BOOK II.

COMMERCIAL ARITHMETIC.

SECTION I.
WEIGHTS, MEASURES, &C.

212. In making the calculations which are necessary in commercial affairs, no more processes are required than those which have been explained in the preceding book. But there is still one thing wanted—not to insure the accuracy of our calculations, but to enable us to compare and judge of their results. We have hitherto made use of a single unit (15), and have treated of other quantities which are made up of a number of units, in Sections II., III., and IV., and of those which contain parts of that unit in Sections V. and VI. Thus, if we are talking of distances, and take a mile as the unit, any other length may be represented,[31] either by a certain number of miles, or a certain number of parts of a mile, and (1 meaning one mile) may be expressed either by a whole number or a fraction. But we can easily see that in many cases inconveniences would arise. Suppose, for example, I say, that the length of one room is ¹/₁₈₀ of a mile, and of another ¹/₁₇₄ of a mile, what idea can we form as to how much the second is longer than the first? It is necessary to have some smaller measure; and if we divide a mile into 1760 equal parts, and call each of these parts a yard, we shall find that the length of the first room is 9 yards and ⁷/₉ of a yard, and that of the second 10 yards and ¹⁰/₈₇ of a yard. From this we form a much better notion of these different lengths, but still not a very perfect one, on account of the fractions ⁷/₉ and ¹⁰/₈₇. To get a clearer idea of these, suppose the yard to be divided into three equal parts, and each of these parts to be called a foot; then ⁷/₉ of a yard contains 2⅓ feet, and ¹⁰/₈₇ of a yard contains ³⁰/₈₇ of a foot, or a little more than ⅓ of a foot. Therefore the length of the first room is now 9 yards, 2 feet, and ⅓ of a foot; that of the second is 10 yards and a little more than ⅓ of a foot. We see, then, the convenience of having large measures for large quantities, and smaller measures for small ones; but this is done for convenience only, for it is possible to perform calculations upon any sort of quantity, with one measure alone, as certainly as with more than one; and not only possible, but more convenient, as far as the mere calculation is concerned.

The measures which are used in this country are not those which would have been chosen had they been made all at one time, and by a people well acquainted with arithmetic and natural philosophy. We proceed to shew how the results of the latter science are made useful in our system of measures. Whether the circumstances introduced are sufficiently well known to render the following methods exact enough for the recovery of astronomical standards, may be matter of opinion; but no doubt can be entertained of their being amply correct for commercial purposes.

It is evidently desirable that weights and measures should always continue the same, and that posterity should be able to replace any one of them when the original measure is lost. It is true that a yard, which is now exact, is kept by the public authorities; but if this were burnt by accident,[32] how are those who shall live 500 years hence to know what was the length which their ancestors called a yard? To ensure them this knowledge, the measure must be derived from something which cannot be altered by man, either from design or accident. We find such a quantity in the time of the daily revolution of the earth, and also in the length of the year, both of which, as is shewn in astronomy, will remain the same, at least for an enormous number of centuries, unless some great and totally unknown change take place in the solar system. So long as astronomy is cultivated, it is impossible to suppose that either of these will be lost, and it is known that the latter is 365·24224 mean solar days, or about 365¼ of the average interval which elapses between noon and noon, that is, between the times when the sun is highest in the heavens. Our year is made to consist of 365 days, and the odd quarter is allowed for by adding one day to every fourth year, which gives what we call leap-year. This is the same as adding ¼ of a day to each year, and is rather too much, since the excess of the year above 365 days is not ·25 but ·24224 of a day. The difference is ·00776 of a day, which is the quantity by which our average year is too long. This amounts to a day in about 128 years, or to about 3 days in 4 centuries. The error is corrected by allowing only one out of four of the years which close the centuries to be leap-years. Thus, a.d. 1800 and 1900 are not leap-years, but 2000 is so.

213. The day is therefore the first measure obtained, and is divided into 24 parts or hours, each of which is divided into 60 parts or minutes, and each of these again into 60 parts or seconds. One second, marked thus, 1″,[33] is therefore the 86400ᵗʰ part of a day, and the following is the

MEASURE OF TIME.[34]

60 seconds are 1 minute 1 m.
60 minutes ” 1 hour1 h.
24 hours ” 1 day1 d.
7 days ” 1 week1 wk.
365 days ” 1 year1 yr.

