A supplement to Hutton's arithmetic: containing the solutions, at full length, of the promiscuous collection of questions proposed in that work. By the author — School-master's guide: or, a complete system of practical arithmetic.

Table of contents

- [titlepage]

- ADVERTISEMENT.

- QUESTION I.

- SOLUTION.

- QUESTION 2.

- SOLUTION.

- QUESTION 3.

- SOLUTION.

- QUESTION 4.

- SOLUTION.

- QUESTION 5.

- SOLUTION.

- QUESTION 6.

- SOLUTION.

- QUESTION 7.

- SOLUTION.

- QUESTION 8.

- SOLUTION.

- QUESTION 9.

- SOLUTION.

- QUESTION 10.

- SOLUTION.

- QUESTION 11.

- SOLUTION.

- QUESTION 12.

- SOLUTION.

- QUESTION 13.

- SOLUTION.

- QUESTION 14.

- SOLUTION.

- QUESTION 15.

- SOLUTION.

- QUESTION 16.

- SOLUTION.

- QUESTION 17.

- SOLUTION.

- QUESTION 18.

- SOLUTION.

- QUESTION 19.

- SOLUTION.

- QUESTION 20.

- SOLUTION.

- QUESTION 21.

- SOLUTION.

- QUESTION 22.

- SOLUTION.

- QUESTION 23.

- SOLUTION.

- QUESTION 24.

- SOLUTION.

- QUESTION 25.

- SOLUTION.

- QUESTION 26.

- SOLUTION.

- QUESTION 27.

- SOLUTION.

- QUESTION 28.

- SOLUTION.

- QUESTION 29.

- SOLUTION.

- QUESTION 30.

- SOLUTION.

- QUESTION 31.

- SOLUTION.

- QUESTION 32.

- SOLUTION.

- QUESTION 33.

- SOLUTION.

- QUESTION 34.

- SOLUTION.

- QUESTION 35.

- SOLUTION.

- QUESTION 36.

- SOLUTION.

- QUESTION 37.

- SOLUTION.

- QUESTION 38.

- SOLUTION.

- QUESTION 39.

- SOLUTION.

- QUESTION 40.

- SOLUTION.

- QUESTION 41.

- SOLUTION.

- QUESTION 42.

- SOLUTION.

- QUESTION 43.

- SOLUTION.

- QUESTION 44.

- SOLUTION.

- QUESTION 45.

- SOLUTION.

- QUESTION 46.

- SOLUTION.

- QUESTION 47.

- SOLUTION.

- QUESTION 48.

- SOLUTION.

- QUESTION 49.

- SOLUTION.

- QUESTION 50.

- SOLUTION.

- QUESTION 51.

- SOLUTION.

- QUESTION 52.

- SOLUTION.

- QUESTION 53.

- SOLUTION.

- QUESTION 54.

- SOLUTION.

- QUESTION 55.

- SOLUTION.

- QUESTION 56.

- SOLUTION.

- QUESTION 57.

- SOLUTION.

- QUESTION 58.

- SOLUTION.

- QUESTION 59.

- SOLUTION.

- QUESTION 60.

- SOLUTION.

- QUESTION 61.

- SOLUTION.

- QUESTION 62.

- SOLUTION.

- QUESTION 63.

- SOLUTION.

- QUESTION 64.

- SOLUTION.

- QUESTION 65.

- SOLUTION.

- QUESTION 66.

- SOLUTION.

- QUESTION 67.

- SOLUTION.

- QUESTION 68.

- SOLUTION.

- QUESTION 69.

- SOLUTION.

- QUESTION 70.

- SOLUTION.

- QUESTION 71.

- SOLUTION.

- QUESTION 72.

- SOLUTION.

- QUESTION 73.

- SOLUTION.

- ERRATUM.

- Lately published,

A PROMISCUOUS COLLECTION of QUESTIONS and their SOLUTIONS.

FINIS.

A SUPPLEMENT TO HUTTON'S ARITHMETIC: CONTAINING THE SOLUTIONS, AT FULL LENGTH, OF THE PROMISCUOUS COLLECTION OF QUESTIONS PROPOSED IN THAT WORK.

BY THE AUTHOR.

LONDON: PRINTED FOR G. ROBINSON AND R. BALDWIN, IN PATER-NOSTER ROW.

M DCC LXXVIII.

ADVERTISEMENT.

THE promiscuous collection of questions proposed in the Arithmetic, was intended for the exercise of pupils after they had gone through the several rules of the book, and has been found very useful to them, by accustoming them to think, and to strike out methods of solution unassisted by other helps. But many gentlemen having expressed a desire to have a publication of the solutions of those questions at full length, that they might have the satisfaction to compare their own solutions with those of the Author, and thereby the opportunity of chusing which they like best, he has herein complied with their request.

These solutions are published apart from the work itself, that the teacher may avail himself of them, and yet keep them from the sight of his pupils if he so chuses, each learner having only the book of arithmetic for the purposes therein mentioned.

