Ingenious COCKER! (Now to Rest thou'rt Gone No Art can Show thee fully but thine own Thy rare Arithmetic alone can show Th' vast Sums of Thanks we for thy Labour Cockers ARITHMETIC, Being A plain and familiar Method suitable to the meanest capacity for the full understanding of that incomparable Art, as it is now taught by the ablest Schoolmasters in City and Country. Composed By Edward Cocker late Practitioner in the Arts of Writing, Arithmetic, and Engraving. Being that so long since promised to the world. Perused And published by john Hawkins Writing Master near St. George's Church in Southwark, by the Authors correct Copy, and commended to the World by many eminent Mathematicians and Writing-Masters in and near London. Licenced Sept. 3. 1677. Roger L' Estrange. London Printed, for T. Passenger at the three Bibles on London-Bridge, and T. Lacie at the Golden Lion in Southwark. And sold by C. Passenger, at the 7 Stars in the New-Buildings upon London-Bridge. 1678. TO his much honoured Friends, Manwering Davies of the Inner Temple Esquire, And Mr. Humphrey Davies of St. Mary Newington Butts in the County of Surry. john Hawkins As an acknowledgement of unmerited favours, humbly Dedicateth this Manuel of Arthmetick. Courteous Reader, I Having the Happiness of an Intimate Acquaintance with Mr. Cocker in his life time, often solicited him to remember his Promise to the world of Publishing his Arithmetic, but (for Reasons best known to himself) he refused it, and (after his Death) the Copy falling accidentally into my hands, I thought it not convenient to smother a work of so considerable a moment, not questioning but it might be as kindly accepted as if it had been presented by his own hand. The Method is Familiar and easy, discovering as well the Theoric as the Practic of that most Necessary Art of Vulgar Arithmetic: and thou mayst speedily expect his Decimal, Logarithmetical, and Algebraical Arithmetic, concerning which thou hast a further account given at the end of this Book. In the mean time Judge favourably of the present undertaking, and thou wilt Oblige him, who is, Thine to Serve thee, John Hawkins. From my School near St. George's Church in Southwark, Nou. 29. 1677. Mr. Edward Cocker's PROEM or PREFACE. BY the sacred influence of Divine Providence, I have been instrumental to the benefit of many, by virtue of those useful Arts, Writing and Engraving: And do now with the same wont alacrity cast this my Arithmetical Mite into the public Treasury; beseeching the Almighty to grant the like blessing to these, as to my former labours. Seven Sciences supremely excellent Are the chief Stars in Wisdom's Firmament: Whereof Arithmetic is one, whose worth In beams of Profit and Delight shines forth. This crowns the rest: this makes man's mind complete; This treats of Numbers, and of this we treat. I have been often desired by my intimate Friends to publish something on this Subject; who in a pleasing freedom have signified to me that they expected it would be extraordinary. How far I have answered their expectations, I know not; but this I know, that I have designed this Work not extraordinary abstruse or profound, but have by all means possible, within the circumference of my capacity, endeavoured to render it extraordinary useful to all those whose occasions shall induce them to make use of Numbers. If it be objected that the Books already published, treating of Numbers, are innumerable; I answer, that's but a small wonder, since the Art is infinite. But that there should be so many excellent Tracts of Practical Arithmetic extant, and so little practised, is to me a greater wonder: knowing that as Merchandise is the Life of the Weal-public; so Practical Arithmetic is the Soul of Merchandise. Therefore I do ingenuously profess, that in the beginning of this undertaking, the numerous concerns of the honoured Merchants first possessed my consideration: and how far I have accommodated this Composure for his most worthy service, let his own profitable experience be judge. Secondly, For your service, most excellent Professors, whose understandings soar to the subli●●●y of the Theory and Practice of this noble Science, was this Arithmetical Tractate composed: which you may please to employ as a monitor to instruct your young Tyroes, and thereby take occasion to reserve your precious moments, which might be exhausted that way, for your more important affairs. Thirdly, For you, the ingenious Offspring of happy Parents, who will willingly pay the full price of Industry and Exercise for those Arts and choice accomplishments which may contribute to the felicity of your future state. For you, I say (ingenious Practitioners) was this Work composed, which may prove the pleasure of your youth, and the glory of your age. Lastly, for you the pretended Numerists of this vapouring age, who are more disingenuously witty to propound unnecessary questions, than ingeniously judicious to resolve such as are necessary. For you was this Book composed and published, if you will deny yourselves so much as to invert the streams of your ingenuity, and by studiously conferring with the Notes, Names, Orders, Progress, Species, Properties, Proprieties, Proportions, Powers, Affections and Applications of Numbers delivered herein, become such Artists indeed, as you now only seem to be. This Arithmetic ingeniously observed, and diligently practised, will turn to good account to all that shall be concerned in Accounts. All whose Rules are grounded on Verity, and delivered with Sincerity. The examples are built up gradually from the smallest consideration to the greatest. All the Problems or Propositions are well weighed, pertinent and clear, and not one of them throughout the Tract taken upon trust; therefore now, Zoilus and Momus lie you down and die, For these inventions your whole force defy. Edward Cocker. Courteous Reader, Being well acquainted with the deceased Author, and finding him knowing and studious in the Mysteries of Numbers and Algebra, of which he had some choice Manuscripts, and a great Collection of Printed Authors in several Languages. I doubt not but he hath writ his Arithmetic suitable to his own Preface, and worthy acceptation, which I thought to certify on a request to that purpose made to him that wisheth thy welfare, and the progress of Arts. John colens. Novemb. 27th. 1677. This Manual of Arithmetic is recommended to the World by us whose names are subscribed, viz. Mr. john colens Math. Mr. james Atkinson Math. Mr. Peter Perkins Math. Mr. Rich. Noble of Guildford Mr. Rich. Laurence, Senior Mr. Eleazar Wigan. Mr. Benj. Williams. Mr. Luke Talbot. Mr. William Norgate Mr. William Mason M. Steph. Thomas Mr. Peter Storey Mr. Benj. Tichbourne Mr. joseph symmond's Mr. jerem. Milles Mr. josiah Cuffley Mr. john Hawkins. A Table of the Contents of this BOOK. Chap. Pag. NOtation of Numbers 1 1 Of the natural division of Integers, and the denomination of their parts 2 18 Of the species or kinds of Arithmetic 3 31 Of Addition of whole Numbers 4 33 Of Substraction of whole Numbers 5 46 Of Multiplication of whole Numbers 6 60 Of Division of whole Numbers 7 75 Of Reduction 8 104 Of Comparative Arithmetic, viz. the relation of Numbers one to another 9 148 The single Rule of 3 Direct 10 153 The single Rule of 3 Inverse 11 185 The double Rule of 3 Direct 12 197 The double Rule of 3 Inverse 13 206 The Rule of 3 composed of 5 Numbers 14 211 Single Fellowship 15 215 Double Fellowship 16 220 Alligation Medial 17 226 Alligation Alternate 18 229 Reduction of vulgar Fractions 19 243 Addition of vulgar Fractions 20 259 Substraction of vulgar Fractions 21 262 Multiplication of vulgar Fractions 22 265 Division of vulgar Fractions 23 267 The Rule of 3 Direct in vulgar Fractions 24 270 The Rule of 3 Inverse in vulgar Fractions 25 274 Rules of Practice 26 276 The Rule of Barter 27 301 Questions in loss and gain 28 304 Equation of Payments 29 309 Exchange 30 316 Single Position 31 324 Double Position 32 326 CHAP. I. Notation of Numbers. I. ARITHMETIC is an Art of Numbering, or Knowledge which teacheth to Number well; (viz.) the Doctrine of Accounting by Numbers. And there are divers species and kinds of Arithmetic and Geometry, the which we do intend to treat of in order, applying the Principles of the one, to the Definitions of the other: For as Magnitude or Greatness is the subject of Geometry, so Multitude or Number, is the subject of Arithmetic; and if so, than their first Principles and chief Fundamentals, must have like Definitions; or at least, a Semblable Congruency. II. Number, is that by which the Quantity of any thing is Expressed or Numbered; as the Unit is a number by which the quantity of one thing is expressed or said to be one, and two by which it is named two, and ½ half by which it is named or called half, and the Root of 3 by which it is called the Root of 3, the like of any other. 3. Hence it is that Unit is Number, for the part is of the same matter that is his whole, the Unit is part of the Multitude of Units, therefore the Unit is of the same matter that is the Multitude of Units; but the matter of the multitude of Units is number, therefore the matter of Unit is number: for else if from a number given no number be substracted, the number given remaineth; let three be the number given, from which number subtract, or take away one (which as some conceive is no number) therefore the number given remaineth, that is to say, there remaineth three, which is absurd. 4. Hence it will be convenient to examine from whence number hath its Rise or Beginning; Most Authors maintain that Unit is the beginning of number, and itself no number; but looking upon the Principles and Definitions in the first rudiments of Geometry, we shall find, that the definition of a point is in no way congruous with the definition of an Unit in Arithmetic; and therefore one, or unit must be in the bounds or limits of Number, and consequently the beginning of number is not to be found in the number one; wherefore to make number and magnitude congruent in Principles, and like in Definitions, we make and constitute a cipher to be the beginning of number, or rather the medium between Increasing and Decreasing numbers, commonly called absolute or whole numbers, and negative or fractional numbers, between which, nothing can be Imagined more agreeable to the definition of a point in Geometry; for as a point is an adjunct of a line and itself no line, so is (0) cipher an adjunct of number and itself no number: And as a point in Geometry cannot be divided or increased into parts, so likewise (0) cannot be divided or increased into parts; for as many points though in number infinite do make no line, so many (0) Ciphers, though in number infinite do make no number. For the line AB cannot be Increased by the addition of the point. C, neither can the number D be increased by the addition of the (0) cipher E, for if you add nothing to 6, the sum will be 6, (0) neither increasing nor diminishing the number 6, but if it be granted that AB be extended or prolonged to the point C, so that AC be made a continued line, then AB is increased by the addition of the point C, In like manner if we grant D 6 be prolonged to E (0) so that DE (60) be a continued number making 60, than 6 is Augmented by the Aid of (0) as to the constituting the number (60) sixty, And furthermore that one or unit is material and a number, and that (0) is the beginning of number, is proved by all Authors although indirectly, for the Tables of Sines and Tangents prove one degree to be a number, because the Sine of 1 degree is 174524 (the Radius being 10000000) and the beginning of that Table is (0) and to it Answereth 00000 etc. 5. Hence it is that number is not quantity discontinued, for all that which is but one quantity, is not quantity disjunct; (60) sixty as it is a number, is one quantity, viz. one number: (60) sixty therefore as it is number, it is not quantity disjunct; for number is some such thing in magnitude, as humidity in water; for as humidity extends itself through all and every part of Water, so Number alated to magnitude, doth extend itself through all and every part of magnitude. Also as to continued Water doth answer continued Humidity, so to a continued Magnitude doth answer a continue● Number. As the continued Humidity or any entire Water, suffereth the same Division and Distinction that his Water doth; so the continued Number suffereth the same Division and Distinction that his Magnitude doth. From all which considerations we might enlarge a farther digression concerning Number and Magnitude, by comparing the Definitions of the one, with the Principles of the other, for having found a (0) cipher to be answerable in Definition to a point in Magnitude, we may very well conclude that number may be congruent to a line; as also the Figurative number to be consonant in definition with a superficies, and solid, etc. in the order of Geometrical Magnitudes. 6. The Characters or Notes by which Numbers are signified, or by which a Number is ordinarily expressed, are these following, (viz) 0 cipher or nothing, 1 One, 2 Two, 3 Three, 4 Four, 5 Five, 6 Six, 7 Seven, 8 Eight, 9 Nine; The cipher which though of itself signifieth nothing, (viz.) expresseth not any certain, or known quantity, but is the beginning, Radix or Root of Number, and the other nine Figures or Characters are called significant Figures or Digits. 7. In Numbers of any sort, two things are to be considered, (viz.) Notation and Numeration. 8. Notation teacheth how to describe any Number by certain Notes and Characters, and to declare the value thereof being so described, and that is by Degrees and Periods. 9 A degree consists of three figures, (viz.) of three places comprehending Units Ten and Hundreds, so 365 is a degree, and the first figure (5) on the right hand, stands simply for its own value, being Units or so many ones (viz.) five; the second in order from the right, signifies as many times ten, as there are units contained in it, (viz.) sixty; the third in the same order signifies so may hundreds as it contains units so will the expression of the Number be, three hundred sixty five, also 789, is seven hundred eighty nine, etc. 10. A period is when a Number consists of more than three figures, or places, and whose proper order is to prick or distinguish every third place, beginning at the right hand, and so on to the left; so the Number 63452 being given it will be distinguished thus, 63.452 and expressed thus sixty three thousand four hundred fifty two, likewise 4.578.236.782 being distinguished as you see will be expressed thus, four thousand five hundred seventy eight millions, two hundred thirty six thousand, seven hundred eighty two. 11. Number is either Absolute or Negative. 12. An Absolute, or Entire, whole, Increasing Number, is that which by annexing of another figure or cipher it becomes ten times as much as it stood for before; and if two figures or cyphers be annexed, it makes it a hundred times more than it stood for before, etc. as if you annex to the figure 6 a cipher, than it will become (60) sixty: so if two Ciphers be annexed, than it will be (600) six hundred; and if you do annex to it a (4) four, than it will be (64) sixty four; and if you annex (78) seventy eight, it will be then (678) six hundred seventy eight, and so on: By annexing more figures or cyphers, it will increase in a decuple proportion ad Infinitum. 13. A Negative, or Broken, Fractional, Decreasing Number, is that which by prefixing a point or prick towards the left hand its value is decreased from so many units, to so many tenth parts of any thing; and if a point and (0) cipher, or a digit be prefixed, it will be then so many hundred parts, and if a point, and two Ciphers, or digi●s be prefixed, its value is decreased to be so many thousandth parts; as if you would prefix before the figure 3 a point (.) or prick thus (.3) it is then decreased from 3 Units or Integers, to (3) three tenth parts of a Unit or Integer, and if you prefix a point and cipher thus (.03) it is decreased from 3 Integers to 3 hundredth parts of an Integer, and by this means 5 l. Absolute by prefixing of a point will be decreased to 5 l. Negative which is 5 tenth parts of a pound, equal in value to ten shillings; And so by prefixing of more Ciphers or Digits, its value is decreased in a decuple proportion ad infinitum. As in the following Scheme or rather order of Numbers, we have placed (0) cipher in its due place and order, as it is both the beginning and medium of Number; for going from (0) towards the left hand you deal with Entire, Absolute, Whole, Increasing Numbers. Increasing Numbers. Decreasing Numbers 29 876 543 256 21 0 12 345 678 976 3 mm mmm mmm mmm CX U XC mmm mmm mmm m mm mmm mmm CX XC mmm mmm m mm mmm CX XC mmm m mm CX XC m X But going from (0) the place of Units towards the Right hand, you meet with broken, Negative, Fractional and Decreasing Numbers. And hence it follows that Multiplication increaseth the product in Absolute Numbers, but decreaseth the product in Negative Numbers; Also Division decreaseth the quotient in whole Numbers, and increaseth it in Negative, or Fractional Numbers. 14. An Absolute, Entire, whole, Increasing number, hath always a point annexed towards the right hand and therefore 15. A Negative, Broken, Decimal, decreasing number, hath always a point prefixed before it towards the left hand. When we express Integers, or whole number, as 5 pounds, 5 feet, 26 men, we usually annex a point or prick after the number thus 5.5.26.347. l. feet. men. Inch. But when we express Decimals, or Numbers that are denied to be Entire, as decreasing numbers, we do commonly prefix a point or prick before the said decimal, or decreasing number, thus (.3) that is 3 tenths, or 3 primes, .03, that is 3 hundredths or 3 seconds. 16. A whole or absolute number is an unit or a Composed multitude of units, and it is either a prime or else a compounded number. 17. Prime numbers amongst themselves are those which have no multitude of units for a common measurer as 8 and 7 or 10 and 13 because not any multitude of units can equally measure or divide them without a Remainder. 18. Compound numbers amongst themselves are those which have a multitude of units for a common measurer, as 9 and 12 because 3 measures them exactly and abbreviates them to 3 and 4. 19 A Broken number commonly called a fraction, is a part or parts of a whole number; viz. a part of an Integer, as 1/●; one third, is one third part of an unit. 20. A broken number, or fraction, consists of 2 parts, viz. the Numerator and the Denominator. 21. The Numerator and Denominator of a fraction, are set one over the other, with a line between them; and the Numerator is set above the line, and expresseth the parts therein contained. 22. The Denominator of a fraction is the Inferior number placed below the line, and expresseth the number of parts into which the unit or Integer is divided; as let ¾ be the fraction given, so shall 3 be the numerator, and doth express or number the multitude of parts contained in this fraction, for ¾ is a fraction composed of fourth's, or quarters, and the figure 3 in numbering shows us that in that fraction there are 3 of those fourth parts or quarters; also in the same fraction ¾, 4 is the denominator and doth express the Quality of the fraction, viz. that the whole, or integer, is here divided into 4 equal parts. 23. A broken number is either Proper or Improper; viz. Proper, when the numerator is lesser than the denominator; so ¾ is a perfect proper fraction: But an Improper fraction hath its numerator greater, or at least equal to the denominator; thus 13/8 is an Improper fraction: the Reason is given in the definition. 24. A proper broken number, is either Simple, or Compound; viz. Simple, when it hath one Denomination, and Compound when it consisteth of divers Denominations. If ¾ l. 6/12 l. 25/100 l. were given, we say they are either of them single, or simple fractions because they consist but of one numerator and one denominator; but if ¾ of 6/12 of 25/100 of a pound sterling were given, we say that it is a compound broken number, or fraction, because the expression and representation, consisteth of more denominations than one; and such by some are called fractions of fractions; and they have always this particle (of) between them. 25. When a single broken number or fraction, hath for his denominator a number consisting of a Unit in the first place towards the left hand, and nothing but Ciphers from the Unit towards the right hand, it is then the more aptly and rightly called a decimal fraction; under this head are all our decreasing numbers placed, and in our 13th definition called Negative, and by that order there prescribed we order them to be Decimals, by signing a point or prick before them, or the numerator rejecting the denominator: Therefore according to our last Rule, 5/10 5/100 25/100 25/1000 are said to be Decimals; and a Decimal fraction may be expressed without its denominator (as before) by prefixing a point or prick before the numerator of the said fraction, and then shall the former fraction 5/10 and 25/100 stand thus .5 and .25. But oftentimes as in the second and 4th fractions 5/100 and 25/1000 a prick or point will not do without the help of a cipher or cyphers prefixed before the significant figures of the numerator, and therefore when the numerator of a decimal fraction, consisteth not of so many places, as the denominator hath cyphers, fill up the void places of the numerator with prefixing Ciphers before the significant figures of the numerator, and then sign it for a decimal, so shall 5/100 be .05 and 25/1000 will be .025 and 72/10000 will be .0072. Now by this we may easily discover the denominator having the numerator; for always the denominator of any decimal fraction, consists of so many Ciphers as the numerator hath places, with a unit prefixed before the said Ciphers, viz. under the point or prick. 26. A Decimal number or fraction, is that which is expressed by Primes, Seconds, Thirds, Fourths, etc. and is Number decreasing. Here instead of Natural and Common fractions, as ¾ of a thing, we order the thing or Integer into Primes, Seconds, Thirds, Fourths, Fifths, etc. that our expression may be consonant to our former order. 27. In Decimal Arithmetic, we always imagine (and it would be very commodious if it were really so) that all entire units, Integers, and things are divided first into ten equal parts, and these parts so divided we call Primes; and secondly, we divide also each of the former primes into other ten equal parts, and every of these divisions we call seconds; and thirdly, we divide each of the said seconds into ten other equal parts, and those so divided we call Thirds, and so by decimating the former and sub-decimating these latter, we run on ad infinitum. 28. Let a pound sterling, Troy weight, Averdupois weight, Liquid measure, Dry measure, Long measure, time, dozen or any other thing, or Integer be given to be decimally divided; in this notion premised, we ought to let the first division be Primes, the next division Seconds, the next Thirds, etc. So one pound sterling being 20 shillings, which divided into ten equal parts the value of each part will be two shillings; therefore one Prime of a pound sterling will stand thus (.1) which is in value 2 shillings: Three Primes will stand thus (.3) and that is in value 6 shillings. Again, a Prime or .1 being divided into ten equal parts, each of those parts will be one Second, and is thus expressed, (.01) and its value will be found to be 2d. farthing and 6/10 of a farthing; and so will .05 signify one shilling or five Seconds. And if .01 be divided into ten other equal parts, each of those parts so divided will be Thirds, and will stand thus .001, and its value will be found to be .96 of a farthing, or 96/100 of a farthing; and .009 Thirds will be 2d. and .64 of a farthing, or 64/100 of a farthing, etc. So that .375 l. will be found to Represent 7 s. and 6 d.; for the 3 Primes are 6 shillings, and the 7 Seconds are 1 s. 4 d. and 8/10 of a penny, and the five Thirds are 1 penny and 2/10 of a penny, both which added together make 7 s. 6 d. 29. If you put any bulk, or body, representing an Integer; if it be decimally divided, than the parts in the first decimation are Primes, the next Seconds, and the next decimation is Thirds, the next Fourths, etc. As let there be given a bullet of Lead, or such like, whose weight let be 50 l. Troy, this call an unit, Integer, or thing; then with the like weight and matter make 10 other, the which together will be equal to 50 l. and will weigh each of them 5 l. a piece; take of the same matter, and equal to 5 l. make 10 more, than each of those will weigh 6 ounces a piece; also if again you take 6 ounces and thereof make 10 other small bullets each of them will weigh 12 penny weight Troy; and thus have you made Primes, Seconds, and Thirds, in respect of the Integer containing 50 l. Troy weight: So that 5 Primes is equal to the half mass, and 2 Primes and 5 Seconds, is a quarter of the mass; and therefore 1 of the first division, 2 of the second division, and 5 of the third division, will be equal in weight to ½ a quarter of the mass and contain 6 l. and 3 ounces. 30. When a decimal fraction followeth a whole number, you are to separate or part the decimal from the whole number by a point or prick; so if .75 followed the whole number 32, set them thus 32.75. You will find that divers Authors have divers ways in expressing mixed numbers, as thus, 3275 or 32 75/100 or 32.75 but you will find that 32.75 thus placed and expressed is fittest for Calculation. 31, A mixed Number hath 2 parts, the whole and the broken; the whole is that which is composed of Integers, and the broken is a fraction annexed thereunto. So the mixed Number 36 8/12 being given we say that 36 is the whole Number, which is composed of Integers, and the 8/12 is the broken Number annexed, which showeth that one of the former Integers (of that 36) being divided into 12 parts, this 8/12 doth express 8 of those 12 parts more belonging to the said 36 Integers. 32. Denominative Numbers are of one, or of many, and those are of divers sorts and kinds, viz. Singular called unit, as 1; and Plural called multitude as 2, 3, 4, 5: Single of one kind only, called digits, as 1, 2, 3, 4, 5, 6, 7, 8, 9, and Compound of many, as 10, 11, 12, etc. 102, 367, etc. Proportional as single, Multiple, Double, Triple, Quadruple, etc. Denominate as Pounds, Shillings, Pence; Undenominate as 1, 2, 3, etc. Perfect as 6, 28, 496, 8128, 130816, 2096128, etc. Whose parts are equal to the numbers; Imperfect, unequal and more in the sum as 12 to 1, 2, 3, 4, 6. Imperfect, unequal and less than the sum, as 8 to 1, 2, 4. Numbers Commensurable and Incommensurable, as 12 and 9 are Commensurable because three measures them both. triangular array of dots square array of dots CHAP. II. Of the Natural Division of Integers, and the several Denominations of their parts. 1. BEfore we come to Calculation or the ordering of Numbers to operate any Arithmetical Question proposed, we will lay down Tables of the Denomination of several Integers; and after that (having mentioned the several Species or kinds of Arithmetic) we shall immediately handle the Species of Numeration, which are the main Pillars upon which the whole Fabric of this Art is built. Of Money, Weights, etc. 2. The least Denomination or Fraction of money used in England is a farthing, from whence is produced the following Tables, called the Tables of Coin, (viz.) and therefore 1 farthing make 1 farthing l. s. d. qrs 4 farthings 1 Penny 1 20 12 4 12 Pence 1 shilling 1 20 240 960 20 Shillings 1 Pound 1 12 48 1 4 The first of these Tables viz. that on the left hand is plain and easy to be understood, and therefore wants no directions. In the second Table above the line you have 1 l. 20 s. 12 d. 4 qts. whereby is meant that 1 pound is equal to 20 shillings, and one shilling is equal to 12 pence, and one penny equal to 4 farthings, under the line is 1 l. 20 s. 240 d. 960 qts. which signifies one pound to contain 20 shillings or 240 pence, or 960 farthings; in the second line below that is 1 s. 12 d. 48 qts. the first standing under the denomination of shillings, whereby is to be noted that one shilling is equal to 12 pence, or 48 farthings, and likewise that below, that one penny is equal in value to four farthings; understand the like reason in all the following Tables of weight, measure, time, motion, and dozen. Troy weight. 3. The least Fraction or Denomination of weight used in England is a grain of wheat gathered out of ●●e middle of the ear, and well dried▪ from whence are produced these following ●●bles of weight, called Troy weight, 32 Grains of wheat make 24 artificial grains 24 Artificial grains 1 Penny weight 20 Peny-weight 1 Ounce 12 Ounces 1 Pound Troy weight And Therefore. l. owned. dra. grains. 1 12 20 24 1 12 240 5760 1 20 480 1 24 Troy Weight serveth only to weigh Bread, Gold, silver, and Electuaries; it also regulateth, and prescribeth a form how to keep the money of England at a certain standard. The Goldsmiths have divided the ounce Troy weight into other parts, which they generally call mark weight, the denominative parts thereof are as followeth, viz. A mark (being an ounce Troy) is divided into 24 equal parts, called Carects and each Carect into 4 grains, so that in a mark are 96 Grains; by this weight they distinguish the different finess of their Gold, for if to the finest of Gold be put 2 Carects of Alloy (which is of Silver, Copper, or other base mettle, with which they use to mix their Gold or silver to abate the finess thereof) both making when cold but an ounce, or 24 Carects, than this Gold is said to be 22 Carects fine, for if it come to be Refined the 2 Carects of alloy will fly away and leave only 22 Carects of pure Gold, the like to be considered of a greater or lesser quantity; And as the finess of gold is estimated by Carects, so the finess of silver is distinguished by ounces, for if a pound of it be pure, and looseth nothing in the Refining, such silver is said to be twelve ounces fine, but if it looseth any thing, it is said to contain so much fineness as the loss wanteth of 12 ounces, as if it lose an ounce, it is said to be 11 ounces fine, and if it lose one ounce 14 penny weight, than it is said to be 10 ounces 6 penny weight fine, and that which loseth two ounces four penny weight 16 grains, is said to be nine ounces 15 penny weight 8 grains fine, etc. the like of a greater or lesser quantity. Apothecary's weights. 4. The Apothecaries have their weights deduced from Troy weight, a pound Troy being he greatest Integer, a Table of whose division and sub-division followeth, viz. 1 pound makes 12 ounces And therefore 1 ounce 8 drams l. owned. dram scrup. gr. 1 dram 3 scruples 1 12 8 3 20 1 scruple 20 grains 1 12 96 218 5760 1 8 24 480 1 3 60 1 20 5. Thus much concerning Troy weight, and its derivative weights (which as was said before) serveth to weigh Bread, Gold, Silver, and Electuaries; now besides Troy weight there is another kind of weight used in England, commonly known by the name of Averdupois weight, (a pound of which is equal to 14 ounces 12 penny weight Troy weight) and it serveth to weigh all kinds of Grocery wares, as also butter, Cheese, Flesh, Wax, Tallow, resin, Pitch, Led, and all such kind of garbel, the Table of which weight is as followeth. The Table of Averdupois weight. 4 quarters of a dram makes one dram 16 drams one ounce 16 ounces one pound 28 Pounds 1 quarter of a hundred 4 quarters 1 hund. weight, or 112 20 Hundred 1 Tun. And therefore. Tun C. qts. l. owned. does. qts. 1 20 4 28 16 16 4 1 20 80 2240 13440 215040 860160 1 4 112 1792 28672 114688 1 28 448 7168 28672 1 16 256 1024 1 16 64 1 4 Wool is weighed with this weight, but only the divisions are not the same; A Table whereof followeth. A Table of the denominative parts of Wool-weight. 7 Pounds make 1 Clove 2 Cloves 1 Stone 2 Stone 1 Todd 6 Todd 1 Stone 1 Wey 2 Weighs 1 Sack 12 Sacks 1 Last And Therefore Last Sacks Weigh Todd Stone Cloves l. 1 12 2 6 ½ 2 2 7 1 12 24 156 312 624 4368 1 2 13 26 52 364 1 6 ½ 13 26 182 1 2 4 28 1 2 14 1 7 Note that in some Country's the Wey is 256 l. Averdupois, as is the Suffolk Wey; But in Essex there is 336 l. in a Wey. 6. The least Denominative part of Liquid measure is a Pint, and it is taken from Troy weight, because 1 pound of wheat Troy weight makes 1 Pint liquid: A Table of which Measure followeth. The Table of Liquid Measure. 1 pound of wheat Troy make 1 pint 2 pints 1 quart 2 quarts 1 pottle 2 pottles 1 gallon 8 gallons 1 firk. of ale, soap, or herr. 9 gallons 1 firkin of beer 10 gallons and a half 1 firk. of Salmon or Eeles 2 firkins 1 Kilderkin 2 Kilderkins 1 barrel 42 gallons 1 Tierce of wine 63 gallons 1 hogshead 2 hogsheads 1 pipe or butt 2 pipes or butts 1 Tunn of wine And Therefore. Tun▪ pipes hhds gall. pts grains of wh. 1 2 2 63 8 7680 1 2 4 252 2016 15482880 1 2 126 1008 7741440 1 63 504 3870720 1 8 61440 1 7680 Thus you see that according to the standard of England a pint doth contain 7680 grains of Natural wheat, but whosoever shall try the wheat growing in Norfolk, shall find a Concave inch (that is a hollow, or hole made just the bigness of an inch) to contain 280 kernels of wheat, and if you admit of 28 ⅞ solid Inches in a pint, which according to the Judgement of Artists, is the content of a pint wine-measure (the least of all measures) than a pint of such measure will contain 8085 grains of wheat; which is very different from the standard, but these things Quere. 7. The least Denominative part of dry measure is also a pint, and this is likewise taken from Troy weight. The Table of whose division followeth. The Table of Dry Measure. 1 pint make 1 pint 2 pints 1 quart 2 quarts 1 pottle 2 pottles 1 gallon 2 gallons 1 peck 4 pecks 1 bushel 4 bushels 1 Comb 2 Combs 1 quarter 4 quarters 1 Chalder 5 quarters 1 Wey 2 Wey's 1 Last And therefore. last weigh qts come. bush. peck gall. pints 1 2 5 2 4 4 2 8 1 2 10 20 80 320 640 5120 1 5 10 40 160 320 2560 1 2 8 32 64 512 1 4 16 32 256 1 4 8 64 1 2 16 1 8 8. The least Denominative part of Long Measure is a Barley-corn well dried and taken out of the middle of the ear: whose Table of parts followeth. The Table of Long Measure. 