Theory of Gunnery, independent of -the Conic Sections

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EXERCISES FOR

Young Proficients^^ '%

I N T H E »* r

MATHEMATICKS.. 7

Containing, I. A large Variety of Algebraical Problems with their Solutions.

II. A choice Number of Geometrical Problems with their Solutions both Algebraical and Geome trical.

HE The. Theory of Gunnery, independent of -the Conic Sections.

IV. A new and very comprehensive Method for finding the Roots of E- quations in Numbers.

V. A short Account of the Nature and first Prin ciples of Fluxions.

VI. The Valuation of Annuities for fingle and joint Lives, with a Set of new Tables, far more ex tensive than any extant.

By THOMAS SIMPSON, F. R. S.

LONDON:

Printed for J. Nourse, opposite Ketharint-Street in;

the Strand. M.DCC.LH.

TO

JOHN BACON,

Of Newtoncafi, Esq; F. R. S.

Sir,

HEN Gentlemen of your

Station and Figure become

the Patrons of Science it is a Benefit to the Publick, their Expecta tions of farther Improvements having then the best Foundation. And All who have the Pleasure of your Acquaint

ance, and know your Attachment to

polite and useful Learning, in which

a Knowledge of the Mathematicks

may be justly included, will be sen sible of my Happiness in being thus permitted to address You.

Believe

Believe me, Sir, whatever may be the Fate os these Sheets, I mall, at all Times, consider this Use of your Name as a singular Honour to,

Sir,

Tour most obedient , and mojl

Humble Servants

T. Simpson.

Woolwich, May t, in2'

THE

PREFACE.

HE ensuing Work, or at least the greatest Part of it, was originally composed for my civn Use in the Royal Academy: And it is upon a "Presumption that It may also be os Service to Others, especially 'Those employed in a like Publick Way os Teaching, that It now appears in the World.

The Work itself coiistjls offix distinB Parts, or Trails ; each of which I sjall here give some Account of.

The frft Part contains a Number of Alge braical Problems, with their . Solutions ; de signed as proper Exercises for Young Beginners.

In the Course of these Problems and Solutions (whereof the greater Part will appear to be new ) the Art of managing Equations, and the various Methods of Substitution are taught and illustrated.

a The

ii The PREFACE.

The second Part comprehends a Variety oj Geometrical Problems with their Solutions, both by Algebra and aljo independent of it, from Principles purely Geometrical. In this Part the Learner will find a large Field to exercise his Industry in : He will moreover have the Opportunity of comparing the two Methods of Solution together, and from thence observing, that sometimes the One has the Ad vantage, and sometimes the Other-, that, in some Cases 'They both proceed upon the very fame Properties, and in others, upon quite different Ones : Aud it may be further re marked from hence (which will be of some Use to know) that, when Quantities are given in Magnitude only, the Algebraic Method generally claims the Preference, in Point of Ease and Expedition, at least ; whereas the Advantage is almost always on the Side of the Geometrical EffeSlion, when the Positions of Points and Lines, and the Quantities of Angles are given.

There is, however, one Particular, or Two, in this Part, that may be thought to stand in need of some Apology.

In the first Place, thefrequent Use of Sym bols, common to the Algebraic Notation, may, perhaps, be look'd upon as repugnant to the

Rigour

The PREFACE. iii Rigour and Strictness of Geometry. But it is not the Use of Symbols (which Some, more scru pulous than discerning, have condemned) but

the Ideas annexed to them, that render the Con sideration Geometrical, or Ungeometrical. In pure Geometry regard is always had to the absolute Quantity ofsome One of the three Kinds of Extension, abstractedly considered j and,

•whatever Symbols are used Here, are to be considered as expressive of the Quantities them selves, and not as any Measures, or numerical

Values of them. 'Thus by A X B, taken _ in a geometrical Sense, we have an Idea, not of the Product of two Numbers (as in the Algebraic Notation), butof a real, rectangular, Space comprehended under two Right-lines, repre sented by A and B, and two Others equal to

them. So, likewise, - * - is not to be under stood here in • the Light of an Algebraic

Fraction, but as a Right-line, which is Fourth Proportional to three other Right-lines, repre sented by A, B, and C. — These Distinctions

are absolutely necessary to Those who would have an accurate Idea os the Subject .

The second Particular, above hinted at, re lates to the Quotations ; wherein I have re ferred to my own' Elements of Geometry, and

not

iv The PREFACE.

not to those of Euclid, so universally known and established. But for this there were two Reasons : Firfl, those Persons, for whose In struction these Sheets are, in a more particular

manner, designed, are taught the Firs Prin ciples of Geometry from other Elements than those of Euclid ; and, secondly, a Number of Propositions are used Here that are only to be met with in modern Authors.

In the third Part the Theory of Gunneryt or the Motion of Projectiles, is considered, ex clusive of the Conk Sections ; and the practical Solutions of the several Cafes depending on the Theory (as well Those where the Object is ele vated or depressed as where it% is situate in the Plain of the Horizon) are given, at large, by Plane Trigonometry.

Tlje fourth Part exhibits a new, and very comprehensive Method for extracting the Roots of algebraical Equations ; whereby the Num ber sought may be determined, to any proposed Degree of Exactness, without the Trouble' of repeating the Operation, as in the common Way, by Converging Series's.

The fifth Part gives some Account of the Nature of Fluxions, together with the Investi gation of the Fundamental Rules ; and may

be of Use, not only to Beginners, but also to Such)

The PREFACE. v Such, who, though tolerably well versed in the Practice and Application of Fluxions, have nevertheless but an imperfect Idea of the First Principles of this difficult Branch of Science.

The sixth, and last, Part, is concerned about the Valuation of Annuities, on Jingle, and joint Lives ; wherein, befdes a new Set of Tables, far more comprehensive than any yet published, are given the Solutions of up

wards of Forty different Problems, on the most important and intricate Cafes of the Subject : Many of which are quite new\ and are, be sides, such as actually occur in Business, being, most of them, taken from real Cafes, proposed to the Author's Consideration, by Gentlemen in the Law, and Others.

This, fxth, Part, which the Reader will perceive is upon a different Plan from the fve preceding Ones, was designed as a Supplement to my Doctrine of Annuities and Reversions, printed in 1 7 42 j but, being thought too small to publisld alone, it is inserted Here. This, if a Fault, will not, perhaps, be look'd upon as inexcusable : Though, as to the Performance itself, 1 do not in the least doubt but that it will be depreciated by Some, on Account of the Ob servations whereon the Calculations are ground- . cd. I am sensible that there neither is, nor can

vi The PREFACE.

can be, a 'Table of Observations on the De grees of Mortality of Mankind, but what may

be objected to ; and that those Persons who make a very despicable Figure, when They come to Calculations, seldom fail os Dis playing their Talents, and being illustrious

Here; where, gratifying themselves in the Liberty which the Nature of the Subject allows them, They can boldly lanch out, without ha ving to do with Science and Demonstration.

But, though the London Bills of Mortality, whereon I build, appear to me to be the best Foundation, at least, for this Place ; yet I have no Inclination to enter the Lists with any of these Gentlemen. The Examples, given hereafter, are indeed wrought according to the London Bills; but the Solutions themselves are general, and may be apply'd, with equal Facility and Advantage, to any Table of Ob servations.

PART

-

PART I.

CONTAINING

A seleSi Number of Algebraical

Problems, with their Solutions.

DESIG N'D

As proper Exercises for young Beginners*

Q.U ESTION L

7JTHAT Number is that, which being doubled anJ 'y 1 6 added to the ProduEt, the Sum Jhall U 188 ?

Let represent the required Number ; then 2x will denote the Double thereof;

and so 2*+i6=188, by the Question.

Therefore 2x=z 1 88- 16=17 1, by Transpoftion*

And x=z- —86, by Divijitn.

Q_U ESTION II.

To find that Number, which being added to 56, the Treble of the required Number Jhall be produced.

If x be put for the Number sought, then yt will be the Treble thereof:

And therefore %x—x-\- $6, by the ^ucstion.

Hence lx—^6, by Transpoftion.

And *\=— =28, by Division.

B QUESTION

2 Algebraical Problems, Q_U E S T I O N III.

The Sum of 155 /. was raised (for a certain Purpose) by three different Persons, A, B, and C; whereof B ad vanced 15 /. more than A ; and C, 20/. more than B : How much did each contribute ?

Let x be the Number of Pounds advanced by A:

Then x-\-i$, is the Number of Pounds advanced by B, and x-\-25 tn< Number of Pounds advanced by C.

Therefore 3x4-50.^155, by the ^uestion.

Whence 3*=. 105, and x— 35.

From which it also appears that B contributed 50 /. and C'70/.

Q,U E S T I O N IV. '

A Gentleman (ly Will) left 550/. to be divided among four Servants A, B, C, and D ; whereof B was to have twice as much as A; C as much as A and B ; and D as much as C and B. How much had Each ? Let * be the Number of A's Pounds.