214. The second having been obtained, a pendulum can be constructed which shall, when put in motion, perform one vibration in exactly one second, in the latitude of Greenwich.[35] If we were inventing measures, it would be convenient to call the length of this pendulum a yard, and make it the standard of all our measures of length. But as there is a yard already established, it will do equally well to tell the length of the pendulum in yards. It was found by commissioners appointed for the purpose, that this pendulum in London was 39·1393 inches, or about one yard, three inches, and ⁵/₃₆ of an inch. The following is the division of the yard.

MEASURES OF LENGTH.

The lowest measure is a barleycorn.[36]

3 barleycornsare1 inch1 in.
12 inches1 foot1 ft.
3 feet1 yard1 yd.
5½ yards1 pole1 po.
40 poles or 220 yards1 furlong1 fur.
8 furlongs or 1760 yards1 mile1 mi.
Also 
6 feet1 fathom1 fth.
69⅓ miles1 degree1 deg. or 1°.

A geographical mile is ¹/₆₀th of a degree, and three such miles are one nautical league.

In the measurement of cloth or linen the following are also used:

2¼inches are1 nail1 nl.
4nails1 quarter (of a yard)1 qr.
3quarters1 Flemish ell1 Fl. e.
5quarters1 English ell1 E. e.
6quarters1 French ell1 Fr. e.

215. MEASURES OF SURFACE, OR SUPERFICIES.

All surfaces are measured by square inches, square feet, &c.; the square inch being a square whose side is an inch in length, and so on. The following measures may be deduced from the last, as will afterwards appear.

144square inches are1 square foot1 sq. ft.
9square feet1 square yard1 sq. yd.
30¼square yards1 square pole1 sq. p.
40square poles1 rood1 rd.
4roods1 acre1 ac.

Thus, the acre contains 4840 square yards, which is ten times a square of 22 yards in length and breadth. This 22 yards is the length which land-surveyors’ chains are made to have, and the chain is divided into 100 links, each ·22 of a yard or 7·92 inches. An acre is then 10 square chains. It may also be noticed that a square whose side is 69⁴/₇ yards is nearly an acre, not exceeding it by ⅕ of a square foot.

216. MEASURES OF SOLIDITY OR CAPACITY.[37]

Cubes are solids having the figure of dice. A cubic inch is a cube each of whose sides is an inch, and so on.

1728cubic inches are1 cubic foot1 c. ft.
27cubic feet1 cubic yard1 c. yd.

This measure is not much used, except in purely mathematical questions. In the measurements of different commodities various measures were used, which are now reduced, by act of parliament, to one. This is commonly called the imperial measure, and is as follows:

MEASURE OF LIQUIDS AND
OF ALL DRY GOODS.

4gills  are1 pint1 pt.
2pints1 quart1 qt.
4quarts1 gallon1 gall.
2gallons1 peck[38]1 pk.
4pecks1 bushel1 bu.
8bushels1 quarter1 qr.
5quarters1 load1 ld.

The gallon in this measure is about 277·274 cubic inches; that is, very nearly 277¼ cubic inches.[39]

217. The smallest weight in use is the grain, which is thus determined. A vessel whose interior is a cubic inch, when filled with water,[40] has its weight increased by 252·458 grains. Of the grains so determined, 7000 are a pound averdupois, and 5760 a pound troy. The first pound is always used, except in weighing precious metals and stones, and also medicines. It is divided as follows:

AVERDUPOIS WEIGHT.

27¹¹/₃₂grainsare1 dram1 dr.
6 drams, or drachms1 ounce[41]1 oz.
16 ounces1 pound1 lb.
28 pounds1 quarter1 qr.
4 quarters1 hundred-weight1 cwt.
20 hundred-weight1 ton1 ton.

The pound averdupois contains 7000 grains. A cubic foot of water weighs 62·3210606 pounds averdupois, or 997·1369691 ounces.