The solutions are all delivered in what have been judged to be the best forms. Sometimes two methods are given, where they seemed necessary. In such of them as contain any proportions, or rule-of-three statings, the fourth term is generally put down in the form of a vulgar fraction, placing the product of the second and third terms, as numerator, to be divided by the first as a denominator; the fraction is then abbreviated as much as it can be, before the multiplication and division are performed; by which means generally these two operations are intirely saved, or at least much shortened and facilitated, by bringing them to such small numbers as can be easily multiplied and divided mentally, and the answer thence discovered without any other intermediate figures or operations to be written down. So that there is never occasion for more writing than appears in the operations as here printed, where it will be found that, to avoid the long form of division, the large divisors are separated into their component parts, and the divisions performed separately by them, which are always done by writing down only the quotients. But where the numbers are so large as to make long divisions, or multiplications, unavoidable, those operations are always placed down at the bottom of the solution, that so none of the necessary part of the writing in the full solution, might be omitted.

QUESTION I.

A WAS born when B was 21 years of age: how old will A be when B is 47; and what will be the age of B when A is 60? — Ans. A 26, B 81.

SOLUTION.

- From 47

- take 21

- Ans.——26=A's age.

- To 60

- add 21

- 81=B's age.

QUESTION 2.

What difference is there between twice five and twenty, and twice twenty-five? — Ans. 20.

SOLUTION.

- From 50=2 × 25=twice 25,

- take 30=2 × 5 + 20=10 + 20=twice 5 and 20, or=20 and twice 5,

- remains 20=the answer.

QUESTION 3.

What number taken from the square of 48 will leave 16 times 54? — Ans. — 1440.

SOLUTION.

〈 math 〉

OTHERWISE.

〈 math 〉 the answer.

QUESTION 4.

What number added to the thirty-first part of 3813 will make the sum 200? — Ans. 77.

SOLUTION.

〈 math 〉

OTHERWISE.

〈 math 〉

QUESTION 5.

What number deducted from the 23d part of 29440 will leave the 64th part of the same? — Ans. 820.

SOLUTION.

〈 math 〉

OTHERWISE.

〈 math 〉

QUESTION 6.

The remainder of a division is 325, the quotient 467, the divisor is 43 more than the sum of both; what is the dividend? — Ans. 390270.

SOLUTION.

The answer or dividend here is to be found by multiplying the quotient by the divisor, and to the product adding the remainder. That is 〈 math 〉 the dividend required.

〈 math 〉 sum is 390270 the dividend or answer.

QUESTION 7.

A person, at the time of his out-setting in trade, owed 350l. and had in cash 5307l. 10s. in wares 713l. 7d. and in good debts 210l. 5s. 10d. Now after having traded a year he owed 703l. 17s. and had in cash 4874l. 9s. 4d. in bills 350l. in wares 1075l. 14s. 3½d. and in recoverable debts 613l. 13s. 10½d. What was his real gain that year? — Ans. 329l. 4s. 1d.

SOLUTION.

〈 math 〉

QUESTION 8.

Two persons depart from the same place at the same time, the one travels 30, the other 35 miles a day: how far are they distant after 7 days if they travel both the same road, and how far if they travel in contrary directions? — Ans. 35 and 455 miles.

SOLUTION.

Here 〈 math 〉 the answer in the 1st case.

And 〈 math 〉 miles in the other.

QUESTION 9.

A gentleman's daily expence is 4l. 8s. 1 19/365d. and he saves 500l. in the year: What is his yearly income?

— Ans. 2107l. 12s.

SOLUTION.

Here the daily expence must be multiplied by 365 the days in year, and the 500l. added to the product. But 365 is=73 × 5, therefore multiply by 73 and by 5, as below.

〈 math 〉

QUESTION 10.

Having a piece of land 11 poles in breadth, I demand what length of it must be taken to contain an acre, when 4 poles in breadth require 40 poles in length to contain the same? — Ans. 14 pls. 3 yds.

SOLUTION.

As 〈 math 〉.

QUESTION 11.

If a gentleman, whose annual income is 1000l. spend 20 guineas a week, whether will he save or run in debt, and how much in the year? — Ans. 92l. debt.

SOLUTION.

〈 math 〉

QUESTION 12.

In the latitude of London, the distance round the earth, measuring in the parallel of latitude, is about 15550 miles; now as the earth turns round in 23 hours 56 minutes, at what rate per hour is the city of London carried by this motion from west to east?

— Ans. 649 259/359 miles an hour.

SOLUTION.

〈 math 〉

QUESTION 13.

In order to raise a joint stock of 10000l. A, B, and C, together subscribe 7950l. and D the rest: now A and B are known together to have set their hands to 5800l. and A has been heard to say that he had undertaken for 550l. more than B. What did each proprietor advance?

Ans. A 3175, B 2625, C 2150, D 2050.

SOLUTION.

〈 math 〉

QUESTION 14.

A tradesman increased his estate annually by 100l. more than ¼ part of it, and at the end of 4 years found that his estate amounted to 10342l. 3s. 9d. What had he at out-setting? — Ans. 4000l.

SOLUTION.

Here the amount at the end of each year, is equal to 100l. more than ¼ of what he had at the beginning of the same year; therefore subtract 100l. and there remains 5/4 of what he had at the beginning of that year, consequently subtracting ⅕ of this remainder from itself, there will at last remain the sum at the beginning of the year. And this operation must be made 4 times for the 4 years, as here follows.

〈 math 〉

QUESTION 15.

Paid 1012l. 10s. for 750l. taken in 7 years ago; at what rate per cent. per ann. did I pay interest?

— Ans. 5l.

SOLUTION.

〈 math 〉

Then as 750∶100 ∷ 37½ ∶ 75/15=5 per cent. the answer.

Or as 15 ∶ 2 ∷ 37½ ∶ 75/15=5 per cent. the answer.