Then 2x is the Number of B's Pounds, also 3* is the Number of C's Pounds, and 5* the Number of D's Pounds : Therefore 11 *.=550, by Question.

And, consequently, x = =^=50.

From which the rest of the Shares are easily determined;

Q.U E S T I O N V.

'Tis required to divide the Number 92 into four such Parts, that the First may exceed the Second by 10, the Third 1 1 8, end the Fourth by 24.

Let x be the first Part.

Then ) *-18 > will be the other Part*.

t *-24 3

And

with their Solutions. j And 4^-52=92, by the Question.

Hence 4*= 144;

and x — ^ = 36.

Q_U ESTION VI.

A certain Sum of Money was Jhared among five Persons, A, B, C, D, and E; whereof B received 10 L less than A; C 16/. more than B ; D 5/. ///j than C ; and E 1 5 /. 7nsr* than D : Moreover it appeared that E received as much as both A B. What was the whole Sum Jhared, and how much did Each receive ? Let x be the Share of A.

Therefore x-\-i6—2x— 10, by the Question.

Whence 26=*.

From which it appears that 26, 16, 32, 27, and 4a (Pounds) were the respective Shares ; and that the whole Sum was 143/.

Q_U ESTION VII.

To find that Number, whereof the Double increased by 24, Jhall as much exceed 80, as the Number itself it below 100.

Let x be the required Number.

Then 2*+24— 80- ioq—*, by the Question. ...

Whence 2x-\-x ^100— 24+80, that is 3* = 156 ;

and therefore x — ^=52.

3

B2 QUESTION

4 Algebraical Problems, Q_U E S T I O N VIII.

What two Numbers are those, whereof the Difference it

« and the Sum 33.

Let x be the lesser Number, then * -+- 7 will be the greater, and 2x + 7 = 33.

Therefore 2*= 3 3—7=26, and x= — —13= the Lesser, Whence *4-7=2Cc= the Greater.

Q.U E S T I O N IX.

To divide the Number 75 into two such Parts, that 3 times the greater may exceed 7 times the lejfer by 15.

If x be the greater Part,

then 7 c — x will be the Lesser ? , , ~

13 \ by the Question.

and 3* = 75-* x 7 + 15 J that is, 3* = 525 — 7*+ 1 5 i therefore 1 0*^: 540, and *=54.

From whence the lesser Part (75-*) is found=ai.

Q_U E S T I O N X.

A, after winning 10 Guineas ofB, had as much Money as B and 6 Guineas more; and betwixt them both they hadforty Guineas : What Money had Each at first ? Let x be the Guineas that A began with ;

then 40-*, are the Guineas that B began with : Therefore, after Play,

A had, *+ 10 Guineas ; and B, 30— x, Guineas.

Whence *4- 10^30-,* +6 (by the Question.) Therefore 2x—26.

And

Q.UESTION ' 5

•with their Solution*.

Q.U E S T I 0 N XI.

The Sum of 500 /. was divided among four Persons, ft that the First and Second, between them, had 280 / ; the First and Third 260 / ; and the First and Fourth 220 /. How many Pounds had Each ?

If x be the Number of Pound* the First had, fad i C 280-x

then the .} 3d vhad < 260-x I 4th j C 220-Jf The Sum of all which,

being 760—2*, is = 500, by the Question.

Whence x _.7oo SSB^jy) ,

Therefore the four Shares were 130, 150, 133, and 90 Pounds, respectively.

QUESTION XII.

'Tts proposed to divide 60 into two such Parts, that the Difference between the Greater and 64, may be equal to twice the Difference between the Lesser and 38.

If * be the greater Part ; then 60-x will be the lesser :

Also 64— x, will be the first mentioned Difference, and 38—6o4-*, or x— 22, will be the Second.

Therefore 64-^=2 x *-22, by the Question.

that is, 64—jr=2*-44.

Whence 108=3*, an£* 3,^—x'

Q_U E S T I O N XIII.

Jfter 34 Gallons had been drawn out of one of two, equal, Cajks, and 80 Gallons out of the Other, there remained just twice as much Liquor in the One as in the Other :

What did each Cajk contain when full?

Let x be the Number of Gallons sought ; then #-34 will be what remained in the first Calk,

B 3 and

6 Algebraical Problems, and *-8o, what remained in the Second.

Hence Ar-34=2XJr-8o, by the Question.

Or, #-34. n 2*- 1 60.

Therefore 126 =x.

Q_U E S T I O N XIV.

A Son, a/king his Father how old he was, received the following Answer : Your Age four Years ago, says the Father, was only I of mine, at that Time ; but now your Age is jufl \ of mine : What was the Age of Each ? Let * be the Age of the Son,

then 3* will be that of the Father :

Also x—4 will be the Age of the Son, four Years be fore the Time in Question ; and 3^-4 will be the cor responding Age of the Father : which, by the £htestiont is equal to 4 times *-4 : Hence we have this Equation 4*-i6=3*-4.

Therefore *=i2, and 3*=36 j which are the two Ages required.

Q_U ESTION XV.

What Number is that, whose i exceeds its I Part by id?

Let x be the required Number j Then, its ] Part being — , and its I Part — ,

3 4

VTC have — — = 16, by the Question.

3 4 ;

Hence 4*-3*=i 92, by Reduction j that is, #=192.

QUESTION

•with their Solutions. 7 Q.U E S T I O N XVI.

In a Mixture of Wine and Cyder, one half of the whole -f-25 Gallons was Wine ; and, \ Part — 5 Gallons, Cyder. How many Gallons were there of Each ? If be put for the Number of Gallons in the whole Mixture, the Gallons of Wine will be expressed by

— +25, and those of Cyder by 5: Which to-

2 ' 3

gether being equal to the whole, we therefore have — -f- 2 5 +— — 5 = x j

2 3

_ XX

Ut, 20 = * :

2 3

Hence 120= 6x— 3*- 2x;

Or 1 20 = x. From which it appears that the Mix ture consisted of 85 Gallons of Wine, and 35 of Cyder.

Q_U E S T I O N XVII.

In a Lottery, consisting of 100000 Tickets, half the Number of Prizes added to ^ of the Number of Blanks^

was 35000. How many Prizes were there in the Lottery ?

If x be the Number of Prizes ;

then iooc00-at, will be the Number of Blanks.

A , r x 1 00000— *

And so, 1- —— =35000.

Hence 3*+ 200000 — 2# = 210000:

And x =10000= the Number sought.

1 4 QUESTION

t l

8 Algebraical Problems,

QUESTION XVIII.

Tq the Compoftion of a certain Quantity of Gunpowder, j of the whole ifeight 4-6 Ib. os Saltpetre was necessary, the Sulphur used was j of the whole — $lb and the Charcoal \ of the whole — 3 Ib. How many Pounds of each of the three Ingredients were there taken ?

Let x be the Number of Pounds in the Whole: Then

there were

r te

— -f- 6, Pounds of Saltpetre.

—— 5, Pounds of Sulphur.

, Pounds of Charcoal.

x , x , x

X X X

And therefore H — by the Question.

« 3 4

Whence 12*+ 8* + 6* — 48 = 24*:

48

And consequently x = — = 24. , 2

Therefore there were taken 181b. of Saltpetre, 3 lb. of Sulphur, and 3 lb. of Charcoal.

Q.U E S T I O N XIX.

A General, after having lost a Battle, found that he had only I of hi, Army 4- 360* left, fit for Atlion ; | of his Men 4- 600 being wounded, and the Rest, which were -f of the whole Army, either slain, taken Prisoners, or miffing. What was the whole Number of the Army ? The Number sought being denoted by the Numr her of Men left unhurt will be —I- 2600.

2 3

And the Number of the Wounded - -•+. 600 : o

To which adding, the Number of the Slain and Pri soners, we have the Number of the whole Army j

Of

with their Solutions.

9 Or — + 3600 +4 +600*+ — =*.

205

rT ( X X X \ IX M

HenCC 4aool = x~ ~T~

Therefore 168000 (=15*' — 8*) =jx : And 24000 =#.

Q_U E S T I O N XX.

^ Pr/z* of 2000 /. war divided between two Persons, A B ; whofe Shares therein were in proportion as 7 to 9 : What was the Share of Each ?

Let * denote the Share of A ; then 2000-tf, will be That of B.

But x : 2000-* : : 7 : 9, by the Question.

From whence (as the Rectangle of the two Extremes, of any four proportional Numbers, is equal to the Rec tangle of the two Means] we get this Equation, 9 x x = 2000 - x x 7 ;

that is, gx =: 14000-7*.

Hence i6x = 14000*

aim x — °7V

Therefore the Share of A was 875 /. and that of B,

Q.U ESTION XXI.

To divide 44 into two such Parts, that the Greater in creased by 5, may bt to the Lesser increased by 7, as 4 is to 3.