3 barley corns make an Inch 12 Inches 1 foot 3 feet 1 yard 3 feet 9 inches or a yard and quart. 1 Ell English 6 feet 1 fathom 5 yards and an half 1 pole or perch 40 poles or perches 1 furlong 8 furlongs 1 English mile And Therefore. mile furl. poles yards feet Inches barl. corns 1 8 40 5 ½ 3 12 3 1 8 320 1760 5280 63360 190080 1 40 220 660 7920 23760 1 5 ½ 16 ½ 198 594 1 3 36 108 1 12 36 1 3 And note that the yard, as also the ell, is usually divided into 4 quarters, and each quarter into 4 Nails. Note also, that a Geometrical pace is 5 feet; and there are 1056 such paces in an English mile. 9 The parts of the Superficial measures of land, are such as are mentioned in the following Tables, viz. The Table of Land Measure. 40 Square Poles or Perches make 1 Rood or quarter of an Acre. 4 Roods 1 Acre. By the foregoing Table of long Measure, you are Informed what a pole, or (which is all one) perch is; and by this that 40 square perches are 1 Rood. Now a square perch is a Superficies very aptly resembled by a square Trencher, every side thereof being a Perch or 5 Yards and a half in length, 40 of them is a Rood, and 4 Roods an Acre. So that a Superficies that is 40 perches long and 4 broad is an Acre of land the Acre containing in all 160 square Perches. 10. The least Denominative part of Time is a Minute, the greatest Integer being a Year; from whence is produced this following Table. The Table of Time. 1 Minute make 1 Minute 60 Minutes 1 Hour 24 Hours 1 Day natural 7 Days 1 Week 4 Weeks 1 Month 13 months 1 day & 6 hours 1 Year. But the year is usually divided into 12 unequal Calendar Months, whose names and the number of Days that they Contain, follow, viz. days january 31 February 28 March 31 April 30 May 31 june 30 july 31 August 31 September 30 October 31 November 30 December 31 So that the year containeth 365 days, and 6 hours, but the 6 hours is not reckoned but only every 4 year, and then there is a day added to the latter end of February, and then it containeth 29 days, and that year is called Leap-year, and containeth 366 days. And here Note, that as the Hour is divided into 60 Minutes, so each Minute is subdivided into 60 Seconds, and each Second into 60 Thirds, and each Third into 60 Fourths, etc. The Tropical year by the exactest observations of the most accurate Astronomers, is found to be 365 Days, 5 Hours, 49 Minutes, 4. Seconds and 21 Thirds. CHAP. III. Of the Species or Kind's of Arithmetic. 1. Arithmetic is either Natural, Artificial, Analiticall, Algebraical, Lineal, or Instrumental. 2. Natural Arithmetic, is that which is performed by the Numbers themselves; and this is either Positive or Negative. Positive which is wrought by certain infallible numbers propounded, and this is either single or Comparative; Single, which considereth the nature of numbers simply by themselves; and Comparative which is wrought by numbers as they have Relation one to another. And the Negative part relates to the Rule of False. 3. Artificial (by some called Logarithmetical) Arithmetic is that which is performed by Artificial or borrowed numbers invented for that purpose, and are called Logarithmes. 4. Analiticall Arithmetic, is that which shows from a thing unknown, to find truly that which is sought; always keeping the Species without Change. 5. Algebraical Arithmetic, is an obscure and hidden art of Accounting by numbers in resolving of hard Questions. 6. Lineal Arithmetic, is that which is performed by lines, fitted to proportions as also Geometrical projections. 7. Instrumental Arithmetic, is that which is Performed by Instruments, fitted with Circular and Right lines of proportions, by the motion of an Index or otherwise. 8. The parts of single Arithmetic are Numeration and the Extraction of Roots. 9 Numeration is that which by certain known numbers propounded, we discover another Number unknown. 10. Numeration hath four Species; viz. Addition, Subtraction, Multiplication, and Division, CHAP. IU. Of Addition of whole Numbers. 1. ADdition is the Reduction of two, or more numbers of like kind together into one Sum or Total. Or it is that by which divers numbers are added together, to the end that the Sum or Total value of them all may be discovered. The first number in every addition is called the Addable number, the other, the number or numbers added, and the number invented by the Addition is called the Aggregate or Sum, containing the value of the Addition. The Collation of the numbers, is the right placing of the numbers given respectively to each denomination; And the Operation is the Artificial adding of the numbers given together in order to the finding out of the Aggregate or Sum. 2. In Addition, place the numbers given respectively the one above the other, in such sort, that the like degree, place, or denomination may stand in the same Series, viz. Units under Units, Ten under Ten, Hundreds under Hundreds, etc. Pounds under Pounds, Shillings under Shillings, Pence under Pence, etc. Yards under Yards, Feet under Feet, etc. 3. Having thus placed the numbers given (as before) and drawn a line under them, Add them together, beginning with the lesser denomination, viz. at the right hand and so on, subscribing the sum under the line Respectively; as for Example. Let there be given 3352 and 213 and 133 to be added together, I set the Units in each particular number under each other, and so likewise the Ten under the Ten, etc. and draw a line under them as in the margin, than I begin at the place of Units and add them together upwards saying, 3 and 3 are 6 and 2 make 8, which I set under the line, and under the same figures added together; then I proceed to the next place, being the place of Ten, and add them up in the same manner as I did the place of Units, saying 3 and 1 are 4 and 5 are 9, which I likewise set under the line Respectively; then I go to the place of Hundreds, and add them up as I did the other, saying 1 and 2 are 3 and 3 are 6, which I also set under the line; and lastly I go to the place of Thousands, and because there are no other figures to add to the 3, I set it under the line in its respective place, and so the work is finished; and I find the sum of the 3 given numbers to be 3698. 4. But if the sum of the figures of any Series exceeds ten, or any number of ten, subscribe under the same the Excess above the ten, and for every ten carry one to be added to the next Series towards the left hand, and so go on until you have finished your addition; always remembering, that how great soever the sum of the figures of the last Series is, it must all be set down under the line Respectively. So 3678 being given to be added to 2357, I set them down as is before directed and as in the margin, with a line drawn under them, than I begin and add them, saying 7 and 8 are 15 which is 5 above ten, I set 5 under the line, and carry 1 for the ten to be added to the next Series, saying 1 that I carried and 5 is 6 and 7 are 13 I set down 3 and carry 1 for the ten, then to the next Series, I say 1 that I carried and 3 are 4 and 6 are ten, now because it come to just 10 and no more, I set 0 under the line and carry 1 for the ten to the next, and say 1 that I carried and 2 are 3 and 3 are 6 which I set down in its Respective place, thus the addition is ended, and the total Sum of these numbers is found to be 6035, several Examples of this kind follow. 5. If the Numbers given to be added are contained under divers denominations, as of Pounds, Shillings, Pence and Farthings; or of Tuns, Hundreds, Quarters, Pounds, etc. Then in this case having disposed of the numbers, each denomination under other of like kind; begin at the least denomination, (minding how many of one denomination do make an Integer in the next) and having added them up, for every Integer of the next greater denomination that you find therein contained, bear a unit in mind to be added to the said next greater denomination, expressing the excess respectively under the line, proceed in this manner until your addition be finished, the following Examples will make the Rule plain to the learner. Thus these several sums being given to be added, viz. 136 l. 13 s. 4 d. 2 qrs. and 79 l. 07 s. 10 d. 3 qrs. and 33 l. 18 s. 09 d. 1 qrs. also 15 l. 9 s. 5 d. 0 qrs. The numbers being disposed according to order will stand as in the Margin. Then I begin at the denomination of Farthings l. s. d. qrs. 136 13 04 2 79 07 10 3 33 18 09 1 15 09 05 0 265 09 05 2 and add themup, saying 1 and 3 are 4 & 2 make 6, now I consider that 6 Farthings, is 1 penny and 2 farthings, wherefore I set down the 2 Farthings in its place under the line, and keep 1 in mind to be added to the next denomination of Pence; then I go on, saying 1 that I carried and 5 are 6 and 9 are 15 and 10 are 25 and 4 are 29, now I consider that 29 pence are 2 shillings and 5 pence, wherefore I set the 5 pence in order under the line and keep 2 in mind for the 2 shillings, to be added to the shillings; then I go on saying, 2 that I carried, and 9 are 11, and 18 are 29, and 7 are 36, and 13 are 49; then I consider that 49 shillings are 2 Pounds and 9 shillings, wherefore I set the 9 shillings under the line, and carry 2 for the 2 pounds, to the next and last denomination of pounds, and proceed saying, 2 that I carried, and 5 make 7. and 3 are 10, and 9 are 19, and 6 are 25; I then set down 5 and carry 2 for the 2 ten and proceed saying 2 that I carry and 1 is 3, and 3 are 6, and 7 are 13, and 3 make 16; I set down 6 and carry 1 for the ten, and go on saying 1 that I carried and 1 are 2 which I set in its place under the line, and the work is finished, and thus I find the Sum of the foresaid Numbers to be 265 l. 09 s. 05 d. 2 qrs. This to the Ingenious practitioner is sufficient, but I shall (for the further illuminating of weaker apprehensions) explain the operation of another Example in Troy weight; and here the Learner must take notice of the Table of Troy weight, mentioned or set down in the third Section of the second Chapter. The numbers given in this Example, are 38 l. 07 oz. 13 p.w. 18 gr. And 50 l. 10 oz. 10 p.w. 12 gr. And 42 l. 08 oz. 05 p.w. 16 gr. And in order to the Addition thereof, I place them as you see, and proceed to operation; saying 16 and 12 are 28, and 18 are 46; now because 24 grains l. oz. p.w. gr. 38 07 13 18 50 10 10 12 42 08 05 16 132 02 09 22 make 1 penny weight, 46 grains are 1 penny weight and 22 grains; wherefore I set down 22, and carry 1 for the the penny weight, and going on I say o●e that I carry and 5 make 6, and 10 are 16, and 13 are 29, which is 1 ounce and 9 penny weight; I set down 9 in its place under the line, and carry 1 to the ounces, saying 1 that I carry and 8 are 9, and 10 are 19, and 7 are 26, and because 26 ounces make 2 pound 2 ounces, I set down 2 for the ounces, and carry 2 to the pounds; going on, 2 that I carry and 2 are 4, and 8 make 12, that is 2 and go 1; then 1 I carry and 4 are 5, and 5 are 10, and 3 are 13, which I set down as in the Margin, and the work is finished, and I find the sum of the said numbers to amount to 132 l. 02 oz. 09 p.w. 22 gr. This is sufficient for the understanding of the following Examples, or any other that shall come to thy view. The way of proving these or any sums in this Rule is showed Immediately after the ensuing Examples. Addition of English money. l. s. d. qrs. l. s. d. qrs. 436 13 07 1 48 15 11 1 184 09 10 3 76 10 07 3 768 17 04 2 18 00 05 3 564 11 11 0 24 19 09 2 1954 12 09 2 168 06 10 1 Addition of Troy weight. l. owned. p.w. gr. l. owned. p.w. gr. 15 07 13 12 145 09 12 18 18 06 04 20 726 08 14 10 11 10 16 18 380 07 06 13 09 04 10 22 83 10 16 20 19 11 18 04 130 00 10 12 22 00 00 00 74 07 15 00 97 05 04 04 1541 08 16 01 Addition of Apothecary's weights. l. owned. dr. scr. gr. l. owned. dr. scru. gr. 48 07 1 0 14 60 03 4 0 10 74 05 5 2 10 48 10 6 0 14 64 10 7 1 16 34 08 2 1 15 17 08 1 0 11 18 11 2 2 11 34 09 6 1 09 160 07 1 2 15 35 02 5 1 07 240 05 6 1 00 358 07 7 0 12 Addition of Averdupois weight. Tun C. qrs. l. i. owned. dr. 75 13 1 15 36 10 12 48 07 3 21 22 11 13 60 11 1 17 11 07 04 21 07 0 25 15 04 10 12 16 0 11 20 00 09 218 16 0 05 106 03 00 Addition of Liquid Measure. Tun Pipe hhd. gall. Tun hhds gall. pts. 45 1 1 48 30 3 40 4 15 0 1 17 12 0 28 6 38 0 0 47 47 5 60 5 12 1 0 56 57 3 22 3 21 1 1 18 17 0 00 0 133 1 1 60 166 1 26 2 Addition of Dry Measure. Chald. qrs. bush. pec. qrs. bush. pec. gall. 48 3 7 3 17 3 1 1 13 1 4 0 50 1 3 0 54 0 6 2 14 5 3 1 16 3 6 1 40 2 0 1 40 1 0 1 30 0 3 0 173 3 0 3 152 5 3 1 Addition of Long Measure. yds' qrs. na. else qrs. na. 35 3 3 56 1 3 14 1 2 13 3 2 74 2 3 48 2 1 48 0 1 50 1 0 30 1 0 74 0 2 15 0 0 17 1 0 218 1 1 260 2 0 Addition of Land Measure. Acre Rood per. Acr. Rood Perch. 12 3 18 86 1 36 14 0 24 47 3 24 30 2 19 73 2 18 48 3 30 60 0 07 28 1 38 04 2 08 50 3 26 14 1 14 185 3 35 286 3 27 The proof of Addition. 6. Addition is proved after this manner, when you have found out the sum of the Numbers given, then separate the uppermost line from the rest, with a stroke or dash of the pen, and then add them all up again as you did before, leaving out the uppermost line, and having so done add this new invented Sum to the uppermost line you separated, and if the Sum of those two lines be equal to the Sum first sound out, than the work was performed true, otherwise not, As for Example, let us prove the first example of Addition of money whose sum we found to be 265 l. 9 s. 5 d. 2 qrs, and which we prove thus, having separated the l. s. d. qts. 136 13 04 2 79 07 10 3 33 18 09 1 15 09 05 0 265 09 05 2 128 16 01 0 265 09 05 2 uppermost number from the rest, by a line as you see in the margin, than I add the same together again, leaving out the said uppermost line, and the sums thereof, I set under the first Sum, or true sum, which doth amount to 128 l. 16 s. 01 d. 0 qrs. then again I add this new Sum to the uppermost line that before was separated from the rest, and the Sum of these two is 265 l. 9 s. 05 d. 2 qrs. the same with the first Sum, and therefore I conclude that the operation was rightly performed. 7. The main end of Addition in Questions Resolvable thereby to know the sum of several debts, parcels, Integers, etc. some Questions may be these that follow Quest. 1. There was an old man whose age was required, to which he replied, I have seven sons each having two years between the birth of each other, and in the 44 year of my age my eldest son was born, which is now the age of my youngest; I demand what was the old man's age? Now to Resolve this Question, first set down the father's age at the birth of his first child, which was 44, than the difference between the eldest and the youngest, which is 12 years, and then the age of the youngest which is 44, and then add them all together, and their sum is 100, the complete age of the Father. Quest. 2. A man lent his friend at several t●mes, these several sums, (viz.) at one time 63 l. at another time 50 l. at another time 48 l. at another time 156 l. now I desire to know how much was lent him in all. Set the sums lent one under another, as you see in the margin, and then add them together, and you w●ll find their sum to amount to 317 l. wh●ch is the Total of all the several sums lent, and so much is due to the Creditor. Quest. 3. From London to Ware is 20 miles, thence to Huntingdon 29 miles, thence to Standford 21, thence to Tuxford 36 miles, thence to Wentbridge 25 miles, from thence to York 20 miles. Now I desire to know how many miles it is from London to York according to this Reckoning? Now to answer this Question, set down the several distances given as you see in the margin, and add them together, and you will find their sum to amount to 151, which is the true distance in miles between London and York. Quest. 4. There are 2 numbers the least whereof is 40, and their difference is 14, I desire to know what is the greater number, and also what is the sum of them both? First set down the least viz. 40, and 14 the difference, and add them together, and their sum is 54 for the greatest number, than I set 40, (the least) under 54, (the greatest) and add them together, and their sum is 94, equal to the greatest and least numbers. CHAP. V. Of Subtraction of whole Numbers. 1. Subtraction is the taking of a lesser number out of a greater of like kind, whereby to find out a third number, being or declaring the Inequality, excess, or difference between the numbers given, or Subtraction is that by which one number is taken out of another number, given to the end that the residue, or remainder may be known, which remainder is also called the rest or difference of the numbers given. 2. The number out of which Subtraction is to be made, must be greater, or at least equal with the other number given, the higher or superior number is called the major number, and the lower or inferior is called the minor number, and the operation of Subtraction being finished, the rest or remainder is called the difference of the numbers given. 3. In Subtraction place the numbers given respectively, the one under the other, in such sort as like degrees, places, or denominations may stand in the same Series, viz. Units under units▪ Ten under Ten, etc. Pounds under Pounds, etc. Feet under Feet, and Parts under Parts, etc. This being done, draw a line underneath, as in Addition. 4. Having placed the numbers given as is before directed, and drawn a line under them, Subtract the lower number, (which in this case must always be lesser than the uppermost) out of the higher number, and subscribe the difference, or remainder, respectively below the line; and when the work is finished, the number below the line will give you the Remainder; As for Example, let 364521 be given to be Subtracted from 795836, I set the lesser under the greater as in the margin, and draw a line under them, then beginning at the Right hand, I say 1 out of 6 and there Remains 5, which I set in order under the line; then I proceed to the next, saying 2 from 3 rests 1, which I note also under the line, and thus I go on until I have finished the work, and then I find the Remainder or difference to be 431315. 5. But if it so happen (as commonly it doth) that the lowermost number or figure is greater than the uppermost; then in this case, add ten to the uppermost number, and Subtract the said lowermost number from their sum, and the remainder place under the line, and when you go to the next figure below, pay a unit by adding it thereto for the 10 you borrowed before, and subtract that from the higher number or figure: And thus go on until your Subtraction be finished. As for Example; Let 437503 be given, from whence it is required to subtract 153827, I dispose of the numbers as is before directed, and as you see in the margin; then I begin, saying 7 from 3 I cannot, but (adding 10 thereto I say) 7 from 13 and there Remains 6 which I set under the line in order; then I proceed to the next figure, saying 1 that I borrowed and 2 is 3 from 0 I cannot, but 3 from 10 and there remains 7, which I likewise set down as before; then one that I borrowed and eight is nine, from five I cannot, but nine from fifteen and there remains six; then one I borrowed and three is four, from seven and there remains three; then five from three I cannot, but five from thirteen and there remains eight; then one I borrowed and one are two, from four and there rests two; And thus the work is finished, and after these numbers are Subtracted one from another, the inequality, remainder, excess, or difference is found to be 283676. Examples for thy further experience may be these that follow. From 3475016 From 3615746 Take 738642 Take 5864 Rests 2736374 Rests 3609882 6. If the Sums or Numbers to be Subtracted, are of several Denominations, place the lesser Sum below the greater, and in the same Rank and order as is showed in Addition of the same Numbers; then begin at the Right hand, and take the lower number out of the uppermost, if it be lesser; but if it be bigger than the uppermost, then borrow a Unit from the next greater Denomination, and turn it into the parts of the lesser Denomination, and add those parts to the uppermost Number, and from their Sum subtract the lowermost, noting the remainder below the line; then proceed and pay 1 to the next Denomination for that which you borrowed before, and proceed in this order until the work be finish●ed, An Example of this Rule may be thi● that followeth; let 375 l. 13 s. 07 d. 1 qrs be given, from whence let it be required to Subtract 57 l. 16 s. 03 d. 2 qrs. In order whereunto I place the numbers as you see i● the margin, and thus I begin at the leas● Denomination saying l. s. d. qr● 375 13 07 1 57 16 03 2 317 17 03 ●3 two from one I cannot, therefore I borrow one penny from the next denomination and turn it into farthings which is four, and adding four 〈◊〉 one which is five I say, but two from fiv● and there remains three, which I put und●● the line; then going on, I say one that I borrowed and three is four, from 7 and there Rests three; then going on, I say sixteen from thirteen I cannot, but (borrowing one pound and turning it into twenty shillings; I add it to thirteen and that is thirty three, wherefore I say) sixteen from thirty three, and there remains seventeen, which I set under the line and go on, saying one that I borrowed and seven is eight, from five I cannot, but eight from fifteen and there remains seven; then one that I borrowed and five is six, from seven there Rests one, and nothing from three Rests three, and the work is done; And I find the remainder or difference to be 317 l. 17 s. 03 d. 3 qrs. Another Example of Troy weight may be this, I would Subtract 17 l. 10 oz. 11 p.w. 20 gr. from 24 l. 05 oz. 00 p.w. 08 gr. I place the numbers according to Rule and begin, saying twenty l. oz. p.w. gr. 24 05 00 08 17 10 11 20 06 06 08 12 from eight I cannot, but borrow one penny weight which is twenty four grains, and add them to eight and they are thirty two, wherefore I say twenty from thirty two Rests twelve; then one that I borrowed and eleven is twelve, from 00 I cannot, but twelve from twenty (borrowing an ounce which is twenty penny weight) and there remains eight; then one that I borrowed and ten is eleven, from five I cannot, but eleven from seventeen and there rests six; then one that I borrowed and seven is eight, from four I cannot but eight from fourteen and there Rests six; then one that I borrowed and one is two from two and there rests nothing; so that I find the Remainder or difference to be 6 l. 6 oz. 8 p.w. 12 gr. 7. It many times happeneth that you have many Sums or Numbers to be Subtracted from one number; as suppose a man should lend his friend a certain Sum of Money, and his friend had paid him part of his debt at several times, then before you can conveniently know what is still owing you are to add the several Numbers or Sums of Payment together, and Subtract their Sum from the whole Debt, and the Remainder is the Sum due to the Creditor, as Suppose A dareth to B 564 l. 13 s. 10 d. and B hath Repaid him 79 l. 16 s. 08 d. at one time, and 163 l. 18 s. 11 d. at another time, and 241 l. 15 s. 08 d. at another time; and you would know how the Account standeth between them, or what more is due to A. In order whereunto I first set down the Sum which A lent, and draw a line underneath it, then under that line set the several Sums of payment as you see in the margin; and having brought the several l. s. d. Lent 564 13 10 paid at several paym. 79 16 08 163 18 11 241 15 08 paid in all 485 11 03 Remaines 79 02 07 Sums of payment into one Total by the 5th. Rule of the 4th. Chapter foregoing, I find their Sum amounteth to 485 l. 11 s. 3 d. which I Subtract from the sum first lent by A by the 6th Rule of this Chapter, and I find the Remainder to be 79 l. 02 s. 07 d. And so much is still due to A. When the Learner hath good knowledge of what hath been already delivered, in this and the foregoing Chapter, he will with ease understand the manner of working the following Examples. Subtraction of Money. l. s. d. l. s. d. qrs. Borrowed 374 10 03 700 10 11 2 Paid 79 15 11 9 03 11 3 Remaines 294 14 04 691 06 11 3 l. s. d. l. s. d. qrs. Borrowed 1000 00 00 711 03 00 0 Paid 19 00 06 11 13 00 1 Rem. due 980 19 06 699 09 11 3 l. s. d. qrs. Borrowed 3300 00 00 0 Paid at several payments. 170 10 00 0 361 13 10 1 590 03 04 3 73 04 11 3 Paid in all 1195 12 02 3 Remain due 2104 07 09 1 Subtraction of Troy weight. l. oz. p.w. gr. Bought 174 00 13 00 Sold 78 04 16 15 Remaines 95 07 16 09 l. oz. p.w. gr. Bought 470 10 13 00 Sold at several Times 60 00 00 00 35 10 18 00 16 07 09 08 48 04 00 00 61 11 19 23 23 00 00 00 Sold in all 245 10 07 07 Rem. unsold 225 00 05 17 Subtraction of Apothecary's weight. l. oz. dr. scr. gr. l. oz. dr. scr. gr. Bought 12 04 3 0 00 20 00 1 0 07 Sold 8 05 1 1 15 10 00 1 2 12 Remains 3 11 1 1 05 9 11 7 0 15 Substraction of Averdupois weight. C. qrs. l. Tun C: qrs. l. oz. dr. Bought 35 0 15 5 07 1 10 10 05 Sold 16 1 20 3 17 1 16 09 13 Rem. 18 2 23 1 09 3 22 00 08 Subtraction of Liquid Measure. Tun hhd gall. Tun hhd gall. pts. Bought 40 1 30 60 3 42 4 Sold 16 1 40 15 3 46 6 Remains 23 3 53 44 3 58 6 Substraction of Dry Measures. Chal. qrs. bush. pec. Ch. qrs. bush. pec. Bought 100 0 00 0 73 2 3 2 Sold 54 1 04 3 46 2 3 3 Remains 45 2 03 1 26 3 7 3 Substraction of Long Measure. Yards qrs. Nails Yards qrs. Nails. Bought 160 1 0 344 0 1 Sold 64 1 2 177 1 3 Remains 95 3 2 166 2 2 Subtraction of Land Measure. Acres Rood perch. Acre Rood perch. Bought 140 2 13 600 0 00 Sold 70 3 22 54 0 16 Remains 69 2 31 545 3 24 The Proof of Subtraction. 8. When your Subtraction is ended, if you desire to prove your work, whether it be true or no, then add the remainder to the minor number, and if the aggregate of these two be equal to the major number, than was your operation true, otherwise false; thus let us prove the first Example of the fifth rule of this Chapter, where after Subtraction is ended the numbers stand as in the Margin, the remainder or difference being 283676 now to prove the work, I add the said remainder 283676 to the minor number 153827, by the fourth Rule of the foregoing Chapter, and I find the sum or aggregate to be 437503 equal to the major number, or number from whence the lesser is Subtracted, behold the work in the Margin. The proof of another Example, may be of the first Example, of the sixth Rule of this Chapter where it is required to Subtract 57 l. 16 s. 03 d. 2 qrs. from 375 l. 13 s. 07 d. qrs. and by the Rule I find the Remainder to be 317 l. 17 s. 03 d. 3 qrs. now to prove it, I add the said Remainder 317 l. 17 s. l. s. d. qrs 375 13 07 1 57 16 03 2 317 17 03 3 375 13 07 1 03 d. 3 qrs. to the minor number 57 l. 16 s. 03 d. 2 qrs. and their sum is 375 l. 13 s. 07 d. 1 qrs. equal to the major number which proves the work to be true, but if it had happened to have been either more or less than the said major number, than the operation had been false. 9 The general effect of Subtraction is to find the difference or excess between two numbers, and the Rest of a payment made of a greater sum, the date of Books printed, the age of any thing by knowing the present year, and the year wherein they were made, created or built; and such like. The Questions appropriated to this Rule, are such as follow. Quest. 1. What difference is there between one thing of 125 foot long and another of 66 foot long? To resolve this Question, I first set down the major or greater number 125 and under it the minor or lesser number 66, as is directed in the third rule of this Chapter, and according to the fourth Rule of the same, I Subtract the minor from the major, and the Remainder, excess, or difference I find to be 59, see the work in the Margin. Quest. 2. A Gentleman oweth a Merchant 365 l. whereof he hath paid 278 l. what more doth he owe? To give an Answer to this Question, I first set down the major number, 365 l. and under it I place 278 the minor, and subtract the one from the other, and thereby I discover the excess, difference, or remainder to be 87, and so much is still due to the Creditor. Quest. 3. An obligation was written, book printed, a child born, a Church built, or any other thing made, in the year of our Lord 1572, and now we Account the year of our Lord 1677, The Question is to know the age of the said things, that is how many years are passed since the said things were made; I say if you subtract the lesser number 1572, from the greater 1677, the Remainder will be 105▪ and so many years past are since the making of the said th●ngs. Quest. 4. There are 3 Towns lie in a straight line (viz.) London, Huntingdon, and York, now the distance between the f●rthest of these towns, viz. London and York is 151 miles, and from London to Huntingdon, is 49 miles, I demand how far it is from Huntingdon to York. To Resolve this Question, subtract 49 the distance between London and Huntingdon, from 151 the distance between London and York, and the remainder is 102, for the true distance between Huntingdon and York. CHAP. VI Of Multiplication of whole Numbers. 1. Multiplication is performed by two numbers, of like or unlike kind, for the production of a third, which shall have such reason to the one, as the other hath to unite, and in effect is a most brief and artificial compound Addition of many equal numbers of like kind into one sum. Or Multiplication is that by which we Multiply two or more numbers, the one into the other, to the end that their product may come forth, or be discovered. Or Multiplication is the increasing of any one number by another, so often as there are unites in that number, by which the other is increased, or by having two numbers given to find a third, which shall contain one of the Numbers as many times as there are unites in the other. 2. Multiplication hath three parts, first the Multiplicand, or number to be Multiplied, Secondly, the Multipliar, or number given, by which the multiplicand is to be multiplied, and Thirdly, the product or number produced by the other two, the one being multiplied by the other, as if 8 were given to be multiplied by 4, I say 4 times 8 is 32, here 8 is the multiplicand, and 4 is the multipliar, and 32 is the product. 3. Multiplication is either single by one figure, or compound that consists of many. Single multiplication is said to consist of one figure, because the multiplicand and multipliar consist each of them of a digit, and no more, so that the greatest product that can arise by single multiplication is 81, being the square of 9; and Compound multiplication is said to consist of many figures, because the multiplicand or multipliar consist of more places than one; as if I were to multiply 436 by 6, it is called compound, because the multiplicand 436 is of more places than one, (viz) 3 places. 4. The Learner ought to have all the varieties of single multiplication by heart before he can well proceed any further in this Art, it being of most excellent use, and none of the following Rules in Arithmetic but what have their principal dependence thereupon, and they may be learned by the following Table. Multiplication Table. 1 2 3 4 5 6 7 8 9 2 4 6 8 10 12 14 16 18 3 6 9 12 15 18 21 24 27 4 8 12 16 20 24 28 32 36 5 10 15 20 25 30 35 40 45 6 12 18 24 30 36 42 48 54 7 14 21 28 35 42 49 56 63 8 16 24 32 40 48 56 64 72 9 18 27 36 45 54 63 72 81 The use of the precedent Table is this, in the uppermost line or Collume you have expressed all the digits from 1 to 9, and likewise beginning at 1 and going downwards in the side-colume you have the same; so that if you would know the product of any two single numbers multiplied by one another, look for one of them (which you please) in the uppermost Collume, and for the other in the side Collume, and running your eye from each figure along the respective Collumes, in the common Angle (or place) where these two Collumes meet there is the product required. As for Example, I would know how much is 8 times 7, first I look for 8 in the uppermost Collume, and 7 in the side Collume, then do I cast my eye from 8 along the Collume downwards from the same, and likewise from 7 in the side Collume, I cast my eye from thence towards the right hand, and find it to meet with the first Collume at 56, so that I conclude 56 to be the product required, it would have been the same if you had looked for 7 in the top, and 8 on the side, the like is to be understood of any other such numbers. The learner being perfect herein, it will be necessary to proceed. 5. In Compound Multiplication, if the Multiplicand consists of many places, and the multipliar of but one figure; first set down the multiplicand, and under it place the multipliar in the place of unites and draw a line underneath them, then begin and multiply the multipliar into every particular figure of the multiplicand, beginning at the place of units, and so proceed towards the left hand, setting each particular product under the line, in order as you proceed, but if any of the products exceed 10 or any number of ten, set down the excess, and for every 10 carry a unite to be added to the next product, always remembering to set down the Total product of the last figure; which work being finished, the sum or number placed under the line shall be the true and total product required. As for example, I would multiply 478 by 6, first I set down 478, and underneath it 6 in the place of units, and draw a line underneath them as in the Margin, than I begin saying 6 times 8 is 48, which is 8 above four ten, therefore I set down 8 (the excess) and bear 4 in mind for the four ten, than I proceed saying 6 times 7 is 42 and 4 that I carried is 46, I then set down 6 and carry 4, and go on saying 6 times 4 is 24, and 4 that I carried is 28, and because it is the last figure, I set it all down, and so the work is finished, and the product is found to be 2868, as was required. 