For the precious metals and for medicines, the pound troy, containing 5760 grains, is used, but is differently divided in the two cases. The measures are as follow:

TROY WEIGHT.

24grainsare1 pennyweight1 dwt.
20pennyweights1 ounce1 oz.
12ounces1 pound1 lb.

The pound troy contains 5760 grains. A cubic foot of water weighs 75·7374 pounds troy, or 908·8488 ounces.

APOTHECARIES’ WEIGHT.

20grainsare1 scruple℈
3scruples1 dramʒ
8drams1 ounce℥
12ounces1 poundlb

218. The standard coins of copper, silver, and gold, are,—the penny, which is 10⅔ drams of copper; the shilling, which weighs 3 pennyweights 15 grains, of which 3 parts out of 40 are alloy, and the rest pure silver; and the sovereign, weighing 5 pennyweights and 3¼ grains, of which 1 part out of 12 is copper, and the rest pure gold.

MEASURES OF MONEY.

The lowest coin is a farthing, which is marked thus, ¼, being one fourth of a penny.

2farthings are1 halfpenny ½d.
2halfpence1 penny1d.
12pence1 shilling1s.
20shillings1 pound[42] or sovereign£1
21shillings1 guinea.[43]

219. When any quantity is made up of several others, expressed in different units, such as £1. 14. 6, or 2cwt. 1qr. 3lbs., it is called a compound quantity. From these tables it is evident that any compound quantity of any substance can be measured in several different ways. For example, the sum of money which we call five pounds four shillings is also 104 shillings, or 1248 pence, or 4992 farthings. It is easy to reduce any quantity from one of these measurements to another; and the following examples will be sufficient to shew how to apply the same process, usually called Reduction, to all sorts of quantities.

I. How many farthings are there in £18. 12. 6¾?[44]

Since there are 20 shillings in a pound, there are, in £18, 18 × 20, or 360 shillings; therefore, £18. 12 is 360 + 12, or 372 shillings. Since there are 12 pence in a shilling, in 372 shillings there are 372 × 12, or 4464 pence; and, therefore, in £18. 12. 6 there are 4464 + 6, or 4470 pence.

Since there are 4 farthings in a penny, in 4470 pence there are 4470 × 4, or 17880 farthings; and, therefore, in £18. 12. 6¾ there are 17880 + 3, or 17883 farthings. The whole of this process may be written as follows:

II. In 17883 farthings, how many pounds, shillings, pence, and farthings are there?

Since 17883, divided by 4, gives the quotient 4470, and the remainder 3, 17883 farthings are 4470 pence and 3 farthings (218).

Since 4470, divided by 12, gives the quotient 372, and the remainder 6, 4470 pence is 372 shillings and 6 pence.

Since 372, divided by 20, gives the quotient 18, and the remainder 12, 372 shillings is 18 pounds and 12 shillings.

Therefore, 17883 farthings is 4470¾d., which is 372s. 6¾d., which is £18. 12. 6¾.

The process may be written as follows:

EXERCISES.

A has £100. 4. 11½, and B has 64392 farthings. If A receive 1492 farthings, and B £1. 2. 3½, which will then have the most, and by how much?—Answer, A will have £33. 12. 3 more than B.

In the following table the quantities written opposite to each other are the same: each line furnishes two exercises.

£15 . 18 . 9½15302 farthings.
115ˡᵇˢ 1ᵒᶻ 8ᵈᵚᵗ663072 grains.
3ˡᵇˢ 14ᵒᶻ 9ᵈʳ1001 drams.
3ᵐ 149 yds 2ᶠᵗ 9 in195477 inches.
19ᵇᵘ 2ᵖᵏˢ 1 gall 2 qᵗˢ1260 pints.
16 ʰ 23ᵐ 47ˢ59027 seconds.

220. The same may be done where the number first expressed is fractional. For example, how many shillings and pence are there in ⁴/₁₅ of a pound? Now, ⁴/₁₅ of a pound is ⁴/₁₅ of 20 shillings; ⁴/₁₅ of 20 is

4 × 20, or 4 × 4 (110), or 16 ,
1533

or (105) 5⅓ of a shilling. Again, ⅓ of a shilling is ⅓ of 12 pence, or 4 pence. Therefore, £⁴/₁₅ = 5s. 4d.