If x be the greater Part, 44-*, will be the Lesser,

Therefore (by multiplying Extremes and Means) we have 3*+ 15=204-4*:

QUESTION

io Algebraical Problems,"

Q_U E S T I O N XXII.

to find two Numbers in the proportion of i to 2 ; ft that, 1 2 being added to Each, the Sums Jhall be in proportion as 5 is to j.

Let x be the lesser Number ; then 2*, will be the Greater :

Hence *-f-12 : 2*+ 12 : : 5 : 7, by the Question.

Therefore 2*4- 12x5 =*•+ 12x75 that is io* -f- 60 = 7* +84.

From which 3*— 24 ; and jc = 8.

So that the two Numbers are 8 and 1 6.

Q_U E S T I O N XXIII.

A, at Play, first won 5 Guineas of B, and had then as much Money as B ; but B, upon winning back his own Money and 5 Guineas more, had 5 times as much Money gi A : What Money had Each, at first ?

If x be the Number of A's Guineas, at first ; then it is plain, from the Question, that* +10 will be the Number of B's Guineas :

Whence x-\- 10+ 5 (=^-5x5) —5X-25, that is, 40 =4* :

Therefore * = 10 = the Number of A's Guineas ; and x 4- 10 =20= the Number of B's Guineas.

Q.U E S T I O N XXIV.

A Grocer, with 56 Ib. of fine lea, at 20 Shillings a Pound, would mix a coarser fort, of 14 Shillings, so as to afford the whole, together, at 18 Shillings, per Pound; What Quantity of the latter Sort must he

take ? i

Let x be the Number of Pounds required j then 1 120 is the value of finest Sort,

and 14* that of the Coarsest.

Moreover)

with their Solutions. ir Moreover, the Number of Pounds of both Sorts together being 56 x, it is evident

that 56-f-* x18, or 1008 4- 18*, is the Value of the whole Mixture. And therefore

1 120 4- 14*= 1008+ 18*:

Whence 112 = 4-*-:

And consequently 28 — x.

QUESTION XXV.

A Farmer would mix Wheat, at 4 Shillings a BuJhel, with Rye, at is. 6d. a Bushel ; so that the whole Mixture may constjl of 90 BuJJjels, and be afforded at 31. 4 d.

a Bushel : 'Tis required to find how many BuJhds of each Sort mujl be taken.

If the Number of Bufliels of Wheat be x, Those of Rye will be 90-*.

Moreover, the Value of the Wheat will be 48*, Pence, and the Value of the Rye 90 —x x 30, Pence :

Whence 48* 4- 90—x x 30 =. 90x38 (by the Que/litm.) that is, 48* 4- 2700 — 30* =3420 :

Therefore 18*. = 720 : And consequently x = •

Q^U E S T I O N XXVI.

A Workman was hiredfor 40 Days, at 3 s. 4 d. per Day, for every Day he worked ; but with this Condition, that, for every Day he played, he was to forfeit is. \d:

. And it so happened, that, upon the whole, he had 3/. 3/, ifd. to receive. The Question is, to find how manj Days, of the 40, he work d, and how many he play'd.

Let x he the Number of Days he work'd ; then 40— * will be the Number of Days he play'd : Moreover, as he was to receive 40 Pence, for every Day he work'd, and to forfeit 16 Penc« for every Day he play'd,

12 Algebraical Problems,

we have 40*= the Number of Pence earn'd by Work, and 40-Jf x 1 6 = the Number of Pence forfeited by Play : Whence 40* — 40-* x 16 = 760, by the Question ; that is, 40-f — 640 -+- i6x — 760 :

Therefore 56* = 1400;

and consequently x = 25.

Fn m which it is evident that he work'd 25 Days, and play'd 15.

Q_U E S T I O N XXVII.

A Bill ef 120/. was paid in Guineai and Moidores, and the Number of Pieces used of both Sorts was just 100:

How many were there of Each ?

If x be put for the Number of Guineas ; then 100—* will be the Number of Moidores : And so, 21* -f- 100 — x x 27 = 120 X 20 i by the

Or, 2 1 a- -f- 2700 — 27* = 2400 \ Question.

Hence 300 = 27*— 21* r_ 6x.

And consequently * = 222.— e0, 6

Q_U E S T I O N XXVm.

One bought 30 Pounds of Sugar, of two different Sorts, and paidfor the whole 19 Shillings ; the best Sort cost 10 d. per Pound, and the worst 7 d. How many Pounds were there of Each ?

Let x stand for the Number of Pounds of the best Sort, and then 30-* will express the Pounds of the other Sort ; Therefore *x 10 -f- 30-* x 7 = 19 X 12, by the Question;

that is, ioa--(- 210— yx = 228 : Whence 3A; 228-210 = 1 8,

and * = 6. Therefore there were 6 Pounds of the best Sort, and 24 of the Worst.

QUESTION

with their Solutions. j j Q_U E S T I O N XXIX.

A Lady gave a Guinea, in Charity, among a Number of Poor, consisting of Men, Women, and Children : Each Man had 12 d, each Woman 6 d. and each Child "id.

Moreover there were twice as many Women as Men, wanting 2 j and 3 times as many Children as Woment wanting 4 : How many Persons were there relieved?

Let * be the Number of the Men ; then %x-a, will be the Number of Women, and 6x— 10, the Number of Children.

Hence xx 12 + 2x — 2x6 + 6x— 10x3 = 21x12 ; that is, 12* -f- 12X—12 -f- i8x— 30 = 252,

or, 42* = 294 : Whence x— j.

Therefore there were 7 Men, 12 Women, and 3a Children; in all 51 Persons.

Q.U E S T 1 O N XXX.

To find that Number, which being divided, either into three, or four, equal Parts, the continual Product of all the Parts7 in both Cases, Jhall be exactly thesame.

Let x be the required Number ;

so shall the continual Product of the three equal Parti be— x — x —ss— » and that of the four equal

3 3 3 27

_ x x K * **

Parts — x — x — x—'=—r.

4 4 4 4 256

Whence —; = by the Question.

256 27, ^'-/

Therefore 27*= 256 j and*=9--*

QUESTION

14 Algebraical Problems, Q_U E S T I O N XXXI.

la find two Numbers, in the Proportion of'3/9 4, whose Sum is to the Sum of their Squares, as 7 to 50.

Let 3* denote the lesser Number : Then 4* will express the greater.

And we (hall have 3* -(-4*- : qv* 4- i6*z : : 7 : 50, or, yx : 25** : : 7 : 50, by the Question.

Therefore 25^x7 =7^x50, or, 25** = 50* ;-

whence x =— =2 : So that 6 and 8 are the two Nura- bers that answer the Question.

Q_U E S T I O N XXXII.

To find two Numbers in the Proportion of 9 to 7 ; so that the Square of their Sum, and the Cube of their Difference, Jhall be equal.

If 9* be put for the greater Number;

then 7* will be the lesser ;

And so 16*1* = 2*|3, by the Question, that is, 256** =8*3.

Hence * = — % = 32 : o

Therefore 288 and 224, are the two Numbers sought.

Q_U E S T I O N XXXHI.

To find two Numbers whofe Difference is 4 and the Dif ference of their Squares 120.

Let * be the lesser Number ; then x + 4, will be the greater : Also xx will be the Square of the lefler, and xx -f- 8* 16 that of the greater : Whence 8* -f 16=: 120, by the Question:, Therefore %x — 104 ; and * = 1 3 :

So that 13 and 17 are the two Numbers that were t»

be found. QUESTION

ivith their Solutions. 1$

Q_U E S T I O N XXXIV.

To divide 100 into two such Parts, that the Difference of their Squares may be iooo.

If x be the greater Part, loo-*, will be the lesser :

Therefore xx — 100 — jTj* = iooo ; that is, ** — ioooo + 200* — ** = iooo : Whence 200* = i iooo ;

and consequently x = — = 55.

Q_U E S T I O N XXXV.

To divide 100 into two Parts, so that the Square of their Difference may exceed the Square of twice the lesser Part by 2000.

The lesser Part being denoted by x, the greater will be expressed by 100-*, and the Difference by 100-2*.

Therefore, by the Problem, 100— 7.xf — 2*)* -\- 2000, that is, 1 0000 — 400* + \xx — 4** + 2000,

or, 1 0000 — 2000 = 400*.

Hence x = ^00Q = 20; and 100-* = 80.

400

Q_U E S T I O N XXXVI.

A and B make a joint Stock of 500 /. by which they gain 160 /. whereof A, for his Share, had 32 /. more than B : What did each Person bring into Stock ?

. If * be the Number of Pounds advanced by A ; then it will be, as 500 [the whole Stock) is to 160 (the whole Gain) so is x (the Stock of A) to the

500, Gain of A: Whence the Gain of B being 160 .

we

16 Algebraical Problems, we have — = 160 — — + 32, by the Question :

5° 50

Therefore — iqZ, 9t ——(/;

S0 " 50 and consequently * = 50 x 6 = 300.

Q_U ESTION XXXVTI.

A Sum of Money was divided between two Persons, A and B, so that the Share os A was to That osB, as 5 to 3, and exceeded -J- of the whole Sum by 50 /. What was the Share of each Person ?