6. When in Compound Multiplication the Multipliar consisteth of divers places, then begin with the figure in the place of units in the Multipliar, and Multiply it into all the figures of the Multiplicand, placing the product below the line as was directed in the last Example, then begin with the figure in the second place of the Multipliar (viz.) the place of ten, and Multiply it likewise into the whole Multiplicand (as you did the first figure) placing its product, under the product of the first figure, do in the same manner by the third, fourth, and ●●fth, etc. until you have Multiplied all the figures of the Multipliar particularly into the whole multiplicand, still placing the product of each particular figure under the product of its precedent figure; herein observing the following Cau●ion. In the placing of the product of each particular figure of the Multipliar, A Caution. you are not to follow the 2 Rule of the fourth Chapter, viz. ●ot to place units under units, and ten under ten, etc. but to put the figure or cipher in the place of Units of the second line under the second figure or place of Ten in the line above it, and the figure or cipher in the place of units of the third line under the place of Ten in the second line, etc. Observing this order till you have finished the work, viz. still placing the first figure of every line or product under the second figure or place of Ten in that which is above it, and having so done, draw a line under all these particular products, and add them together; so shall the sum of all these products be the total product Required. As if it were Required to Multiply 764 by 27, I set them down the one under the other with a line drawn underneath them; then I begin saying seven times four is 28, than I set down 8 and Carry 2, then say 7 times 6 is 42 and 2 that I carried is 44, that is 4 and go 4; then 7 times 7 is 49, and 4 that I carry is 53, which I set down because I have not another figure to Multiply; Thus have I done with the 7; then I begin with the 2 saying 2 times 4 is 8, which I set down under the 4 the second figure or place of ten in the line above it, as you may see in the margin; Then I proceed, saying 2 times 6 is 12 that is 2 and carry one, than two times seven is fourteen, and one that I carry is fifteen, which I set down because 'tis the product of the last figure; so that the product of 764 by 7 is 5348, and by 2 is ●528 which being placed the one under the other as before is directed, and as you see in the margin, and a line drawn under them, and they added together Respectively, make 20628 the true product Required, being equal to 27 times 764. Another Example may be this; Let it be Required to Multiply 5486 by 465, I dispose of the Multiplicand and Multiplier according to Rule, and begin Multiplying the first figure of the Multiplier, which is five, into the whole Multiplycand, and the product is 27430; then I proceed and Multiply the second figure (6) of the Multiplier into the Multiplicand and find the product to amount to 32916 which is subscribed under the other product Respectiuly, then do I Multiply the third and last figure (4) of the Multiplier into the Multiplicand, and the product is 21944, which is likewise placed under the second line Respectively; then I draw a line under the said products (being placed the one under the other according to Rule) and add them together and the sum is 2550990 the true product sought being equal to 5486 times 465, or 465 times 5486. More Examples in this Rule are these following. 7. Although the foormer Rules are sufficient for all Cases in multiplication, yet because in the work of multiplication many times great labour may be saved, Si e numeris propositis unus vel uterque adjunctos habeat ad dextram circulos; omissis circulis siat ipsorum numerorum multiplicatio, & facto demu●r tot in●uper integrorum loci accenseanturquot sunt omissi circuli in utroque factore. Cl●vis Math: c. 4.3. , I shall acquaint the Learner therewith, viz. If the multiplicand or multiplier, or both of them end with Ciphers, then in your multiplying you may neglect the Ciphers, and multiply only the significant figures; and to the product of those significant figures; add so many Ciphers as the Numbers given to be multiplied did end with; that is, annex them on the Right hand of the said product, so shall that give you the true product Required. As if I were to multiply 32000 by 4300, I set them down in order to be multiplied as you see in the margin, but neglecting the Ciphers in both numbers, I only multiply 32 by 43 and the product I find to be 1376, to which I annex the 5 Ciphers that are in the multiplicand and multiplier and than it makes 137600000 for the true product of 32000 by 4300. 8. If in the multiplier, Ciphers are placed between significant figures, then multiply only by the significant figures neglecting the cyphers, Si intermedio multiplicantis loco circulus suerit, ille negl gitu. Alsted: Cap. 6. de Arithm. but here special notice is to be taken of the true placing of the first figure after the neglect of such cipher or Ciphers, and therefore you must observe in what place of the multiplier the figure you multiply by standeth, and set the first figure of that product under the same place of the product of the first figure of your multiplier; As for Example, let it be Required to multiply 371568 by 40007, first I multiply the multiplicand by seven and the product is 2600976, then neglecting the Ciphers, I multiply by 4 and that product is 1486272 now I consider that four is the fifth figure in the multiplier, therefore I place two (the first figure of the Product by four) under the fifth place of the first Product by seven, and the rest in order, and having added them together, the total product is found to be 14865320976. other Examples in this Rule are these following. 9 If you are to multiply any Number by a unit with Ciphers, (viz.) by 10, 100▪ 1000, etc. Then prefix so many Ciphers before the multiplicand, and that Numbe● when the Ciphers are prefixed is the Pro●duct Required; as if you would multiply 428 by 100, annex two Ciphers to 428 and it is 42800; If it were Required t● multiply 102 by 10000, annex 4 Cypher● and it gives 1020000 for the Product Required. The Proof of Multiplication. 10. Multiplication is Proved by Division and to speak truth all other ways are false; and therefore it will be most convenient in the first place, Neque est quod aliam expectes examinandi viam; nam aliae vulgares & falsae sunt, & nullo inn●xae fundamento. Gem. ●ris. to learn Division and by that to prove Multiplication. There is a way (at this day generally used in Schools) to Prove multiplication, which is this, first add all the Figures in the multiplicand together, as if they were simple Numbers, casting away the Nine as often as it comes to so much, and noting the Remainder at last, which in this case cannot be so much as 9, Cast likewise the Nine out of the multiplier as you did out of the multiplicand, and note that Remainder; then multiply the Remainders the one by the other, and cast the Nine out of that Product, observing the Remainder, and lastly, Cast the Nine out of the total product, and if this Remainder be equal to the Remainder last found, than they conclude the work to be Rightly performed; but there may be given a thousand (nay infinite) false Products in a multiplication, which after this manner may be Proved to be true, and therefore this way of Proving doth not deserve any Example; but we shall defer the Proof of this Rule till we come to Prove Division, and then we shall Prove them both together. 11. The general effect of Multiplication is contained in the definition of the same which is to find out a third Number, so often containing one of the two given Numbers as the other containeth unit. The second effect is by having the length and breadth of any thing (as a pararellogram, or long plain) to find the superficial content of the same, and by having the superficial content of the base and the length, to find the solidity of any parallelepipedon, Cylinder are other solid figures. The third Effect is by the contents price, value, buying, selling, expense, wages, exchange, simple Interest, gain, or loss of any one thing, be it Money, Merchandise, etc. to find out the value, price, expense, buying, selling, exchange, or Interest of any Number of things, of like Name, Nature and Kind. The fourth Effect is (not much unlike the other (by the Contents, Value, or price of one part of any thing Denominated, to find out the Content, Value or Price of the whole thing, all the parts into which the whole is divided, multiplying the price of one of those parts. The Fifth Effect is, to aid, to compound and to ma●e 〈◊〉 Rules, as chiefly the Rule of 〈◊〉, called th● Golden Rule▪ 〈…〉 of three; also by 〈◊〉 things of 〈…〉 are reduced to another. 〈…〉 any Number of Integers by the price of the Integer, the Product will discover the price of the Quantity, or Number of Int●gers given. In a Rectangular Solid, if you multiply the breadth of the base by the depth, and that Product by the length, this last Product will discover the Solidity or content of the same Solid. Some Questions proper to this Rule may be these following. Quest. 1. What is the content of a square piece of ground, whose length is 28 perches, and breadth 13 perches. Answer, 364 square perches, for multiplying 28 the length, by 13 the breadth, the Product is so much. Quest. 2. There is a square battle whose Flank is 47 men, and the files 19 deep, what Number of men doth that battle contain? Facit 893, for multiplying 47 by 19, the Product is 893. Quest. 3. If any one thing cost 4 shillings, what shall 9 such things cost? Answer, 36 shillings; for multiplying 4 by 9, the Product is 36. Quest. 4. If a piece of Money or Merchandise be worth or cost 7 shillings, what shall 19 such pieces of Money or Merchandise cost? Facit 133 shillings, which is equal to 6 l. 13 s. Quest. 5. If a Soldier or Servant get or spend 14 s. per month, what is the Wages or Charges of 49 Soldiers or Servants for the same time? multiply 49 by 14 the Product is 686 s. for the Answer. Quest. 6. If in a day there are 24 hours, how many hours are there in a year, accounting 365 days to constitute the year? Facit 8760 hours to which if you add the 6 hours over and above 365 days as there is in a year, than it will be 8766 hours, now if you multiply this 8766 by 60 the Number of Minutes in an hour, it will produce 525960, for the Number of Minute's in a Year. CHAP. VII. Of Division of whole Numbers. 1. DIVISION is the Separation, or Parting of any Number, or Quantity given, into any parts assigned; Or to find how often one Number is Contained in another; Or from any two Numbers given to find a third that shall consist of so many Units, as the one of those two given Numbers is Comprehended or contained in the other. 2. Division hath three Parts, or Numbers Remarkable, viz. First the Dividend, Secondly the Divisor, and Thirdly the Quotient. The Dividend is the Number given to be Parted or Divided. The Divisor is the Number given, by which the Dividend is divided; Or it is the Number which showeth how many parts the Dividend is to be divided into. And the Quotient is the Number Produced by the Division of the two given Numbers the one by the other. So 12 being given to be divided by 3, or into three equal parts, the Quotient will be 4 for 3 is contained in 12 four times, where 12 is the Dividend; and 3 is the Divisor, and 4 is the Quotient. 3. In Division set down your Dividend, and draw a Crooked line at each end of it, and before the line at the left hand, place the Divisor, and behind that on the right hand, place the figures of the Quotient, as in the margin, where it is required to divide 12 by 3; First I set down 12 the Dividend, and on each side of it do I draw a crooked line, and before that on the left hand do I place 3 the Divisor; then do I seek how often 3 is contained in 12, and because I find it 4 times, I put 4 behind the Crooked line on the Right hand of the Dividend, denoting the Quotient. 4. But if the Divisor being a single Figure, the Dividend consisteth of two or more places, than (having placed them for the work as is before directed) put a point under the first Figure on the left hand of the Dividend, provided it be bigger then, (or equal to) the Divisor, but if it be lesser than the Divisor, than put a point under the second Figure from the left hand of the Dividend, which Figures as far as the point goeth from the left hand are to be Reckoned by themselves, as if they had no dependence upon the other part of the Dividend, and for distinction sake may be called the Dividual, then ask how often the Divisor is contained in the Dividual, placing the answer in the Quotient; then multiply the Divisor by the Figure that you placed in the Quotient▪ and set the product thereof under the Dividual; then draw a line under that product, and Subtract the said Product from the Dividual, placing the Remainder under the said line, than put a point under the next figure in the Dividend, on the Right hand of that which you put the point before, and draw it down, placing it on the Right hand of the Remainder, which you found by Subtraction; which Remainder with the said Figure annexed before it, shall be a new dividual; then seek again how often the divisor is contained in this new dividual, and put the Answer in the Quotient on the Right hand of the Figure there before, then multiply the divisor by the last Figure that you put in the Quotient, and subscribe the Product under the dividual, and make Subtraction, and to the Remainder draw down the next Figure from the grand dividend, (having first put a point under it) and put it on the right hand of the Remainder for a new dividual as before, etc. Observing this general Rule in all kind of Division, first to seek how often the divisor is contained in the dividual; then (having put the answer in the quotient) multiply the Divisor thereby, and Subtract the Product from the dividual. An Example or two will make the Rule plain. Let it be Required to divide 2184 by 6, I dispose of the Numbers given as is before directed, and as you see in the margin, in order to the work, than (because 6 the divisor is more than 2 the first Figure of the dividend) I put a point under 1 the second Figure, which make the 21 for the Dividual, then do I ask how often 6 the divisor, is contained in 21 and because I cannot have it more than 3 times, I put 3 in the Quotient, and thereby do I multiply the divisor 6 and the product is 18, which I set in order under the dividual, and Subtract it therefrom, and the Remainder (3) I place in order under the line as you see in the Margin. Then do I make a point under the next Figure of the dividend being 8, and draw it down, placing it before the Remainder 3, So have I 38 for a new dividual, then do I seek how often 6 is contained in 38, and because I cannot have more than 6 times, I put 6 in the quotient, and thereby do I multiply the divisor 6, and the product (36) I put under the dividual (38) and Subtract it therefrom, and the remainder 2 I put under the line as you see in the Margin. Then do I put a point under the next (and last) Figure of the dividend (being 4) and draw it down to the remainder 2, and putting it on the Right hand thereof it maketh 24 for a new dividual; then I seek how often 6 is contained in 24, and the Answer is 4, which I put in the quotient, and multiply the divisor (6) thereby and the product (24) I put under the dividual, and Subtract it from it, and the Remainder is 0, and thus the work is finished, and I find the quotient to be 364, that is, 6 is contained in 2184 just 364 times, or 2184 being divided into 6 equal parts, 364 is one of those parts. If it were Required to divide 2646 by 7 or into 7 equal parts, the quotient would be found to be 378, as by the following operation appeareth. So if it were Required to divide 946 by 8, the Quotient will be found to be 118 and 2 〈◊〉 after Division is ended. The work followeth. Many times the dividend cannot exactly be divided by the divisor, but something will Remain, as in the last Example; where 946 was given to be divided by 8, the quotient was 118 and there Remaineth 2 after the division is ended; Now what is to be done in this case with the Remainder, the Learner shall be taught when we come to treat of the Reducing, or Reduction of Fractions. And here Note, that if after your Division is ended, any thing do Remain, it must be lesser than your divisor, for otherwise your work is not Rightly performed. Other Examples are such as follow. 5. But if the divisor consisteth of more place than one, then choose so many Figures from the left side of the dividend for a dividual, as there are Figures in the divisor, and put a point under the farthest Figure of that dividual to the Right hand, and seek how often the first Figure on the Left side of the divisor, is contained in the first Figure on the Left side of the dividual, and place the Answer in the quotient, and thereby multiply your divisor, placing the product under your dividual, and Subtract it therefrom, placing the Remainder below the line; then put a point under the next Figure in the dividend, and draw it down to the said Remainder, and annex it on the Right side thereof, which makes a new dividual, and proceed as before, etc. And if it so happen that after you have chosen your first dividual (as is before directed) you find it to be lesser than the divisor, than put a point under a Figure more, nearer to the Right hand, and seek how often the first Figure on the Left side of the divisor, is contained in the two first Figures on the Left side of the dividual, and place the answer in the quotient, by which multiply the divisor and place the product thereof in order under the dividual, and Subtract it therefrom, and proceed etc. Always remembering, that (in all the cases of division) if after you have multiplied your divisor by the Figure last placed in the quotient, the product be greater than the dividual, than you must cansel that Figure in the quotient, and instead thereof put a Figure lesser by a unit (or one) and multiply the divisor thereby, and if still the product be greater than the dividual, make the Figure in the quotient yet lesser by a unit, and thus do until your product be lesser than the dividual, or at the most equal thereto, and then make Subtraction, etc. So if you would divide 9464 by 24, the quotient will be found to be 394, I first put down the given Numbers as before is directed, now because my divisor consisteth of two Figures, I therefore put a point under the second Figure from the Left hand in my dividend, which here is under four, wherefore I seek how often two the first figure (on the Left side of the divisor) is contained in nine (the like first in the dividual) the answer is four, which I put in the quotient and thereby multiply all the divisor and find the product to be ninety six, which is greater than the dividual ninety four, wherefore I cancel the four in the quotient, and instead thereof I put three (a unit lesser) and by it I multiply the divisor twenty four, and the product is seventy two, which I Subtract from ninety four the dividual, and the Remainder is twenty two, then do I make a point under the next Figure six in the dividend, and draw it down and place it on the Right side of the Remainder twenty two, and it makes 226 for a new dividual, now because the dividual 226 consisteth of a Figure more than the divisor, therefore I seek how often 2 (the first Figure of the divisor) is contained in 22 (the two first of the dividual) I say nine times, wherefore I put nine in the quotient, and thereby multiply the divisor 24, the product 216 I place under the dividual 226, and Subtract it from it, and there Remaineth 10. Then I go on and make a point under the next and last Figure (4) in the dividend, and pull it down to the Remainder 10, and it maketh 104, for a new dividual, which is also a Figure more than the divisor, and therefore I seek how often two is contained in ten, I answer five, but multiplying my divisor by five, the product is 120, which is greater than the divisor, and therefore I make it but four, and by it multiply the divisor, and the product is 96 which being placed under, and Subtracted from the Dividual, there Remaineth 8, and thus the whole work of this Division is ended, and I find that 9464 divided by 24, or into 24 equal parts the Quotient, or one of those equal parts, is found to be 394 as was said before, and there Remaineth 8. Another Example may be this, let there be Required the quotient of 1183653 divided by 385, first I dispose of the Numbers in order to their dividing, and because 118 the three first Figures of the dividend is lesser than the divisor, I therefore make a point under the fourth Figure, which is 3, and seek how often 3 is contained in 11? the answer is (3) which I put in the quotient, and thereby multiply the divisor, and the Product is 1155 which I subtract from the dividual 1188 and there Remaineth 28. Then (as before) I pull down the next Figure, which is six, and place it before the Remainder 28, so have I 286 for a new dividual, and because it hath no more Figures than the divisor I seek how often 3 (the first Figure in the divisor) is contained in 2 (the first Figure of the dividual) and the answer is 0, for a greater Number cannot be contained in a lesser, wherefore I put 0 in the the quotient and thereby (according to Rule) I should multiply my Divisor, but if I do the product will be 0, and 0 Subtracted from the dividual 286 the remainder is the same; wherefore I pull down the next figure (5) from the dividend and put it before the said Remainder 286 so have I 2865 for a new dividual, and because it consisteth of four places (viz.) a place more than the divisor, I seek how often 3 (the first Figure of the divisor) is contained in 28 (the two first of the dividual) and I say there is 9 times 3 in 28, but multiplying my whole divisor (385) thereby I find the product to be 3465, which is greater than the dividual 2865, wherefore I choose eight which is lesser by a Unit than nine, and thereby I multiply my divisor 385, and the product is 3080, which still is greater than the said dividual, wherefore I choose another Number yet a Unit lesser, viz. 7; and having multiplied my divisor thereby the product is 2695 lesser than the dividual 2865, wherefore I put seven in the quotient, and Subtract 2695 from the dividual 2865, and there remains 170, than I pull down the last figure 3 in the dividend and place it before the said Remainder 170, and it makes 1703 for a new dividual, then (for the reason abou-said) I seek how often three is contained in 17, the answer is 5, but multiplying the divisor thereby, the product is 1925 greater than the dividual, wherefore I say it will bear four, a Unit lesser, and by it I multiply the divisor 385, and the product is 1540, which is lesser than the dividual, and therefore I put four in the quotient, and Subtract the said product from the dividual▪ and there Remaineth 163, and thus the work is finished, and I find that 1183653 being divided by 385, or into 385 equal shares or parts, the Quotient or one of those parts is 3074, and besides there is 163 Remaining. And thus the Learner being well versed in the method of the foregoing Examples, he may be sufficiently qualified for the dividing of any greater sum or number into as many parts as he pleaseth, that is, he may understand the method of dividing by a Divisor consisting of 4 or 5 or 6 or any greater number of places, the method being the same with the foregoing Examples in every Respect. Other Examples of Division. So if you divide 47386473 by 58736 you will find the Quotient to be 806, and 45257 will Remain after the work is ended. In like manner if you would Divide 3846739204 by 483064, the Quotient will be 7963 and the Remainder after Division will be 100572. Compendiums in Division. 1. IF any given Number be to be Divided by another Number that hath Ciphers prefixed on the Right side thereof, (omitting the Ciphers) you may cut off so many Figures from the Right hand of the Dividend, Et si Divisor adjunctos sibi habeat Circulos ad dextram, omissis circulis & abseissis totidem ultimis figuris dividendi, in numeris Reliquis fiat divisio in fine autem divisionis Restituendi sunt tum omissi circuli, tum figurae abscissae. Aught. Cla. Math. Cap. 5.3. as there are Ciphers before the Divisor, and let the Remaining numbers in the Dividend, be divided by the Remaining number or numbers in the Divisor; observing this Caution, that if after your Division is ended, any thing Remain; you are to prefix the number or numbers that were cut off from the Dividend before the Numbers Remaining; and such new found Number shall be the Remainder. As for Example; Let it be Required to divide 46658 by 400 now because there are two cyphers before the divisor, I cut off as many Figures from before the Dividend, viz. 58, so that then there will Remain only 466 to be divided by 4, and the Quotient will be 116, and there will Remain 2, before which I prefix the two Figures (58) which were cut off from the Dividend and it makes 258 for the True Remainder, so that I conclude 46658 divided by 400 the Quotient will be 116, and 258 Remaineth after the work is ended; as by the work in the Margin. 2. And hence it followeth that if the divisor be (1) or a unite with Ciphers prefixed, you may cut off so many figures from before the dividend, Divisurus quemcunque numerum per 10, Aufer ex dextera parte unicam, eamque primam figuram; Reliquae enim figurae productum ostendunt; ablatum Residuum, etc. Gem. Fris. Arith. par. 1. as there are Ciphers in the divisor, and then the figure or figures that are on the left hand, will be the Quotient, and those on the right hand will be the Remainder, after the Division is ended: as thus, if 45783 were to be divided by 10, I cut off the last figure (3) with a dash thus (45783) and the work is done, and the quotient is 4578 (the number on the left hand of the dash,) and the Remainder is 3 (on the right hand,) In like manner if the same Number 45783 were to be divided by 100, I cut off 2 figures from the end thus (45783) and the quotient is 457, and the remainder 83. And if I were to divide the same by 1000, I cut off 3 figures from the end thus (45783) and the quotient is 45, and 783 the Remainder, etc. 6. The General effect of Division is contained in the definition of the same, (th●● is) by having two unequal numbers given to find a third number in such proportion to the dividend, as the divisor hath to unite, or 1, It also discovers what reason, or proportion there is between numbers, so if you divide 12 by 4 it quotes 3, which shows the reason, or proportion of 4 to 12 is triple. The second effect is by the superficial measure or content, and the length of any oblonge, Rectangular paralelogram, or square plane known, to find out the breadth thereby; or chose by having the superficies, and breadth of the said figure, to find out the length thereof. Also by having the solidity and length of a solid to find the superficies of the base, & Contra. The third effect is, by the contents, Reason, price, value, buying, selling, expenses, wages, exchange, interest, profit, or loss of any things (be it Money, Merchandise, or what else) to find out the contents, reason, price, value, buying, selling expense, wages, exchange, interest, profit, or loss, of any one thing of like kind. The fourth effect is to aid, to compose, and to make other Rules, but principally the Rule of proportion, called the Golden Rule, or Rule of three; and the Reduction of Moneys, weight and measure, of one denomination into another, by ●t also fractions are abbreviated by finding a common measurer, unto the numerator and denominator, thereby discovering commensurable numbers. If you Divide the value of any certain quantity by the same quantity, the Quotient 〈◊〉 the Rate or value of ●he Integer, as 〈◊〉 eight yards of Cloth cost 96 shillings, here if you divide (96) the value, or price of the given quantity by (8) the same quantity, the Quotient will be 12 s, which is the value or price of 1 of those yards, & contra. If you divide the value, or price of any unknown quantity, by the value of the Integer, it gives you in the quotient that unknown quantity, whose price is thus divided; as if 12 shillings were the value of 1 yard, I would know how many yards are worth 96 shillings, here if you divide (96) the price or value of the unknown quantity, by (12) the rate of the Integer, or one yard, the quotient will be 8, which is the number of yards worth 96 shillings. Some Questions answered by Division may be these following. Quest. 1. If 22 things cost 66 shillings, what will 1 such like thing cost? facit 3 shillings, for if you divide 66 by 22 the Quotient is 3 for the Answer; so if 36 yards or els of any thing be bought or sold for 108 l. how much shall 1 yard or ell be bought or sold for? facit 3 l. for if you divide 108 l. by 36 yards the Quotient will be 3 l. the price of the Integer. Quest. 2. If the expense, charges, or wages of 7 years' amounts to 868 l. what is the expense, charges, or wages of one year? facit 124 l. for if you divide 868 (the wages of 7 years,) by 7 (the number of years) the Quotient will be 124 l. for the Answer, see the work. Quest. 3. If the content of a superficial foot be 144 Inches, and the breadth of a board be 9 Inches, how many Inches of that board in length will make such a foot? facit 16 Inches; for by dividing 144 (the number of square Inches in a square foot,) by 9 (the Inches in the breadth of the board) the Quotient is 16 for the number of Inches in length of that board, to make a superficial foot. Quest. 3. If the content of an Acre of Ground be 160 square Perches, and the length of a furlong (propounded) be 80 Perches, how many Perches will there go in breadth to make an Acre, facit 2 Perches for if you divide 160 (the number of Perches in an Acre) by 80 (the length of the furlong in Perches) the Quotient is 2 Perches; and so many in breadth of that furlong will make an Acre. Quest. 5. If there be 893 men to be made up into a battle, the front consists of 47 men, what Number must there be in the File? Facit 19 deep in the File: For if you divide 893 (the Number of men) by 47 (the number in front) the Quotient will be 19 the file in depth, the work followeth. Quest. 6. There is a Table whose Superficial Content is 72 feet, and the breadth of it at the end is 3 feet, now I demand what is the Length of this Table? Facit 24 feet long; for if you divide 72 (the content of the Table in feet) by 3 (the breadth of it) the Quotient is 24 feet for the length thereof which was Required. See the operation as followeth. The proof of Multiplication and Division. Multiplication and Division Interchangeably prove each other; for if you would prove a sum in Division, whether the operation be Right or no, Multiply the Quotient by the Divisor; and if any thing Remain after the Division was ended, add it to the Product, which Product (if your sum was Rightly divided) will be equal to the Dividend; And chose if you would prove a sum in Multiplication, divide the Product by the Multipliar, and if the work was Rightly performed, the Quotient will be equal to the Multiplicand. See the Example where the work is done and undone; Let 7654 be given to be Multiplied by 3242, the product will be 24814268 as by the work appeareth. And than if you Divide the said Product 24814268 by 3242 the Multipliar, the Quotient will be 7654 equal to the given Multiplicand. In like manner (to prove a Sum or Number in Division) If 24814268 were Divided by 3242 the Quotient would be found to be 7654; then for proof, if you Multiply 7654 the Quotient by 3242 the Divisor, the Product will amount to 24814268, equal to the Dividend. Or you may prove the last or any other Example in Multiplication thus, viz. Divide the Product by the Multiplicand, and the Quotient will be equal to the Multipliar see the work. From whence ariseth this Corollary, that any operation in Division may be proved by Division; for if after your Division is ended, you divide the Dividend by the Quotient, the new Quotient thence ariseing will be equal to the Divisor of the first operation; for Trial whereof let the last Example be again Repeated. For proof whereof divide again 24814268 by the Quotient 7654, and the Quotient thence will be equal to the first Divisor 3242 see the work. But in proving Division by Division, the Learner is to observe this following Caution, that if after his Division is ended there be any Remainder, before you go about to prove your work, Subtract that Remainder out of your Dividend, and then work as before, as in the following Example, where it is Required to divide 43876 by 765, the Quotient here is 57 and the Remainder is 271; See the work following. Now to prove this work Subtract the Remainder 271 out of the Dividend 43876 and there Remaineth 43605 for a new Dividend to be divided by the former Quotient 57, and the Quotient thence arising is 765 equal to the given Divisor, which proveth the operation to be Right. Thus have we gone through the four Species of Arithmetic, viz. Addition, Hae sunt igitur quatuor illae species arithmetices per qua● omnia quaecunque deinceps dicenda sunt vel quae per numeros fier● possbile est, absolvantur. Quare ea●, quisquises, ante omnia perdisces. Gem. Fris. Arith. par. 1. Subtraction, Multiplication, and Division; upon which all the following Rules and all other operations whatsoever that are possible to be wrought by numbers have their Immediate dependence, and by them are Resolved. Therefore before the Learner make a further step in this Art, let him be well acquainted with what hath been delivered in the foregoing Chapters. CHAP. VIII. Of Reduction. 1. REDUCTION, is that which brings together 2 or more numbers of different denominations, into one denomination; or it serveth to change or alter Numbers, Hills Arith. Ch. 13.152. Money, Weight, Measure, or Time, from one Denomination to another; and likewise to abridge fractions to their lowest Terms. All which it doth so precisely, that the first Proportion Remaineth without the least jot of Error or Wrong Committed. So that it belongeth as well to Fractions as Integers, of which in its proper place. Reduction is generally performed, either by Multiplication or Division; from whence we may gather, that 2, Reduction is either Descending or Ascending. 3. Reduction Descending, is when it is Required to Reduce a Sum or Number of a greater Denomination, into a lesser; which Number, when it is so reduced, shall be equal in value to the Number first given in the greater Denomination; as if it were Required to know how many shillings, Wing. Arith. Ch. 7.2, 3, 4. pence or farthings are equal in value to a hundred pounds? or how many ounces are contained in 45 hundred weight; or how many days, hours, or minutes, there are in 240 Years, etc. And this kind of Reduction is generally performed by Multiplication. 