Also, ·23 of a day is ·23 × 24 in hours, or 5ʰ·52; and ·52 of an hour is ·52 × 60 in minutes, or 3ᵐ·2; and ·2 of a minute is ·2 × 60 in seconds, or 12ˢ; whence ·23 of a day is 5ʰ 31ᵐ 12ˢ.

Again, suppose it required to find what part of a pound 6s. 8d. is. Since 6s. 8d. is 80 pence, and since the whole pound contains 20 × 12 or 240 pence, 6s. 8d. is made by dividing the pound into 240 parts, and taking 80 of them. It is therefore £⁸⁰/₂₄₀ (107), but ⁸⁰/₂₄₀ = ⅓ (108); therefore, 6s. 8d. = £⅓.

EXERCISES.

⅖ of a day is9ʰ 36ᵐ
 ·12841 of a day3ʰ 4ᵐ 54ᔆ·624[45]
 ·257 of a cwt.28ˡᵇˢ 12ᵒᶻ 8ᵈʳ·704
£·149362ˢ 11ᵈ 3ᶠ·3856

221, 222. I have thought it best to refer the mode of converting shillings, pence, and farthings into decimals of a pound to the Appendix ([See Appendix On Decimal Money]). I should strongly recommend the reader to make himself perfectly familiar with the modes given in that Appendix. To prevent the subsequent sections from being altered in their numbering, I have numbered this paragraph as above.

223. The rule of addition[46] of two compound quantities of the same sort will be evident from the following example. Suppose it required to add £192. 14. 2½ to £64. 13. 11¾. The sum of these two is the whole of that which arises from adding their several parts. Now

¾d. + ½d. = ⁵/₄d. = £0 . 0 . 1¼(219)
11d. + 2d. = 13d. = 0 . 1 . 1
13s. + 14s. = 27s. = 1 . 7 . 0
£64 + £192 = = 256 . 0 . 0
The sum of all of which is£257. 8 . 2¼

This may be done at once, and written as follows:

Begin by adding together the farthings, and reduce the result to pence and farthings. Set down the last only, carry the first to the line of pence, and add the pence in both lines to it. Reduce the sum to shillings and pence; set down the last only, and carry the first to the line of shillings, and so on. The same method must be followed when the quantities are of any other sort; and if the tables be kept in memory, the process will be easy.

224. Subtraction is performed on the same principle as in (40), namely, that the difference of two quantities is not altered by adding the same quantity to both. Suppose it required to subtract £19 . 13. 10¾ from £24. 5. 7½. Write these quantities under one another thus:

Since ¾ cannot be taken from ½ or ²/₄, add 1d. to both quantities, which will not alter their difference; or, which is the same thing, add 4 farthings to the first, and 1d. to the second. The pence and farthings in the two lines then stand thus: 7⁶/₄d. and 11¾d. Now subtract ¾ from ⁶/₄, and the difference is ¾ which must be written under the farthings. Again, since 11d. cannot be subtracted from 7d., add 1s. to both quantities by adding 12d. to the first, and 1s. to the second. The pence in the first line are then 19, and in the second 11, and the difference is 8, which write under the pence. Since the shillings in the lower line were increased by 1, there are now 14s. in the lower, and 5s. in the upper one. Add 20s. to the upper and £1 to the lower line, and the subtraction of the shillings in the second from those in the first leaves 11s. Again, there are now £20 in the lower, and £24 in the upper line, the difference of which is £4; therefore the whole difference of the two sums is £4. 11. 8¾. If we write down the two sums with all the additions which have been made, the process will stand thus:

225. The same method may be applied to any of the quantities in the tables. The following is another example:

From 7 cwt. 2 qrs. 21 lbs. 14 oz. Subtract 2 cwt. 3 qrs. 27 lbs. 12 oz.

After alterations have been made similar to those in the last article, the question becomes:

From7 cwt. 6 qrs. 49 lbs. 14 oz.
Subtract3 cwt. 4 qrs. 27 lbs. 12 oz.
The difference is4 cwt. 2 qrs. 22 lbs.  2 oz.