QUESTION 16.

What is the interest of 720l. for 73 days, or ⅕ of a year, at 3l. per cent. per annum? Ans. 4l. 6s. 4d. 3⅕ q.

SOLUTION.

〈 math 〉 the answer.

QUESTION 17.

Part 1200 acres of land among A, B, and C, so that B may have 100 more than A, and C 64 more than B?

Ans. A 312, B 412, C 476.

SOLUTION.

〈 math 〉

QUESTION 18.

Divide 1000 crowns, give A 120 more and B 95 less than C. — Ans. A 445, B 230, C 325.

SOLUTION.

〈 math 〉

QUESTION 19.

To how much amounts the order, for which my factor, at the rate of 2½ per cent. receives 22l. 10s? — Ans. 900l.

SOLUTION.

As 〈 math 〉. the answer.

QUESTION 20.

What sum of money will amount to 132l. 16s. 3d. in 15 months, at 5 per cent. per annum simple interest?

— Ans. 125l.

SOLUTION.

As 12 ∶ 15, or as 4 ∶ 5 ∷ 5 ∶ 25/4=6¼=the interest of 100l. for 15 months.

And therefore 106¼=425/4=the amount of 100l. for the same time.

Hence 〈 math 〉 the answer.

QUESTION 21.

Laid out 165l. 15s. in wine at 4s. 3d. a gallon; some of which receiving damage in carriage, I sold the rest at 6s. 4d. gallon, which produced only 110l. 16s. 8d. What quantity was damaged? — Ans. 430 gal.

SOLUTION.

〈 math 〉

Ans. 430 gallons unsold, or damaged.

QUESTION 22.

A father divided his fortune among his sons, giving A 4 as often as B 3, and C 5 as often as B 6; what was the whole legacy, supposing A's share were 5000l.

— Ans. 11875l.

SOLUTION.

A having 4 for B's 3, is the same as A 8 for B 6; and C had 5 for B's 6. Therefore

As 〈 math 〉 the answer.

QUESTION 23.

A stationer sold quills at 10s. 6d. a thousand, by which he cleared ⅓ of the money; but growing scarce, raised them to 12s. a thousand; what did he clear per cent. by the latter price? — Ans. 71l. 8s. 6 6/7d.

SOLUTION.

〈 math 〉

Therefore as 〈 math 〉 the answer.

QUESTION 24.

If 1000 men, besieged in a town, with provisions for 5 weeks, allowing each man 16 oz a day, were reinforced with 500 men more; and hearing that they cannot be relieved till the end of 8 weeks; how many ounces a day must each man have, that the provision may last that time? — Ans. 6⅔ oz.

SOLUTION.

〈 math 〉. the answer.

QUESTION 25.

If a quantity of provisions serve 1500 men 12 weeks, at the rate of 20 ounces a day for each man; how many men will the same provisions maintain for 20 weeks, at the rate of 8 oz. a day for each man? — Ans. 2250 men.

SOLUTION.

〈 math 〉 men, the answer.

QUESTION 26.

In what time will the interest of 72l. 12s. equal that of 15l. 5s. for 64 days, at any rate of interest?

— Ans. 13 161/363 days.

SOLUTION.

Here 〈 math 〉.

And 〈 math 〉.

Then as 〈 math 〉 days, the answer.

QUESTION 27.

A person possessed of ⅜ of a ship, sold ⅔ of his share for 1260l. what was the reputed value of the whole at the same rate? — Ans. 5040l.

SOLUTION.

First ⅔ of 〈 math 〉 the part sold.

Then ¼ ∶ 1 ∷ 1260 ∶ 1260 × 4=5040l. the value of the whole ship.

QUESTION 28.

What sum of money at 4½ per cent. will clear 29l. 15s. in a year and a half's time? — Ans. 440l. 14s. 9 7/9d.

SOLUTION.

- First 4½=9/2,

- and 29l. 15s.=29¾=119/4l.

- also 1½=3/2.

Then 〈 math 〉 the principal required.

QUESTION 29.

What number is that, to which if 2/7 of 5/9 be added, the sum will be 1? — Ans. 53/63.

SOLUTION.

First 2/7 of 〈 math 〉.

Then 1 − 10/63=63/63 − 10/63=53/63 the answer.

QUESTION 30.

A father dying, left his son a fortune, ¼ of which he ran through in 8 months; 3/7 of the remainder lasted him a twelve-month longer, after which he had bare 410l. left: What did his father bequeath him?

— Ans. 956l. 13s. 4d.

SOLUTION.

After spending ¼ he had ¾ remaining.

And after spending 3/7 of the remainder he had 4/7 of that remainder left.

Therefore 4/7 of ¾=3/7 of the whole left at last, the value of which is 410l.

Hence 〈 math 〉 the whole sum bequeathed.

QUESTION 31.

Bought a quantity of goods for 250l. and 3 months after sold it for 275l. How much per cent. per annum did I gain by them? — Ans. 40.

SOLUTION.

Here 275 − 250=25 the gain of 250 for 3 months.

Therefore as 〈 math 〉

QUESTION 32.

A guardian paid his ward 3500l. for 2500l. which he had in his hand 8 years: What rate of interest did he allow him? — Ans. 5 per cent.

SOLUTION.

Here 3500 − 2500=1000 the interest of 2500 for 8 years.

Therefore 〈 math 〉

QUESTION 33.