Let 5* express the Share of A, and 3* the Share of B j

then fix will be the whole Sum, and will be — thereof :

9 9

Therefore 5* —^L— 50, by the Question ; 9

that is, — = 50, or *• = go.

9

Hence 450 /. and 270 /. are the two Share* required.

Q.U ESTION XXXVIII.

A and B began to play together with equal Sums of Money } A first won 20 Guineas , but afterwards lost back the Half of all he then had ; and thereupon had only half as much Money as B. (What Money did each begin with ? Let x be the Number of Guineas required :

Then A, aster winning 20 Guineas, had .v -f- 20 ; the Half of which, or — + 10, is therefore what he had at

2

last : And this deducted from 2* (the whole Sum betwixt Both) leaves lx —— 10 = what B had at last.

2

Therefore

with their Solutions. 17 Therefore 2x —— - 10= a- 4- 20 ;

2

whence 2*-20= 2a.- + 40, and consequently x =z 60.

QUESTION XXXIX.

A Gentleman left his whole EJiate among his four Sons ; whereof the Eldest had $ wanting 800 /. the Second i and 120/. ever ; the Third had half es much as the Eldest ; and the Youngest \ of what the Second had.

What was the whole Estate ? and how much had Each?

Let x be the whole Estate ; then The First had — 800.

1 2

The Second— 4- i*o, 4

The Third -1— 400.

4

The Fourth^ 80.

ia

The Sum of all which is, consequently, equal to the Whole,

Or, J U — 1000 = *.

2 4 T 4 T 6

But it is plain (without Reduction) that JT4T41 — = *.

Hence our Equation becomes

— 1000=0, or— =1000. . .

6 6

Therefore the whole Estate wai 6000/. whereof the eldest Son had 2200/. the Second 1620/. the Third

1 1 00/. and the Youngest 1080/.

C QUESTION

18 Algebraical Problems,

QUESTION XL.

One being ajk'd his Age, reply d; If \ of my Years ht multiply'd by 3, and 4 of them be added to the Product, the Amount will be 115. What was his Age?

If x be the required Number os Years, then x3 + — =H5, by the Question \

\ 3

„. . ox , x that J9, — 4. =115.

5 3

Therefore 18* + 5*= 15x1 15, or 23*= 15x1 15 :

Consequently x = = 15x5 = 75.

QUESTION XLI.

A Person being ajk'd the Hour of the Day, answered thus : If \ of the Number of Hours remaining till Midnight be multiply' d by 4, the Product will as much exceed 12 Hours , as Half the present Hour from Noon is below 4 : What was the Hour after Noon ?

Let x be the required Hour ;

then 12-*, will be the Hours till Midnight, and *—f- x 4 - 1 2 = 4 , by the Question.

8 2

That is, -l^z3f— 12=4-

2 2

whence 36- 3* -24= 8 -a-j and consequently a- = 2.

QUESTION

<witb their Solutions. 19 Q.U E S T I O N XLII.

* •

A Market-woman bought in a certain Number of Eggs, at the Rate of 5 for two Pence ; one half of which Jht fold out again at 1 a Penny, and the remaining Half at 3 a Penny ; and cleared 4 Pence, by so doing : What Number of Eggs hadJhe?

Let lx be the Number sought ; then, by the Question,

5 : 2 : : 2.x : the Number of Pence the Eggs cost:

But the Number of Pence They were sold for, again, X X ...

is —I— : Therefore we have this Equation,

2 3

xx 4*

viz. — — = 4 :

* 3 5

From whence 15*+ 10 x— 24*= 120, or x= 120.

Q.U E S T I O N XLIII.

A certain Sum of Money, put out at Interest, amounts, in 8 Months, to 297 /. 12 s. And, in 1 5 Months its Amount ( computed according to fimpk Interest) is 306/.

What is that Sum ? And what the Rate of Interest ? Let x be the Number of Pounds in the required Sum : Then, the Interest thereof for 8 Months being 297, 6-x, and for 1 5 Months 306-*, we have,

as 8 : 15 : : 297,6-.*: 306 — at:

Whence, by multiplying Extremes and Means, we get 2448 — 8* =4464— 15*.

Therefore 7*:= 2016,

and consequently x = 288 /. the Sum required.

For the Rate of Interest, it will be,

as 288/. x 15 : 100/. x 12 : : 18/. (the Interest of 288/.

for 15 Months) to 5/. the required Interest of 100/. for 12 Months.

li

C 2 QUESTION

2o Algebraical Problems, Q_U E S T I O N XLIV.

A Waterman finis by Experience, that he eon, with the Advantage of a common Tide, row from London to Greenwich, which is 5 Miles, in 3 Quarters of an Heur ; and that, to return to London, against an equal Tide, though he rows hack along-Jhore, where the Stream is only half as strong as in the Middle, takes him a full Hour and Half "Us required to find, from hence, at what Rate, per Hour, the Tide runs in the Middle where it is Jlrongejl.

In the first Place, it will be

3:4:: 5:6} = Dist^row'd per Hour with the Tide, 6:4:: 5:3 -5 = Dist. row'd per Hour against Tide.

If now the former of these two Distances ( b\) be put=rtf, and the latter (3 \)—b; and * be assumed to express the required Distance run, per Hour, by the Stream in the Middle of the River ; then a — x will be the real Effect of his Rowing, per Hour, in going from London, the Motion of the Tide being deducted ; and 44-— will be the like Effect in his Return :

And so, these two Quantities being equal to each other, we have bA— = a-x :

2

Whence ib -+- x = ia — zx ; o.a-7.b m and consequently *=.—-— as 3$.

Q_U E S T I O N XLV.

To divide 36 (z) into 3 such Parts, that \ of the Fir/7, 4 of the Second, and \ of the Third, may be equal to each other.

If be put for the first Part, x

then is—= \ of the second Part, (by the Quejlion.) And

with their Solutions. 21 And so — = second Part.

2

Moreover — being=x| of the third Part, therefore (or 2*)= third Part.

Hence x -f 2— + 2x+a, 2

and 2x 4- 3* -f 4* = %a.

Consequently x = 2fL- 2X3&

9 9

From which the second Part (— )

appears to be = 12, and the Third (2x) =16.

Q.U E S T I O N XLVI.

To divide the Number 90 into 4 such Parts, that, if the firji be increased by 5, the second diminijhed by 4, the third multiply'd by 3, and the fourth divided by 2, the Resultt in each Case, Jhall be exactly the same.

Let x be the fourth, or last, Part : Then, three times the third Part being =- - the third Part will be 4

Moreover, the second Part— 4 being, also, =—

. 2,

the second Part will be -^--f- 4 :

And, by adding all the Parts thus found together, 6.

And, the first Part -f- 5 being = *- the first Part, alone, will be—— 5.

2

we have x+~+JL+ 4 44-5=90;

c3 that

22 Algebraical Problems, that is, 2* -f-£— i = 90.

D

Whence 13*1=91x6; and *=42.

Therefore the four required Parts are 16, 25, 7, and 24, respectively.

Q_U E S T I O N XLVII.

Two Workmen A and B were employed together for 50 Days, at 5 Jhillings per Day, Each ; during which Tim: A, by spending only Sixpence a Day less than B, had saved twice as much as B, besides the expence of 1 Days over : What did each Person expend a Day f Let x be the Pence A spent per Day ;

then 60-x, will be what he faved per Day, and 54— .v, what B faved.

Therefore 3000—50* are A's whole favings, and 2700-50* those of B.

Hence 3000-50* = 2 x 2700-50* -f- 2* , Or, 3000-50* =5400 -98*:

From which 48 v= 2400, and *=5o.

Q_U ESTION XLVIII.

Two Persons, A and B, have both the same Income ;.

A lays by ' of his ; but B, ly spending 60 /. per Ann.

mere than A, at the End of three Years finds himself 1 00 /. in debt. What did Each receive, and expends per Annum ?

Let * be the yearly Income of Each ; then JtfLis the Sum expended by A, per Ann.

5

and -~-\-6o, That expended by B.

Therefore -^--\-6o — *, is what B runs in Debt.

Consequently ^-+6o-*X3 = 100,

or

•with their Solutions. 23 or {- 180- 7x= 100 ;

5

that is, 180 2iL=ioo.

Whence 900 — 3* = 500, and * =-122.= 1 w. 6j. %d.

3 "

Therefore A expended 106/. 13 j. ifd. and B 166/

13s. 4^. Annum.

Q_U E S T I O N XLIX.

j/ Grazier bought in as many Sheep, of different Sorts, as tost h'm 33/. -js. 6d. For the first Sort, which were } of the whole, he paid qs. 6d. a-piecc, for the second Sort, which were. £ of the whole, he paid 1 1 s.

each ; andfor the rest, 12 s. 6d. each : What Num*

ter of Sheep did he buy in all ? If be the whole Number of Sheep ;

then, the Number of the first Sort being-1, and of the second Sort —, the Number of the remaining Sort (at lit.