4. Reduction Ascending, is when it is Required to Reduce or bring a Sum or Number of a smaller Denomination into a Greater, which shall be equivalent to the given number; As suppose it were Required to find out how many Pence, Shillings or Pounds, are equal in value to 43785 Farthings; or how many Hundreds are equal to (or in) 3748 l. pounds, etc. and this kind of Reduction is always performed by Division. 5. When any Sum or Number is given to be Reduced into another Denomination, you are to consider whether it ought to be Resolved by the Rule Descending or Ascending, viz. by Multiplication or Division; If it be to be performed by Multiplication, consider how many parts of the denomination into which you would Reduce it, are contained in a Vnit or Integer of the given Number, and multiply the said given Number thereby, and the Product thereof will be the Answer to the Question. As if the Question were, in 438 l. pounds how many shillings? here I consider that in one pound are 20 shillings, and that the number of shillings in 38 l. pounds will be 20 times 38 wherefore I multiply 38 l. by 20, and that product is 760, and so many shillings are contained in 38 pounds. But when there is a Denomination, or Denominations between the Number given and the Number Required, you may (if you please) Reduce it into the next inferior Denomination, and then into the next lower than that, etc. until you have brought it into the denomination Required; As for Example, let it be demanded in 132 pounds, how many Farthings? First I multiply 132 (the Number of pounds given) by 20 to bring it into shillings, and it makes 2643 shillings, then do I Multiply the shillings (2640) by 12, to bring them into pence, and it produceth 31680, and so many pence are contained in 2640 shillings, or 132 pounds; then do I Multiply the pence, viz. 31680 by 4 to bring them into Farthings (because 4 farthings is a penny) and I find the product thereof to be 126720, and so many farthings are in equal value to 132 pounds, the work is manifest in the Margin. And if the number propounded to be Reduced, is to be divided, or wrought by the Rule Ascending; consider how many of the given numbers are equal to an Unit or Integer in that denomination to which you would reduce your given number, and make that your divisor, and the given number your dividend; and the Quotient thence arising will be the number sought or required; As for Example, Let it be Required to Reduce 2640 shillings into pounds'; here I consider that 20 shillings are equal to one pound, wherefore I divide 2640 (the given number) by 20, and the Quotient is 132 and so many pounds are contained in 2640 shillings. In Reduction descending and ascending the Learner is advised to take particular notice of the Tables delivered in the second Chapter of this Book, where he may be informed what Multipliars, or Divisors to make use of in the reducing of any number to any other denomination whatsoever, Especially English Moneys, Weights, Measures, Time and Motion, but in this place it is not convenient to meddle with Foreign Coins, Weights or Measures. But if in Reduction Ascending it happen that there is a denomination, or denominations between the number given and the number Required, than you may Reduce your number given into the next Superior denomination, and when it is so Reduced, bring it into the next above that, and so on until you have brought it into the Denomination Required. As for Example, Let it be demanded in 126720 farthings, how many pounds? First I divide my given number (being farthings) by 4, to bring them into pence, (because 4 farthings make one penny) and they are 31680 pence, than I divide 31680 pence by 12, and the Quotient giveth 2640 shillings, and then I divide 2640 shillings by 20, and the Quotient giveth 132 pounds, which are equal in value to 126720 farthings. See the whole work as it followeth. 6. When the number given to be Reduced, consisteth of divers denominations, as pounds, shillings, pence, and farthings; or of hundreds, quarters, pounds, and ounces, etc. then you are to Reduce the highest (or greatest) denomination into the next Inferior, and add thereunto the number standing in that denomination which your greatest or highest number is Reduced to; then Reduce that Sum into the next Inferior denomination, adding thereto the number standing in that denomination; do so until you have brought the number given into the denomination proposed. As if it were Required to Reduce 48 l. 13 s. 10 d. into pence, first I bring 48 l. into shillings, by multiplying it by 20 and the product is 960 shillings, to which I add the 13 shillings and they make 973, than I multiply 973 by 12, to bring the shillings into pence, and they make 11676 pence, to which I add the 10 pence, and they make 11686 pence for the Answer, see the whole work done 7. If (in Reduction Ascending) after Division is ended, any thing Remain, such Remainder is of the same denomination with the Dividend. Examples in Reduction of Coyn. Quest. 1. In 438 l. how many shillings? Facit 8760 shillings, for Multiplying 438 by twenty, the Product amounteth to so much. Quest. 2. In 467 l. how many pence? First multiply the given number of pounds 467 by 20, to bring it into shillings, and it makes 9340 shillings, then multiply the shillings by 12 and it produceth 112080 pence, thus Or it may be Resolved thus, viz. multiply the given number of pounds 467, by 240 the Number of Pence in a Pound, and the Product is the same, viz. 112080 pence, Quest. 3. In 5673, l. how many Farthings? First Multiply the given Number by 20 to bring it into shillings, and it produceth 113460, then multiply that product by 12 to bring it into pence, and it produceth 1361520; then lastly, Multiply the pence by 4, and it produceth 5446080 farthings. See the operation. Or this Question might have been thus Resolved, viz. Multiply 5673 the given Number of pounds by 960 the Number of farthings in a pound, and it produceth the same Effect. Otherwise thus, First bring the given Number 5673 l. into shillings, and multiply the shillings by 48, the Number of Farthings in a shilling, and the same effect is thereby likewise produced, viz. These various ways of Operating are expressed to Inform the Judgement of the Learner with the Reason of the Rule; more ways may be shown, but these are sufficient even for the meanest capacities. Quest. 4. In 458 l. 16 s. 07 d. 3 qrs. how many farthings? To Resolve this question consider the 6 Rule of this Chapter, and work as you are there directed, and you will find the aforesaid given number to amount to 440479 farthings, viz. This last Question (or any other of this kind, viz. where the number given to be Reduced consisteth of several denominations) may be more concisely Resolved thus viz. when you multiply the pounds by 20 to bring them into shillings, to the product of the first Figure, add the Figure standing in the place of Units in the denomination of shillings, but because the first Figure in the Multipliar is (0) I say 0 times 8 is nothing, but 6 is 6, which I put down for the first Figure in the product, then because this Multipliar is 0, I go on no further with it, for if I should, the whole product would be 0, but proceed, and when I come to multiply by the second Figure in the multipliar, to the Product of it, and the first Figure of the Multiplicand, I add the Figure standing in the place of ten in the denomination of shillings, which is (1) saying 2 times 8 is 16 and (the said Figure) 1 is 17, than I set down 7 and carry a Unit to the product of the next Figure as is directed in the 5 Rule of the 6 Chapter foregoing; and finish the work. So that you now have the whole product and sum of shillings at one operation; and when you multiply the shillings by 12 to bring them into pence, after the same manner add to the product, the number standing in the denomination of pence, and so when you multiply the pence by four to bring them into Farthings, add to the product the number standing under the denomination of Farthings, see the last Question thus wrought. After the method last prescribed (which if Rightly considered, differeth not any thing from the sixth Rule of this Chapter) are all the following Examples that are of the same Nature wrought and Resolved. Quest. 5. In 4375866 farthings I demand how many Pounds, Shillings, Pence and Farthings. To Resolve this Question, first I divide the given number of Farthings by 4, and the Quotient is 1093966 pence, and there Remaineth 2 after the Division is ended, which (by the Seventh Rule foregoing) is two Farthings; then I divide 1093966 pence by 12, and the quotient is 91163 shillings, and there Remaineth 10 after division which by the said 7th Rule is so many Pence, viz. 10 d. then I divide 91163 shillings by 20, and the Quotient is 4558 l. and there Remaineth 3 shillings; so the work is finished, and I find that in 4375866 farthings there are 4558 l. 03 s. 10 d. 2 qrs. See the Operation. Quest. 6. In 4386 l. I demand how many Groats? To Resolve this Question, I Reduce the given number of pounds into shillings, and they are 87720 shillings, now I consider that in a shilling are 3 Groats, therefore I multiply the shillings by 3, and it produceth 263160 Groats; See the work. This Question might have been otherways Resolved thus, viz. considering that in a pound (or 20 shillings) there are three times 20 Groats which makes 60, by which I multiply the number of pounds given, and it produceth the same effect at one operation as followeth. Quest. 7. In 43758 three pences I desire to know many pounds? To Resolve this (many such like) Question, first I divide my given number of 3 pences by 4 because 4 three pences are in a shilling, and the Quotient is 10939 shillings; and there Remaineth 2 after division is ended, which is 2 three pences (by the seventh Rule of this Chapter) which are equal in value to 6 d. then I divide 10939 shillings by 20, and the quote giveth 546 l. and 19 shillings Remain; so that I conclude in 43758 pieces of three pence per piece, there are 546 l. 19 s. 06 d. as by the work appeareth. This question might have been otherwise This question might have been otherwise Resolved thus, viz. first multiply the given number of three pences 43758, by 3 the number of pence in three pence, and the product viz. 131274 is the number of pence equal to the given number of three pences, which number of pence may be brought into pounds by dividing by 12 and by 20, and the quotient you will find to be equal to the former work, viz. 546 l. 19 s. 06 d. Or thus, divide the given number of 3 pences by the number of 3 pences in 1 l. or 20 shillings (which you will find to be 80 if you multiply 20 s. by 4, the number of three pences in a shilling) and you will find the quote to be 546 l. as before, and a Remainder of 78 three pences, and if you divide those 78 three pences by 4, you will find the quote to be 19 s. and 2 three pences Remain, which are equal to 6 d. Quest. 8. In 4785 l. 13 s. how many pieces of 13 d. 1/● per piece? This Question cannot be Resolved by Reduction, descending, or ascending, absolutely, (because 13 d. 1/● is no even part of a pound) but rather by them both 〈◊〉 viz. by Multiplication and Division; for if you bring the number given into half pence, and divide the half pence by the half pence in 13 d. ½ viz. 27, the quotient will be the Answer; for having brought 4785 l. 13 s. into half pence, I find it makes 2297112, which I divide by 27, because there are so many half pence in 13 d. ½ and the quote gives 85078 pieces of 13 d. ½ and 6 halfpences Remain over and above, observe the work following. It would have produced the same answer if you had Reduced your given number into farthings, and divided by the farthings in 13 d. ½ viz. 54, (for always the dividend and the Divisor must be of one denomination) and then you would have had a Remainder of 12 farthings which are equal in value to the former Remainder of 6 halfpences. Quest. 9 In 540 Dollars at 4 s. 4 d. per dollar, how many pounds sterling. First bring your given number of Dollars into Pence, and then your pence into Pounds according to the former directions. Thus in 4 s. 4 d. (viz. a dollar) you will find 52 pence, by which multiply 540 dollars, and it produceth 28080 pence, which if you divide by 240 (the pence in one pound) the quotient will give you 117 l. which are equal in value to 540 dollars at 4 s. 4 d. per dollar; observe the operation. The foregoing question might have been otherwise wrought, thus, viz. Multiply your given number of dollars by 13 the number of groats in a dollar, or 4 s. 4 d. and it produceth 7020 groats, which divide by 60 (the groats in one pound or twenty shillings) and the quote is 117 l. as before. Quest. 10. In 547386 pieces of 4 d. ½ per piece, I demand how many Pounds, Shillings, and Pence. First bring your given number of four pence half pentes all into half pence, which you will do if you multiply by 9 the number of halfpences in 4 d. ½, and the product is 4926474 halfpences, which are brought into pounds if you divide them by 24 (the halfpences in a shilling) and 20 (the shillings in a pound, and it makes 10263 l. 09 s. 9 d. as by the work. Quest. 11. In 4386 l. I demand how many pieces of 6 d of 4 d. and of 2 d. of each an equal Number; that is to say, what Number of Six-pences, Groats, and two Pences, will make up 4386 l. and the Number of each equal. The way to Resolve questions of this nature is to add the several pieces (into which the given Number is to be brought) into one sum and to Reduce the given Number into the same denomination with their Sum, and to divide the said given Number (so Reduced) by their Sum, and the Quotient will give you the exact number of each piece. And after the same method will we proceed to Resolve the present Question, viz. So that I conclude by the operation that 87720 six-pences, and 87720 groats, and 87720 twopences are Just as much as, (or equal to) 4386 l. or if you admit of 5 s, to be thus divided, it is equal to 5 six pences, and 5 four pences or groats, and 5 two pences. For if two Right lines (or two Numbers) be given, and one of them be divided into as many Parts, or Segments as you please, the Rectangle, (or Product) comprehended under the two whole Right lines, (or Numbers given) shall be equal to all the Rectangles (or Products) contained under the whole line (or Number) and the several Segments, (or Parts) into which the other line (or Number) is divided, Eucl. 2.1. Another Question of the same Nature with the last may be this following, viz. Quest. 12. A Merchant is desirous to Change 148 l. into pieces of 13 d. ½, of 12 d. of 9 d. of 6 d. and of 4 d.; and he will have of each sort an equal Number of pieces, I desire to know the Number? Do as you were taught in the last question, viz. add the several pieces together, and Reduce it into half pence, then Reduce the sum to be changed, viz. 148 l. into the same denomination, and divide the greater by the lesser, and in the Quotient you will find the Answer, viz. 798 is the Number of each of the pieces Required and ●8 doth Remain, as by the following work appeareth. The truth of the 2 foregoing operations will thus be proved, viz. multiply the Answer by the parts, or pieces into which the given Number was Reduced, and ●aving added the several products together, If the●r sum be equal to the given Number, the Answer is Right, otherwise not. So the Answer to the 11 Question was 87720, which is proved as followeth, viz. l. 87720 Six-pences make 2193 Four-pences make 1462 Twopences make 731 The total Sum of them 4386 which was the Sum given to be Changed. The Answer to the 12th question was 798, and 18 halfpence Remained after the work was ended, now the truth of the work may be proved as the former was, viz. d. l. s. d. 798 Pieces of 13 1/● make 44 17 09 Pieces of 12 make 39 18 00 Pieces of 9 make 29 18 06 Pieces of 6 make 19 19 00 Pieces of 4 make 13 06 00 and 18 halfpences or 9 d. rem. 00 00 09 The total Sum of them 148 00 00 which Total Sum is equal to the Number that was first given to be changed, and therefore the operation was Rightly performed. Reduction of Troy weight. We come now to give the Learner some Examples in Troy weight, wherein we shall be brief, having given so large a Taste of Reduction in the foregoing Examples of Coin, and now the Learner must be mindful of the Table of Troy weight delivered in the second Chapter of this Book. Quest. 13. In 482 l. 07 oz. 13 p.w. 21 gr. how many Grains? Multiply by 12● by 20, and by 24, taking in the Figures standing ●n the several denominations, according to the direction given in the 6 Rule of this Chapter, and you will find the Product to be 2780013 Grains, which is the Number Required, or Answer to the Question. Quest. 14. In 2780013 grains I demand how many pounds, ounces, penny weights, and grains? This is but the foregoing question Inverted, and is Resolved by dividing by 24, by 20, and by 12, and the Answer is 482 l. 07 oz. 13 p.w. 21 gr. Quest. 15. A Merchant sent to a Goldsmith 1● Ingots of S●lver, each containing in weight 2 l. 4 ●z. and ordered it to be made into Bowls of 2 l. 8 oz. per bowl, and Tankards of 1 l. 6 oz. per piece, and Salts of 10 oz. 10 p.w. per Salt, and Spoons of 1 oz. 18 p.w. per Spoon; of each an equal Number I desire to know how many of each sort he must make? This Question is of the same Nature with the 11 and 12 Questions foregoing, and may be answered after the same method viz. First add the weight of the several vessels (into which the silver is to be made) into one sum, and Reduce it to one denomination, and they make 1248 penny weights than Reduce the weight of the Ingot into the same denomination, viz. penny weights, (and it makes 560 penny weights) and multiply them by the Number of Ingots, viz. 16, and the product will give you the weight of the 16 Ingots, viz. 8960, then divide this Product by the weight of the vessels, viz. 1248, and the quotient giveth you the Answer to the question, viz. 7, and 224 p.w. remaining over and above. The Proof of the work is as followeth, Viz. l. oz. p.w. l. oz. p.w. 7 Bowls of 2 08 00 per bowl is 18 08 00 Tank. of 1 06 00 per tank. is 10 06 00 Salts of 0 10 10 per salt is 06 01 10 Spoons of 0 01 18 per spoon is 01 00 06 224 penny weight Remaining is 00 11 04 Total Sum 37 03 00 So that you see the Sum of the weights of each vessel together with the Remainder is 37 l. 03 oz. which is equal to the weight of the 16 Ingots delivered. Reduction of Averdupois Weight. In Reducing Averdupois weight, the Learner must have Recourse to the Table of Averdupois weight delivered in the second Chapter foregoing. Quest. 16. In 47 C. 1 qrs. 20 l. how many ounces? Multiply by 4 by 28 and by 16, and the last product will be the Answer, viz. 84990 ounces. Quest. 17 In 84992 ounces I demand ho● many C. qrs. l. and ounces? This is ●he foregoing Question Inverted, and will be Re●olved if you divide by 16, by 28, and by 4, and the Answer 47 C. 1 qrs. 20 l. equal to the given number in the foregoing question. Quest. 18. A Chapman buyeth of a Grocer 4 C. 1 qrs. 14 l. of Pepper, and ordered it to be made up into parcels of 14 l. of 12 l. of 8 l. of 6 l. and of 2 l. and of each parcel an equal Number now I would know the Number of each parcel. This Example is of the same nature with the 11. and 12, and 15 questions foregoing and after the same manner is Resolved, see the operation as followeth. Reduction of Liquid Measure. Quest. 19 In 45 Tun of wine, how many Gallons. Multiply by 4, and by 63, th● product is 11340 Gallons for the Answer▪ Quest. 20. In 34 Roundlets of wine each containing 18 Gallons, I demand how many Hogsheads? First find how many Gallons is in the 34 Roundlets, which you may do if you multiply 34 by 18 the content of a Roundlet and the Product is 612 Gallons, which you may Reduce into Hogsheads if you divide them by 63, and the Quote will be 9 Hogsheads and 45 Gallons; See the work. Quest. 21. In 12 Tun how many Roundlets of 14 Gallons per Rundlet? Reduce your Tuns into Gallons, and divide them by 14, the Gallons in a Roundlet, and the Quotient 216 is your Answer; See the work following. Reduction of Long Measure. Quest. 22. I demand how many Furlongs, Poles, Inches, and Barley Corns, will Reach from London to York, it being accounted 151 Miles? Quest. 23. The Circumference of the Earth (as all other Circles are) is divided into 360 Degrees, and each degree into 60 Minutes, which (upon the Superficies of the Earth) are equal to 60 miles; now I demand how many miles, Furlongs, perches, yards, feet, and Barley-corns will Reach round the Globe of the Earth? Facit 4105728000 Barley Corns; And so many will reach Round the World, the whole being 21600 Miles, so that if any Person were to go Round, and go 15 Miles every Day, he would go the whole Circumference in 1440 Days, which is 3 Years, 11 Months, and 15 Days. Reduction of Time. Quest. 24. In 28 years 24 weeks 4 days 16 hours 30 minutes, how many Minutes? Note that in Resolving the last question after the method expressed there is lost in every year 30 Hours, for the year consisteth of 365 Days and 6 Hours, but by multiplying the years by 52 weeks, which is but 364 days you lose 1 day and 6 hours every year▪ wherefore to find an exact Answer, bring the odd week's days, and hours into hours, and then multiply the years by the Number of hours in a year, viz. 8766, and to the Product add the hours contained in the odd time, and you have the exact time in hours, which bring into Minutes as before. See the last Question thus Resolved So you see that according to the method first used to Resolve this Question the hours contained in the given time are 248752, but according to the last, best, or true method, they are 249592 which exceeds the former by 840 hours. But for most occasions it will be sufficient to multiply the given years by 365, and to the product add the days in the odd time if there be any, and then there will be only a loss of 6 hours in every year, which may be supplied by taking a fourth part of the given years, and adding it to the contained days, and you have your desire. Quest. 25. In 438657540 Minutes, how many years? Facit 834 years, 4 days 19 hours. Quest. 26. I desire to know how many hours and minutes it is since the birth of our Saviour Jesus Christ, to this present year, being accounted 1677 years? This Question is of the same Nature with the 24th foregoing, and after the same manner is Resolved; viz. Multiply the given Number of years by 8766, the Product is 14700582 hours, and that by 60 and the product is 882034920 minutes; See the work. Note that as Multiplication and Division do Interchangeably prove each other, so Reduction Descending, and Ascending, prove each other by Inverting the Question, as the 13 and 14 and likewise the 16 and 17 Questions foregoing, by Inversion, do Interchangeably prove each other, the like may be performed for the proof of any Question in Reduction whatsoever? Thus far have we discoursed concerning single Arithmetic, whose Nature, and parts are defined in the second, eighth, ninth, and tenth definitions of the third Chapter of this book, for although Reduction is not reckoned or defined among the parts of single Arithmetic, yet considered Abstractly, it is the proper effect of multiplication and division; and as for the extraction of Roots (which ought to be handled in the next place as parts of single Arithmetic) we shall omit, until the Learner is made acquainted with the Doctrine of Decimals, and Immediately enter upon Comparative Arithmetic. CHAP. IX. Of Comparative Arithmetic, viz. The Relation of Numbers one to another. 1. COmparative Arithmetic is that which is wrought by Numbers as they are considered to have Relation one to another, Boetius Arith. lib. 1. cap. 21. and this consists either in quantity, or in quality. 2. Relation of Numbers in Quantity, is the Reference or Respect, that the Numbers themselves have one to another: vide Wing: Arith. ch. 34. where the Terms, or Numbers propounded are always two, the first called the Antecedent and the other the Consequent. 3. The Relation of Numbers in quantity, consists in the differences, or in the Rate or Reason 〈◊〉 is found betwixt the Terms propounded the difference of two Numbers, being the Remainder found by Substraction, Alsted Mathemat. lib. 2. cap. 11. & 12. but the Rate or Reason betwixt two Numbers is the Quotient of the Antecedent divided by the Consequent. So 21 and 7 being given the difference betwixt them will be found to be 14, but the rate or reason that is betwixt 21 and 7 will be found to be Triple Reason, for 21 divided by 7 quotes 3, the reason or rate. 4. The Relation of Numbers in Quality, (otherwise called Proportion) is the Reference or Respect that the Reason of Numbers have one unto another; therefore the Terms given ought to be more than two. Now this Proportion or Reason between Numbers relating one to another, Alsted. Math. lib. 2. cap. 12. is either Arithmetical or Geometrical. 5. Arithmetical proportion (by some called Progression) is when divers Numbers differ one from another by equal Reason, that is, have equal differences. So this Rank of Numbers, 3, 5, 7, 9, 11, 13, 15, 17, differ by equal Reason, viz. by 2 as you may prove. 6. In a Rank of Numbers that differ by Arithmetical proportion, the Sum of the first and last term, being multiplied by half the number of terms, the Product is the total Sum of all the Terms. Or if you multiply the Number of the terms by the half Sum of the first and last terms, the Product thereof will be the total sum of all the Terms. So in the former Progression given, 3 and 17 is 20, which Multiplied by 4 (viz. half the Number of terms) the Product gives 80, the Sum of all the terms, or multiply 8, (the Number of terms) by 10 (half the sum of the first and last terms) the Product gives 80, as before. So also 21, 18, 15, 12, 9, 6, 3, being given, the Sum of all the terms will be found to be 84; for here the Number of terms is 7, and the Sum of the first and last (viz. 21 and 3) is 24, half whereof (viz. 12) multiplied by 7, produceth 84 the sum of the terms sought. 7. Three numbers that differ by Arithmetical proportion, the double of the mean (or middle number) is equal to the sum of the Extremes. So 9, 12 and 15 being given, the double of the mean 12 (viz. 24) is equal to the sum of the Extremes, 9 and 15. 8. Four numbers that differ by Arithmetical proportion (either continued or interrupted) the sum of the two means is equal to the sum of the two Extremes. So 9, vide Wing. Arithm. chap. 35. 12, 18, 21, being given, the sum of 12 and 18 will be equal to the sum of 9 and 21, viz. 30; also 6, 8, 14, 16, being given, the Sum of 8 and 14, is equal to the sum of 6 and 16, viz. 22. etc. 9 Geometrical Proportion (by some called Geometrical Progression) is when divers numbers differ according to like Reason. So 1, 2, 4, 8, 16, 32, 64, etc. differ by double Reason and 3, 9, 27, 81, 243, 729, differ by Triple Reason, 4, 16, 64, 256, etc. differ by quadruple Reason, etc. 10. In any numbers that increase by Geometrical proportion, if you multiply the last term by the Quotient of any one of the terms, divided by another of the terms, which being less, is next unto it, and having deducted, or subtracted the first term out of that product, divide the Remainder by a number that is an unit less than the said Quotient, the last quote will give you the sum of all the Terms. So 1, 2, 4, 8, 16, 32, 64, being given, first I take one of the terms, viz. 8 and divide it by the term which is less and next to it, (viz. by 4) and the Quotient is 2, by which I multiply the last term 64, and the Product is 128, from whence I subtract the first term, (viz. 1) the Remainder is 127 which divided by the quotient 2 made less by 1 (viz. 1.) the quote is 127 for the sum of all the given terms, as by the work in the margin. So if 4, 16, 64, 256, 1024 were given, the Sum of all the terms will be found to be 1364. For first I divide 64 one of the terms by his next lesser term, and the Quotient is 4, by which I multiply the last term 1024 and it produceth 4096; from whence I subtract the first term 4, and the Remainder is 4092, which I divide by the quote less 1 (viz. 3) and the quote is 1364 for the total sum of all the terms, as per margin. So likewise if 2, 6.18, 54, 162, 486, were given the sum or total of all the terms will be found to be 728, see the work. 11. Three Geometrical proportionals given, the square of the mean is equal to the Rectangle, or Product of the Extremes. So 8, 16, 32 being given, the square of the mean, viz. 16 is 256, which is equal to the Product of the Extremes 8 and 32, for 8 times 32 is equal to 256. 12. Of 4 Geometrical proportional numbers given, the Product of the two means, is equal to the Product of the two Extremes. So 8, 16, 32, 64, being given, I say that the Product of the two means, viz. 16 times 32 which is 512 is equal to 8 times 64 the product of the Extremes. Also if 3, 9, 21, 63, were given (which are Interrupted) I say 9 times 21 is equal to 3 times 63. From hence ariseth (that precious Gem in Arithmetic, which for the excellency thereof is called) the Golden Rule, or Rule of 3. CHAP. X. The single Rule of three Direct. 1. THE Rule of Three (not undeservedly called the Golden Rule) is that by which we find out a fourth number, in proportion unto three given numbers, so as this fourth number sought may bear the same Rate, Reason, or Proportion to the third (given) number, as the second doth to the first, from whence it is also called the Rule of Proportion. 2. Four Numbers are said to be Proportional, when the first containeth, or is contained by the second, as often as the third containeth, or is contained by the fourth. Vide Wingates Arith. Chap. 8. Sect. 4. So these Numbers are said to be Proportionals, viz. 3, 6, 9, 18, for as often as the first Number is contained in the second, so often is the third contained in the fourth, viz. twice. Also 9, 3, 15, 5, are said to be Proportional, for as often as the first Number containeth the second, so often the third Number containeth the fourth, viz. 3 times. 3. The Rule of Three is either simple, or composed. 4. The simple (or single) Rule of three, consisteth of 4 Numbers, that is to say, it hath 3 Numbers given to find out a fourth; and this is either Direct, or Inverse. Vide Alsted. Math. lib. 2. Cap. 13. 5. The single Rule of three direct, is when the Proportion of the first term is to the second, as the third is to the fourth; or when it is Required that the Number sought (viz.) the fourth Number must have the same Proportion to the second, as the third hath to the first. 6. In the Rule of three, the greatest difficulty is (after the question is propounded) to discover the order of the 3 terms, viz. which is the first, which is the second, and which the third, which that you may understand observe, That (of the three given Numbers) two are always of one kind, and the other is of the same kind with the Proportional Number that is sought; as in this Question, viz. If 4 yards of Cloth cost 12 shillings, what will 6 yards cost at that Rate? here the two Numbers of one kind are 4 and 6, viz. they both signify so many yards; and 12 shillings is the same kind with the Number sought, for the price of 6 yards is sought. Again observe, that of the 3 given numbers, those two that are of the same kind, one of them must be the first, and the other the third, and that which is of the same kind with the number sought, must be the second number in the Rule of three; and that you may know which of the said numbers to make your first, and which your third, know this, that to one of those two Numbers there is always affixed a demand, and that Number upon which the demand lieth must always be Reckoned the third Number. As in the forementioned Question, the demand is affixed to the Number 6, for it is demanded what 6 yards will cost? and therefore 6 must be the third Number and 4 (which is of the same denomination (or kind) with it) must be the first, and consequently the Number 12, must be the second, and then the Numbers being placed in the forementioned order will stand as followeth, viz. yards s yards 4 12 6 7. In the Rule of three Direct (having placed the Numbers as is before directed, the next thing to be done will be to find out the fourth Number in proportion which that you may do) Multiply the second Number by the third, and divide the Product thereof by the first, or (which is all one) multiply the third term (or Number) by the second and divide the Product thereof by the first, and the Quotient thence arising is the fourth number in a direct proportion, and is the Number sought or Answer to the Question, As thus let the said Question be again Repeated, viz. If 4 yards of Cloth cost 12 shillings, what will 6 yard's cost? Having placed my Numbers according to the 6 Rule (of this Chapter) foregoing, I multiply the second number 12, by the third number 6, and the product is 72, which I divide by the first number 4, and the quotient thence arising is 18, which is the fourth Proportional or number sought, viz. 18 shillings, which is the Price of the 6 yards, as was Required, see the work following. Quest. 2. Another question may be this, viz. If 7 C. of Pepper cost 21 l. how much will 16 C. cost at that Rate? To Resolve which question, I consider that (according to the 6 Rule of this Chapter) the terms or numbers ought to be placed thus, viz. the demand lying upon 16 C. it must be the third number, and that of the same kind with it must be the first, viz. 7 C. and 21 l. (being of the same kind with the number sought) must be the second number in this Question, than I proceed according to this 7 Rule, and multiply the second number by the third, viz. 21 by 16, and the product is 336 which I divide by the first number 7, and the Quotient is 48 l. which is the value of 16 C. of Pepper at the Rate of 21 l. for 7 C. See the work as followeth. 