In this example, and almost every other, the process may be a little shortened in the following way. Here we do not subtract 27 lbs. from 21 lbs., which is impossible, but we increase 21 lbs. by 1 qr. or 28 lbs. and then subtract 27 lbs. from the sum. It would be shorter, and lead to the same result, first to subtract 27 lbs. from 1 qr. or 28 lbs. and add the difference to 21 lbs.

226. EXERCISES.

A man has the following sums to receive: £193. 14. 11¼, £22. 0. 6¾, £6473. 0. 0, and £49. 14. 4½; and the following debts to pay: £200 . 19. 6¼, £305. 16. 11, £22, and £19. 6. 0½. How much will remain after paying the debts?

Answer, £6190. 7. 4¾.

There are four towns, in the order A, B, C, and D. If a man can go from A to B in 5ʰ 20ᵐ 33ˢ, from B to C in 6ʰ 49ᵐ 2ˢ and from A to D in 19ʰ 0ᵐ 17ˢ, how long will he be in going from B to D, and from C to D?

Answer, 13ʰ 39ᵐ 44ˢ, and 6ʰ 50ᵐ 42ˢ.

227. In order to perform the process of Multiplication, it must be recollected that, as in (52), if a quantity be divided into several parts, and each of these parts be multiplied by a number, and the products be added, the result is the same as would arise from multiplying the whole quantity by that number.

It is required to multiply £7. 13. 6¼ by 13. The first quantity is made up of 7 pounds, 13 shillings, 6 pence, and 1 farthing. And

1 farth. × 13 is 13 farth. or£0 . 0 . 3¼(219)
6 pence × 13 is 78 pence, or 0 . 6 . 6
13 shill. × 13 is 169 shill. or 8 . 9 . 0
7 pounds × 13 is 91 pounds, or91 . 0 . 0
The sum of all these is £99 . 15 . 9¼

which is therefore £7. 13. 6¼ × 13.

This process is usually written as follows:

228. Division is performed upon the same principle as in (74), viz. that if a quantity be divided into any number of parts, and each part be divided by any number, the different quotients added together will make up the quotient of the whole quantity divided by that number. Suppose it required to divide £99. 15. 9¼ by 13. Since 99 divided by 13 gives the quotient 7, and the remainder 8, the quantity is made up of £13 × 7, or £91, and £8. 15. 9¼. The quotient of the first, 13 being the divisor, is £7: it remains to find that of the second. Since £8 is 160s., £8. 15. 9¼ is 175s. 9¼d., and 175 divided by 13 gives the quotient 13, and the remainder 6; that is, 175s. 9¼d. is made up of 169s. and 6s. 9¼d., the quotient of the first of which is 13s., and it remains to find that of the second. Since 6s. is 72d., 6s. 9¼d. is 81¼d., and 81 divided by 13 gives the quotient 6 and remainder 3; that is, 81¼d. is 78d. and 3¼d., of the first of which the quotient is 6d. Again, since 3d. is ¹²/₄, or 12 farthings, 3¼d. is 13 farthings, the quotient of which is 1 farthing, or ¼, without remainder. We have then divided £99. 15. 9¼ into four parts, each of which is divisible by 13, viz. £91, 169s., 78d., and 13 farthings; so that the thirteenth part of this quantity is £7. 13. 6¼. The whole process may be written down as follows; and the same sort of process may be applied to the exercises which follow:

Here, each of the numbers 99, 175, 81, and 13, is divided by 13 in the usual way, though the divisor is only written before the first of them.

EXERCISES.