6d. each) must be x- i— * = I 3*- 5f :

.3 4 '2 12

Whence, by the conditions of the Problem, we have

*. , x , CAT

— X19 + — X22+^X 25 = 1335;

3 ' 4

^18,121+^14-^51=1335.

3 4 12

Let each Term of this Equation be now raultiply'd by 12, and it will become

76* + 66*4- 125* = 1 6020, or, 267*= 16020.

Therefore x = 60.

C 4 QUESTION

24 ALGEBRAICAL PROBLEMS, Q_U E S T I O N L.

A Draper, of a Piece of Cloth, [standing him in 3 S. id.

per Yard, fold ? Part, at 4 s. per Yard ; I at 3 j. 8</.

per Yard; \ at 3*. 6d. per Yard ; and the Remnant at 3 j. ^.d. a Yard: And his Gain upon the Whole, was 1 5 j. 2 d. How many yards did the Piece contain ? If the Number sought be denoted by x;

then the Number of yards in the Remnant

will be r- - * — — — 6oy 1 2°x - 1 5* " 1 lx — 1 3*-

- 3 4 5 60 60

Therefore, by the Question, we haye

:* x484-^X44 + yX42+-|^-x4o-38^i82, Or, x6*+u*+-±2i+— -38*= 182,

5 3

that is-i^-^-— - 11*= 182 : Whence i26*,-f-i3PJt'-165*'=273o:

And therefore x =—^-22_= 30.

9*

Q_U E S T I O N LI. - •

J Distiller proposes to mix Foreign Brandy, standing him 'tn 8 Shillings a Gallon, with Britijh Spirits of 3 Shillings per Gallon, in such Proportion that he may gain 30 per Cent by selling out the Compound at 9 s.

it Gallon. What is that Proportion ?

Suppose, that, with a Gallons of Brandy, he mixes

*• Gallons of Spirits ; then, the Brandy, standing him in 8a (Shillings) and the Spirits in 3* (Shillings), the true value of the whole Mixture will be 8« 4-3* : But the Value of a 4- * Gallons, at 9 Shillings per Gallon, is 9*7+9*: Therefore, by laying out 8«4-3*»

he gains a \ 6x : And so

we have 2<f-\-2x : : : 1P0 : 30, by the Question.

V_ , ' Consequently

tuith their Solutions. 25 Consequently 1 00a+ 600x = 240a -f go* ;

whence 510*= 1400, and*=s-^.

From which it appears that, to every 51 Gallons of Brandy, there must be taken 14 Gallons, of Spirits.

Q.U E S T I O N LII.

To find two Numbers in the Proportion of' 4 to 5, from which two other (required) Num;ers, in the proportion of 6 to 7 being, respectively, deducted, the Remainders Jhall be in the proportion of 2 to 3, and their Sum equal

to 20.

Let 4* and 5*- be the 2 first Numbers, and 6y and jy, the other 2 Numbers.

Then 4*-6y : 5»-7/ : : * : 3 I h tbt

And qx—i-ry = 20 J J ^ J

From which Proportion, by multiplying Extremes and Means, we have 12*-i8y:= 10* -147, and

therefore x-=.2j\ which substituted in the above Equation gives 1 8y — 1 37 = 20 ;

20

whence y =—=4; and x (—iy)~ 8.

Therefore the 2 first Numbers are 32 and 40 ; and the

•ther Two, 24 and 28.

Q_U E S T I O N Lffl.

A Farmer soli, at one time, 30 BuJhels of IVheat and 40 of B'dr ley, and for the whole received 13/. 10s;

and, at another time, he fold 50 Bujhels of IVheat and 30 of Barley, at the same Prices as before, and for the whole received 17/. the Question is, to find what each Sort of Grain was fold at per BujJhel.

Let x and y express the Numbers of Shillings, re spectively, that the Wheat and Barley were sold at per Bushel ; and then, from the Conditions of the Question^

We foall have the two following Equations, viz.

3°*

26 Algebraical Problems, 30* +40.7=270,

50* + 3py= 340.

From 4 times the Second of which Equations let 3 times the First be subtracted, and there will remain

110^=550.

Therefore x =.15-= e :11 J And ^=iZZL3£j=3.

Q-U E S T I O N LIV.

A Farmer, with 28 BuJhels of Barley, at is. 4*/. per Bushel, would mix Rye, at is. per BuJhel, and Wheats at 4 s. per BuJhel ; so that the whole Mixture may con- sijl of 100 BuJhels, and he worth 3*. ^d. a BuJhel:

How many BuJhels of Rye, and how many of Wheat must he mingle with the Barley ?

Let x be the Number of Bushels of Rye, and y those of the Wheat : Then, the value of the Barley being 784 (Pence), of the Rye 36* (Pence), and of the Wheat 487 (Pencej, we have

784+ 3^+48^=4000 J j he gueJ}i and 28 + *+.¥= 100 S ^* J

From the first of which Equations, take 36 times the second, and there results,

784-36x28+ i2y=4oo, that is, -224+ 127=400.

Therefore izy — 624,

and consequently 7 = ^i= 52.

Whence x (=100-28-7) =20-

QUESTION

with their Solutions. 27

Q.U ESTION LV.

A and B, working together on the same Work, can earn 40 Shillings in 6 Days ; A and C together can earn 54 Shillings in g Days ; and B and C, 80 Shillings in 15 Days: 'Tis required to find what each Person, alone, can earn per Day.

Let x, y and z express the Numbers of Shillings in the three required Values, respectively.

s 6*+ 67 = 40 S

Then 1 gx 4- qz = 54 >£y t&< Question.

.■ C*+.y=6! 7

And < * -|~ z — 6 [• Division, Ly + z = Ki

Hence y—z by subtracting the ld Equationfrom \st.

And 2y — 6, by adding the two last.

Consequently y = 3 : From which

wehave#(=6f —y) =33- = 3 s. %d. and z (= 57 ->J 5324= is. $d.

Q_U ESTION LVL

To find three Numbers, so that f the First, \ of the Second, and i of the Third, Jhall together be equal to 62. ; also 7 of the First, \ of the Second, and | of the Third, equal to 47 ; and, lastly, $ of the First, -\ of the Second, and I of the Third, equal /0 38.

Put a— 62, b=4j, and ^=38; and let the three re quired Numbers be denoted by x, y and z, respectively ; then the Conditions of the Problem will be expressed, in the three following Equations, viz.

2 3 4

3 T4 r 5 .1+

4 5 9

7 * which,

28 Algebraical Problems, Which, clear'd of Fractions, become

1 2x + 8y + 6 % —

20* + 15y+ i2z = 6oi <

30*+ 24y+ 20z = 120s.

Now (in order to exterminate z) let the Second of these Equations be taken from the Double of the First, and also the Treble of the Third from the Quintuple of the Second ; and there results

4*+.y = 4%a-6ob ; and io*+3J' = 30o£ ~3^oc '-

Whence, by deducting the second of These from the Treble of the Former, and dividing by 2, there comes out

x=j2a— 24.0b 180^ = 24.

From which y (=480 - 60b - \x) is also found = 60, and z (=«-f*- iyx 4) = 120.

Q_U E S T I O N LVH.

To divide the Number 90 (a) into three Parts, so that, the Double of the first Part + 40 (b) ; the Treble os the Second + 20 (c) ; and the Quadruple os the Third + 10(d), may be all equal to one another.

Let x, y, and z represent the three required Parts, respectively ; then, from the Conditions of the Problem, we shall have

2* + * = 37 + c, 2x -f- b = 42: -j- d.

Now, in order to exterminate^ and z, let 12 times the first Equation, 4 times the Second, and 3 times the Third, be added all together ; and you will have 26*+ 12Jr + 12z + -jb = 12«+ 12J>+ 12z + 4C+ 3</.

Therefore 26*= 12a + 4s + 3^- -jbt and x =JL£i±4L+_3^.= 3S.

26 J

Whence 7[=^i-^ZfJ= 30, and z (= a-x-y) =25.

QUESTION

with their Solutions. 29 Q_U E S T I O N LVIII.

To find three Numbers, so that the First with half the other Two, the Second with \ os the other Two, and the Third with \ of the other Two, may be the same, and amount to 51 in each Case.

Vita— 51, and let x, y and z denote the three re quired Numbers ; then, by the Question,

Which, cleared of Fractions, become 2.x-\-y-\-z =2<7,

* + 3)' + a: = 3<7,

From whence, by taking the Second from the Third, and the First from the Double of the Second, there re sults - 2y + 3z = a,

and $y 4- z = 4a.

And, by deducting the former of These, from the Treble of the latter, we have ijy= 11 a.

Therefore y = = 33*

17

2 (=4*— 5^=-^y=39, and * (= 3* - 3/ - x) —-^-= 1 5>

QUESTION

30 Algebraical Problems, Q_U E S T I O N LIX.

A certain Sum of Money was divided between three Per sons, A, B and C ; so that, A's Share exceeded t of

the Shares of B and C by 30 / ; also the Share of B exceeded | of the Shares of A and C by 30/ ; and the Share of C likewise exceeded % of the Shares of A and B, by 30 /. Question is, to find the Share of each Person.