8. If when you have divided the product of the second and third numbers by the first, any thing Remain after Division is ended, such Remainder may be multiplied by the parts of the next Inferior denomination, that are equal to a unit (or Integer) of the second number in the question, and the product thereof divide by the first number in the question, and the quotient is of the same denomination with the Parts by which you multiplied the Remainder, and is part of the fourth number which is sought. And furthermore, if any thing Remain, after this last Division is ended multiply it by the parts of the next inferior denomination equal to a unit of the last quotient, and divide the product by the same divisor, (viz. the first number in the question) and the quote is still of the same denomination with your Multipliar; follow this method unt●ll you have Reduced your Remainder into the lowest denomination, etc. An Example or two will make the Rule very plain which may be this following. Quest. 3. If 13 yards of Velvet (or any other thing) cost 21 l. what will 27 yards of the same cost at that Rate? Having ordered, and wrought, my Numbers according to the 6 and 7 Rules of this chapter, I find the quotient to be 43 l. and there is a Remainder of 8, so that I conclude the price of 27 yards to be more than 43 l. and to the intent that I may know how much more I work according to the foregoing Rule, viz. I multiply the said Remainder 8, by 20 s. (because the second number in the Question was pounds) and the product is 160, which divided by the first number, viz. 13, it quotes 12, which are 12 shillings, and there is yet a Remainder of 4, which I multiply by 12 pence (because the last Quotient was shillings) and the product is 48. which I divide by 13 (the first number) and the quotient is 3 d. and yet there Remaineth 9, which I multiply by 4 farthings, and the product is 36, which divided by 13 again, it quotes 2 farthings, and there is yet a Remainder of 10, which (because it cometh not to the value of a farthing) may be neglected, or Rather set (after the 2 farthings) over the Divisor with a line between them, and then (by the 21 and 22 definitions of the first Chapter of this book) it will be 10/13 of a farthing; So that I conclude, that if 13 yards of Velvet cost 21 l. 27 yds. of the same will cost 43 l. 12 s. 3 d. 2 10/13 qrs. which Fraction is ten thirteenths of a farthing. See the operation as followeth. Quest. 4. Another Example may be this following, viz. If 14 l. of Tobacco cost 27 s. what will 478 l. cost at that rate? Work according to the last Rule, and you will find it to amount to 921 s. 10 d. 1 2/14 qrs. and by the 5 Rule of the 8 Chapter 921 s. may be Reduced to 46 l. 01 s. So that then the whole worth or value of the 478 l. will be 46 l. 01 s. 10 d. 1 2/14 qrs. the whole work followeth. 9 In the Rule of 3 it many times happeneth that although the first and third numbers be Homogeneal (that is, of one kind) as both money, weight, measure, etc. yet they may not be of one denomination, or perhaps they may both consist of many denominations, in which case you are to reduce both numbers to one denomination; and likewise your second number (if it consisteth of divers denominations) must be reduced to the least name mentioned, or lower if you please which being done, multiply second and third together and divide by the first, as is directed in the 7 Rule of this Chapter. And note that always the Answer to the question is in the same denomination that your second number is of, or is Reduced to. Quest. 5. If 15 ounces of Silver be worth 3 l. 15 s. what are 86 ounces worth at that Rate? In this question the numbers being ordered according to the 6 Rule of this Chapter, the first and third numbers are ounces, and the second number is of divers denominations, viz. 3 l. 15 s. which must be Reduced to Shillings, and the shillings multiplied by the third number, and the Product divided by the first gives you the Answer in shillings, viz. 430 s. which are reduced to 21 l. 10 s. See the work. In resolving the last question the work would have been the same if you had reduced your second Number into pence, for then the Answer would have been 5160 pence, equal to 21 l. 10 s. or if you had Reduced the Second number into farthings the Quotient, or Answer, would have been 20640 farthings equal to the same, as you may prove at your leisure. Quest. 6. If 8 l. of Pepper cost 4 s. 8 d. what will 7 C. 3 qrs. 14 l. cost? In this Question the first number is 8 l. and the third is 7 C. 3 qrs. 14 l. which must be reduced to the same denomination with the first, viz. into pounds, and the second number reduce into pence, then Multiply and Divide, according to the 7 Rule foregoing and you will find the answer to be 6174 pence which is reduced into 25 l. 14 s. 6 d. Quest. 7. I● 3 C. 1 qrs. 14 l. of Raisins cost 9 l. 9 s. what will 6 C. 3 qrs. 20 l. of the same cost? Here the first and third numbers each consist of divers denominations, but must be brought both into one denomination etc. as you see in the operation which followeth; the Answer is 388 s. which is Reduced into 19 l. 8 s. Quest. 8. If in 3 weeks 4 days I spend 13 s. 4 d. how long will 53 l. 06 s. last me at that Rate? Answer, 2238 days equal to 6 year● 48 days. See the work. Quest. 9 Suppose the yearly Rent of a house, a yearly pension, or wages be 73 l. I desire to know how much it is per day? Here you are to bring the year into days, and say, If 365 days Require 73 l. what will 1 day Require? Now when you come to multiply 73 by 1 the product is the same, for 1 neither multiplieth nor divideth, and 73 cannot be divided by 365 because the divisor is bigger than the dividend, wherefore bring the 73 l. into shillings, and they make 1460, which divide by the first number 365, and the quote is 4 shillings for the Answer, as you see in the work. Quest. 10. A Merchant bought 14 pieces of broad-cloath, each piece containing 28 yards, for which he gave after the Rate of 13 s. 6 d ½ per yard, now I desire to know how much he gave for the 14 pieces at that Rate. First find out how many yards are in the 14 pieces which you will do if you multiply the 14 pieces by 28 the number of yards in a piece, and it makes 392, then say If 1 yard cost 13 s. 6 d. ½ what will 392 yard's cost? work as followeth and the Answer you will find to be 127400 halfpences, which Reduced make 265 l. 8 s. 4 d. For after you have multiplied your second and third numbers together the product is 127400, which (according to the Rule) should be divided by the first number, but the first number is 1, which neither multiplieth nor divideth, and therefore the quotient, or fourth number is the same with the Product of the second and third, which is in halfpences because the second number was so Reduced. See the work. Quest. 11. A Draper bought 420 yards of broadcloth, and gave for it after the Rate of 14 s· 10 d. ¾ per el English, now I demand how much he paid for the whole at that Rate? Bring your ell into quarters and your yards given into quarters, the ell is 5 quarters and in 420 yards are 1680 quarters, then say, If 5 quarters cost 14 s. 10 d. ¾ or 715 farthings, what will 1680 quarters cost? facit 250 l. 05 s. 00 d. Quest. 12. A Draper bought of a Merchant 50 pieces of Kerseys each piece containing 34 els flemish, (the Ell Flemish being 3 quarters of a yard) to pay after the Rate of 8 s. 4 d. per el English, I demand how much the 50 pieces cost him at that Rate? First find how many els Flemish are in the 50 pieces by multiplying 50 by 34, the product is 1700 which bring into quarters by 3 it makes 5100 quarters, then proceed as in the last question, and the Answer you will find to be 102000 pence or 425 l. Quest. 13. A Goldsmith bought a wedge of Gold which weighed 14 l. 3 oz. 8 p.w. for the sum of 514 l. 4 s. I demand what it stood him in per ounce? Answer 60 shillings or 3 l. See the work. Quest. 14. A Grocer bought 4 hhds of Sugar, each weighing neat 6 C. 2 qrs. 14 l. which cost him 2 l. 8 s. 6 d. per C. I demand the value of the 4 hhds. at that Rate? First find the weight of the 4 hhds, which you may do by Reducing the weight of one of them into pounds, and multiply them by 4 (the number of hhds) and they make 2968 l. then say, If 1 C. or 112 l. cost 2 l. 8 s. 6 d. what will 2968 l. cost? Facit 64 l. 5 s. 3 d, as by the operation. Quest. 15. A Draper bought of a Merchant 8 Packs of Cloth, each pack containing 4 parcels, and each parcel 10 pieces, and in each piece 26 yards, and gave after the Rate of 4 l. 16 s for 6 yards, now I desire to know how much he gave for the whole, Answer 6656 l. First find out how many yards there were in the 8 packs, as by the following work, you will find there are 8320 yards; then say, If 6 yards cost 4 l. 16 s. what will 8320 yard's Cost, etc. By this time the Learner is (surely) well Exercised in the Practic and Theoric of the Rule of 3 Direct, but at his leisure he may look over the following Questions, whose Answers are given, but the operation purposely omitted as a Touchstone for the Learner, thereby to try his ability in what hath been delivered in the former Rules. Quest. 16. If 24 l. of Raisins cost, 6 s. 6 d, what will 18 Frails cost, each weighing Neat 3 qrs. 18 l.? Answer 24 l. 17 s. 03 d. Quest. 17. If an ounce of Silver be worth 5 shillings, what is the price of 14 Ingots, each Ingot weighing 7 l. 5 oz. 10 p.w.? Answer 313 l. 05 s. Quest. 18. If A piece of Cloth cost 10 l. 16 s. 8 d. I demand how many els English there are in the same, when the Ell at that Rate is worth 8 s. 4 d. Answer 26 els English. Quest. 19 A Factor bought 84 pieces of Stuffs, which cost him in all 53 l. 12 s. at 5 s. 4 d. per yards, I demand how many yards there were in all, and how many els English were contained in a piece of the same? Answer 2016 yards in all, and 19 ⅕ els English per piece. Quest. 20. A Draper bought 240 yards of Broadcloth, which cost him in all 254 l. 10 s. for 86 yards of which he gave after the Rate of 21 s. 4 d. per yard, I demand how much he gave per yard for the Remainder? Answer 20 s. 10 64/156 d. per yards. Quest. 21. A Factor bought a certain quantity of Serge, and Shalloon, which together cost him 226 l. 14 s. 10 d. the Quantity of Serge he bought was 48 yds. at 3 s. 4 d. per yd. and for every 2 yds. of Serge he had 5 yards of 〈◊〉▪ I demand how many yards of Shalloon he had, and how much the Shalloon cost him per yard? Answer 120 yards of Shalloon at 1 l. 16 s. 05 58/120 d per yard. Quest. 22. An Oylman bought 3 Tun of Oil, which cost him 151 l. 04 s. and it so chanced that it leaked out 85 gallons, but he is minded to sell it again so as that he may be no loser by it, I demand how he must sell it per gallon? Answer at 4 s. 6 54/671 d. per gallon. Quest 23. Bought 6 packs of cloth, each pack containing 12 clothes, which at 8 s. 4 d. per el Flemish cost 1080 l. I demand how many yards there were in each Cloth? Answer 27 yards in each Cloth. Quest. 24. A Gentleman hath 536 l. per annum, and his expenses are one day with another 18 s. 10 d. 3 qrs. I desire to know how much he layeth up at the years end? Answer 191 l. 03 s. 00 d. 1 qrs. Quest. 25. A Gentleman expendeth daily one day with another 27 s. 10 d. ½ and at the years end layeth up 340 l. I demand how much is his yearly Income? Answer 848 l. 14 s. 4 d. ½. Quest. 26. If I sell 14 yards for 10 l. 10 s. 00 d. how many els Flemish shall I sell for 283 l. 17 s. 06 d. at that Rate? Answer 504 2/3 els Flemish. Quest. 27. If 100 l. in 12 month's gain 6 l. Interest, how much will 75 l. gain in the same time, and at the same Rate? Answer 4 l. 10 s. Quest. 28. If 100 l. in 12 month's gain 60. Interest, how much will it gain in 7 months at that Rate? Answer 3 l. 10 s. Quest. 29. A certain Usurer put out 75 l. for 12 Month, and Received principle and Interest 81 l. I demand what rate per Cent. he Received Interest? Answer, 8 l. per Cent. Quest. 30. A Grocer bought 2 Chests of Sugar the one weighed neat 17 C. 3 qrs. 14 l. at 2 l. 6 s. 8 d. per C. the other weighed neat 18 C. 1 qrs. 21 l. at 4 d ½ per l. which he mingleth together, now I desire to know how much a C. weight of this mixture is worth? Answer 2 l. 4 s. 3 d. 2 530/●●● qrs. Quest. 31. Two men viz. A and B, departed both from one place, the one goes East, the other West, the one travaileth 4 miles a day, the other 5 miles a day, how far are they distant the 9th day after their departure? Answer 91 miles. Quest. 32. A flying every day 40 miles, is pursued the 4th day after by B, Moor. Arithm. Chap. 8. Quest. 7. posting 50 miles a day, now the Question is in how many days, and after how many miles Travail will A be overtaken? Answer, B overtakes him in 12 days, when they have 〈◊〉 600 miles. 11. The General Effect of the Rule of three Direct, is contained in the definition of the same, that is, to find a fourth number in proportion consisting of two equal Reasons, as hath been fully shown in all the foregoing Examples. The second Effect is, by the price or value of one thing to find the price or value of many things of like kind? The third Effect is by the price or value of many things to find the price of 1, or by the price of many things (the said price being 1) to find the price of many things of like kind. The fourth Effect is, by the price or value of many things, to find the price or value of many things of like kind. The fifth Effect is, thereby to Reduce any Number of moneys, weight or measure the one sort into the other, as in the Rules of Reduction contained in the 8 Chapter foregoing Examples of its various Effects have been already Answered. 12. The Rule of 3 Direct is thus proved, viz. Multiply the first Number by the fourth, The proof of the Rule of Three Direct. and note the product, then multiply the second Number by the third, and if this product is equal to the product of the first and fourth, than the work is Rightly performed, otherwise it is Erroneous. So the first Question of this Chapter (whose Answer, or fourth Number we found to be 18 s.) is thus proved, viz. the first Number is 4, which multiplied by 18 (the fourth) produceth 72. And the second and third Numbers are 12 and 6, which multiplied together produce 72, equal to the product of the first and fourth, and therefore I conclude the work to be Rightly performed. Always observing, that if any thing Remain after you have divided the product of the second and third Numbers, by the first, such Remainder in proving the same, must be added to the product of the first and fourth Numbers, whose sum will be equal to the product of the second and third, (the second number being of the same denomination with the fourth, and the first with the same denomination of the third.) So the Fourth Question of this Chapter, being again repeated, viz. If 14 l. of Tobacco cost 27 s. what will 478 l. cost at that Rate? the Answer (or 4th number) was 46 l. 01 s. 10 d. 1 qrs. 2/14 which is thus proved, viz. bring the fourth number into farthings and it makes 44249 which multiplied by the first number 14 produceth 619488 (the 2 which remained being added thereto) then (because I Reduced my fourth number into farthings) I Reduce my second (viz. 27 s.) into farthings and they are 1296 which multiplied by the third number 478, their product is 619488 equal to the product of the first and fourth Numbers. Wherefore I conclude the operation to be true. This is an Infallible way to prove the Rule of 3 Direct, and it is deduced from the twelfth Sect. of the 9th. Chap. of this Book. Thus much concerning the single Rule of 3 Direct, and I Question not but by this time the Learner is sufficiently qualified to Resolve any Question pertinent to this Rule, not Relying upon fractions, or Geometrical magnitudes. Those that are desirous to see the demonstration of this Rule, let them Read the 6th. Chapter of (the Ingenious) Mr. Kerseys Appendix to Wingates Arithmetic. Or the 6th. Chapter of Mr. Oughtreds (Incomparable) Clavis Mathematicae: By both which Authors this Rule is largely demonstrated, being grounded upon the 19 Prop. of the 7th, and the 19 Prop. of the 9th. of Euclids Elem. CHAP. XI. The single Rule of three Inverse. 1. THE Golden Rule or Rule of 3 Inverse is when there are 3 Numbers given to find a 4th. in such Proportion to the 3 given Numbers, so as the 4th. proceeds from the second, according to the same Rate, Reason, or Proportion that the first proceeds from the third, or the Proportion is As the third number is in proportion to the second, Alsted. Math. lib. 2. cap. 14. so is the first to the fourth. So if the 3 numbers given were 8, 12, and 16, and it were Required to find a 4th. number in an inverted proportion to these, I say that as 16 (the third Number) is the double of the first term or number (8) so must 12 (the second number) be the double of the 4th; so will you find the 4th. term or number to be 6. And as in the Rule of 3 Direct, you multiply the second and third together, and divided their product for a 4th Proportional number, So 2. In the Rule of 3 Inverse, you must Multiply the second term by the first (or first term by the second) and divide the product thereof by the 3d. term, so the Quotient will give you the 4th. term sought in an Inverted Proportion. The same order being observed in this Rule as in the Rule of 3 Direct, for placing or disposing of the given numbers, and after your numbers are placed in order, that you may know whether your Question be to be Resolved by the Rule Direct or Inverse, observe the general Rule following. 3. When your Question is stated, and your numbers orderly disposed, Consider in the 1st. place whether the 4th. term or number sought, aught to be more, or less than the second term; which you may easily do; And if it is required to be more, or greater than the second term, than the lesser Extreme must be your divisor, but if it require less, than the biggest Extreme must be your Divisor, (In this Case the 1st and 3d numbers are called Extremes in Respect of the second) and having found out your Divisor, you may know whether your Question belong to the Rule Direct or Inverse, for if the 3d. term be your Divisor, than it is Inverse, but if the first term be your Divisor, than it is a Direct Rule. As In the following Questions. Quest. 1. If 8 Labourers can do a certain piece of work in 12 days in how many days will 16 Labourers do the same? Answer in 6 days. Having placed the numbers according to the 6 Rule of the 10th. Chapter. I consider that if 8 men can finish the work in 12 days, 16 men will do it in lesser (or fewer days than 12) therefore the biggest Extreme must be the Divisor which is 16 and therefore it is the Rule of 3 Inverse, wherefore I multiply the first and second numbers together viz. 8 by 12 and their product is 96, which divided by 16 Quotes 6 days for the Answer, and in so many days will 16 Labourers perform a piece of work when 8 can do it in 12 days. Quest. 2. If when the measure (viz. a peck) of wheat cost 2 shillings the penny Loaf weighed (according to the Standard, Statute, or Law of England) 8 ounces, I demand how much it will weigh when the peck is worth 1 s. 6 d. according to the same Rate or Proportion? Answer 10 oz. 13 p.w. 8 grs. Having placed and reduced the given Numbers according to the 6 and 9 rules of the 10th Chapter I consider that at 1 s. 6 d. per Perk, the penny Loaf will, weigh more than at 2 s. per Perk, for as the price decreaseth, the weight Increaseth, and as the price Increaseth so the weight diminisheth, wherefore because the term Requireth more than the second, the lesser Extreme must be the Divisor, viz. 1 s. 6 d. or 18 pe●●e, and having finished the work I find the Answer to be 10 oz. 13 p.w. 8 gr. and so much will the penny Loaf weigh, when the peck of wheat is worth 1 s. 6 d. according to the given Rate of 8 ounces, when the peck is worth 2 shillings, the work is plain in the following operation. Quest. 3. How many pieces of money or Merchandise at 20 s. per piece, are to be given, or Received for 240 pieces, the value or price of every piece being 12 shillings? Answer 144. For if 12 s. Require 240 pieces, than 20 shillings will Require less; therefore the biggest Extreme must be the Divisor, which is the 3 number, etc. See the work. Quest. 4. How many yards of 3 quarters broad are Required to double, or be equal in measure to 30 yard's, that are 5 quarters broad? Answer 50 yards. For say If 5 quarter wide Require 30 yards long, what length will three quarters broad require? here I consider that 3 quarters broad will Require more yards than 30, for the narrower the cloth is, the more in length will go to make equal measure with a broader piece. Quest. 5. At the Request of a friend I lent him 200 l. for 12 months, promising to do me the like Courtesy at my necessity, but when I came to Request it of him he could let me have but 150 l. now I desire to know how long I may keep this money to make plenary Satisfaction for my former kindness to my Friend? Answer, 16 Months. I say, If 200 l. Require 12 Months, what will 150 l. Require? 150 l. will Require more time than 12 Months, therefore the lesser extreme (viz. 150) must be the Divisor, Multiply and Divide, and you will find the 4th inverted Proportional to be 16, and so many Months I ought to keep the 150 l. for satisfaction. Quest. 6. If for 24 s. I have 1200 l. weight carried 36 M. how many M. shall 800 l. be carried for the same money? Answer, 24 M. Quest. 7. If for 24 s. I have 1200 l. carried 36 Miles how many pound weight shall I have carried 24 miles for the same money? Answer, 800 l. weight. Quest 8. If 100 workmen in 12 days finish a piece of work or service, how many workmen are sufficient to do the same in 3 days? Answer 400 workmen. Quest. 9 A Colonel is besieged in a Town in which are 1000 Soldiers, with provision of Victuals only for 3 Months, the Question is how many of his Sould●ers must he dismiss that his Victuals may last the Remaining Soldiers 6 Months? Answer, 500 he must keep, and dismiss as many. Quest. 10. If wine worth 20 l. is sufficient for the ordinary of 100 men, when the Tun is sold for 30 l. how many men will the same 20 pounds worth suffice when the Tun is worth 24 l. Answer, 125 men. Quest. 11. How much plush is sufficient to line a Cloak which hath in it 4 yards of 7 quarters wide, when the Plush is but 3 quarters wide? Answer 9 ⅓ yds. of Plush. Quest. 12. How many yards of Canvas that is Ell wide will be sufficient to line 20 yards of Say that is 3 quarters wide? Answer, 12 yards. Quest. 13. How many yards of matting that is 2 foot wide, will cover a floor that is 24 foot long, and 20 foot broad, Answer, 240 foot. Quest. 14. A Regiment of Soldiers consisting of 1000 are to have new Coats, and each coat to contain 2 yds. 2 qrs. of Cloth that is 5 qrs. wide, and they are to be lined with Shalloon that is 3 quarters wide, I demand how many yards of Shalloon will line them? Answer, 16666 ⅔ yards? Quest. 15. A Messenger makes a Journey in 24 days when the day is 12 hours long, I desire to know in how many days he will go the same when the day is 16 hours long? Answer, in 18 days. Quest. 16. Borrowed of my friend 64 l. for 8 Months, And he hath occasion another time for to borrow of me for 12 Months, I desire to know how much I must lend to make good his former kindness to me? Answer 42 l. 13 s. 4 d. 4. The General Effect of the Rule of 3 Inverse is contained in the Definition of the same, that is, to find a fourth term in a Reciprocal Proportion, inverted to the Proportion given. The second Effect, is by two prizes or values of two several pieces of money or Merchandise known, to find how many pieces of the one price is to be given for so many of the other. And consequently to Reduce or Exchange one sort of Money, or Merchandise, into another. Or chose to find the price unknown of any piece given to Exchange, in Reciprocal Proportion. The third Effect, is by two differing prizes of a measure of wheat bought or sold, and the weight of the Loaf of bread, made answerable to one of the prizes of the measure given, to find out the weight of the same Loaf, answerable to the other price of the said measure given. Or chose by the two several weights of the same prized Loaf, and the price of the measure of wheat answerable, to one of those weights given, to find out the other price of the measure answerable to the other weight of the same Loaf. The Fourth Effect, is by two lengths and one breadth of two Rectangular planes known, to find out another breadth unknown▪ Or by two breadths and one length given, to find out another length unknown in an inverted Proportion. The fifth Effect, is by double time and a Capital sum of money borrowed or lent, to find out another Capital sum answerable to one of the given times, Or chose, by two Capital sums, and a time answerable to one of them given, to find out a time answerable to the other Capital sum in Reciprocal Reason. The sixth Effect is, by two differing weights of Carriage, and the distance of the places in Miles or in Leagues given, to find another distance in miles answerable to the same price of payment, Or otherwise by two distances in miles and the weight answerable to one of the distances (being carried for a certain price) to find out the weight answerable to the other distance for the same price. The seventh Effect▪ is by double workmen, and the time answerable to one of the numbers of workmen given, to find out the time answerable to the other number of workmen, in the performance of any work or service. Or chose, by double time, and the workmen answerable to one of those times given, to find out the number of workmen answerable to the other time, in the performance any work or service. Also by a double price of provision, and the number of men, or other Creatures nourished for a certain time, answerable to one of the prizes of provision given, to find out another number of Men or other Creatures answerable to the other price of the provision for the same time. Or chose by two numbers of Men or other Creatures nourished, and one price of provision answerable to one of the numbers of Creatures given, to find out the other price of the same provision answerable to the other number of Creatures, both being supposed to be nourished for the same time, etc. As in the foregoing Examples is fully declared. To prove the operation of the Rule of 3 Inverse, multiply the third and fourth terms together, and note their product, then multiply the first and second together, and if their product is equal to the product of third and fourth, then is the work truly wrought, but if it falleth out otherwise then it is erroneous. As in the first Question of this Chapter the product of 16 (the third number) multiplied by 6 (the 4th. number) is 96, and the product of 8 (the first number) multiplied by 12 (the second number) is 96, equal to the first product which proves the work to be Right. And Note, that if in Division any thing Remain, such Remainder must be added to the product of third and 4th. terms, and if the sum be equal to the product of the first and second (the homogeneal terms being of one denomination) the work is Right. CHAP. XII. The double Rule of three Direct. WE have already delivered the Rules of Single Proportion, and we come now to lay down the Rules of Plural Proportion. 1. Plural Proportion, is when more operations in the Rule of 3 than one, are Required before a Solution can be given to the Question propounded. Therefore in Questions that Require Plurality in Proportion, there are always given more than three numbers. 2. When there are given 5 numbers, and a sixth is Required in Proportion thereunto, than this sixth Proportional is said to be found out by the double Rule of 3, as in the Question following, viz. If 100 l. in 12 months' gains 6 l. Interest, how much will 75 l. gain in 9 Months? 3. Questions in the double Rule of 3 may be Resolved either by two single Rules of three, or by one single Rule of 3 compounded of the 5 given Numbers. 4. The double Rule of three is either Direct or else Inverse? 5. The double Rule of 3 Direct is when unto 5 given Numbers a sixth Proportional may be found out by two single Rules of three Direct. 6. The 5 given Numbers in the double Rule of three, consist of two parts, viz. first a Supposition, and secondly of a Demand. The Supposition is contained in the three first of the 5 given Numbers, and the Demand lies in the two last; as in the Example of the second Rule of this Chapter, viz. If 100 l. in 12 months' gain 6 l. Interest, what will 75 l. gain in 9 Months? here the Supposition is expressed in 100, 12, and 6, for it is said if (or suppose) 100 l. in 12 months' gain 6 l. Interest; and the Demand lieth in 75 and 9; for it is demanded how much 75 l. will gain in 9 Months? 7. When your Question is stated, the next thing will be to dispose of the given Numbers in due order and place, as a preparative for Resolution; which that you may do, first observe which of the given Numbers in the Supposition is of the same denomination with the Number Required; for that must be the second Number (in the first operation) of the single Rule of 3, and one of the other Numbers in the Supposition (it matters not which) must be the first Number, and that Number in the demand which is of the same denomination with the first, mu●t be the third Number, which three Numbers being thus placed, will make one perfect Question in the single Rule of three, as in the forementioned Example, first I consider, that the Number Required in the Question is the Interest or Gain, of 75 l. therefore that Number in the Supposition which hath the same name (viz. 6 l. which is the Interest or Gain of 100 l.) must be the second Number in the 1st operation, and either 100 or 12 (it matters not which) must be the first Number, but I will take 100, and then for the 3● number, I put that number in the Demand which hath the same denomination w●th 100, which is 75, (for they both signify pounds principal) and then the Numbers will stand as you see in the margin. But if I had for the first number put the other number in the Supposition viz. 12, which signifieth 12 Months, than the third Number must have been 9, which is that number in the Demand which hath the same Denomination with the first, viz. 9 Months, and then they will stand as you see in the Margin. There yet Remains two numbers to be disposed of, and those are, one in the supposition, and another in the Demand, that which is of the supposition, I place under the first of the three Numbers and the other which is in the demand I place under the 3d number, and then 2 of the terms in the Supposition will stand (one over the other) in the first place, and the 2 terms in the Demand will stand (one over the other) in the third place, as in the margin. 8. Having disposed, or ordered the numbers given according to the last Rule, we may proceed to a Resolution, and first I work with the three uppermost numbers which according to the first disposition are 100, 6, and 75, which is as much as to say, If 100 l. Require 6 l. (Interest) how much will 75 l. Require, which by the 3 Rule of the 11 Chapter I find to be Direct, and by the 7 and 8 Rules of the 10th. Chapter I find the 4th. Proportional number, to be 4 l. 10 s. so that by the foregoing single question I have discovered how much 75 l. will gain Interest in 12 M. the operation whereof followeth on the left hand under the Letter A; and having discovered how much 75 l. will gain in 12 Months, we may by another question easily discover how much it will gain in 9 Months, for this 4th. number (thus found) I put in the middle between the two lowest numbers of the five, after they are placed according to the 7th. Rule of this Chapter; and than it will be a second number. In another Question in the Rule of 3, the numbers being the first and third numbers being of one denomination, viz. both Months, and may be thus expressed. If 12 Months Require 4 l. 10 s. Interest, what will 9 Months Require? and by the 3 Rule of the 11 Chapter, I find it to be the Direct Rule, and by working according to the directions laid down in the 7, 8, and 9 Rules of the 10th. Chapter, I find the 4th. Proportional number to the last single Question to be 3 l. 07 s. 06. which is the sixth Proportional number to the 5 given numbers and is the Answer to the General Question. The work of the last single Question is expressed on the Right side of the page under the letter B as followeth. So that by the foregoing operation I conclude that If 100 l. in 12 months' gain 6 l. Interest, 75 l. will gain 3 l. 7 s. 6 d. in 9 Months, after the same Rate. The Answer would have been the same, if the 5 given numbers had been ordered according to the second method, viz. as you see in the margin. For first I say, If 12 months' gain 6 l. what will 9 months' gain, this Question I find to be Direct by the 3 Rule of the 11th. Chapter, and by the 7 and 8 Rules of the 10th. Chapter, I find the fourth Proportional number to these three to be 4 l. 10 s. Thus have I found out what is the Interest of 100 l. for 9 Months, and I have now to find the Interest of 75 l. for 9 Months, to effect which, I make this 4th. number (found as before) to be my second number in the next Question, and say, If 100 l. Require 4 l. 10 s. what will 75 l. Require, this Question I find (by the said 3 Rule of the 11th. Chapter) to be Direct, and by the said 7, 8, and 9 Rules of the 10th Chapter, I find the Answer to be as before viz. 3 l. 7 s. 6 d. This Rule hath been sufficiently explained by the foregoing Example, so that the Learner may be able to Resolve the following (or any other) questions pertinent to the double Rule of 3 Direct, whose answers are there given, but the operation purposely omitted to try the Learners ability in the knowledge of what hath been before delivered. Quest. 2. A second Example in this Rule may be as followeth, viz. A Carrier Receiveth 42 shillings for the Carriage of 3 C. weight 150 Miles▪ I demand how much he ought to Receive for the Carriage of 7 C. 3 qrs. 14 l. 50 Miles at that Rate? Answer 36 s. 9 d. Quest. 3. A Regiment of 936 Soldiers eat up 351 quarters of wheat in 168 days, I demand how many quarters of wheat 11232 Soldiers will eat in 56 days at that Rate? Answer 1404 qrs. Quest. 4. If 40 Acres of Grass be mowed by 8 men in 7 days, how many Acres shall be mowed by 24 men in 28 days? Answer, 480 Acres. Quest. 5. If 48 Bushels of Corn (or other seed) yield 576 Bushels in 1 year, how much will 240 Bushels yield in 6 years at that Rate? that is to say, if there were sowed 240 Bushels every one of the 6 years? Answer 17280 bushels. Quest. 6. If 40 shilli●● 〈…〉 wages of 8 men for 5 days, 〈◊〉 ●hall be the wages of 32 men for 24 days? Answer 768 shillings, or 38 l. 8 s. Quest. 