2 cwt. 1 qr. 21 lbs. 7 oz. × 53 = 129 cwt. 1 qr. 16 lbs. 3 oz.
2ᵈ 4ʰ 3ᵐ 27ˢ × 109=236ᵈ 10ʰ 16ᵐ 3ˢ
£27 . 10 . 8 × 569=£15666 . 9 . 4
£7 . 4 . 8 × 123=£889 . 14
£166 × ₈/₃₃=£40 . 4 . 10⁶/₃₃
£187 . 6 . 7 × ³/₁₀₀=£5 . 12 . 4¾ ²/₂₅
4s. 6½d. × 1121=£254 . 11 . 2½
4s. 4d. × 4260=6s. 6d. × 2840

229. Suppose it required to find how many times 1s. 4¼d. is contained in £3. 19. 10¾. The way to do this is to find the number of farthings in each. By 219, in the first there are 65, and in the second 3835 farthings. Now, 3835 contains 65 59 times; and therefore the second quantity is 59 times as great as the first. In the case, however, of pounds, shillings, and pence, it would be best to use decimals of a pound, which will give a sufficiently exact answer. Thus 1s. 4¼d. is £·067, and £3. 19. 10¾ is £3·994, and 3·994 divided by ·067 is 3994 by 67, or 59⁴¹/₆₇. This is an extreme case, for the smaller the divisor, the greater the effect of an error in a given place of decimals.

EXERCISES.

How many times does 6 cwt. 2 qrs. contain 1 qr. 14 lbs. 1 oz.? and 1ᵈ 2ʰ 0ᵐ 47ˢ contain 3ᵐ 46ˢ?

Answer, 17·30758 and 414·367257.

If 2 cwt. 3 qrs. 1 lb. cost £150. 13. 10, how much does 1 lb. cost?

Answer, 9s. 9d. ¹³/₃₀₉.

A grocer mixes 2 cwt. 15 lbs. of sugar at 11d. per pound with 14 cwt. 3 lbs. at 5d. per pound. At how much per pound must he sell the mixture so as not to lose by mixing them?

Answer, 5d. ¾ ¹⁵³/₉₀₅.

230. There is a convenient method of multiplication called Practice. Suppose I ask, How much do 153 tons cost if each ton cost £2. 15. 7½? It is plain that if this sum be multiplied by 153, the product is the price of the whole. But this is also evident, that, if I buy 153 tons at £2. 15. 7½ each ton, payment may be made by first putting down £2 for each ton, then 10s. for each, then 5s., then 6d., and then 1½d. These sums together make up £2. 15. 7½, and the reason for this separation of £2. 15 . 7½ into different parts will be soon apparent. The process may be carried on as follows:

1. 153 tons, at £2 each ton, will cost

£306 0 0

2. Since 10s. is £½, 153 tons, at 10s. each, will cost £15³/₂, which is

76 10 0

3. Since 5s. is ½ of 10s., 153 tons, at 5s., will cost half as much as the same number at 10s. each, that is, ½ of £76 . 10, which is

38 5 0

4. Since 6d. is ⅒ of 5s., 153 tons, at 6d. each, will cost ⅒ of what the same number costs at 5s. each, that is, ⅒ of £38 . 5, which is

3 16 6

5. Since 1½ or 3 halfpence is ¼ of 6d. or 12 halfpence, 153 tons, at 1½d. each, will cost ¼ of what the same number costs at 6d. each, that is, ¼ of £3 . 16 . 6, which is

0 19 1½

The sum of all these quantities is 425 10 7½

which is, therefore, £2 . 15 . 7½ × 153.

The whole process may be written down as follows:

or what
153 tons
would
cost at
£153 0 0 £1 per ton.
£2 is 2 × £1306 0 02 0 0 
10s. is ½ of £176 10 00 10 0
10s. is ½ of £176 10 00 10 0
5s. is ½ of 10s.38 5 00 5 0 
6d. is ⅒ of 5s.3 16 60 0 6 
1½d. is ¼ of 6d.0 19 1½0 0 1½
Sum£425 10 7½0 0 1½

ANOTHER EXAMPLE.