Let <?=30; and let x,y, and z be assumed to express the three required Numbers j then by the Conditions of the Problem,

x-M±Al=a, 7

-3x4- 3z y — -2—1 J— = a,

9

Whence, by Reduction, 7*—-W—4z =7*»

—3*+8^—3z =8fi, -— 2a-— 2y+9z T^ga.

Now, to get rid of y (which, because of the even Coef ficients, is the easiest to be exterminated) let the Double of the first Equation and the Quadruple of the Third be, successively, added to the Second ; by means where of we have

11*— 11z =22tf,

— "*4-33z— 44*-

Moreover, by adding these two last Equations together, we have 22z — 66a.

Therefore z = 3a = 90 ;

whence x (— 2a-j-z) =2 $a— 150, and y ^—a 4- *4/z^=4«=120.

Q.UESTION

with their Solutions. 31 Q.U E S T I O N LX.

If A and B together can perform a Piece of Work in 8 Days ; A and C together in 9 Days ; and B and C in 10 Days : How many Days will it take each Person^

alone; to perform the same Work ?

Let the whole Work be represented by a, and let x, y, and z stand for the Parts thereof performed by A, B,

and C in one Day, respectively.

~~C8*+8j=« 1

WI" liqy+io*=a)

J 8

And ^ *-f-z=— )by Division, a

y+z — I0J

Whence y—z(=^— *- ) =^v

And 2y = — •

10 72 720

Consequently y =-^1^-: From which * ^=-^--j> j is found =_125,

720

and — v ]=:-2I£.

V 10 - - 720

Now, the Part of the Work (a) performed by each Per son in one single Day being thus assigned, the Number of Days it will take any one of them to do the Whole, will be found by dividing the Whole by the assigned Part.

Thus, ) a — .I2,0a-— 72?L— 1 7-2! is the Number

720 ' 41a 41 ' 41

of Days in which B, alone, can do the Whole. And, in like

32 Algebraical Problem?,

like manner, the Number of Days in which A, orC, can do the Whole, appears to be

or 23-2-, respectively.

49 3i

Q_U E S T I O N LXI.

Is A, B and C can, together, finifl) a Piece of Work in g Days ; A, B and D together, in io Days ; A, C and D together, in 1 1 Days ; and B, C and D in 1 2 Z)*yr .«

/« Aaw long Time can they all Four, together, finiJh It ? Here, denoting the given Numbers by a, h, c, and d, and putting «, x, y, and z, for the Parts of the whole Work (g) done by Each in one Day, respectively, we shall, by the Question, have these Equations,

■ axu+x + y=gl cxu + y-\-z=£

«+ *+y=ll

The Sum of all which, divided by 3, by Division.

gives « + x £-lZ+£+-*-

3 - a 0 c «

Therefore, seeing the Work done by all the Four, in one Day, is expressed by— x— +4-+— +4, the

3 * h c d

Whole Work g, divided hereby, will consequently give

•with their Solutions. 33

give M , - 3 -

a + b + c + d a + b + c + d

. 3 g*f* _ 7 122L , for the Number bed + acd + abd + abc '2289

of Days required. ,

Q_U E S T I O N LXII.

To fnd that Number whose square Root is to its Cube Root, in the Proportion of 5 to 2.

Let x6 express the required Number : Then x3 will be its square Root, and x* its Cube Root.

Now x3 : x* : : 5 : 2, by the Question.

Therefore 2*3=5*% or 2*=5 ; or, lastly, x—^~-=. 2.5.

2

Whence x6 =244.140625= the Number sought.

Q.U E S T I O N LXIII.

To find two Numbers in the Proportion of 3 to 5 ; where of the fifth Power of the First Jhall be to the third Power of the Second, as 972 to 125.

If 3* be put for the first Number,

then 5* will express the Second ; and we shall have j*) s : 5*|3 : : 972 : 125, by the $veftunt that is, 243*s : 125*3 : : 972 : 125 :

Hence 243*' x 125 as 125*3 x 972, Or, 243** = 972.

Therefore x* 4 ; _ 243

and # = ^4 =2: So that 6 and 10 are the tw»

Numbers that were to be found.

D QUESTION

34 Algebraical Problems,

QUESTION LXIV.

To find three Numbers in the Ratio of f, \, and | f vihereof the Sum of all the Squares shall be 549.

Let * denote the first Number,

then it will be f : 7 : : * : -^-=p the second Number.

3

And \ : | : : *- : —= the third Number.

2

Hence x1 -\ J -)— =549, by the Question^

3 1 2

that is, *1 -f-4^—Y~— 549.

9 4

or, 36*1 + 1 6*1 + 9*1 = 36 x 549 : From which X--36^W _ 36xg,

and * = 6x3 = i8. Therefore 18, 12 and 9, are the three required Numbers.

Otherwise,

By reducing the given Fractions §, 4, and |, to the fame Denomination, they will appear to be in the Pro portion as 6, 4 and 3. If, therefore, the first of the three Numbers sought be denoted by 6x, the other Two will be expressed by 4*, and 3*, respectively : And so we shall have

36*1 + 1 6*1 + 9*1 ac 549.

Whence x = 3, and the Numbers sought, as before.

Q.U ESTION LXV.

Having given the Difference of two Numbers =6, and their Product'=720 j r» find the Numbers.

Let the Lester of them be denoted by x ; then the Greater will be *-f- 6 ; and so, by the Question, we shall have xv-f- 6* ac 720.

But in order to the Resolution of this Equation (in which both the first and second Powers of x are in

volved)

hsoith their Solutions." 35 volvcd) let Half the Coessicient of *, which (in this Case) is 3, be taken and squared, and let that Square be added to both Sides of the Equation : By which Means it becomes xx 4- 6* + 9 = 729.

Whereof the former Part being now a cotnpleat Square, its Root may, therefore, be extracted ; and will be expressed by the faid Half Coessicient joined to x with its proper Sign ; that is, by x -|f- 3 (as may be very easily proved by the Multiplication of x -)- 3 intd itself ; whence xx -+- 6x -+- 9, the very Quantity above, is produced). Hence it is evident, that x-\- 3 = ^729 = 27 (for equal Quantities have equal square Roots) ; and consequently x — 24.

Q_U E S T I O N LXVI.

The Sum of two Numbers being given .=. 60, and the Stim of their Squares = 1872 ; to find the Numbers.

Let * be the Greater of them ; then 60-* will be the Lesser : And therefore x* 4- 60—x\* = 1872 ; Or ^+3600— 120* + ** = 1872 : Whence 2**— 120* =^—1728,

and x%—60* =—864 :

From which, by compleating the Square (as in the Idji Problem) we get x*~6ex -f- 900 (=-8644-900) =36 : And consequently, by taking the Root, x~ 30=^36=6.

Therefore ^ = 6+30= 36; and 60-* = 24: Which are the two Numbers that were to be found.

But, to solve the Problem in a more general manner (by Letters) put the Sum of the two Numbers =.a, the Sum of their Squares =£, and the greater Number =*, es before.

Then will **-f a* -lax + x*=zb, by the Question.

Hence 2x*-2ax=b-a*,

A* b a2

and x —ax — —

2 2

D 2 Where,

36 Algebraical Problems,

Where, Half the Coessicient of the second Term beings- » the compleated Square will therefore be

, a% r b a\ «» \ b <?

xt-ax-\ = + )= :

'4V 2 2^4/ 2 4

From which, by extracting the Root, we have ,

a /b aa

* 2 V 2 4"

And therefore x —%S 1 : Which, if a be

v 2 4 2

taken = 60, and b = 1872, will come out 5= 36, very same as before.

Q.U EST.ION LXVIL

To divide the Number 60 (a) into two such Parts, that their Product may be 864 (b).

If x be put for the greater Part, the Lesser will be de noted by a—x ; and we (hall therefore have

ax—xx-z= b; by the Conditions of the Question.

This Equation, by changing the Signs of all its Terms (in order to have the highest Pewer of x assirmative) bf- comes xx — ax z=— b.

Whence, by compleating the Square, . aa 1 , aa we have xx-ax -j zz-o-\-. ;

4 4

and consequently x—(^~—\/~~—*'

Therefore *=VX — — * +—=36, the greater Part ;

* 4 ^

and a-x— 24, the Lesser.

QUESTION

with their Solutions. 37 Q_U E S T I O N LXVIII.

To divide the Number 60 (a) into two Parts, so that the Square of the Greater multiply d by the Lejser, added to the Square of the Lejser multiply 'd by the Greater, may amount to 51840 (b).

If * be the greater Part ; then a-x will be the lesser ; and *—x -f- a—xr xxzzib,

that is, ax*-x3 -\-azx-2ax*-\-x1=ibJ or, a*x — b :

Whence x* — ax — —— ;

and, consequently, xzz^f — — - +"~= 3&

Q_U E S T I O N LXIX.