7. If 14 Horses eat 5● bushels of provender in 16 days, how many Bushels will 20 horses eat in 24 days? Answer 120 Bushels. Quest. 8. If 8 Cannons in 1 day spend 48 barrels of Powder, I demand how many barrels 24 Cannons will spend is 12 days at that Rate? Answer, 1728 barrels. Quest. 9 If in a Family consisting of 7 Persons, there are drank out 2 Kilderkins of beer in 12 days, how many Kilderkins will there be drank out in 8 days, by another Family consisting of 14 persons? Answer 48 Gallons, or 2 Kilderkins and 12 Gallons. Quest. 10. A Usurer put 75 l. out to Receive Interest for the same, and when it had continued 9 Months, he Received for Principal and Interest 78 l. 7 s. 6 d. I demand at what Rate per Cent. per annum, he Received Interest? Answer, at 6 l. per Cent. per annum. CHAP. XIII. The Double Rule of three Inverse. 1. The Double Rule of 3 Inverse, is when a question in the Double Rule of 3 is Resolved by 2 Single Rules of 3, and one of those Single Rules falls out to be Inverse, or requires a 4th. number in Proportion Reciprocal, (for both the questions are never Inverse.) 2. In all questions of the Double Rule of 3 (as well Inverse as Direct) you are (in the disposing of the 5 given numbers) to observe the 7th. Rule of the 12 Chapter, and in resolving of it by two single Rules, observe to make choice of your Numbers for the first, and second, single questions, according to the directions given in 8 Rule of the same Chapter, as in the Example following, viz. Quest. 1. If 100 l. Principal in 12 months' gain 6 l. Interest, what Principal will gain 3 l. 7 s. 6 d. in 9 Months? This Question is an Inversion of the first question of the 12 Chapter, and may serve for a proof thereof. In order to a Resolution, I dispose of the 5 given numbers according to the 7th. Rule of the last Chapter, and being so disposed will stand as followeth. 12 100 9 l. s. d. 6 3 7 6 Or thus l. s. d. 6 100 3 7 6 12 9 Here observe, that according to the 8th. Rule of the 12 Chapter, the first question (if you take it from the 5 numbers as they are ordered or placed first) will be, If 12 Months Require 100 l. Principal, what will 9 Months Require? (to make the same Interest) this (according to the 3 Rule of the 11 Chapter) is Inverse, and the Answer will be found (by the 2 Rule of the 11th. Chapter) to be 133 l. 6 s. 8 d. the second question than will be, If 6 l. Interest, Require 133 l. 6 s. 8 d. Principal, how much Principal will 3 l. 7 s. 6 d. require, this is a direct Rule, and the Answer in a direct Proportion is 75 l. See the work. So that by the foregoing work I find that if 6 l. Interest be gained by 100 l. in 12 Months, 3 l. 7 s. 6 d. will be gained by 75 l. in 9 Months. But if the Resolution had been found out by the numbers as they are Ranked in the second place, than the second question in the single Rule would have been Inverse, and the first Question Direct, and the conclusion the same with the first method; viz. 75 l. Quest. 2. If a Regiment consisting of 936 Soldiers can eat up 351 quarters of wheat in 168 days, how many Soldiers will eat up 1404 quarters in 56 days at that Rate? Answer, 11232 Soldiers. Quest. 3. If 12 Students in 8 weeks spend 48 l. I demand how many Student's will spend 288 l. in 18 weeks? Answer, 32 Students. Quest. 4. If 48 l. serve 12 Students 8 weeks, how many weeks will 288 l. serve 4 Students? Answer, 144 weeks. Quest. 5. If when the Bushel of wheat cost 3 s. 4 d. the penny Loaf weigheth 12 ounces, I demand the weight of the Loaf worth 9 pence when the Bushel cost 10 s.? Answer, 3 l. or 36 ounces. Quest. 6. If 48 pioners in 12 days, cast a Trench 24 yards long, how many pioners will cast a Trench 168 yards long in 16 days? Answer, 252 pioners. Quest. 7. If 12 C. weight being Carried 100 miles cost 5 l. 12 s. I desire to know how many C. may be carried 150 miles at that Rate? Answer, 18 C. Quest. 8. If when wine is worth 30 l. per Tun, 20 pounds worth is sufficient for the ordinary of 100 men, how many men will 4 l. worth suffice when it is worth 24 l. per Tun? Answer 25 men. Quest. 9 If 6 men in 24 days mow 72 Acres in how many days will 8 men mow 24 Acres? Answer in 6 days. Quest. 10. If when the Tun of wine is worth 30 l. 100 men will be satisfied with 20 l. worth, I desire to know what the Tun is worth when 4 l. worth will satisfy 25 men at the same Rate? Answer 24 l. per Tun. CHAP. XIV. The Rule of three Composed of 5 Numbers. 1. THE Rule of three Composed, is when questions (wherein there are 5 Numbers given to find a 6 in Proportion thereunto) are Resolved by one single Rule of 3 Composed of the 5 given Numbers. 2. When questions may be performed by the double Rule of 3 Direct, and it required to resolve them by the Rule of 3 Composed, (first Order or Rank your Numbers according to the 7 Rule of the 12th. Chapter, then) The Rule is Multiply the terms (or numbers) that stand one over the other in the first place, the one by the other, and make their Product the first term in the Rule of 3 Direct, then Multiply the terms that stand one over the other in the third place, and place their product for the third term in the Rule of 3 Direct, and put the middle term of the 3 uppermost for a second term, then having found a fourth Proportional, direct to these 3, than this 4th. Proportional so found, shall be the Answer Required. So the first question of the 12 Chap. (viz. If 100 l. in 12 months' gain 6 l. Interest, what will 75 l. gain in 9 Months? the numbers being Ranked (or placed) as is there Directed and done. Then I multiply the 2 first terms 100 and 12, the one by the other, and their Product is 1200 (for the first term) than I multiply the two last terms 75 and 9 together, and their Product is 675 for the 3d. term. Then I say, As 1200 is to 6, so is 675 to the Answer, which by the Rule of 3 Direct will be found to be 3 l. 7 s. 6 d. as before. 3. But if the Question be to be Answered by the double Rule of 3 Inverse, than (having placed the 5 given terms as before) multiply the lowermost term of the first place, by the uppermost term of the third place, and put the Product for the first term; then multiply the uppermost term of the first place, by the lowermost term of the third place, and put the Product for the third term, and put the second term of the 3 highest numbers for the middle term to those two, then if the Inverse Proportion is found in the uppermost 3 numbers, the 4th. Proportional direct to these 3 shall be the Answer; so the first question of the 13th. Chapt. being stated, viz. If 100 l. Principal in 12 months' gain 6 l. Interest, what Principal will gain 3 l. 7 s. 6 d. in 9 Months? State the numbers as is there Directed in the first order, viz. M. l. M. 12 100 9 l. l. s. d. 6 3 7 6 then Reduce the 6 l. and 3 l. 7 s. 6 d. into pence, the 6 l. is 1440 d. and 3 l. 7 s. 6 d. is 810 d. then multiply 1440 by 9, the product is 12960 for the first term in the Rule of 3 Direct, and multiply 810 by 12, the product is 9720 for the third term, than I say, As 12960 is to 100 l. so is 9720 to the Answer, viz. 75 l. as before. But if the terms had been placed after the second order, viz. l. l. l. s. d. 6 100 3 7 6 M. M. 12 9 than the Inverse Proportion is found in the lowest numbers, and having composed the numbers for a single Rule of 3 as in the second Rule foregoing, than the Answer must be found by a single Rule of 3 Inverse, for here it falls out to multiply 810 by 12 for the first number, and 1440 by 9 for the third number, and then must say, As 9720 is to 100 l. so is 12960 to the Answer, which by Inverse Proportion will be found to be 75 l. as before. The questions in the 12 and 13 Chapters may serve for thy further experience? CHAP. XV. Single Fellowship. 1. FELLOWSHIP is that Rule of Plural Proportion, whereby we balance Accounts depending between divers Persons having put together a general Stock, so that they may every man have his Proportional part of gain, or sustain his Proportional part of loss. 2. The Rule of Fellowship is either single, or it is double. 3. The single Rule is when the Stocks propounded are single numbers, without any Respect or Relation to time. 4. In the single Rule of Fellowship, the Proportion is, As the whole Stock of all the Partners, is in Proportion to the total Gain or Loss, so is each man's particular share in the Stock, to his particular share in the Gain or Loss. Therefore take the total of all the Stocks for the first term in the Rule of 3, and the whole Gain or Loss for the second term, and the particular Stock of any one of the partners for the third term, then Multiply and Divide according to the 7 Rule of the 9th. Chapter, and the 4th. Proportional number is the particular Loss or Gain of him whose Stock you m●de your second number, wherefore repeat the Rule of 3 as often as there are particular Stocks, or Partners in the question, and the 4th. terms produced upon the several operations, are the Respective Gain or Loss of those particular Stocks given, as in the Examples following. Quest. 1. Two Persons, viz. A and B bought a tun of wine, for 20 l. of which A paid 12 l. and B paid 8 l. and they gained in the Sale thereof 5 l. now I demand each man's share in the Gains according to his Stock? First I find the sum of their Stocks, by adding them together, viz. 12 l. and 8 l. which are 20 l. then according to this Rule I say first, If 20 l. (the sum of their Stocks) Require 5 l. the total gain how much will 12 l. (the Stock of A) Require? Multiply and Divide by the seventh Rule of the ninth Chapter, and the Answer is 3 l. for the 〈◊〉 of A in the gains; then again I say, If 20 l. Require 5 l. what will 8 l. Require? the Answer is 2 l. which is the gain of B. So I conclude that the share o● A in the gain is 3 l. and the share of B in the gain is 2 l. Quest. 2. Three Merchants, viz. A, B and C, enter a joint Adventure; A, put into the the common stock 78 l. B, put in 117 l. and C, put in 234 l. and they find (when they make up their Accounts) that they have gained in all 264 l. now I desire to know each man's particular share in the gains? First I add their Particular stocks together, and their sum is 429 l. then say, If 429 l. gain 264 l. what will 78 l. gain? and what 117 l. and what will 234 l. (the stocks of A, B, and C) gain? work by 3 several Rules of 3 and you will find that Quest. 3. Four Partners, viz. A, B, C, and D, between them build a Ship, which cost 1730 l. of which A paid 346 l. B 519 l. C 692 l. and D 173 l. and her Freight for a certain Voyage is 370 l. which is due to the Owners, or builders, I demand each man's share therein according to his charge in building her? Answer Quest. 4. A, B, and C, enter Partnership for a certain time, A put into the common stock 364 l. B put in 482 l. C put in 500 l. and they gained 867 l. now I demand each man's share in the gain, Proportionable to his Stock? Answer. l. s. d. A 234 09 3 354/1346 B 310 09 5 ●●●/1346 C 322 01 3 930/1346 Sum 867 00 0 5. To prove the Rule of Single Fellowship, add each man's Particular gains or loss together, and if the sum total is equal to the general gain or loss, The proof of the Rule 〈◊〉 Single Fellowship. then is the work Rightly perfomed, but otherwise it is erroneous. Example in the first question of this Chapter, the Answer was that the gain of A was 3 l. and the gain of B 2 l. which added together make 5 l. equal to the total gain given. If in finding out the Particular shares of the several Partners, any thing Remain after Division is ended, such Remainders must be added together (they being all Fractions of the same denomination) and their sum divide by the Common divisor in each question (viz. the total stock) and the Quotient add to the Particular gains, and then if the Total sum is equal to the total gain the work is Right otherwise not. As in the last question, the Remainders were 354, 62, and 930 which added together make 1346 which divided by 1346 (the sum of their Stocks) the quotient is 1 d. which I add to the pence etc. and the sum of their shares is 867 equal to the Total gain; wherefore the work is Right. CHAP. XVI. Double Fellowship. 1. DOUBLE Fellowship, is when several persons enter into Partnership for unequal time, that is, when every man's Particular stock hath Relation to a Particular time. 2. In the Double Rule of Fellowship, multiply each Particular stock by its Respective time, and having added the several products together make the●r sum the first number (or term) in the Rule of 3, and the Total gain or loss the second number, and the Product of any ones particular stock by his time, the third term and the 4th number in Proportion thereunto is his Particular gain or loss, whose Product of stock and time is your third number. Then Repeat (as in single Fellowship) the Rule of 3 as often as there are products (or partners) and the 4th. terms Invented are the numbers Required. Example. Quest. 1. A and B enter Partnership, A put in 40 l. for 3 months, B put in 75 l. for 4 Months, and they gained 70 l. now I demand each man's share in the gains, proportionable to his stock and time? Answer A 20 l. B 50 l. To Resolve this question I first multiply the stock of A (viz. 40 l.) by its time (3 Months) the product is 120, than I multiply the stock of B by its time (viz. 75 by 4) and it produceth 300, which I add to the Product of A his stock and time, the sum is 420. Then by the Rule of 3 I say; As 420 (the sum of the products is to 70 (the total gain) so is 120 (the product of A his Stock and time) to 20 l. the share of A in the gains. Then I say again, As 420 is to 70, so is 300 to 50 l. the share of B in the gains. Quest. 2. A, B, and C, make a Stock for 12 Months, A put in at first 364 l. and 4 Months after that he put in 40 l. B put in at first 408 l. and at the end of 7 Months he took out 86 l. C put in at first 148 l. and 3 Months after he put in 86 l. more, and 5 M. after that he put in 100 l. more, and at the end of 12 Months their gain is found to be 1436 l. I desire to know each man's share in the gains according to his stock and his time? First I consider, that the whole time of their Partnership is 12 Months. Then I proceed to find out the several products of stock and time as followeth. A, had at first 364 l. for 4 Months wherefore their product is 1456 Then he put in 40 l. which with the first sum makes 404 l. which continued the Remainder of the time, viz. 8 Months, and their product is 3232 The sum of the products of the stock and time of A is 4688 B, had 408 l. in 7 Months whose product is 2856 And then took out 86 l., therefore he left in Stock 322 l. which Continued the Rest of the time viz. 5 Months, whose product is 1610 The Sum of the Products of the stock and time of B is 4466 C▪ put in 148 l. for 3 Mon. whose product being multiplied, is 444 Then he put in 86 l. which added to the first (viz. 148) makes 234 l. which lay in stock 5 mon. their product is 1170 Then he put in 100 l. more, so than he had in stock 334 l. which continued the Remainder of the time (viz. 4 months) which multiplied together produce 1336 The sum of the product of the money and time of C is 2950 B 4466 A 4688 The Total sum of all the product is 12104 Then I say, As 12104 is to 1436 (the total gain) so is 2950 to the share of A in the gains, etc. go on as in the foregoing Examples, and you will find their shares in the gain to be as followeth, viz. Answer. l. s. d. The share of A is 556 03 6 6192/121●4 B 529 16 9 5496/21●4 C 349 19 8 416/2104 1436 00 0 Quest. 3. Three Graziers A, B, and C, Take a piece of Ground for 46 l. 10 s. in which A put 12 Oxen for 8 months, B put in 16 Oxen for 5 months, and C put 18 Oxen for 4 months, now the question is, what shall each man pay of the 46 l. 10 s- for the Ground? Answer. l. s. A shall pay 18 00 B 15 00 C 13 10 46 10 3. The Proof of this Rule is the same with that of Single Fellowship, laid down in the 5th. Rule of the 15th. Chapter, and note that If a loss be sustained instead of gain amongst Partners, every man's share to be be born in the loss, is to be found after the same method as their gain, whether their stocks be for equal or unequal time. CHAP. XVII. Allegation Medial. 1. THE Rule of Allegation is that Rule in Plural proportion, by which we Resolve questions, wherein is a composition or mixture of divers simples, as also it is useful in the Composition of medicines both for quantity, quality, and price. And its species are two, viz. Medial and Alternate. 2. Alligation Medial is when having the several quantities, and prizes of several simples propounded we discover the mean price, or Rate of any quantity of the mixture compounded of those simples and the proportion is As the sum of the simples to be mingled is to the Total value of all the simples, so is any part, or quantity of the Composition or mixture, to its mean Rate or Price. Quest. 1. A Farmer mingleth 20 bushels of wheat at 5 s. per bushel and 36 bushels of Rye at 3 s. per bushel, with 40 bushels of barley at 2 s. per bushel, now I desire to know what one bushel of that mixture is worth? To Resolve this question, add together the given quantities and also their values, which is 96 bushels, whose total value is 14 l. 8 s. as appeareth by the work following, for bush. l. s. 20 of wheat at 5 s. per bush. is 05 00 36 of Rye at 3 s. per bush. is 05 08 40 of barley at 2 s. per bush. is 04 00 The sum of the given quantities is 96 and their value is 14 08 Then say by the Rule of 3 Direct. If 96 Bushels cost (or is worth) 14 l. 8 s. what is 1 Bushel worth? Quest 2. A Vintner mingleth 15 gallons of Canary at 8 s. per gallon with 20 gallons of Malligo at 7 s. 4 d. per gallon, with 10 gallons of Sherry at 6 s. 8 d. per gallon● and 24 gallons of white wine at 4 s. per gallon▪ now I demand what a gallon of that mixture is worth? work as in the last Quest. and you will find the Answer to be 6 s. 2 d. 2 qrs. 4●/6●. Quest. 3. A Grocer hath mingled 3 C. of Sugar at 56 s. per C. with 4 C. of the Sugar at 3 l. 14 s. 8 d. per C. and with 6 C. at 1 l. 17 s. 4 d. per C. I desire to know the price of a hundred weight of that mixture? Answer. 2 l. 13 s. 1 d. 7/13. 3. The proof of this operation is by the price of any quantity of the mixture to find out the total value of the whole composition, The proof of Alleg. Medial. and if it is equal to the total value of the several simples, the work is right, otherwise not. As in the first example, the Answer to the quest. was that 3 s. is the price of one bushel, wherefore I say by the rule of proportion, If 1 bushel be 3 shillings, what is 96 bushels. Answ. 14 l. 8 s which is the total value of the several simples, wherefore the work is right. CHAP. XVIII. Allegation Alternate. 1. ALlegation alternate is when there are given the particular prizes of several simples, and thereby we discover such quantities of those simples, as being mingled together shall bear a certain rate propounded. 2. When such a question is stated, place the given prizes of the simples one over the other, and the propounded price of the Composition against them in such sort that it may represent a root and they so many branches springing from it as in the following example. Quest. 1. A certain Farmer is desirous to mix 20 bushels of wheat at 5 s. or 60 d. per bushel with rye at 3 s. or 36 d. per bush. and with Barley at 2 s. or 24 d. per bushel, and Oats at 1 s. 6 d. per bushel, and desireth to mix such a quantity of Rye, Barley and Oats with the 20 bushels of wheat as that the whole composition may be worth 2 s. 8 d. or 32 d. per bushel. The prizes of the simples being placed according to the last rule, with the price of the composition propounded as a root to them will stand as followeth, 3. Having thus placed the given numbers you are to link or combine the several rates of the simples the one to the other, by certain arches in such sort that one that is lesser than the root, (or mean rate) may be linked or coupled to another that is greater than the mean rate, so the question last propounded will stand, 4. Then take the difference between the root and the several branches, and place that difference against the number or branch, with which it is coupled, or linked, and having taken all the differences and placed them as aforesaid, than those differences so placed, will show you the number of each simple to be taken to make a composition to bear the mean rate propounded. So the branches of the last question being linked together as in the first manner, I say the difference between 32 and 60 is 28 which I put against 18 because 60 is linked with 18, than the difference between 32 and 36 is 4 which, I put against 24 because 36 is linked or coupled with 24, than I say the difference between 32 and 24 is 8 which I place against 36 (for the reason aforesaid) than I say the difference between 32 and 18 is 14 which I place against 60; and then the work will stand as you see in the margin. So I conclude that a composition made of 14 bushels of wheat at 60 d. per bushel, and 8 bushel of Rye at 36 d. per bushel and 4 bushels of Barley at 24 d. per bushel, and 28 bushels of Oats at 18 d. per bushel, will bear the mean price of 32 d. or 2 s. 8 d. per bushel. And here observe that in this composition there is but 14 bushels of wheat, but I would mingle 20 bushels, and this kind (or rather case) of Allegation alternate (viz. when there is given a certain quantity of one of the simples, and the quantities of the rest sought to mingle with this given quantity that the whole may bear a price propounded) is called alternation partial. And the proportion to find out the several quantities to be mingled with the given quantity is as followeth, viz. As the difference annexed to the branch that is the value of an integer of the given quantity is to the other particular differences, so is the quantity given, to the several quantities required. So here, to find out how much Rye; barley, and Oats must be mingled with the 20 bushels of wheat, I say by the single rule of three direct, if 14 bushels of wheat require 8 bushels of Rye, what will 20 bush. of wheat require? answer 11 6/14 bushels of rye. Again, if 14 bushels of wheat require 4 bushels of barley what will 20 bushels of wheat require? answer 5 10/14 bushels of barley. Again, I say, if 14 bushels of wheat require 28 bushels of Oats, what will 20 bushels of wheat require? answer 40 bushels of barley. And now I say that 20 bushels of wheat mingled with 11 6/14 bushels of rye, and 5 10/14 bushels of barley, and 40 bushels of oats each bearing the rates as aforesaid, will make a composition or heap of Corn that may yield 32 d. per bushel. But if the branches had been coupled according to the second order, or manner, than the differences would have been thus placed, viz. the differences between 32 and 60 is 28 which I set against 24 because 60 is linked thereto, and the difference between 32 and 36 is 4 which I set against 18, and the difference between 32 and 24 is 8, which I set against 60, than the difference between 32 and 18 is 14 which I set against his yoke-fellow 36, and then I conclude that if you mix 8 bushels of wheat with 14 bushels of Rye, 28 bushels of barley, and 4 bushels of oats each bearing the foresaid prizes the whole mixture may be sold for 32 d. per bushel as by the work in the margin. You see by this work we have found how many bushels of rye, barley, and oats, aught to be mixed with 8 bushels of wheat, and to find out how many of each aught to to be mixed with 20 bushel of wheat I say, As 8 is to 14 so is 20 to 35 bushels of rye. As 8 is to 28 so is 20 to 70 bushels of barley. As 8 is to 4 so is 20 to 10 bushels of oats. whereby I conclude that if to 20 bushels of wheat I put 35 bushels of rye, 70 bushels of barley, and 10 bushels of oats bearing each the foresaid prizes per bush. that then a bush. of this mixture will be worth 32 d. or 2 s. 8ds. And if the branches had been linked as you see in the third place, where each branch bigger than the root, is linked to two that are lesser than the root, then in this case you must have placed the several differences between the root and branches, against those 2 with which each is coupled, as first the difference between 32 and 60 is 28 which I Put against 24 and 18 because it is coupled with them both, than the difference between 32 and 36 is 4 which I set likewise against 24 and 18, because 36 is linked to to them both then the differenee between 32 and 24 is 8 which I put against 60 and 36 because 24 is linked to them both, than the difference between 32 and 18 is 14 which I put against 60 and 36 the yoke fellows of 18. Lastly, I draw a line behind the differences and add the differences which stand against each branch and put the sum behind the said line against its proper branch as you see in the margin. And now by this work I find that 22 bushels of wheat mingled with 22 bushels of rye, and 32 bushels of barley, and 32 bushels of oats each bearing the said price will make a mixture bearing the mean rate of 32 d. per bushel. And to find how much of each of the rest must be mingled with 20 bushels of wheat I say As 22 is to 22 so is 20 to 20 bush of rye. As 22 is to 32 so is 20 to 29 2/22 bush. of barley. As 22 is to 32 so is 20 to 29 2/32 bushels of Oats. Whereby you see the questions of Allegation alternate will admit of more true answers then one for we have found 3 several answers to this first question. Question of alternation partial are proved the same way with questions in Allegation medial, which you may see in the 3 rule of the 17 Chap. The proof of alternation partial. Quest. 2. A Grocer hath 4 sorts of Sugar, viz. of 12 d. per l. of 10 d. per l. of 6 d. per l. and of 4 d. per l. and he would have a composition worth 8 d. per pound the whole quantity whereof should contain 144 l. made of these 4 sorts, I demand how much of each he must take? Questions of this nature are resolved by that part of all●gation alternate called by Arithmeticians alternation total, viz. where there is given the sum, and prizes of several simples, to find out how much of each simple aught to be taken to make the said sum, or quantity, so that it may bear a certain rate propounded. To resolve this question I place the several prizes of the simples, and mean rate propounded, and link them together as is directed in the 2 and 3 rules of this Chapter, and place the differences between the root and branches according to the 4 rule of this Chapter, which will then stand one of these 3 ways, viz. 5. Then add the several● differences together which I have done, and the sums of the first and second order are 12 l. and of the third 24 l. as you may see above, but it required that there should be 144 l. of the composition, therefore to find the quantity of each simples to make the whole composition 144 l. observe this general rule, viz. As the sum of the differences, is to the several differences, so is the total quantity of the composition, to the quantity of each simple. So to find how much of each sort of Sugar I ought to take to make 144 l. at 8 d. per l. I say l. As 12 is to 4, so is 144 to 48 l. at 12 d. per l. As 12 is to 2, so is 144 to 24 l. at 10 d. per l. As 12 is to 2, so is 144 to 24 l. at 6 d. per l. As 12 is to 4, so is 144 to 48 l. at 4 d. per l. Whereby I find that 48 l. at 12 d. per l. and 24 l. at 10 d. per l. and 24 l. at 6 d per l. and 48 l. at 4 d. per l. will make a composition of Sugar containing 144 l. worth 8 d. per l. But as the branches are linked in the second order, the Answer will be 24 l. at 12 d. per l. and 48 l. at 10 d. per l. and 48 l. at 6 d. per l. and 24 l. at 4 d. per l. to make the said quantity, and to bear the said price· And if you had worked as the branches are linked after the third order than you would have found the quantity of 36 l. of each. Quest. 3. A Vintner hath 4 sorts of wine, viz. Canary at 10 s. per gallon, Mallaga at 8 s. per gallon; Rhenish wine at 6 s. per gallon, and white wine at 4 s. per gallon; and he is minded to make a Composition of them all of 60 gallons that may be worth 5 shillings per gallon, I desire to know how much of each he must have? The numbers or terms being ranked according to the second Rule of this Chapter, the branches will be linked as followeth, and will admit of no other manner of coupling, because there is but one branch that is lesser than the Root, therefore all the rest must be linked unto it; and the differences between the Root and the three 1. branches, viz. 10, 8, and 6, which are 5, 3, and 1 must be set against 4 because they are all coupled wit●●●t, and the difference between the Roose (viz. 5) and ● which is 1 must be set against the 3 other because it is linked to them all, so I find 1 gallon of Canary, 1 gallon of Mal●aga, 1 gallon of Rhenish wine, and 9 gallons of White-wine, prised as above, being mingled together will be worth 5 s. per gallon, the Sum being 12 gallons but there must be 60 gallons wherefore I say As 12 is to 1, so is ●0 to 5 gallons of Can. As 12 is to 1, so is 60 to 5 gallons of M●ll. As 12 is to 1, so is 60 to 5 gallons of 〈◊〉. As 12 is to 9, so is 60 to 45 gallons of white wine. So that 5 gallons of Canary, 5 gallons of Malliga, 5 gallons of Rhenish, and 45 gallons of white wine mingled together, will be in all 60 gallons, worth 5 s. per gallon, which was Required. Quest. 4. A Goldsmith hath Gold of 4 several sorts of finess, viz, of 24 Carects fine, Read Chap. 2. deff. 2. of this book. and of 22 Carects fine, of 20 Carects fine, and of 15 Carects fine. And he would mingle so much of each with alloy that the whole mass of 28 ounces of gold so mingled may bear 17 Carects fine? the second and third Rules of this Chapter being observed (for instead of the alloy I put 0 because it bears no finess, but it makes a branch in the operation) the terms may be Allegated and the differences added any of these 4 ways following. More ways may be given for the Allegating, or linking of the terms in this question, but these are sufficient, and it shall also suffice to give an Answer to the question as the terms are linked the first way, not doubting but the ingenious practitioner will be able at his leisure, to find Answers to the other 3 ways, viz, oz. p.w. Car. As 56 is to 17, so is 28 to 8 10 of 24 As 56 is to 2 so is 28 to 1 00 of 22 As 56 is to 19, so is 28 to 9 10 of 20 As 56 is to 8, so is 28 to 4 00 of 15 As 56 is to 10, so is 28 to 5 00 of alloy. Thus much well practised and understood, is sufficient for the understanding of Allegation. In questions of Alternation Total, the Answer is given true, when the sum of each of the quantity of simples found, The proof of Alternation Total. agrees with the Sum or Quantity propounded; as in the last Question, the Answer was 8 oz. 10 p.w. of 24 Carrects fine, 1 oz. of 22 Carrects fine, 9 oz. 10 p.w. of 20 Carrects fine, 4 oz. of 15 Carrects fine, and 5 oz. of Alloy, which added together make 28 oz. the quantity propounded. CHAP. XIX. Reduction of Vulgar Fractions. 1. WHat a Vulgar fraction is, and its parts, and several kinds, hath been already showed in the 19, 20▪ 21, 22, 23, 24 and 31 definitions of the first Chapter of this book, which the Learner is desired diligently to observe before he proceed. 2. To Reduce a Vulgar fraction (which discovereth the Principal knowledge of fractions, and therefore ought greatly to be Regarded) we shall discover plainly under these 8 several heads (or Rules) following, viz. 1. To Reduce a mixed number into an improper Fraction. 2. To Reduce a whole number into an Improper Fraction. 3. To Reduce an Improper fraction into its equivalent whole (or mixed) number. 4. To Reduce a Fraction into its lowest terms equivalent to the Fraction given. 5. To find the value of a Fraction in the known parts of Coin, Weight, Measure, etc. 6. To Reduce a Compound Fraction to a simple one of the same value. 7. To Reduce divers Fractions having unequal denominators, to fractions of the same value having equal denomination. 8. To Reduce a Fraction of one denomination to another of the same value. I. To Reduce a mixed Number to an Improper fraction. The Rule is vide Chap. 1. defin. 31. Multiply the Integral part (or whole Number) by the denominator of the Fraction, and to the product add the Numerator, and that sum place over the denominator, so this new fraction shall be equal to the mixed number given. As for Example 1. Reduce 18 3/7 into an improper fraction, multiply the whole number 18 by 7 the denominator, and to the product add the numerator 3 the sum is 129 which put over the denominator 7 and it makes 129/7 for the answer as per margin. 2. Reduce 183 5/12 to an improper fraction facit. 2201/1●. 3. Reduce 56 13/21 to an improper fraction facit 1189/21. II. To Reduce a whole number to an Improper fraction. The Rule is Multiply the given number, Vide Ch. 1. defin. 23. by the intended denominator and place the product for a numerator over it, as for example. 1. Let it be required to reduce 15 into a fraction whose denominator shall be 12. To effect which I multiply 15 by the intended denominator (12) the product is 180 which I place over 12 as a numerator and it makes 180/12 which is equal to 15 which was required as per margin. 2. Reduce 36 into an improper fraction whose denominator shall be 20 facit 936/26. 3. Reduce 135 into an improper fraction whose denominator shall be 16 facit 2160/16. III. To Reduce an improper fraction into its equivalent, whole or mixed number. The rule is, Divide the numerator by the denominator, and the quotient is the whole number equal to the given fraction, and if any thing remain put it for a numerator over the divisor, example. 1. Reduce 436/8 into its equivalent mixed number, divide the numerator 436 by the denominator 8 and the quotient is 54 and 4 remains which put for a numerator, over the divisor 8 the answer 54 4/8 as per margin. 2. Reduce 3476/15 to a mixed number, facit 231 11/15. 3. Reduce 15576/136 to a mixed number, facit 114 72/136. FOUR To Reduce a fraction into its lowest terms▪ equivalent to the fraction given. The Rule is. 1. If the numerator and denominator are even numbers take ½ of the one and half of the other as often as may be and when either of them falls out to be an odd number, then divide them by any number that you can discover will divide both numerator and denominator without any remainder; and when you have thus proceeded as low as you can reduce ●hem then this new fraction so found out shall be the fraction you desire, and will be in value equal to the given fraction, example. 