Bought a quantity of goods for 150l. ready money, and sold it again for 200l. payable at the end of 9 months; what was the gain in ready money, supposing rebate to be made at 5 per cent. — Ans. 42l. 15s. 5 5/83d.

SOLUTION.

As 〈 math 〉 the interest of 100l. for 9 months.

And therefore 100 15/4=415/4=its amount for that time.

Then 〈 math 〉 the present worth of the 200l.

Consequently 〈 math 〉 the gain in ready money.

〈 math 〉

QUESTION 34.

A person being asked the hour of the day, said, The time past noon is equal to ⅘ths of the time till midnight: What was the time? — Ans. 20 min. past 5.

SOLUTION.

Here the one part of the 12 hours, which are contained between noon and midnight, being ⅘ of the other, the two parts are in the ratio of 4 to 5.

Hence as 〈 math 〉 20 min. the time past noon required.

QUESTION 35.

A person, looking on his watch, was asked what was the time of the day, who answered, It is between 4 and 5; but a more particular answer being required, he said that the hour and minute hands were then exactly together: What was the time? — Ans. 21 9/11 min. past 4.

SOLUTION.

As the minute hand goes once round while the hour hand goes but 1/12 part, in every revolution of the former, it goes 11/12 more than the latter.

Now when the first is at 12, the latter is at 4, and therefore the next time the former overtakes the latter, it will have gone 4 parts of the 12 more than this other.

Then state the increases proportional to the distances, as here below.

As 11 ∶ 4 ∷ 60 min.: 〈 math 〉 min. past 4, the time sought.

QUESTION 36.

With 12 gallons of canary at 6s. 4d. a gal. I mixed 18 gal. of white-wine at 4s. 10d. a gal. and 12 gal. of cyder at 3s. 1d. a gal. At what rate must I sell a quart of this composition so as to clear 10 per cent. Ans. 1s. 3 5/7d.

SOLUTION.

- 12gal. × 6s. 4d.=76s. Then as 100 ∶ 10 ∷ 200s. ∶ 20s gain.

- 18gal. × 4s. 10d.=87s. Theref. the 42g. or 168q. must sell for 220s.

- 12gal. × 3s. 1d.=37s. Conseq. as 168 ∶ 220 ∷ 1 ∶ 220/168=55/42=Theref. 42 gal. cost 200s. 9s.2d./7 = 1s. 3 5/7 d. per quart, so as to gain 10 per cent.

QUESTION 37.

Suppose that I have 3/16 of a ship worth 1200l. what part of her have I left after selling ⅖ of 4/9 of my share, and what is it worth? — Ans. 37/240 worth 185l.

SOLUTION.

〈 math 〉 the part of his share, or of 3/16 which is sold. But when 8/45 of any thing is deducted, there remains 37/45 of the same thing. Therefore 〈 math 〉 is the part of the ship remaining.

OTHERWISE.

〈 math 〉 the part of the ship sold.

Therefore 〈 math 〉 the part of the ship remaining, the same as before.

Then, as 〈 math 〉 the value of the part remaining.

QUESTION 38.

What length must be cut off a board 8⅜ inches broad, to contain a square foot, or as much as 12 inches in length and 12 in breadth? — Ans. 17 13/67 inches.

SOLUTION.

As 〈 math 〉 inches in length to be cut off.

〈 math 〉

QUESTION 39.

What sum of money will produce as much interest in 3¼ years, as 210l. 3s. can produce in 5 years and 5 months? — Ans. 350l. 5s.

SOLUTION.

- First 3¼=13/4,

- and 5 y. 5 m.=5 5/12=65/12,

- also 210l. 3s.=210 3/20=4203/20.

Then as 〈 math 〉 the sum required.

QUESTION 40.

There is gained by trading with a ship 120l. 14s. Now suppose that ¼ of her belongs to S, ⅜ to T, ⅛ to V, and the rest to W; what must each have of the gain? — Ans. S 30l. 3s. 6d. T 45l. 5s. 3d. V 15l. 1s. 9d. W 30l. 3s. 6d.

SOLUTION.

First 〈 math 〉 the sum of S, T, and V's.

Therefore 1 − 6/8=2/8=W's share, which is the same as that of S. Also their respective shares are proportional to the numerators of the fractions, viz. to the numbers 2, 3, 1, 2, the sum of which is 8. Then as 8 ∶ 120l. 14s. or as 1 ∶ 15l. 1s. 9d. ∷

- 2 ∶ 30l. 3s. 6d.=S's share

- 3 ∶ 45 5 3=T's

- 1 ∶ 15 1 9=V's

- 2 ∶ 30 3 6=W's

their sum is 120.14.0=the sum given.

QUESTION 41.

If 100l. in 5 years be allowed to gain 20l. 10s. in what time will any sum of money double itself at the same rate of interest? — Ans. 24 16/41 years.

SOLUTION.

Here it is only to find the time in which 100l. will gain 100l. which is thus.

As 20l. 10s.=20½l.=41/2 ∶ 100 ∷ 5 years ∶ 〈 math 〉, the answer.

〈 math 〉

QUESTION 42.

What difference is there between the interest of 350l. at 4 per cent. for 8 years, and the discount of the same sum, at the same rate, and for the same time?

— Ans. 27l. 3 1/33s.

SOLUTION.

First 4 × 8=32 is the interest of 100l. for the 8 years.

Then 〈 math 〉 the disc. of 500.

And 〈 math 〉 the interest of 350.