The Sum of two Numbers being given =s 20 (a), and the Sum of their Cubes 3= 2240 (b) ; to determine tht Numbers.

Let x be the Greater ; then a-x will be the Lesser ; and therefore x3 -\- a—x3 — b,

that is, x3 -\. a3 - jfif x -\- ^ax* — x3 — b ; Whence yix* — yi* x — b-a3,

, a b a*

and xz-ax— .

3« 3

Therefore, by compleating the Square,

1 aa r b a* \_b ^ a* ,

4 V 3« 3 4/ 3a~I2 '_

And, by extracting the Root, x— —=./ * 5

2 Tr 2a 12

Consequently x = =7 +

v/37l-33T=I2-

D 3 QUESTION

3$ Algebraical Problems, Q_U E S T I O N LXX.

To divide the Number 240 (a) into two such Parts, that the greater Part divided by the Lejfer, may be to the,

\ejser Part divided by the Greater, in the Proportion of 147 to 75 (or of mto n).

If the greater Part he denoted by *, the Lesser will be expressed by a-x ; and we shall have

x a—x

—— : — : '. n '• ft.

a-x x

Hence 2L-»*<>-*

a—x x

From which, by extraaing the square Root, on both

But, in the Case above proposed, y/ ~ being =3

QUESTION

with their Solutions. 39 Q_U E S T I O N LXXI.

Twt Workmen A and B ivere employ' d, by the Day, at different Rates ; A, at the End of a certain Number of Days, had 96 Shillings to receive ; but B, who play'd 6 of those Days, received, only, 54 Shillings : But, had B worked the whole Time, and A play'd 6 Days, They would have received exactly alike. 'Tis proposed to find the Number of Days Tliey were employ d; and

what Each had a Day.

Let x be the Number of Days that A work'd ; then x-6 will be the Days that B work'd.

Moreover-2^., will be ihe Wages of A per Day ; and -51- , the Wages of B per Day.

x-b

Therefore-51- x *, is what B would have earn 'J, had As— 6

he work'd the whole Time :

And ~ x x-6, what A would have earn'd had he play'd 6 Days. Which two Values being equal, by the Question, wehave5iI=9^Ei

*-q x

Whence, by Reduction, 54*1= 96 x*— 6j\

Or, 2fL —x^6\r : Therefore, by extracting the squar*

Root on both Sides, 3HL~x-6i and consequently *c=24.

4

From which it is evident, that A had 4 Shillings, and B

^ Shillings, a Day.

D4 QUESTION

40 Algebraical Problems, Q.U E S T I O N LXXIL

From two Places, at the Distance of 320 (a) Miles, twa Persons, A and B, set out, at the same Time, in order to meet each other ; A travelled 8 (b) Miles a Day more than B ; and the Number of Days in which They met was equal to Half the Number of Miles B went in a Day : 'Tis required to find how far Each travelled tt meet the Other.

Let * be the Number of Days in which They met ; ' q£&£{ S+i H be the Number of Milesi a \ went a Day.

Therefore, by multiplying each of These by (x) the Number of Days, we have 2xx, and 2xx-\-bx, for the whole Number of Miles travelled by B and A, respectively :

And consequently j^xx-\-bx=.a.

tt , bx , bb a . bb

Hence xx -\ K-= Y7-*

.4 64 4 64

and

Therefore 2xx— 128 the Miles travelled by B } and 2xx-\-bx= 192 = Those travelled by A.

C^U E S T I O N LXXIII.

Two Messengers, A and B, were dispatched at the same Time, to a Place, at the Distance of 90 (a) Miles j the former of whom, by riding one Mile an Hour more than the Other, arrived at the End of his fourney one Hour before him : The Shiejiion is, to find at what Rate Each travelled per Hour.

If x be the Miles that A rode per Hour ;

then x-i will be the Miles which B rode per Hour:

Moreover-^- will be the Number of Hours in which A x

performed

with their Solutions. 41 performed the whole Journey j and—^— will be the Num ber of Hours wherein B performed it.

And therefore— —— 1, by the Question.

x x-i

whence, by Reduction ax-az=.ax—xl -\- x, Or x1 — x — a:

Therefore xl-x 4- I = a -f- 1 (by compleating the Square) and consequently x =^/a-\-% -4-j =9s + l SS10.

Q.U E S T I O N LXXIV.

To find two Numbers, so that their Sum multiply' d by tht Greater may produce 100 times the Leffer, and being

• multiply''d by the Lejser may produce 64 times tht Greater.

Let x denote the greater Number, and y the Lesser.

Then x-\-y x x ss iooy, And x -4-y x y =64*.

NoWj the first of these Equations being multiply'd by jr, and the Second by x, they become both alike ;

and so we have iocy1 = 64*1 :

Therefore, by taking the square Root, iqy = 8r, and consequently y =-^- : Which Value substituted in the second Equation, gives x

Or, -£LX JL=64.

Whence x = 25x16 =44A{ and> =35-5-.

9 9 9

QUESTION

42 Algebraical Problems, Q_U E S T I O N LXXV.

To find three Numbers, so that their continual Produfl divided, by the Sum os each two of them, may quote given Numbers ; or (which is the fame Thing) to deter

mine the Values of x, y and z, in the underwritten Equations.

xyz xyz xyz

—-—=200, -^—=150, —4—=120.

First, by Multiplication,

*yz=200*4- 2oqy=i50#4- 150z=1207 -f- 120z:

Therefore, by Reduction,

50* + 200y = 150z 1 C x + 4y = 3z, 200* + = 120z j * + 2y — 3z«

Hence x 4- 4y = S* + ty* an(* therefore j>=2* :

Which, substituted in 3z = x -f- 4^, gives 3z=5*;

and z — 3*.

And, by substituting for both y and z, in the Equation

xyz ^xx2xx-ix

—4—= 200, we get :—-i— — 200 ;

x+y & a-4-2*

that is, 2*1= 200.

Consequently *•= 10, y = 20, and z = 30.

Q_U E S T I O N LXXVI.

7i yfrttf /A* i?«//o of two Numbers, whose Rectangle is equal to the Square of their difference.

Let the lefler Number be to the Greater as 1 is to * ; then, if the said Teller Number be denoted by z, the Greater will be expressed by xz, and we shall have xzxz = xz — z]1, or xz^—xz z*-2xz2-\-z* (by the Question). Whence, dividing the whole by z», there re sults x ss **— 2x -4- 1 :

Therefore xz—$x =- 1 ; '

and consequently x

Hence

•with their Solutions. 43 Hence the Rafio of any two Numbers, whose Rectangle is equal to the Square of their Difference, must be that of 2. 618 &c. to Unity.

QUESTION LXXVn.

To find two Numbers, whofe Product is 300 (a) ; so that, if 10 (b) be added to the Lesser, and 8 (c) subtracted from the Greater, the Product of the Sum and Re

mainder Jhall, also, be equal to 300 (a).

Let the greater Number be denoted by *, and the Lesser by y ;

then will \ = I by the Problem.

I x-c x y+b=a J

By the last of which *y + bx — cy — cb — a.

From whence, the first Equation being subtracted, there rests bx-cy — cb = o:

Therefore bx.=*c b-\-cy, and x— c^ ~jV-? . Which sub- stituted in the first Equation, gives ——\———a.

From which we have y* -f- by=^j 1

And, by compleating the Square, y1+*y+- ^~==^+^"

and x ^=-y-)= 20.

Q.U E S T I O N LXXVIII.

J> divide 100 into two such Parts, that their Difference may be to their Sum, as their Rectangle to the Dif ference of their Squares.

"Let a represent half the given Number, and * half the Difference of its two Parts ; then., the Greater of them

44 Algebraical Problems,

them being expressed by a-\-x, and the Lesser by a—x, it will be 2x : za : : a+x x a—x : a+xfr—a-x]* ; that is, by Reduction, x : a : : aa-xx : 40 x : Hence ifix* = a x aa-Ar*,

Or 4*1 = a1 - *1,

and therefore x =y/ -^-=-22,36.

So that, the greater Part is 72,36 j and the Lesser 27,64.

Q_U E S T I O N LXXIX.

To divide the Number 60 (a) into two such Paris, that their Product may be to the Sum of their Squares, in the Ratio of 2 (va) to 5 (n).

Let * be the greater Part : ' Then, the lesser Part will be a-xt

the Product ax-xx,

and the Sum of the Squares a1— 2ax-\-2xz.

Therefore m: n : : ax—xx : a*—2ax-\-2x* j and so, 2mxz—2max-\-ma*=nax—nx*.

Whence, 2mxz-\-nxz—2max—nax=—maz ; that is, 2m-\-n xxz — 2m-^-n x ax —-maz j

r\ , maa

Or, xz-ax = -c —.

2J»4-»

Therefore xz-ax-\ ^ — J —— x —— %

4 V4 2OT-J-71/ 4 H4-2OT*

and x = — . / —: — =40.