1. Let it be required to reduce 192/336 into its lowest terms. 192 96 48 24 12 4 336 168 84 42 21 7 First I take the half of the numerator 192 and it is 96 then half of the denominator and it is 168, so that now it is brought to 96/168 and next to 48/84, and by halving still to 24/42 and their half is 12/21 and now I can no longer half, it because 21 is an odd number, wherefore I try to divide them by 3, 4, 5, 6, etc. and I find 3 divides them both without any remainder, and brings them to 4/7 as per margin. So I conclude 4/7 thus found to be equal in value to the given fractions 192/336. 2. What is 103●/1184 in its lowest terms? answer 7/8. 3. What is 1342/1586 in its lowest terms? answer 11/13. There is yet another way more excellent than the former to reduce a fraction into its lowest terms, Vide Ought. cla. Math. cap. 7. and that is by finding a common measurer, viz. the greatest number that will divide the numerator and denominator without any remainder, and by that means reduce a fraction to its lowest terms at the first work, and to find out this common measurer, divide the denominator by the numerator, and if any thing remains divide your divisor thereby, and if any thing remains then divide your last divisor by it, do so until you find nothing remains, than this last divisor shall be the greatest common measurer, which will divide both numerator denominator and reduce them into their lowest terms at one work. Example. 4. Reduce 228/304 into its lowest terms by a common measurer. To effect which I divide the denominator 304 by the numerator 228 and there remains 76, than I divide 228, (the first divisor) by 76 (the remainder) and it quotes 3 and remains nothing wherefore the last divisor 76 is the common measurer, by which I divide the numerator of the given fraction, viz 228 it quotes 3 for a new numerator, than I divide the denominator 304 by 76 and it quotes 4 for a new denominator, so that now I have found ¾ equal to 228/304. 5. Reduce 6048/7392 into its lowest terms by a common measurer, facit 9/11. 6. Reduce 3081/20●82 into its lowest terms by a common measurer facit 13/6. A Compendium. Note that if the numerator and denominator of a fraction end each with a cipher or Ciphers then cut of as many cyphers from the one as from the other and the remaining figures will be a fraction of the same value, viz. 3400/710● will be found to be reduced to 34/71 by cutting of the 2 cyphers from the numerator and denominator thus, 34/1700 1700/00 and 460/700 will be 46/70 thus 4●/700/0 etc. V. To find the value of a fraction in the known parts of Coin, weight, etc. The rule is. Multiply the numerator by the parts of the next inferior denomination that are equal to an unit of the same denomination with the fraction, then divide that product by the denominator, and the quote gives you its value, in the same parts you multiplied by▪ and if any thing remain multiply it by the parts of the next inferior denomination, and divide as before, do so till you can bring it no lower and the several quotients will give you the value of the fraction as was required, and if any thing at last remain place it for a numerator over the former denominator, example. 1. What is the value of 27/29 l. Sterling? To Answer this question I multiply the numerator 27 by 20 (the shillings in a pound) the product is 540 which I divide by 29 (the denominator) and the quotient is 18 s. and there Remains 18 which I multiply by 12 pence and the product (216) I divide by the denominator 29 the quotient is 7 d. and 13 Remains, which I multiply by 4 farthings, the product is 52, which I still divide by 29, the quotient is 1 farthing, and there Remaineth 23, which I put for a Numerator over the denominator 29, so I find the value of 27/29 l. to be 18 s. 7 d. 1 qrs. 23/29 as per the following operation. 2. What is the value of 11/15 l. Sterling? facit 14 s. 8 d. 3. What is the value of 28/137 l. Sterling? facit 4 s. 1 d. 7/137. 4. What is 16/21 C. weight? facit 3 qrs. 1 l. 5 oz. 7/21. 5. What is 136/371 l. Troy weight? facit 4 oz. 7 p.w. 23 gr. 179/371. 6. What is 41/50 of a year? Answer 299 day. 7 hour. 12 min. VI To Reduce a Compound Fraction to a simple one of the same value. What a compound Fraction is, hath been showed in Chap. 1. Definition 24, and to Reduce it to a simple Fraction of the same value. The Rule is, Multiply the Numerators continually and place the last product for a new Numerator, then multiply the denominators continually, and place the last product for a new denominator. So this single Fraction shall be equal to the compound fraction given. Example. 1. Reduce ⅔ of ⅗ of ⅝ to a Simple Fraction. Multiply the Numerators 2, 3, and 5 together, they make 30, for a new Numerator; then I multiply the denominators 3, 5, and 8 together and their product is 120 for a denominator, so the simple Fraction is 30/120 and cutting off the Ciphers it is 3/●12 equal to ¼. 2. What is 7/10 of 5/9 of 4/7 of 11/12 Answer 1540/7560 or 154/756 or 77/373. 3. What is 11/12 of 13/14 of 21/29 Answer 3003/4872. By this you may know how to find the value of a Compound Fraction, viz. first Reduce it to a simple one, and then find out his value by the 5 Rule foregoing. 4. What is the value of ¾ of ⅚ of 9/10 of a pound Answer 11 s. 3 d. VII. To Reduce Fractions of unequal denominators to Fractions of the same value, having equal Denominators. The Rule is, Multiply all the denominators together, and the product shall be the Common denominator? Then multiply each Numerator into all the denominators except its own, and the last product put for a Numerator over the denominator found out as before; So this new Fraction is equal to that fraction whose Numerator you multiplied into the Denominators. Do so by all the Numerators given, and you have your desire. Example, 1. Redu●●●/4 ⅘ ⅚ and ⅞ into a common Denomination. Multiply the Denominators 4, 5, 6, and 8 together continually, and the product is 960 for the common Denominator; then multiply the Numerator 3 into the Denominators 5, 6, and 8, and the product is 720, which is a Numerator to 960 (found as before) so 720/96● is equal to the first fraction ¾, than I proceed to find a new Numerator to the second fraction, viz. 4, and I multiply 4 (into all the denominators except its own; viz.) into 4, 6, and 8, which produceth 768/960 equal to ⅘, then multiply the Numerator 5, into the denominators 4, 5, and 8, the product is 800/960 equal to ⅚. Then multiply the Numerator 7 into the denominators 4, 5, and 6, the product is 840/960 equal to ⅞ and the work is done, so that for ¾ ⅘ ⅚ and ⅞ I have 720/960, 768/960 800/960 and 840/960. 2. Reduce 1114/1223 and 19/21 into a common denominator, faciunt 5313/5796 3528/5796 and 5244/5796. VIII. To Reduce a fraction of one Denomination to another. 1. This is either Ascending, or Descending. Ascending when a fraction of a smaller is brought to a greater Denomination; and Descending when a fraction of a greater Denomination is brought lower. 2. When a fraction is to be brought from a lesser to a greater Denomination, then make of it a Compound fraction, by comparing it with the intermediate Denominations between it, and that you would have it Reduced to, then (by the 6 Rule foregoing) Reduce your compound to a simple fraction, and the work is done. Example. Quest. 1. It is Required to know what part of a pound sterling 5/7 of a penny is? To Resolve this, I consider that 1 d. is 1/12 of a shilling, and a shilling is 1/20 of a pound; wherefore 5/7 d. is 5/7 of 1/12 of 1/20 of a pound, which by the said 6th. Rule I find to be 7/1680 l. Quest. 2. What part of a pound Troy weight is ⅘ of a penny weight? Answer ⅘ of 1/20 of 1/12 l. equal to 4/1200 l. Troy. 3. When a fraction is to be brought from a greater to a lesser denomination, then multiply th● Numerator by the parts contained in the several denominations betwixt it, and that you would reduce it to, then place the last product over the denominator of the given fraction. Example, Quest. 3. I would reduce ⅗ l. to the Fraction of a penny? to do which I multiply the Numerator 3 by 20 and 12 the product is 720 which I put over the denominator 5 it makes 720/5 of a penny, equal to ⅗ l. Quest. 4. What parts of an ounce Troy is ⅚ l.? Answer 60/●6 oz. CHAP. XX. Addition of Vulgar Fractions. 1. IF your Fractions to be added have a common Denominator, then add all the Numerators together, and place their sum for a Numerator to the common Denominator, which new Fraction is the sum of all the given Fractions; and if it be Improper, Reduce it to a whole, or mixed Number, by the 3 Rule of the 19 Chap. Quest. 1. What is the sum of 7/24 9/24 16/24 and 14/24? The Denominators are equal, viz. 24, wherefore add the Numerators together, viz. 7, 9, 16, and ●4▪ their sum is 46, which put over the Denominator 24, it makes 4●/21 the sum of the given Fractions, which will be Reduced to the mixed Number 1 22/24 or 1 11/12. 2. But if the Fractions to be added have unequal Denominators, then Reduce them to a common Denominator, by the 7th. Rule of the 19th. Chap. and then add the Numerators together, and put the sum over the common Denominator, etc. as before. Quest. 2. What is the sum of ⅗ ⅞ 9/10 and 11/12? The fractions Reduced to a common Denominator are 2880/4800 4200/4800 4320/4800 and 4400/480●, the sum of their Numerators is 15800 which put over the common Denominator makes 15800/4800 or 158/48 equal to the mixed Number 3 14/48 or 3 7/24 for the sum Required. Quest. 3. What is the sum of 13/17 21/49 and 3/47? Answer 1 37555/3915●. 3. If you are to add mixed Numbers together, then add the fractional parts as before, and if their sum be an Improper Fract on Reduce it to a mixed Number, and add its Integral part to the Integral parts of the given mixed Numbers, and the work is done. Quest. 4. What is the sum of 13 ¾ and 24 ⅝? First add the fractions ¾ and ⅝, sum is 1 12/32 then add this Integer 1, to 13 and 24 their sum is 38, and put after it the fraction 12/32 it is 38 12/32 for the Answer. Quest. 5. What is the sum of 48 3/7 64 ⅝ and 130 2/●? Facit 243 180/224. 4. If any of the fractions to be added is a compound fraction, it must first be Reduced to a Simple Fraction by the 6th. Rule of Chapter 19, and then add it to the Rest according to the 2d Rule of this Chapter. Example, Quest. 6. What is the sum of ⅗ 4/6 and ⅞ of ¾ of ⅚? Reduce 7/● of ¾ of ⅚ into a simple Fraction, and it is 105/192 which Reduced with the other two, and added are 1 2766/5760. Quest. 7. What is the sum of 11/12 and ¾ of ⅘ of 5/●? Answer 1 4/12. 5. If the Fractions to be added are not of one denomination, they must be so Reduced, and then proceed as before. Quest. 8. What is the sum of ¾ l. and ⅚ s.? Here of the given Fractions, one is of a pound, and the other the fraction of a shilling; and before you can add them together, you must Reduce ⅚ s. to the fraction of a pound as the other is by the 8 Rule of Chapter 19, and it makes 5/120 l. then ¾ l. and 5/120 l. will be found to be 380/480 l. or 38/●8 l. by the 7th. Rule of Chapter 19, and in its lowest terms 19/24 l. by the 4 Rule of Chapter 19 It would have been the same, if (by the latter part of the 8th. Rule of Chapter 19) you had Reduced ¾ l. to the fraction of a shilling, which you would have found to have been 60/4 s. which added to ⅚ s. by the said 7th. Rule of the last Chapter the sum is 15 s. 30/24 which is equal to the sum found as before, viz. 19/24 l. for (by the 5th. Rule of Chapter 19) the value of 19/24 l. will be found to be 15 s. 10 d. and so will 15 s. 20/24 be found to be just as much. Quest. 9 What is the sum of ⅗ l. ⅗ s. and ⅗ d.? Answer 379500/600000 or 3795/6000 l. CHAP. XXI. Subtraction of vulgar fractions. 1. THe rules in Addition for reducing the given fractions to one denomination are here to be observed; for before Subtraction can be made, the fractions must be reduced to a common denominator, then Subtract one numerator from the other, and place the remainder over the common denominator, which fraction shall be the excess or difference between the given fractions, example. Quest. 1. What is the difference between ¾ and 5/7? the given fractions are reduced to 21/●8 and 20/●8, then Subtract the numerator 20 from the numerator 21, and there remains 1 which put over the denominator 28 makes 1/28 for the answer, or difference between ¾ and 5/7. Quest. 2. What is the difference between ⅚ and ⅗ of ⅝? Reduce the compound fraction ⅗ of ⅝ to a simple fraction, then proceed as be-before the answer is 110/240 equal to 11/24. 2. When a fraction is given to be Subtracted from a whole number subtract the numerator from the denominator and put the remainder for a numerator to the given denominator, and Subtract a unit (for that you borrowed) from the whole number, and the remainder place before the fraction found as before which mixed number is the remainder or difference sought, example. Quest. 3. Subtract 7/10 from 48? Answer, 47 3/10 for if you Subtract 7 (the numerator) from 10 (the denominator) remains 3, which put over 10 is 3/10 and 1 (I borrowed) from 48 rests 47, to which join 3/10 makes 47 3/10 for the excess. Quest. 4. Subtract 13/21 from 57? remains 56 8/●1. 3. If it is required to Subtract a fraction from a mixed number, or one mixed Number from another reduce the 〈◊〉 fractions to a common denominator, and if the fraction to be Subtracted be lesser than the other, then Subtract the lesser numerator from the greater and that is ● numerator for the common denominator; then Subtract the lesser integral 〈◊〉 from the greater, and the remainder 〈◊〉 the remaining fraction annexed is the difference required between the two given mixed numbers, example. Quest. 5. Subtract 26 3/● from 54 ⅚? First, Subtract 3/● viz. 18/●● from ●/6 viz. 3●/4● the remainder is 17/●2 than 26 from 54 remaineth 28 to which annex 17/4● it makes 28 17/42 for the Answer. 4. But if the fraction to be Subtracted is greater than the fraction from whence you subtract, then take the numerator of the greater fraction out of the denominator, and add the remainder to the numerator of the lesser fraction, and their sum is a new numerator to the common denominator, which fraction note, than (for the unit you borrowed) add 1 to the integral part to be subtracted, and subtract it from the greater number, and annex the remainder to the fraction you noted before, so this new mixed number shall be the difference sought, Example. Quest. 6. Subtract 14 ¾ from 29 4/7? The fractions reduced are viz. ¾ equal to 21/28 and 4/7 equal to 16/28, now I should subtract 21/2● from 1●/28 but I cannot therefore I subtract 21 from 28 rests 7 which added to 16 (the lesser numerator) makes 23 for a numerator to 28 viz. 23/28 than I come to the Integral parts 14 and 29 and say 1 that I borrowed and 15 from 29 rests 14 to which annexing 23/28 it is 14 23/28 for the remainder or difference between 14 ¾ and 29 4/7. Quest. 7. Subtract 36 9/10 from 74 4/9 facit 37 49/90. CHAP. XXII. Multiplication of Vulgar Fractions. 1. IF the multiplicand and multipliar are simple (or single) fractions, then multiply the numerators together for a new numerator, and the denominators for a new denominator, which new fraction is the product required. Quest. 1. What is the product of 5/7 by 9/11? facit 45/77. For the numerator 5 and 9 multiplied is 45 and the denominators 7 and 11 multiplied is 77. Quest. 2. What is the product of by 21/●7 facit 378/●51. 2. If the fractions to be multiplied are mixed numbers, reduce them to improper fractions by the 1 Rule of the 19 Chap. then proceed as before. Quest. 3. What is the product of 48 ⅗ by 13 ⅚? The given mixed numbers reduced to mixed numbers are 48 ⅗ equal to 243/5 and 13 ⅚ equal to 83/6 now 243/5 multiplied by 83/6 according to the first Rule of this Chapter produceth 21069/30 or 702 9/30. Quest. 4. What is the product of 430 6/1● by 18 3/7? facit 555474/70 or 7935 24/70. 3. If a compound fraction is to be multiplied by a simple fraction, first reduce the compound fraction into a simple fraction, then multiply the one by the other as is taught above. Quest. 5. What is the product of 16/21 by ¾ of 5/7 of ⅘? the compound fraction ¾ of 5/7 of ⅘ reduced is 60/140 or ●6/14 which multiplied by 16/21 produceth 96/294 which in its lowest term is 16/49 for the answer. And if the multiplicand and multipliar are both compound fractions, reduce them both to simple ones, then multiply these new fractions as before so have you the product. Quest. 6. What is the product of ¾ of ●/● by ⅗ of 1/21? Answ. 18/120 in its lowest terms 3/20. Quest. 7. What is the product of ⅔ of ¾ by ⅖ of ⅚? Answ. 60/360 or 6/36 or in it least terms 1/6. 4. If a fraction be to be multiplied by a whole number, put under the given whole a unit for a denominator, whereby it will be an improper fraction, then multiply these fractions as before, example. Quest. 8. What is the product of 24 by ⅔? Answ. 48/3 for 24 by putting a unit under it will be 24/1 & 24/1 by ⅔ produceth 48/3 or 16. Quest. 9 What is the product of 36 by 9/11? Answer 324/11 or 29 5/11. CHAP. XXIII. Division of Vulgar Fractions. 1. IF the dividend and the divisor are both simple fractions, then multiply the numerator of the dividend into the denominator of the divisor and the product is a new numerator, then multiply the denominator of the dividend into the numerator of the divisor and the product is a new denominator, which new fraction thus found is the quotient you desire, example. Q. 1. What is the quotient of ⅝ divided by ⅗? A. 42/2● or 1 1/24 for first I multiply (5) the numerator of the dividend into (5) the denominator of the divisor and the product (25) is a numerator for the quotient, than I multiply (8) the denominator of the dividend into (3) the numerator of the divisor and the product (24) I put in the quotient for a denominator, so I find 25/24 is the quotient sought. Q. What is the quotient of 16/21 divided by ⅔? A. 30/42 equal to 5/7 in its lowest terms. 2. But if you would divide a simple fraction by a compound, or a compound by a simple, first reduce such compound to a simple fraction, then go on as before. Q. 3. What is the quotient of 3/10 divided by ¾ of ⅔? A. 36/5● or ⅗, first reduce ¾ of ⅔ into a simple fraction and it is 6/12, by which 3/10 being divided the quotient is 36/30 equal in its least terms to ⅗. And if the dividend, and divisor be both compound fractions, reduce them both to simple fractions, then divide the one by the other as in Rule 1 before-going. Quest. 4. What is the quote of ⅔ of ¾ divided by ⅖ of ⅚? Answ. 180/1●0 or 18/12 or 1 6/12 or 1 ½. 3. If the dividend, or divisor, or both are mixed numbers reduce them to improper fractions, and perform division as you were taught before, example. Quest. 5. What is the quote of 12 ¾ divided by 21 ⅘? Answ. 225/436 for 12 ¾ is equal to 51/4 and 21 ⅘ is equal to 109/● and the quote of 51/4 divided by 109/5 is as before 225/236. 4. If you divide a fraction by a whole number, or a whole number by a fraction, make the whole number an improper fraction by putting an unit for a denominator to it as was taught in Rule 4 of Chap. 22. and then perform division as before was taught, example. Quest. 6. What is the quote of 8 divided by ⅗? Answ. 40/● or else 13 ●/● which is equal thereto, the work is in the margin. Quest. 7. What is the quote of ⅗ di-by 8? Answer. ●/4● as per margin. CHAP. XXIV. The Rule of three direct in vulgar fractions. 1. AS in the rule of 3 in whole numbers so likewise in fractions, you must see that the fractions of the first and third places be of the same denomination. 2. See that if any of the given fractions be compound, that they be reduced to simple of the same value. 3. If there are given mixed numbers, reduce them to improper fractions by the 1 Rule of Chap. 19 4. If any of the 3 terms is a whole number make it an improper fraction by constituting unit for its denominator. Having reduced your fractions as is directed in the 4 last rules, then proceed to a resolution, which is performed the same way as in whole numbers, respect being had to the rules delivered for the working of fractions, viz. multiply the second and 3 fractions together according to the 1 Rule of Chap. 22 and divide the product by the first fraction, according to the 1 Rule of Chap. 23. and the quotient is the answer. Or, (which is better) 5. Multiply the numerator of the first fraction into the denominators of the second and third, and the product is a new denominator, then multiply the denominator of the first fraction into the numerators of the second and third, and the product is a new numerator; which new fraction is the 4th. proportional, or answer, which (if it is an improper fraction) must be reduced to a whole or mixed number by the 3 Rule of Chap. 19 examples. Quest. 1. If ¾ yds. of Cloth cost ⅝ l. what will 9/10 yds. cost? Having placed the given fractions according to the 6 Rule of Chap. 10. I proceed to resolution, and first I multiply the numerator of the first fraction (3) into 8 and 10 the denominators of the second and third fractions and the product is 240 for a denominator, than I multiply 4 the denominator of the first fraction into 5 and the numerators of the second and third fractions, the product is 180 for a numerator, which numerator 180 and denominator 240 make 180/240 l. for the Answer. equal to ¾ l. or 15 s. Quest. 2. If ⅔ l. buy ⅚ yds. of Cloth, what will 11/12 yds. cost at that rate? Answer. 132/18● l. equal to 11/15 l. or 14 s. 8 d. Quest. 3. If ⅞ l. cost 2/6 s. what will 8/9 s. buy? Answer. 336/144 l. equal to 2 ⅓ l. Quest. 4. If ⅗ of an ell of Holland cost ⅛ of a pound how much will 12 2/● else cost at that rate? Answ. 190/●2 equal to 2 23/36 l. In resolving the last question and the two next observe the 3 Rule of this Chap. foregoing. Quest. 5. If 9/1● of a C. cost 28 ¾ s. what will 7 ½ C. cost at that rate? Answ. 239 7/12 s. or 11 l. 19 s. 7 d. Quest. 6. 3 ¼ yds of Velvet cost 3 ⅝ l. how much will 10 ½ yds cost at that rate. Answ. 11 37/●● l. Quest. 7. If 3 yds of broad Cloth cost 2 ⅘ l. what will 14 3/7 yds' cost? Answ. 13 l. 9 s. 4 d. In working the last question and the 4 next observe the 4 Rule of this Chap. foregoing. Quest. 8. If 14 l. of Pepper cost 14 s· 6 ⅗ d. I demand the price of 72 ¾ l. Answer. 3 l. 15 s. 7 83/280 d. Quest. 9 If 1 l. of Couchenele cost 1 l. 5 s. what will 36 7/10 l. cost? Answer. 45 l. 17 s. 6 d. Quest. 10. If 1 yd. of broadcloth cost 15 5/● s. what will 4 pieces each contain 27 3/7 yds. at that rate? Answer. 85 l. 14 s. 3 3/7 d. Quest. 11. A Mercer bought 3 ½ pcs of silk each piece qt. 24 2/● else at 6 s. 0 ¾ d. per ell I demand the value of the 3 ½ pcs at that rate? Answer. 26 l. 3 s. 4 ¾ d. In solving the 4 next questions observe the 8 Rule of Chap. 19 Quest. 12. If ⅖ of an ounce of Silver cost 2 s. I demand the price of 11 ⅖ l. at that rate? Answer 35 l. Quest. 13. If 1 5/7 l. of gold is worth 61 5/7 l. sterling, what is 1 grain worth at that rate? Answer 1 ½ d. Quest. 14. If ¾ yds. of silk is worth ¾ of ⅚ l. with. is the price of 15 ⅖ else Flemish? Answer. 9 l. 12 s. 6 d. Quest. 15. If 2/● of ¾ of a pound of Cloves cost 6 s. 2 1/7 d. what cost the C. weight at that Rate? Answer. 69 l. 4 s. CHAP. XXV. The Rule of Three Inverse in fractions. 1. IT hath been already taught (in the 3 rule of the 11 Ch.) how to discover when the 4th proportional number (to the 3 given numbers) is to be found out by a Rule of 3 direct, and when by a Rule of 3 inverse to which Rule the learner is now referred. 2. When (In fractions) you find a question to be solved by the Rule of 3 Inverse, viz. when the third term is the divisor, than (having reduced the terms exactly according to the Rules in Chap 24.) multiply the numerator of the 3 fractions into the denominators of the second and first fractions, and the product is a new denominator, then multiply the denominator of the third fraction into the numerators of the second and first fractions, and the product is a new numerator, which new fraction thus found is the Answer to the question. Quest. 1. If ¾ of a yard of Cloth that is 2 yds wide will make a garment, how much of any other Drapery that is 3/5 of a yard wide will make the same garment? Answer. 2 ½ yds. Quest. 2. Lent my friend 46 l. for ⅘ of a year how much ought he to lend me for 7/12 of a year. Answer. 63 5/●5. Quest. 3. If 2/● of a yard of Cloth that is 2 ⅓ yards wide will make any garment what breadth is that Cloth when 1 ¾ yds will make the same garment? Answer. 56/63 of a yd wide. Quest. 4. How many inches in length of a board that is 9 Inches broad, will make a foot square? Answer. 16. Quest. 5. If when the bushel of wheat cost 4 ¾ s. the penny loaf weigheth 10 2/● ounces what will it weigh when the bushel cost 8 9/10 s.? Answer. 5 185/●6● ounces. Quest. 6. If 12 men can mow 24 ½ Acres in 10 ⅔ days in how many days will 6 men do the same? Answer. in 21 ⅓ days. CHAP. XXVI. Rules of Practice. 1. IN the single Rule of 3 when the first of the 3 numbers in the question (after they are disposed according to the 6 Rule of the 10 ch.) happeneth to be a unit (or 1) that question many times may be resolved far more speedily then by the Rule of 3, which kind of operation is commonly called Practice, and indeed it is of excellent use amongst Merchants, Tradesmen and others, by reason of its speediness in finding a resolution to such kind of questions. 2. The chiefest question resolveable by these brief Rules may be comprehended under the seven general heads or cases following (viz.) When the given price of the integer Consists. 1 Of farthings under 4 2 Of pence under 12 3 Of pence and farthings 4 Of shillings under 20 5 Of shill. pence and farthings 6 Of Pounds 7 Of pounds, shillings, pence and farthings. It would be very convenient for the practical Arithmetician, to have by heart the several products of the 9 digits multiplied by 12, for his speedy reducing pence into shillings or shillings into pence, which he may gain by the following Table. 12 Times 1 is 12 2 24 3 36 4 48 5 60 6 72 7 84 8 96 9 108 3. Shillings are practically reduced into pounds thus, viz. cut of the figure standing in the place of units with a dash of the pen and note it for shillings, then draw a line under the given number and take half of the remaining figures (after the first is cut off) and set them under the line, and they are so many pounds, but if the last figure is odd, then take the lesser half and add 10 to the figure so cut off (as before) for shillings, as if I were to reduce 43658 shillings into pounds, first I cut of the last figure (8) for shillings, than I take half of the remaining figures (4365) thus half of 4 is 2, which I put under the line, than ½ of 3 is 1, and because 3 is an odd number I make the next figure 6 to be 16, and go on saying ½ of 16 is 8, and then ½ of 5 is 2 which is the last figure, wherefore because 5 is an odd number I add 10 to the 8 I cut off and it makes 18 s. so that I find it to be 2182 l. 18 s. as per margin. 4. It is likewise Convenient that the Learner be acquainted with the Practical Tables following, the first containing the Aliquot (or even) parts of a shilling, the second containing the Aliquot parts of a pound. d. s. The even parts of a shilling. 6 is ½ 4 ⅓ 3 ¼ 2 ⅙ 1 ½ ⅛ 1 1 ½ s. d. l. The even parts of a pound. 10 00 is ½ 6 08 ⅓ 5 00 ¼ 4 00 ⅕ 3 04 ⅙ 2 06 ⅛ 2 00 1/10 1 08 1/12 1 00 1/20 Case 1. 5. When the price of the Integer is a farthing, then take the sixth part of the given Number which will be so many three half pence, and if any thing Remains it is farthings by the 7th. Rule of Chapter 9, then consider that three halfpences is ⅛ of a shilling, wherefore take the eighth parts of them for shillings, and if any thing Remain, they are so many 3 halfpences, which Reduce into pounds by the 3d Rule foregoing. Example 67486 l. at a farthing per l. First I take ⅙ of 67486, and it is 11247 three halfpences and 4 farthings, or 1 penny, than ⅛ of 11247 is 1405 s. and 7 Remains, which is 7 three half penies or 10 d. ½ which with the 4 farthings before make 11 ½ and 1405 shillings which by the 3 Rule is 70 l. 5 s. In all 70 l. 5 s. 11 d. ½ for the Answer, See the work following. Other Examples follow. 6. When the price of the Integer is 2 farthings, then take the third part of the given Number for so many three halfpences, and the Remainder (if any) is halfpences, then take the eighth part of that for shillings as before, etc. Examples. 7. When the price of the Integer is 3 farthings than take half the given Number for three halfpennies (and if any thing Remain it is 3 farthings, then take the eighth of that for shillings as before, etc. Examples. Case 2. 8. When the given price of the Integer, is a part, or parts of a shilling (viz. pence) divide the given Number of Integers (whose value is sought) by the denominator, of the fraction Representing the even part, and the quote is shillings, (always minding the 7th. Rule of Chapter 9) and those shillings may be Reduced to l. by the 3 Rule of this Chapter. Example, Let it be Required to find the value of 438 l. at 3 d. per l. I consider 3 d. is ¼ of a shilling, and 438 l. will cost so many 3 pences, wherefore I divide 438 by 4 the denominator of ¼ and the quote is 109 shillings, and 2 Remains, which is 2 three pences or 6 d. the whole value is 5 l. 9 s. 6 d. as by the following work appeareth More Examples follow. 9 If the price of the Integer be pence under 12 and yet not an even part, than it may be divided into even parts, and so the parts of the given Number taken accordingly, and added together as if it were 5 d.; which is 3 d. and 2 d. viz. ¼ and ⅙ of a shilling, first take ¼ of the given Number, and then ⅙ thereof, and add them together, and their sum is the Answer in shillings, still observing Rule 7 of Chap. 9 for the Remainders (if any be) then bring the shillings into pounds by the 3 Rule foregoing. Likewise 7 d. is ⅓ and ¼ so 9 d. is ½ and ●/4 and 10 d. is ½ and ⅓ and 11 d. is ⅓ and ⅓ and ¼ of a shilling, as in the following Examples, viz. Case 3. 10. When the price of the integer is penny and farthings, if it make an even part of a shilling work as before, but if they are uneven as penny farthing, penny three farthings, 2 d. 1 qrs. or 2 d. 3 qrs. 3 d. 3 qrs. or the l●ke, than first work for some even part, and then consider what part the rest is of that even part, and divide that quotient thereby, then add them together and reduce them to pounds as before, example 3470 l. at 1 d. 1 qrs. per l. first I work for the penny by dividing 3470 by 12 for 1 d. is 1/1● of a shilling, and the quote is 289 s. 2 d. then I conceive that 1 farthing is ¼ of a penny, and the value at 1 farthing, will be ¼ of the value at 1 penny and therefore I take ¼ of 289 s. 2 d. which is 72 s. 3 d. 2 qrs. and add them together and they are 18 l. 1 s. 5 d. 2 qrs. as by the margin, other examples in the same nature follow. Case 4. 11. When the price of the Integer is 2 s. then cut of the figure in the place of units of the given number, and double it for shillings, and the figures of the other hand are pounds, example 436 yds at 2 s. per yd. cut of the last figure 6 and double it, it makes 12 shill. and the other 2 figures, viz 43 are so many pounds so that their value is 43 l. 12 s. as per margin. 12. Hence it is evident that (when the given price of the integer is an even number of shillings then) if you take half of that (even) number of shillings, and multiply the given number of integers thereby, doubling the first figure of the product and setting it a part for shill. the rest of the product will be pounds, which pounds and shill. is the value sought, example, what cost 536 yds. at 8 s. per yd.? to resolve which I take ½ of 8 s. (the price of a yd.) which is 4 and multiply 536 thereby, saying 4 times 6 is 24 then I double the first figure 4 makes 8 for shill. and carry 2 to the next product etc. I find the rest of the product to be 214 which I note for pounds so the value of 536 yds. at 8 s. per yd. is 214 l. 8 s. as per margin, more examples follow. 56 yds. at 6 s. per yd./ 16 l. 16 s. facit. 420 yds. at 12 s. per yd./ 252 l. facit. 123 yds at 4 s. per yrd./ 24 l. 12 s. facit. 326 yds. at 14 s. per yd./ 228 l. 4 s. facit. 48 els at 8 s. per ell./ 19 l. 4 s. facit. 48 yds. at 16 s. per yd./ 38 l. 8 s. facit. 84 yds. at 10 s. per yd./ 42 l. facit. 52 yds. at 18 s. per yd./ 46 l. 16 s. 13. If the given price of the integer is an odd number of shillings, then work first for the even number of shillings by the last Rule, and for the odd shilling take 1/20 of the given number of integers according to the 3 Rule of this Chapter and add them together and you have your desire, examples follow. 14. Except when the given price of the Integer is 5 s. for than it is sooner answered by taking of the given Number whose value is sought, as in the following Example. Case, 5. 15. When the given price of an integer is shillings and pence, or shillings, pence, and farthings, then if the shillings and pence be an even part of a pound divide the given number of integers whose value you seek by the denominator of that fraction representing that even part, as for example, what is the price of 384 yds at 6 s. 8 d. per yd.? here I consider that 6 s. 8 d. is ⅓ of a pound wherefore I divide 384 by 3 and the quote is the answer viz. 128 l. so that 384 yds. at 6 s. 8 d. per yd. amounts to 128 l. per margin, still observing the 7 Rule of the 9 Chap. More Examples follow. 16. When the given value of the Integer is shillings and pence, and not an even part of a pound, yet many times it may be divided into parts (viz. 6 s. 6 d. is 4 s. and 2 s. 6 d. for the 4 s. work according to the 12 Rules foregoing▪ and for the 2 s. 6 d. take the eighth part of the given Number and add them together their sum is the value Required.) So 8 s. 6 d. will be divided into 6 s. and 2 s. 6 d. and the price of the given Number may be found out as before, etc. Examples follow. 