Therefore 〈 math 〉 the difference required.

QUESTION 43.

If, by selling goods at 50s. per cwt. I gain 20 per cent. what do I gain or lose per cent. by selling at 45s. per cwt.?

— Ans. 8l. gain.

SOLUTION.

As 〈 math 〉 the amount or returns of 100 at the rate of 45 per cwt.

Therefore 108 − 100=8 is the gain per cent.

QUESTION 44.

If, by remitting to Holland, at 34s. 6d. per l. sterling, 4½ per cent. be gained; how goes the exchange, when by remittance I clear 10 per cent.? Ans. 36s. 3 165/209d.

SOLUTION.

- First 34s. 6d.=34½=69/2,

- And 100 + 4½=104½=209/2.

- Then 〈 math 〉. the rate of exchange to gain 10 per cent.

〈 math 〉

QUESTION 45.

Sold goods for 60 guineas, and by so doing, lost 17 per cent. whereas I ought, in dealing, to have cleared 20 per cent. Then how much under their just value were they sold? — Ans. 28l. 1s. 8 20/83d.

SOLUTION.

- First 100 + 20=120,

- and 100 − 17=83,

- their difference is 120 − 83=37;

- also 60 guineas=63l.

- Then 〈 math 〉. the answer.

〈 math 〉

QUESTION 46.

If, by selling goods at 27d. per lb. I gain cent. per cent. what do I clear per cent. by selling for 9 guineas per cwt? Ans. 50 per cent.

SOLUTION.

At 27d. per lb. it is per cwt. 27 × 112d.=9 × 28s.

And 9 guineas=9 × 21s.

Therefore 〈 math 〉 the amount of 100 at the latter price.

Consequently 150 − 100=50=the gain per cent.

QUESTION 47.

If 20 men can perform a piece of work in 12 days, how many will accomplish another thrice as big in onefifth of the time? — Ans. 300.

SOLUTION.

As 1 × ⅕ ∶ 20 ∷ 3 × 1 ∶ 3 × 20 × 5=300 men the answer.

QUESTION 48.

A younger brother received 6300l. which was just 7/9 of his elder brother's fortune: What was the father worth at his death? — Ans. 14400l.

SOLUTION.

As the one was 7/9 of the other, their shares were to each other, as 7 is to 9. Therefore

As 7 ∶ 7 + 9=16 ∷ 6300 ∶ 16 × 900=14400l. the answer.

QUESTION 49.

A person making his will, gave to one child 13/20 of his estate, and the rest to another; and when these legacies came to be paid, the one turned out 600l. more than the other: What did the testator die worth? — Ans. 2000l.

SOLUTION.

As the one had 13/20, the other must have had 7/20, and their shares in the ratio of 13 to 7. Therefore

As 13 − 7=6 ∶ 13 + 7=20 ∷ 600 ∶ 100 × 20=2000l. the whole estate.

QUESTION 50.

A father devised 7/18 of his estate to one of his sons, and 7/18 of the residue to another, and the surplus to his relict for life: the children's legacies were found to be 257l. 3s. 4d. different: Pray what money did he leave the widow the use of? — Ans. 635l. 10 30/49d.

SOLUTION.

- First, 1 − 7/18=11/16=the residue after the 1st share.

- Therefore 7/18 of 11/18=77/324=the 2d son's share.

- And 〈 math 〉 the difference of the sons' shares.

- Also fince 18 − 7=11, we have 11/18 of 11/18 = 121/324 = the proportional share of the relict.

- Conseq. as 49 ∶ 121 ∷ 257l. 3s. 4d. ∶ 635l. os. 10 30/49d.

〈 math 〉

QUESTION 51.

What number is that, from which, if you take 2/7 of 3/•, and to the remainder add 7/16 of 1/20, the sum will be 10?

— Ans. 10 191/2240.

SOLUTION.

First 〈 math 〉.

And 7/16 of 1/20=7/320.

Therefore 〈 math 〉 the answer.

QUESTION 52.

There is a number which, if multiplied by ⅔ of ⅞ of 1½, will produce 1: What is the square of that number?

— Ans. 1 15/49.

SOLUTION.

Here 〈 math 〉 the number.

And theref. 8/7 × 8/7 = 64/49 = 1 15/49 = the sq. of the numb.

QUESTION 53.

A person dying, left his wife with child, and making his will, ordered that if she went with a son, ⅔ of his estate should belong to him, and the remainder to his mother; and if she went with a daughter, he appointed the mother ⅔ and the girl the remainder: but it happened that she was delivered both of a son and daughter; by which she lost in equity 2400l. more than if it had been only a girl: What would have been her dowry had she had only a son? — Ans. 2100l.

SOLUTION.

Since the son's share is to the mother's, as 2 to 1, and the mother's to the daughter's, as 2 to 1; therefore their three shares are respectively as the numbers 4, 2, and 1, the sum of which is 7. Consequently their real shares are 4/7, 2/7, and 1/7.

Now 〈 math 〉.

Theref. 〈 math 〉 the answ.

QUESTION 54.

Three persons purchase together a ship, toward the payment of which A advanced 2/9, and B 2/7 of the value, and C 200l. How much paid A and B, and what part of the vessel had C? — Ans. A 90 10/31l. B 116 4/31l. C 31/63 part.

SOLUTION.

First 1 − 2/9 − 2/7 = 1 − 14/63 − 18/63 = 1 − 32/63 = 31/63 = C's part.