Q.U E S T I O N LXXX.

To find two Numbers whose Product Jhall be 320 (a), and the Difference of their Cubes to the Cube of their Dif ference, as 61 (n) is to Unity.

Let x be the greater Number, and y the Lesser ; then will x y = a,

and x3-f : x—yV : : n : 1, by tht Question :

Which

with their Solutions. 45 Which, by actually involving x-y,

becomes x3—y3 : x3 — ^x*y -f- 3*v* — -in: 1.

From whence (by subtracting the Consequents from their Antecedents)

we get 3*Jy — ycy1 : x3 — 3*Æy 4" 3*y1 —Jf* '• • *—1 : 1 » Or, which is the same, 3*yx x-y : x-y\3 : : n—l : 1.

Whence, dividing by x-y, we have 3*y : x—y\l : : »— 1 : 1 ;

Or, 3a -.If-jl1 : : n-i : 1 (because xy—a).

From whence *— vr — —— ;

»— 1

and consequently x-y —^y^ ~~~~, which put = i ; Then from the Equations xyz=a, and *-y=:£, the Value of x will be found =tl£_+-ifii —20; and That of y= ^*1 + ^-* =l6, Jfc pr<)i< fi5«

The same otherwise.

Let z denote the Half Sum, and x the Half Difference, of the two Numbers ; then the Greater will be expressed by z-f-*, and the Lesser by z—x ; and we shall therefore

s z-\-x x z —x z=.a 1 . j z1 — *l=:i7,

thatis ( 6z1 x+2x3 z=Snx*.

The last of which, divided by 2xt give3 3z1 + x% = 4»at1.

From whence, the Treble of the first being deducted, we have 4*1 = — 3« j and consequently x ss v 4*-4

Hence « ( z=</a+xx) =s 18 i therefore z + * as 20, and

46 Algebraical Problems^

and z-x— 16 ; which arc the two Numbers that wefe to be sound.

Q.U E S T I O N LXXXI.

A Farmer received yl. 4*. for a certain Quantity of Wheat, and an equal Sum, at a Price less by is. 6 d.

a BuJhel, for a Quantity of Barley^ which exceeded That of the Wheat by 16 BuJhels : How many Bushels were there of Each ?

Put a = the total Value of each Sort of Grain, b »= the Difference of the Quantities, c = the Difference of the Prices per Buihel, and *- = the Number of Bushels of the Wheat : Then, dividing the whole Price by the Number of Bushels, we have — for the Price of the Wheat per Bushel : And, in the same manner, the Price of the Barley per Bushel will appear to be -~ .

Therefore— h th* Qutstion.

X x+o

Whence (by Reduction) ax -\- ab - ax mm cxz 4- bcx ; Or,

c

Consequently x \ ^ — — = 32, the Num ber of Bushels of Wheat ; and x + 16 -= 48, the Num ber of Bushel* of Barley.

QUESTION

with their Solution^ 47

- Q.U E S T I O N LXXXH.

One bought two Pieces of Cloth of different Sorts, whereof the Finer cost 4 Shillings a Yard more than the Other ; so that, for the Finejl, he paid 360 Shillings j whereas the Coarsest, which exceeded the Finest by 10 Yards, cost him, only, 320 Shillings : How many Yards were there of each Piece ?

Let x be the Number of Yards of the Finest, and y, the Number of Shillings a Yard :

Then jp+io, will be the Length of the coarsest Piece, and y—4., its Price per Yard.

Hence J *y =_26o } by the Question.

t*+iox>—4=320*

By the last Equation xy-\- loy-A^— 360 } from whence deducting xy «= 360,

we have iqy - 4* = 0.

lx

Therefore 1 qy=4*'; and^ = —. Which, being sub- stituted in the first Equation, we get——=360 : Whence x comes out =4/900 = 30.

Q.U E S T I O N LXXXIII.

There are two Numbers, whofe Rectangle is equal to the Difference of their Squares ; and the Sum of their Squares is also equal to the Difference of their Cubes ; What are thofe Numbers ?

Let x denote the lesser Number, and let the Greater be in Proportion thereto as^ is to Unity } Or, which is th«

fame Thing, let the greater Number be denoted by xy.

Therefore j ~Z\^^x\hy Division.

From

48 Algebraical Problems, From the first of which Equations, y» -y = i;

whence f -y + I = i, and consequently j = 1 - But, by the second Equation =. ^ ^ - = 2 (be- cause > y = y4- 1 ) = — + cs«*--j—J— =——h S£ 5 - 2 (hr multiplying both the Numerator and De-

4

nominator by ^ 5 — 1) —iLJL . Therefore the two z

Quantities sought are f V 5, andJLt-i^i.

4 Q_U E S T I O N LXXXIV.

A sets out from London for York, at the same time as B sets out from York for London ; and the Rate at which They travel is such, that A, 9 Hours after their meet ing arrives at York, and Yi at London, in 16 Hours after. The Question is, to find in what Time each Traveller performs his Journey.

Let x denote the Number of Hours travelled by Each before the Time of their Meeting on the Road.

Then, since A (by the former Part of the Question) goes over the very same Ground in 9 (a) Hours, as B travelled in x Hours, we have, therefore, as a : x : : x 1 -~» the Time wherein B travels a Distance equal to That gone over by A, in x Hours : But (by the latter Part of the Question) B, in 16 (b) Hours, travels the very same Distance as A in x Hours. Hence it is evi dent that ——• and b are equal to each Other:

And consequently that x =y ah = 12.

Therefore

with their Solutions. 49 Therefore A performs the Journey in 2 1 Hours, and B

•whose Shares mere in Geometrical Propertion, and the greatest of them exceeded the least by 50 /. What were all the several Shares f

If * be put for the Least of them, then the Greatest will be *+$o; the Sum of which two, subtracted from (igo) the Whole, leaves 140-2* for the Mean Share:

Therefore, x : 140— 2x : : 140—2* : *+ 5°, h Question.

Therefore * = 225—1_J> - 40 : And so the other two Shares are 60 and 90 bounds.

Two Notes, One of 1 26 /. payable in 6 Months, and the Other of 150/. payable in 9 Months, were discounted for 2 I. 10 s. What Rate of Interest were Tlxy dis

counted at P

Let* denote the Interest of One Pound for 12 Months:

Then the Amount of 1 /. in 6 Months being » astd»

in 28 Hours.

Q.U ESTION LXXXVI.

E in

5© Algebraical Problems,

in 9 Months the Present Value of the Bill, due 4

at the End of 6 Months, will therefore be—~— ; and 7^<7r of the Bill, due at the End of g Months, -A -j— • Whence, we have—^—I l->o s— 120 4- 150- 8.5) = 261,5 (ty the Question)

Which, by Reduction, becomes 120 4- 90*+ ISo + 75*= 261,5 x i4-j* 4-

261,5 x 8 4- 1 0*4- 3**.

or, 2704-165* = u '.— . —.—^—

8

Therefore *7Q+'65*x8 = ,*, 4- 10* 4- 8,

261,5 J 1 ' '

... 2l6o 1720X , , -

that is, _- j—i = 3*2 4- 10* 4- 8 : 261,5 261,5 •

From which we have

*+2*>o x^=Jj6_,

3x523 3XS23

Whence, by cotnpleating the Square, &c.

x is found = / '3^ , 1205 |a 1295 _

1569 1569 1 i569

4/ 1 ?6x 1 S6q 4- 1 2Q s x 1 295 - . 295 = 0,Q5093 Which 1569

multiply'd by 100, gives 5,093/. or /. 5 : 1 : iof , nearly, for the Rate per Cent, at which the Notes were discounted.

QUESTION

with their Solutions. 51 Q_U E S T I O N LXXXVII.

A and B take, in Trade, 5940 per Annum, each j but A, whose Profits, are 2 per Cent. greater than Those of B, clears 100/. per Annum more than B: JVi.at are the Profits of Each, per Cent ? And IVhat do They clear per Annum ?

Let c = 100 (Pounds),

b (= 5940)— the whole Sum taken by Each, d (= 100/.)= the given Difference of their Gains, x = What A gains per Cent.

x—a (~ x-i)— What B gains per Cent.

Now, since A, in taking c-\-x Pounds, gains x Pounds, it will be c4-x : x : : h : — the whole Gain of A :

c+x And, in the lame manner, we have,

c+x-a : x-a : : b ** g- , the whole Gain of B : c-\-x-a

Therefore ~-c+x c+x-a~ =d, by the Question.J Hence abc — x* + %cx — ax—ac + ccxd;

abc . and consequently x* + 2s-ax* = —^-+ac-cc.

Whence, by compleating the Square,

, t t 2s— a? ohc , aa

2 I d 4

and therefore * = _ / ■ — .2,c 10.

- V d ' 4 2

From which it appears that A gained 10 per Cent. and cleared 540 /. per Annum ; and that B gained % per Cent.

and cleared 440 /. per Annum.

E 2 QUESTION