17. When the given price of the Integer is shillings and pence, and you cannot Readily divide them according to the last Rule then multiply the given Number whose value you seek by the Number of shillings in the price of the Integer, and then for the pence, work by the 8th. Rule foregoing, then add the Numbers together, and their sum is the value sought in shillings; as for Example, what is the value of 392 yds. at 6 s. 9 d. per yard? Here 6 s. 9 d. cannot be made any even part, nor indeed can it be divided into even parts of a pound, wherefore I multiply the given Number of yards 392 by 6 for the 6 s. the product is 2352 shillings, then for the 9 d. I divide it into 6 d. and 3 d., and work for them by the 8th. Rule foregoing, and at last add the shillings together they make 2646 s. and by the 3 Rule they are reduced to 132 l. 6 s. the value of 392 yds. at 6 s. 9 d. per yard. See the work following. Other Examples follow. 18. When the given price of the Integer is shillings, pence, and farthings, then multiply the given number of Integers by the Number of shillings contained in the value of the Integer, and for the pence and farthings follow the 10th. Rule of this Chapter Examples. Case 6. 19 When the given value of the Integer is pounds, then multiply the Number of Integers whose value is sought by the price of the Integer and the product is the Answer in pounds. Examples. C. l. C. l. 42 at 2 per C./ 84 l. facit 13 at 8 per C./ 104 l. facit C. l. C. l. 30 at 3 per C./ 90 l. facit 48 at 12 per C./ 576 l. facit. Case 7. 20. If the price of the Integer is pounds and shillings, then for the pounds work as in the last Rule, and for the shillings as in the 12 and 13 Rules before going; then add the Numbers produced from them both, and the sum is the value sought. Examples. 21. When the given price of an Integer consists of pounds' shillings and pence, with farthings, then work for the shillings pence and farthings first, according to the 18. Rule of this Chap and find the total value of the given number, as if there were no pounds, then work with the pounds according to the 19 Rule of this Chapter, and add the numbers thus found, and their sum is the total value Required. Examples of this Rule follow. 22. When there is given the value of an Integer, and it is required to know the value of many such Integers together with ¼ or ½ or 3/● of an integer, than first (by the former rules) find out the value of the given number of Integers▪ and then for ¼ of an Integer take ●/4 of the given value of the Integer, or for 1/● take ½ of the given value of the integer, and for ¾ first take ½ of the given value, and then ½ of that 1/●, setting each part under the precedent, then adding them together their sum will be the Required value of the Integers and their parts. Example, what is the value of 116 ½ yds. at 4 s. 6 d. per yard? To give an Answer, first I work for the value of 116 yds. by the 15 Rule foregoing, and then for the ½ yard I take ½ of 4 s. 6 d. which is 2 s. 3 d. and add to the Rest found as before, then adding them together, I find the total value of 116 ½ yds. at 4 s. 6 d. per yard to amount to 26 l. 04 s. 3 d. as by the work in the Margin. Other Examples follow. Many more questions may be stated, and several other rules of practice may be shown according to the method of divers authors but what have been delivered here are sufficient for the practical Arithmetici-cases whatsoever. CHAP. XXVII. The Rule of Barter. 1. BArter is a Rule amongst Merchants which (in the exchanging of one Commodity for another) informs them so to proportion their Rates as that neither may sustain loss. 2. To resolve questions in barter, it will not be difficult to him that is acquainted with the Golden Rule, or Rule of 3, it being altogether used in resolving such quest●ons. Quest. 1. Two Merchants, viz. A and B barter, A hath 13 C. 3 qrs. 14 l. of Pepper at 2 l. 16 s. per C. and B hath Cotton at 9 d. per l. I demand how much Cotton B must give A for his Pepper? Answer. 9 C. 1 qr. First find by the Rule of 3 how much the Pepper is worth, saying If 1 C. cost 2 l. 16 s. what will 13 C. 3 qrs. 14 l. cost. Answ. 38 l. 17 s. Secondly by the Rule of 3 say, if 9 d. buy 1 l. of Cotton how much will 38 l. 17 s. buy? Answer. 9 ¼ and so much Cotton must B give to A for 13 C. 3 qrs. 14 l. of pepper at 2 l. 16 s. per C. when the Cotton is worth 9 d. per l. Quest. 2. Two Merchants (A and B) barter, A hath Ginger worth 1 l. 17 s. 4 d. per C. but in barter he will have 2 l. 16 s. per C. B. hath Nutmegs worth 5 l. 12 s. per C. now I demand how B must rate his Nutmegs per C. to make his gain in barter equal to that of A? Answer, 8 l. 8 s. Say by the Rule of 3 if 1 l. 17 s. 4 d. require 2 l. 16 s. in barter what will 5 l. 12 s. require in barter? Facit 8 l. 8 s. Quest. 3. A and B barter, A hath 120 yds of Broadcloath worth 6 s. per yd. but in barter he will have 8 s. per yd. B hath shalloon worth 4 s. per yd. Now I demand how many yds. of shalloon B must give A for his broadcloth making his gain in barter equal to that of A? Answer, 180 yds. of shalloon. First (as in the last question) find out how B. ought to sell his shalloon in barter, viz. say If 6 s. require 8 s. what will 4 s. require? Answer. 5 s. 4 d. Thus you see that B must sell his shalloon in barter at 5 s. 4 d. if A sell his broadcloth at 8 s. per yd. It remaineth now to find how much shalloon B must give for 120 yds. of broadcloth, which after the same method used to resolve the first question of this Chap. is found to be 180, and so many yds of shalloon must B give A for the 120 yds. of broad-cloath. Quest. 4. A and B bartered A had 14 C. of Sugar worth 6 d. per l. for which B gave him 1 C. 3 qrs. of Cinnamon, I demand how B rated his Cinnamon per l.? Answer 4 s. per pound. Quest. 5. A and B barter, A hath 4 Tun of Brandy worth, 37 l. 16 s. ready money, but in barter he hath 50 l. 8 s. per tun, and A giveth B 21 C. 2 qrs. 11 1/● l. of Ginger for his 4 tun of Brandy, I desire to know how B sold his Ginger in barter per C. and how much it was worth in ready money? Answer. For 9 l. 6 s. 8 d. in barter, and it was worth 7 l. per C. ready money. Quest. 6. A and B barter A hath 320 dozen of Candles at 4 s. 6 d. per dozen, for which B giveth him 30 l. in money, and the rest in Cotton at 8 d. per I. I demand how much Cotton he must give him more than the 30 l.? Answ. 11 C. 1 qr. Quest. 7. A and B barter, A hath 608 yds. of broadcloth worth 14 s. per yd. for which A giveth him 125 l. 12 s. ready money and 85 C, 2 qrs. 24 l. of Bees-wax, now I desire to know how he reckoned his Wax per C.? Answer. 4 l. 13 s. 4 d. per C. CHAP. XXVIII. Questions in Loss and Gain. Quest. 1. A Merchant bought 436 yds of broadcloth for 8 s. 6 d. per yd. and selleth it again at 10 s. 4 d. per yrd. now I desire to know how much he gained in the sale of the 436 yds? Answ. 39 l. 19 s. 4 d. First find out by the Rule of Three or Practise how much the Cloth cost him at 8 s. 6 d. per yd. viz. 185 l. 6 s. then by the same Rule find out how much he sold it for viz. 225 l. 5 s. 4 d. then subtract 185 l. 6 s. which it cost him, from 225 l. 5 s. 4 d. which he sold it for, and there remaineth 39 l. 19 s. 4 d. for his gain in the sale thereof. Otherwise it may sooner be resolved thus, first find out how much he gained per yd. viz. subtract 8 s. 6 d. which he gave per yd. from 10 s. 4 d. which he sold it for per yd. the remainder is 1 s. 10 d. for his gains per yd. then say. If 1 yd gain 1 s. 10 d. what will 436 yds. gain? Answer. by the Rule of three is 39 l. 19 s. 4 d. as was found before. Quest. 2. A Draper bought 124 yds. of Holland cloth, for which he gave 31 l. I desire to know how he must sell it per yd. to gain 10 l. 6 s. 8 d. in the whole sale of the 124 yds? answ. at 6 s. 8 d. per yd. Add the price which it cost him, (viz. 31 l.) to his intended gain (viz. 10 l. 6 s. 8 d.) the sum is 41 l. 6 s. 8 d. then say. If 124 yds require 41 l. 6 s. 8 d. what will 1 yd. require? by the Rule of Three I find the answer 6 s. 8 d. Quest. 3. A Grocer bought 3 C. 1 qrs. 14 l. of Cloves which cost him 2 s. 4 d. per l. and sold them for 52 l. 14 s. I desire to know how much he gained in the whole? answer 8 l. 12 s. Quest. 4. A Draper bought 86 〈◊〉 for 129 l. I demand how he must sell the● per piece to gain 15 l. in laying out 100 l. at that rate? answer 1 l. 14 s. 6 d. per piece, for As 100 l. is to 115 l. so is 129 l. to 148 l. 7 s. So that by the proportion above, I have found how much he must receive for the 86 Kerseys to gain after the rate of 15 l. per C. then to find how he must sell them per piece I say. As 86 pieces is to 148 l. 7 s. so is 1 piece to 1 l. 14 s. 6 d. which is the number sought. Quest. 5. A Grocer bought 4 ¼ C. of pepper for 15 l. 14 s. 7 d. and (it proving to be damnified) is willing to lose 12 l. 10 s. per C. I demand how he must sell it per l.? answer 7 d. per l. Subtract 12 l. 10 s. the loss of 100 l. from l. and there remains 87 l. 10 s. then say As 100 l. is to 87 l. 10 s. so is 15 l. 17 s. 4 d. to 13 l. 17 s. 8 d. so much as he must sell it all for to lose after the rate propounded, then to know how he must sell it per l. I say. As 13 l. 17 s. 6 d. is to 4 ¼ C. so is 1 l. to 7 d. Quest. 6. A Plummer sold 10 fodder of Lead (the fodder containing 19 ½ C.) for 204 l. 15 s. and gained after the rate of 12 l. 10 s. per 100 l. I demand how much it cost him per C.? answer 18 s. 8 d. To resolve this question add 12 l. 10 s. (the gain per C.) to 100 l. and it makes 112 l. 10 s. then say. As 112 l. 10 s. is to 100 l. so is 204 l. 15 s. to 182 l. Which 182 l. is the sum it cost him in all then reduce your 10 fodders to half hundreds and it makes 390, then say As 390 half hundreds is to 182 l. so is 2 half hundreds to 18 s. 8 d. the price of 2 half hundreds or one C. weight and so much it stood him in per C. Quest. 7. A Merchant bought 8 tuns of Wine, which being sophisticated he selleth for 400 l. and loseth after the rate of 12 l. in receiving 100 l. now I demand how much it cost him per tun? and how he must sell it per gall. to lose after the said rate? answer it cost 56 l. per tun, and he must sell it at 3 s. 11 d. 2 10/21 qrs. per gallon to lose 12 l. in receiving 100 l. To resolve this question I consider in the first place, that in receiving 100 l. he loseth 12 l. therefore 100 comes in, for 112 l. laid out, wherefore to find how much he laid out for the whole I say As 100 l. is to 112 so is 400 l. to 448 l. and so much the 8 tun cost him, then to find how much it cost per tun, I say As 8 is to 448 l. so is 1 to 56 l. the price it cost per tun. Now to find how he must sell it per gall. reduce the 8 tuns into gallons they make 2016, then say As 2016 gall. is to 400 l. so is one gall. to 3 s. 11 d. 2 10/2● qrs. the price he must sell it per gallon to lose as aforesaid. Quest. 8. A Merchant bought 8 tuns of Wine, which being sophisticated he is willing to sell for 400 l. and loseth 12 l. in laying out 100 l. upon the same, now I demand how much it cost him per tun? Here I consider that for 100 l. laid out he receiveth but 88 l. therefore to find what the 8 tuns cost him I say As 88 l. to 100 l. so is 400 l. to 454 6/11 the price it all cost him, then to find how much per tun I say As 8 is to 454 6/11 l. so is one to 56 9/11 or 56 l. 16 s. 4 d. 1 5/11 qrs. per tun. CHAP. XXIX. Equation of Payments. 1. EQuation of payments, is that Rule amongst Merchants whereby to reduce the times for payment of several sums of money to an equated time for the payment of the whole debt without damage to Debtor or Creditor, and The Ruie is 2. Multiply the sums of each particular payment by its respective time, then add the several products together and their sum divide by the total debt, and the quotient thence arising is the equated time for the payment of the will debt. Example. Quest. 1. A is indebted to B in the sum of 130 l. whereof 50 l. is to be paid at 2 months, and 50 l. at 4 months, and the rest at 6 months, now they agree to make one payment of the total sum, the question is what is the equated time for payment without damage to debtor or creditor? To resolve this question I multiply each payment by its time, viz. 50 l. multiplied by 2 months produceth 100 50 l. multiplied by 4 months produceth 200 30 l. multiplied by 6 months produceth 180 The sum of the products is 480 Then I divide 480 (the sum of the products) by 130 (the total debt) and the quotient is 3 9/13 months for the time of paying the whole debt. Quest. 2. A Merchant hath owing him 1000 l. to be paid as followeth, viz. 600 l. at 4 months, 200 l. at 6 months. and the the rest which is 200 l. at 12 months; and he agreeth with his debtor to make one payment of the whole, I demand the time of payment without damage to debtor or Creditor; 600 l. multiplied by 4 months is 2400 200 l. multiplied by 6 months is 1200 200 l. multiplied by 12 months is 2400 The sum of the products is 6000 and the sum of the products (6000) divided by the whole debt 1000 l. quotes 6 mon. for the time of payment of the whole debt. 3. The truth of this Rule is thus manifest, if the interest of that money which is paid (by the equated time) after it is due, The proof of the Rule of Equation of Payments. be equal to the interest of that money which (by the equated time) is paid so much sooner than it is due at any rate per C. then the operation is true, otherwise not. example. In the last question 600 l. should have been paid at 4 months but it is not discharged till 6 months (that is 2 months after it is due) wherefore its interest for 2 months at 6 per C. per annum is 6 l. and then 200 l. was to be paid at 6 months which is the equated time for its payment, therefore no interest is reckoned for it, but 200 l. should have been paid at 12 mon. but it is to be paid at 6 months which is 6 months' sooner than it ought, wherefore the interest of 200 l. for 6 months is 6 l. (accounting 6 l. per Cent. per annum) which is equal to the interest of 600 l. for 2 months wherefore the work is right. Quest. 3. A Merchant hath owing him a certain sum to be discharged at 3 equal payments viz. ⅓ at two months' ⅓ at four mon. and ⅓ at 8 mon. the question is what is the equated time for the payment of the whole debt? In questions of this nature (viz. where the debt is divided into equal or unequal parts) each of the parts is to be multiplied by its time, and the sum of the products is the answer ⅓ multiplied by 2 mon. produceth ⅔ ⅓ multiplied by 4 months produceth 1 ⅓ ⅓ multiplied by 8 months produceth 2 ⅔ The sum of the products is 4 ⅔ which is 4 ⅔ months for the equated time of payment. If instead of the fractions (representing the parts) you had wrought by the numbers themselves (represented by those parts) according to the first and second, examples, it would have been the same answer, as suppose the debt had been 90 l. then ⅓ of it is 30 l. for each payment, viz. at 2, 4 and 8 months, than 30 l. multiplied by 2 mon. produceth 60 30 l. multiplied by 4 mon. produceth 120 30 l. multiplied by 8 mon. produceth 240 The sum of the products is 420 which divided by 90 (the whole debt) quoteth 4 60/90 or 4 ⅔ months as before. Quest. 4. A Merchant oweth a Sum of money, to be paid ½ at 5 Months, and ¼ at 8 Months, and ¼ at 10 Months, and he agreeth with his Creditor to make one total payment; I demand the time, without damage to Debtor or Creditor? Work as in the last Question, and you will find the Answer to be 7 Months. Quest. 5. A is indebted to B 640 l. whereof he is to pay 40 l. present money, and 350 l. at 3 Months, and the rest (viz. 250 l.) at 8 Months, and they agree to make an Equated time for the whole Payment, now I demand the time? In Questions of this Nature (viz. where there is ready money paid) you are (in Multiplying) to neglect the money that is to be paid present, and work with the rest as is before directed, and Divide the sum of the Products by the whole Debt, and the Quote is the Answer: for here 40 l. is to be paid present, and hath no time allowed, and according to the Rule it should be Multiplied by its time, which is (0) therefore 40 times 0 is 0, which neither augmenteth nor diminisheth the Dividend; wherefore (to proceed according to direction) I say, 350 by 3 Months, produceth 1050 250 by 8 Months, produceth 2000 The Sum of the Products is 3050 which Divided by 640, the whole Debt, the Quote is 4 49/64 Months, the time of payment. Quest 6. A is indebted to B in a certain Sum, ½ whereof is to be paid present money, ⅓ at 6 Months, and the rest at 8 Months; now I demand the Equated time for the payment of it all? Answer. 3 ⅓ months is the time of payment. Quest. 7. A is indebted to B 120 l▪ whereof ●/● is to be paid at 3 months, ¼ a● 6 months, and the rest at 9 months; what is the Equated time for the payment of the whole Sum? Answer. At 6 ¼ months. Quest. 8. A is indebted to B 420 l. which is due at the end of 6 months; but A i● willing to pay him 140 l. present, provi●ded he can have the remainder forborn so much the longer to make Satisfaction fo● his kindness, which is agreed upon, I desire to know what time ought to be allotted for the payment of the 280 l. remaining? To resolve this Question, first, find ou● what is the Interest of 140 l. for the time i● was paid before it was due, at 6 per Cent. (or any other rate) (viz. 6 months) and you will find it to be 4 l. 4 s. Then it is evident that the remaining 280 l. must be detained so much longer than 6 months, as the while it may eat out that Interest, viz. 4 l. 4 s. which is thus found out, viz. First, see what is the Interest of 280 l. for a month, or any other time; but here we will take one month, and its Interest for one month is 28 s. Then by the Rule of Three say, As 28 s. is to 1 month, so is 84 s. to 3 months; so that the 280 l. remaining must be kept 3 months beyond its first time of payment, (viz. 6 months) which added thereto, makes 9 months, at the end of which time A aught to make payment of the remainder. CHAP. XXX. Exchange. 1. THE Rule of Exchange informeth Merchants how to Exchange Moneys, Weights, or Measures of one Country, into (or for) the Moneys, Weights, or Measures of another Country, and when the Rate, Reason, or Proportion betwixt the Money, Weights, or Measures of different Country's is known, it will not be difficult for the Practitioner that is well acquainted with the Rule of Proportion (or Rule of Three) to solve any Question wherein it is required to Exchange a given quantity of the one kind, into the same value of another kind. 2. In Questions of Exchange, there is always a Comparison made between the Coins, etc. of two Countries (or kinds) or of more. 3. In Questions where there is a Comparison made between two things (whether they be Moneys, Weights, etc.) of different kinds (or Countries) there may be a solution found by a single Rule of 3, as may appear by the following Example. Quest. 1. A Merchant at London delivered 370 l. Sterling, to receive the same at Paris in French Crowns, the exchange 3 ⅓ French Crowns per pound Sterling. I demand how many French Crowns ought he to receive? In placing the Numbers observe the 6 Rule of the 10 Chapter, which being done, this given Numbers will stand thus, l. Crowns. l. 1 3 ⅓ 370 and being Reduced according to the Rules of the 24 Chapter, will stand thus, As is to so is to So that I conclude he ought to Receive 1233 ⅓ French Crowns at Paris for his 370 l. delivered at London. Quest. 2. A Merchant delivered at Amsterdam 587 l. Flemish, to receive the value thereof at Naples in Ducats, the exchange 4 ⅘ Ducats per l. Flemish: I demand how many Ducats he ought to receive? The Proportion is as followeth. As is to so is to So I find he ought to receive 2817 ⅗ Ducats at Naples for the 587 l. Flemish delivered at Amsterdam. Quest. 3. A Merchant at Florence delivereth 3478 Ducatoons, to receive the value at London in pence, the exchange 53 1/● pence Sterling per Ducatoon; I demand how much Sterling he ought to receive? The Proportion for Resolution is, As is to so is to which is equal to 775 l. 6 1/12 for the Answer. I might here (according to the Custom of Arithmetical Writers) lay down Tables for the Reduction of Foreign Coins to English; but by Reason of their Instability (for they continue not at a constant standard, as our Sterling money doth, but are sometimes raised, and sometimes depressed) I shall forbear. 4. When there is a Comparison made between more than two different Coins, Weights, or Measures, there ariseth ordinarily two different cases from such a Comparison. 1. When it is Required to know how many Pieces of the first Coin, Weight, or Measure are equal in value to a known number of Pieces of the last Coin, Weight, or Measure. 2. When it is Required to find out how many Pieces of the last Coin, Weight, or Measure are equal in value to a given number of the first sort of Coin, Weight, or Measure. An Example of the first Case may be this, VIZ. Quest. 4. If 150 pence at London are equal to 3 Ducats at Naples, and 4 4/● Ducats at Naples make 34 ½ Shillings at Brussels, then how many pence at London are equal to 138 shillings at Brussels? Facit 960 d. This Question may be Resolved at two single Rules of Three; for first I say, If 3/1 Ducats at Naples make 150 pence at London, how many pence will 4 ⅘ Ducats make? Answ. 240 pence. By the foregoing Proportion we have discovered, that 4 ⅘ Ducats at Naples make 240 pence at London: And by the Tenor of the Question we see that 4 ⅘ Ducats at Venice make 34 ½ shillings at Brussels, therefore 240 pence at London are equal to 34 ½ shillings at Brussels, (for the things that are equal to one and the same thing, are also equal to one another,) wherefore we have a way laid open to give a solution to this Question by another Single Rule of Three, whose Proportion is, As 34 ½ shillings at Brussels is to 240 pence at London; so is 138 shillings at Brussels to 960 pence at London, which is the Answer to the Question. An Example of the Second Case may be this, VIZ. Quest. 5. If 40 l. Averdupoize weight at London is equal to 36 l. weight at Amsterdam; and 90 l. at Amsterdam makes 116 l. at Dantzick, then how many pounds at Dantzick are equal to 112 l. of Averdupois weight at London? Answer. 129 23/25 pounds at Dantzick. This Question is likewise answered at two single Rules of Three, viz. First, I say, As 36 l. at Amsterdam is to 40 l. at Lond. So is 90 l. at Amsterdam to 100 l. at Lond. And by the Question you find that 90 l. at Amsterdam is 116 l. at Dantzick; and therefore 100 l. at London is likewise equal thereunto, wherefore again I say, As 100 l. at London is to 116 l. at Dant. So is 112 l. at London to 129 23/25 at Dant. By which I find that 112 23/25 l. at Dantzick are equal to 112 l. Averdupois weight at London. 5. There is a more speedy way to Resolve such Questions as are contained under the two Cases before mentioned, laid down by Mr. Kersey in the third Chapter of his Appendix to Mr. Wingates Arithmetic, where he hath given two Rules for the Resolution of the Questions pertinent to the two said Cases. 6. But I shall lay down a general Rule for the solution of both Cases; and first, let the Learner observe the following Directions in placing of the given terms, viz. 7. Let there be made two Collumes, and in these Collumes so place the given terms one over the other, as that in the same Collume there may not be found two Terms of the same kind one with the other. Having thus placed the Terms, the General Rule is, Observe which of the said Collumes hath the most Terms placed in it, and multiply all the Terms therein continually, and place the last Product for a Dividend; then multiply the Terms in the other Collume continually, and let the last Product be a Divisor; then divide the said Dividend by the said Divisor, and the Quotient thence arising is the answer to the question. So the example of the first of the said cases being again repeated, viz. If 150 pence at London make 3 ducats at Naples, and 4 ⅘ ducats at Naples make 34 ½ shill. at Brussels, then how many pence at London are equal to 138 shillings at Brussels? The terms being placed according to the 7 Rule will stand as followeth. A B Pence at Lond. 150 3 Duc. at Na. Ducats at Na. 4 4/● 34 ½ shill. at Bruss. Shi●l. at Bruss. 138 having thus placed the terms, that in neither collume there is 2 terms of one kind, than I observe that the collume under A hath most terms in it, therefore they must be multiplied together for a dividend; viz. 150 mult. by 4 4/● produceth 3600/5 which multiplied by 138 produc. 106800/5 for a dividend, then in the collume under B there are 3 and 34 ½ which multiplied together produceth 207/2 for a divisor; then having divided 496800/● by 207/2 the quotient is 960 pence for the answer as before. Again, let the example of the second case be again repeated, viz. If 40 l. Averdupois weight at London make 36 l weight at Amsterdam, and 90 l. at Amsterdam make 116 l. at Dantzick, then how many pounds at Dantzick are equal to 112 l. Averdupois weight at London. The terms being disposed according to the 7 Rule foregoing will stand thus A B l. at Lond. 40 36 l. at Amsterdam. l. at Amst. 90 116 l. at Dantzick. 112 l. at London. whereby I find that the terms under B multiplied together produce 467712 for a dividend, and the terms under A viz. 40 and 90 produce 3600 for a divisor and division being finished the quotient giveth 129 3312/3600 pounds at Dantzick for the answer. CHAP. XXXI. Single Position. 1. NEgative Arithmetic called the Rule of False, is that by which we find out a truth, by numbers invented or supposed, and this is either single or double. 2. The Rule of Single Position is when at once, viz. by one False position, or feigned number, we find out the true number sought. 3. In the single Rule of False, when you have made choice of your position, work it according to the tenor of the question, as if it were the true number sought, and if by the ordering of your position you find the result either too much or too little you may then find out the number sought, by this proportion following viz. As the result of your position, is to the position, so is the given number to the number sought. Example. Quest. 1. A Person having about him a certain number of Crowns, said if the fourth and third and sixth of them were added together they would make just 45, now I demand the number of Crowns he had about him? Answer 60 Crowns. To resolve this question I suppose he had 24 Crowns (or any other number that will admit of the like division) now the fourth of 24 is 6, and the third is 8, and the sixth is four, all which parts (viz. 6, 8, and 4,) being added together make but 18, but it should be 45, wherefore I say by the Rule of three. As 18 the sum of the parts, is to the position 24 so is 45 the given number to 60 the true number sought. For the fourth of 60 is 15, and the third of 60 is 20, and the sixth of 60 is 10, which added together make 45. Quest. 2. Three Persons, viz. A, B, C, thus discourse together concerning their age, quoth B to A I am as old, and half as old again as you, than quoth C to B I am twice as old as you, than quoth A to them, and I am sure the sum of all our ages is 165, now I demand each man's age? Answer A 30, B 45, C 90, years of age, which added together makes 165. CHAP. XXXII. Double Position. 1. THe Rule of Double Position is, when 2 False positions are assumed, to give a resolution to the question propounded. 2. When any question is stated in double position, make such a cross as followeth. 3. Then make choice of any number you think may be convenient for your working, which call your first position, and place it at that end of the cross at a, then work with this position (as if it were the true number sought,) according to the nature of your question, then having found out your error, either too much or too little place, it on that side the cross d, then make choice of another number of the same denomination with the first position (which call your second position) and place it on that side of the cross at b, then work with this position as with the former, and having found out your error, either too much or too little, place it on that side of the cross at c, and then the positions will stand at the top of the cross, and the errors at the bottom each under his correspondent position, and then multiply the errors into the positions crosswise, that is to say multiply the first position by the second error, and the second position by the first error, and put each product over its position. 4. Having proceeded so far, then consider whether the errors were both alike, that is, whether they were both too much, or both too little, and if they are alike, then subtract the lesser product from the greater and set the remainder for a dividend, then subtract the lesser error from the greater, and let the remainder be a divisor, than the quotient arising by this division is the answer to the question. 5. But if the errors are unlike, that is one too much and the other too little, then add the products of the positions and errors, together and their sum shall be a dividend, then add the errors together and their sum shall be a Divisor, and the quotient arising hence is the Answer; which two last Rules may be kept in memory by this verse following, viz. When Errors are of unlike kinds Addition doth ensue But if alike, Subtraction finds Dividing work for you. Quest. 1. A, B and C Build a house which cost 76 l. of which A paid a certain sum unknown, B paid as much as A and 10 l. over, and C paid as much as A and B, now I desire to know each man's share in that Charge? Having made a Cross according to the 2 Rule, I come according to the third Rule to make choice of my first position, and here I suppose A paid 6 l. which I put upon the Cross as you see, then B paid 16 l. (for it is said he paid 10 l. more than A) and C paid 22 l. for 'tis said he paid as much as A and B, than I add their parts and they amount to 44, but it is said they paid/ 2 l. wherefore it is 32 too little, which I note down at the bottom of the Cross under its position for the first error. Secondly, I suppose A paid 9 l. then B paid 19 l. and C 28 l. all which added together make 56, but they should make 76, wherefore the error of this position is 20 which I put at the bottom of the Cross under his position, for the second Error, than I multiply the Errors and the Positions crosswise, viz. 32 (the Error of the first Position) by 9 (the second position) and the product is 288. Then I multiply 20 (the Error of the second Position) by 6 (the first Position) and the Product is 120. Then (according to the 4th. Rule) I subtract the lesser product from the greater, (viz. 120 from 288 because the Errors are both alike, viz. too little) and there Remaineth 168 for a dividend, than I subtract 20 (the lesser Error) from 32 (the greater Error) and the remainder is 12 for a Divisor, then divide 168 by 12 and the Quotient is 14 for the Answer, which is the share of A in the payment. 6. Again secondly, If the errors had been both too big it had had the same effect, as appeareth by the following work; for first I suppose A paid 20 l. then B paid 30 l. and C 50 l. which in all is 100 but it should have been no more than 76, wherefore the first Error is 24 too much. Again I suppose A paid 18 l. then B must pay 28 l. and C must pay 46 l. which in all is 92 l. but it should have been but 76 l. wherefore the second Error is 16 too much; then I Multiply 20 (the first Position) by 16 (the second Error, and the product is 320, again I multiply 18 (the second Position) by 24 (the first Error) and the product is 432. Then because the Errors are both too much I subtract 320 (the lesser product) from 432 (the greater product) and there Remaineth 112 for a Dividend, likewise I subtract 16 (the lesser Error) from 24 (the greater Error) and the difference is 8 for a Divisor, then perform Division, and the Quotient is 14 (as before) for the answer. Again Thirdly, If the Errors had been the one too big, and the other too little, Respect being had to the 5th. Rule foregoing, the answer would have been the same; as thus, I take for my first Position 6, and then the error is 32 too little, than I take for my second Position 18, and then the error is 16 too much, than I multiply the Positions and errors crosswise, and the products are 96 and 576, and because the errors are unlike viz. one too big, and another too little, I add the products 96 and 576 together, and their sum is 672 for a Dividend, I likewise add the errors 32 and 16 together, and their sum is 48 for a Divisor, then having finished Division I find the quotient to be 14 which is the answer, as was found out at the 2 several Trials before. For Proof of the work I say l. If A paid 14 Then B paid 14 and 10 (that is) 24 And C paid 14 and 24 (that is) 38 The Sum of all is 76 which is the total value of the building and equal to the given Number. Those who desire to see the demonstration of this Rule, let them Read the 7th. Chapter of Mr. Kersyes Appendix to Wingates Arithmetic, Petiscus in the 5th. Book of his Trigonometria. Or Mr. Oughtred in his Clavis Mathematicae. Quest. 2. Three Persons, A, B, C, thus discoursed together concerning their age; Quoth A I am 18 years of age, quoth B I am as old as A and ½ C; and quoth C, I am as old as you both, if your years were added together, Now I desire to know the age of each Person? Answer A is 18, B is 54, and C is 72, years of age. Quest. 3. A Father lying at the point of Death, left to his 3 Sons, viz. A, B, C, all his Estate in Money, and Divided it as followeth, viz. to A he gave ½ wanting 44 l. to B he gave ⅓ and 14 l. over, and to C he gave ⅙ and 30 l. over, now I demand what was the sum left, and each man's part? Answer, the sum bequeathed was 588 l. whereof A had 250 l. B had 210 l. and C had 128 l. Quest. 4. Two Persons viz. A and B had each in their hands a certain Number of Crowns, and A said to B, If you give me 1 of your Crowns I shall have 5 times as many as you, and said B to him again, If you give me one of yours than we shall each of us have an equal Number; now I demand how many Crowns had each Person? Answer, A had 4 and B had 2 Crowns. Quest. 5. What Number is that unto which if I add ¼ of itself, and from the sum subtract ⅛ of itself the Remainder w●ll be 210? Answer 192. Many more questions may be added but these well understood will be sufficient (even for the meanest Capacity) for the Resolution of any other question pertinent to this Rule. There may be an Objection made because we have not treated particularly upon Interest, and Rebate, but the operation of such questions, being more applicable to Decimals, are omitted, till we come to acquaint the Learner therewith. Laus Deo Soli. FINIS. 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