Consequently as 31 ∶ 200 ∷

- 〈 math 〉 paid by A,

- 〈 math 〉 paid by B.

QUESTION 55.

A and B clear by an adventure at sea, 60 guineas, with which they agree to buy a horse and chaise, of which they were to have the use, in proportion to the sums adventured, which was found to be A 9 to B 8; they cleared 45 per cent. What money then did each send abroad?

— Ans. A 74l. 2s. 4 4/17d. and B 65l. 17s. 7 13/17d.

SOLUTION.

First 〈 math 〉 the sum of the adventures.

Therefore as 8 + 9 = 17 ∶ 140 ∷

- 〈 math 〉

- 〈 math 〉

〈 math 〉

QUESTION 56.

In an article of trade, A gains 18s. 3d. and his adventure was 40s. more than B's, whose share of profit is but 12s. What are the particulars of their stock?

— Ans. A 5l. 16s. 9⅗d. and B 3l. 16s. 9⅗d.

SOLUTION.

The difference of the adventures being 40s. and the difference of the gains = 18s. 3d. − 12s. = 6s. 3d. = 6¼s. = 25/4.

Therefore as 25/4 ∶ 40, or as 25 ∶ 160, or as 5 ∶ 32 ∷

- 〈 math 〉

- 〈 math 〉

QUESTION 57.

Three persons entered joint trade, to which A contributed 240l. and B 210l. they clear 120l. of which 30l. belongs of right to C. Required that person's stock, and the several gains of the other two?

— Ans. C's stock 150l. A gained 48l. and B 42l.

SOLUTION.

First 120 − 30=90=the sum of the gains of A and B.

And 240 + 210=450=the sum of their stocks.

Therefore, as 450 ∶ 90, or as 5 ∶ 1 ∷

- 240 ∶ 240/5=48=A's gain,

- 210 ∶ 210/5=42=B's gain.

Also, as 90 ∶ 450, or as 1 ∶ 5 ∷ 30 ∶ 150=C's stock.

QUESTION 58.

A and B in partnership equally divide the gain; A's money, which was 96l. 12s. lay for 15 months, and B's for no more than 6: What was the adventure of the latter? — Ans. 241l. 10s.

SOLUTION.

Since the two shares of the gain are equal, by the rule of Double-Fellowship it appears that the two products are equal which are made by multiplying each stock by its time, and consequently that the stocks are inversely or reciprocally as the times. Hence

As 6 ∶ 15, or as 2 ∶ 5 ∷ 96l. 12s. ∶ 48l. 6s. × 5=241l. 10s.=the sum adventured by B.

QUESTION 59.

Put out 420l. to interest, and in 6½ years time there was sound to be due 556l. 10s. What was the rate of interest? — Ans. 5 per cent.

SOLUTION.

First 556½ − 420=136½=the interest of 420 for 6½ yrs. Theref. 136½ ÷ 6½=273 ÷ 13=21=its interest for 1 yr. Then as 420 ∶ 21, or as 20 ∶ 1 ∷ 100 ∶ 5. Therefore the rate of interest was 5 per cent.

QUESTION 60.

A clears 12l. in 6 months, B 15l. in 5 months, and C, whose stock was 40l. c•••s 21l. in 9 months: What was the whole stock? — Ans. 125 5/7l.

SOLUTION.

As 21 ∶ 360=40 × 9, or as 7 ∶ 120 ∷

- 〈 math 〉 the prod. of A's stock and time,

- 〈 math 〉 the prod. of B's stock and time,

Then

- 〈 math 〉 A's stock,

- 〈 math 〉 B's stock.

Lastly 240/7 + 360/7 + 40 = 600/7 + 40 = 85 5/7 + 40 = 125 5/7 = the sum of all their stocks, as required.

QUESTION 61.

A had 12 pipes of wine, which he parted with to B at 4½ per cent. profit, who sold them to C for 40l. 12s. advantage; C made them over to D for 605l. 10s. and cleared thereby 6 per cent. How much a gallon did this wine cost A? — Ans. 6s. 8 6640/11077d.

SOLUTION.

The 12 pipes cost D 605l. 10s.

Therefore as 106 ∶ 100, or as 〈 math 〉 the sum they cost C.

Consequently 〈 math 〉 the sum they cost B.

Hence as 〈 math 〉 the sum the 12 pipes cost A.

The 12th part of this is 468720/11077 = the price of 1 pipe or 126 gallons.

Divide now the numerator by 126, or by its component parts 2, 9, and 7; and lastly divide by the denominator, for the answer, thus: 〈 math 〉

QUESTION 62.

A, of Amsterdam, orders B of London to remit to C of Paris, at 52½d. ster. a crown, and to draw on D, of Antwerp, for the value, at 34½s. slem. a l. ster. but as soon as B received the commission, the exchange was on Paris at 53d. a crown: Pray at what rate of exchange ought B to draw on D, to execute his orders, and be no loser?

— Ans. 34s. 2 5/53d.

SOLUTION.

As 〈 math 〉 the answer required.

〈 math 〉

QUESTION 63.

A, with intention to clear 20 guineas, on a bargain with B, rates hops at 15d. a lb. which cost him 10½d. B, apprized of that, sets down malt, which cost 20s. a quarter, at an adequate price: For how much malt did they contract? — Ans. 49 qrs.

SOLUTION.

As 〈 math 〉 the gain per quarter.

Then, as 〈 math 〉 the number of quarters required.

QUESTION 64.

A and B venturing equal sums of money, clear by joint trade 180l. By agreement, A was to have 8 per cent. because he spent time in the execution of the project, and B was to have only 5: What was allotted to A for his trouble? — Ans. 41l. 10s. 9 3/13d.

SOLUTION.

As 〈 math 〉. the answer required.

QUESTION 65.

Laid out in a lot of muslin 500l. upon examination of which, 3 parts in 9 proved damaged, so that I could make but 5s. a yard of the same; and by so doing find I lost 50l. by it. At what rate per ell am I to part with the undamaged muslin in order to gain 50l. upon the whole?

— Ans. 11s. 7 2/7d.

SOLUTION.

In order to gain 50l. by the whole, he must gain 100l. by the undamaged part, because he lost 50l. by the part which was damaged.

Now the part damaged was ⅓, and the rest ⅔; also the whole cost 500l.; the ⅓ of which is 166⅔, and the ⅔ of it is 333⅓. Consequently the damaged part was sold for 166⅔ − 50 or 116⅔; and the sound part must be sold for 433⅓=333⅓ + 100.

But the damaged part sold at 5s. per yard, therefore as 5s. or ¼l. ∶ 116⅔l. ∷ 1 yd. ∶ 116⅔ × 4=466⅔ yards, the quantity which was damaged. And the double of it, or 933⅓ yards was the undamaged part, which must sell for 433⅓l. Therefore as 933⅓ ∶ 1 ∷ 433⅓ ∶ 433⅓ / 933⅓ = (by multiplying the terms both by 3) 1300/2800=13/28l. the price per yard.

And consequently, as 〈 math 〉 the price per ell required.

OTHERWISE.

Since the sound part is the double of the part damaged, and the former must gain a sum just the double of that which was lost by the latter, it is evident that it must be sold at a rate as much above the prime cost, as the other was below it.

Now the loss on ⅓ part was 50l. at which rate the whole 500 would have brought only 350; therefore as 350 ∶ 500, or as 7 ∶ 10 ∷ 5s. ∶ 50/7=7 1/7s. the prime cost per yard.

Hence 7 1/7 − 5=2 1/7=the loss per yard on the damaged part, and 7 1/7 + 2 1/7=9 2/7s. the price per yard of the sound part.

Lastly, as 4 ∶ 5 ∷ 9 2/7 ∶ 9 2/7 × 5/4=65/7 × 5/4=325/28s. the price per ell, the same as before.

QUESTION 66.

A, at Paris, draws on B in London, 1400 crowns, at 56d. ster. a crown, for the value of which B draws again on A at 57d. sterl. a crown, besides reckoning commission ½ per cent. Did A gain or lose by this transaction, and what? — Ans. He gained 17 13/19 crowns.

SOLUTION.

First, 1400 × 56=78400 pence, the value of A's draft on B.

Then as 100 ∶ 100½, or as 200 ∶ 201 ∷ 78400 ∶ 392 × 201d.=the sum that B must draw for at 57d. per crown.

Therefore as 〈 math 〉 crowns which B must draw for.

Consequently 1400 − 1382 6/19=17 13/19 crowns is A's gain.

〈 math 〉

QUESTION 67.

A, B, and C are in company; A put in his share of the stock for 6 months, and laid claim to ⅙ of the profits; B put in his for 9 months; C advanced 500l. for 8 months, and required on the balance ⅗ of the gain: Required the stock of the other two adventurers?

— Ans. A 185l. 3s. 8 4/9d. and B 172l. 16s. 9 13/27d.

SOLUTION.

First ⅙ + ⅗=5/30 + 18/30=23/30 the sum of the shares of the gain of A and C.

Conseq. 1 − 23/30=7/30=the share of B. And the gains of A, B, C are respectively proportional to the numbers 5, 7, 18.

But the gains are as the products of the stocks and times, and the product of C's stock and time is 4000=500 × 8. Therefore as 18 ∶ 4000 or as 9 ∶ 2000 ∷

- 5 ∶ 10000/9=the prod. of A's stock and time,

- 7 ∶ 14000/9=the prod. of B's stock and time,

These being divided by their respective times, which are 6 and 9 months, we have

- 〈 math 〉=A's stock,

- 〈 math 〉=B's stock.

QUESTION 68.

A young hare starts 40 yards before a greyhound, and is not perceived by him till she has been up 40 seconds; she scuds away at the rate of 10 miles an hour, and the dog, on view, makes after her at the rate of 18: How long will the course hold, and what ground will be run over, beginning with the out-setting of the dog?

— Ans. 60 5/22 sec. and 530 yards run.

SOLUTION.

First 60 × 60=3600=the number of seconds in an hour. And 1760 yards are a mile. Therefore as 3600 ∶ 40, or as 90 ∶ 1 ∷ 1760 × 10 ∶ 1760/9=the yards run by the hare before the dog starts. Consequently 〈 math 〉 the distance of the hare before the dog when he starts, and which therefore he must run more than she in order to overtake her.

But in 1 hour or 3600 seconds, the dog runs 8 miles or 8 × 1760 yards more than the hare. Therefore, as 〈 math 〉 seconds, the time of the dog's running.

And consequently as 〈 math 〉. the whole space run by the dog.

QUESTION 69.

If A leave Exeter at 8 o'clock on Monday morning for London, and ride at the rate of 3 miles an hour without intermission; and B set out from London for Exeter at 4 the same evening, and ride 4 miles an hour constantly: Supposing the distance between the two cities be 130 miles, whereabout on the road shall they meet?

— Ans. 69 3/7 miles from Exeter.

SOLUTION.

From 8 o'clock till 4 o'clock, are 8 hours. Therefore 8 × 3=24 are the miles rode by A before B sets out from London. And consequently 130 − 24=106 are the miles to travel between them after that.

Hence, as 7=3 + 4 ∶ 3 ∷ 106 ∶ 318/7=45 3/7 miles more travelled by A at the meeting.

Consequently 24 + 45 3/7=69 3/7 miles from Exeter is the place of their meeting.

QUESTION 70.

A reservoir for water has two cocks to supply it; by the first alone it may be filled in 40 minutes, by the second in 50 min. and it hath a discharging cock, by which it may, when full, be emptied in 25 min. Now, supposing that these 3 cocks are all left open, and that the water comes in; in what time, supposing the influx and efflux of the water to be always alike, would rhe cistern be filled? — Ans. 3 hrs. 20 min.

SOLUTION.

The rates of running are reciprocally as the times of filling. Therefore the rate of increase of the influx over the efflux, is as 〈 math 〉, which rate of increase is also reciprocally as the time of filling.

Therefore the whole time of filling, is 200/1 minutes, or 3 hours 20 minutes=the answer required.

QUESTION 71.

A sets out of London for Lincoln, at the very same time that B at Lincoln sets forward for London, distant 100 miles: After 7 hours they meet on the road, and it then appeared that A had road 1½ miles an hour more than B. At what rate an hour did each of them travel?

— Ans. A 7 25/28, and B 6 11/28 miles.

SOLUTION.

First, 7 × 1½=10½ miles which A travels more than B.

- Hence 〈 math 〉 travelled by A,

- And 〈 math 〉 travelled by B.

Then dividing each distance by 7, the time of travelling, we have

- 55¼ / 7 = 7 25/28 = A's rate of travelling,

- 44¾ / 7 = 6 11/28 = B's rate of travelling.

QUESTION 72.

A and B truck; A has 12½ cwt. of Farnham hops, at 2l. 16s. a cwt. but in barter insists on 3l. B has wine worth 5s. a gal. which he raises in proportion to A's demand. On the balance A received but a hhd. of wine: What had he in ready money? — Ans. 20l. 12s. 6d.

SOLUTION.

First, 12½ × 3=37½l.=37l. 10s. is the amount of the hops.

But as 〈 math 〉 the barter price per gallon of the wine. Therefore 〈 math 〉 is the value of the hogshead of wine.

Consequently the difference, or 37l. 10s. − 16l. 17s. 6d.=20l. 12s. 6d. is the sum given in money.

QUESTION 73.

A, of Amsterdam, owes to B, of Paris, 3000 guilders of current specie, which he is to remit to him, by order, the exchange 91d. Flem. de banco a crown, the agio 4 per cent. but when this was to be negotiated, the exchange was down at 90d. a crown, and the agio 5 per cent. What did B get by this turn of affairs?

— Ans. 5 liv. 12 sol. 8 584/1183 den.

SOLUTION.

First, 3000 guilders=3000 × 40=120000 pence, currency.

And as 104 ∶ 100 ∷ 120000 ∶ 12000000/104=1500000/13 d. banco.

Then as 〈 math 〉 crowns, the amount at the first exchange.

Again, as 105 ∶ 100, or as 21 ∶ 20 ∷ 1200000 ∶ 800000/7d. banco.

Then as 90 ∶ 800000/7 ∷ 1 cr.=80000/63 crowns, the amount by the latter exchange.

The difference, or 〈 math 〉 is the sum gained by B.

〈 math 〉

ERRATUM.

In the note to the equation of payments, (in the Arithmetic) containing Malcolm's rule, the remark concerning a greater number of payments than two, should be omitted, as that method of equating for 3 or more payments, will not give the answer strictly true. But in all such cases, to obtain the just answer, Malcolm's General Principle of Solution ought to be used, viz. making the interests of the sums that are kept till after they are due, equal to the discounts of those which are paid before they are due. The resolution of the resulting equation will indeed require some knowledge in Algebra; but for ordinary purposes, the rule in common use will bring out answers sufficiently near the truth.

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A supplement to Hutton's arithmetic: containing the solutions, at full length, of the promiscuous collection of questions proposed in that work. By the author

School-master's guide: or, a complete system of practical arithmetic.

Hutton, Charles, 1737-1823.

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http://ota.ox.ac.uk/id/4255

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1106012542

ID [isbn13]

9781106012548

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CW3306612051

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0030700600

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— A supplement to Hutton's arithmetic: containing the solutions, at full length, of the promiscuous collection of questions proposed in that work. By the author, School-master's guide: or, a complete system of practical arithmetic., Hutton, Charles, 1737-1823.. 43,[5]p. ; 12⁰. London :: printed for G. Robinson and R. Baldwin,. 1778.. [First published as 'The school-master's guide: or, a complete system of practical arithmetic'.] [Reproduction of original from the British Library.] [English Short Title Catalog, ESTCT53058.] [Electronic data. Farmington Hills, Mich. : Thomson Gale, 2003. Page image (PNG). Digitized image of the microfilm version produced in Woodbridge, CT by Research Publications, 1982-2002 (later known as Primary Source Microfilm, an imprint of the Gale Group).]

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