What do 1735 lbs. cost at 9s. 10¾d. per lb.? The price 9s. 10¾d. is made up of 5s., 4s., 10d., ½d., and ¼d.; of which 5s. is ¼ of £1, 4s. is ⅕ of £1, 10d. is ⅙ of 5s., ½d. is ¹/₂₀ of 10d., and ¼d. is ½ of ½d. Follow the same method as in the last example, which gives the following:

or what
1735 tons
would
cost at
£1735 0 0 £1 per ton.
5s. is ¼ of £1433 15 00 5 0  
4s. is ⅕ of £1347 0 00 4 0  
10d. is ⅙ of 5s.72 5 100 0 10 
½d. is ¹/₂₀ of 10d.3 12 3½0 0 0½
¼d. is ½ of ½d.1 16 1¾0 0 0¼
by addition ...£858 9 3¼£0 9 10¾

In all cases, the price must first be divided into a number of parts, each of which is a simple fraction[47] of some one which goes before. No rule can be given for doing this, but practice will enable the student immediately to find out the best method for each case. When that is done, he must find how much the whole quantity would cost if each of these parts were the price, and then add the results together.

EXERCISES.

What is the cost of

243 cwt. at £14 . 18 . 8¼ per cwt.?—Answer, £3629 . 1 . 0¾.

169 bushels at £2 . 1 . 3¼ per bushel?—Answer, £348 . 14 . 9¼.

273 qrs. at 19s. 2d. per quarter?—Answer, £261 . 12. 6.

2627 sacks at 7s. 8½d. per sack?—Answer, £1012 . 9 . 9½.

231. Throughout this section it must be observed, that the rules can be applied to cases where the quantities given are expressed in common or decimal fractions, instead of the measures in the tables. The following are examples:

What is the price of 272·3479 cwt. at £2. 1. 3½ per cwt.?

Answer, £562·2849, or
£562. 5. 8¼. 66½lbs. at 1s. 4½d. per lb. cost £4. 11. 5¼.

How many pounds, shillings, and pence, will 279·301 acres let for if each acre lets for £3·1076?—Answer, £867·9558, or £867. 19. 1¼.

What does ¼ of ³/₁₃ of 17 bush. cost at ⅙ of ⅔ of £17. 14 per bushel?

Answer, £2·3146, or £2. 6. 3½.

What is the cost of 19lbs. 8oz. 12dwt. 8gr. at £4. 4. 6 per ounce?—Answer, £999. 14. 1¼ ⅙.

232. It is often required to find to how much a certain sum per day will amount in a year. This may be shortly done, since it happens that the number of days in a year is 240 + 120 + 5; so that a penny per day is a pound, half a pound, and 5 pence per year. Hence the following rule: To find how much any sum per day amounts to in a year, turn it into pence and fractions of a penny; to this add the half of itself, and let the pence be pounds, and each farthing five shillings; then add five times the daily sum, and the total is the yearly amount. For example, what does 12s. 3¾d. amount to in a year? This is 147¾d., and its half is 73⅞d., which added to 147¾d. gives 221⅝d., which turned into pounds is £221. 12. 6. Also, 12s. 3¾d. × 5 is £3. 1. 6¾, which added to the former sum gives £224. 14. 0¾ for the yearly amount. In the same way the yearly amount of 2s. 3½d. is £41. 16. 5½; that of 6¾d. is £10. 5. 3¾; and that of 11d. is £16. 14. 7.

233. An inverse rule may be formed, sufficiently correct for every purpose, in the following way: If the year consisted of 360 days, or ³/₂ of 240, the subtraction of one-third from any sum per year would give the proportion which belongs to 240 days; and every pound so obtained would be one penny per day. But as the year is not 360, but 365 days, if we divide each day’s share into 365 parts, and take 5 away, the whole of the subtracted sum, or 360 × 5 such parts, will give 360 parts for each of the 5 days which we neglected at first. But 360 such parts are left behind for each of the 360 first days; therefore, this additional process divides the whole annual amount equally among the 365 days. Now, 5 parts out of 365 is one out of 73, or the 73d part of the first result must be subtracted from it to produce the true result. Unless the daily sum be very large, the 72d part will do equally well, which, as 72 farthings are 18 pence, is equivalent to subtracting at the rate of one farthing for 18d., or ½d. for 3s., or 10d. for £3. The rule, then, is as follows: To find how much per day will produce a given sum per year, turn the shillings, &c. in the given sum into decimals of a pound (221); subtract one-third; consider the result as pence; and diminish it by one farthing for every eighteen pence, or ten pence for every £3. For example, how much per day will give £224. 14. 0¾ per year? This is 224·703, and its third is 74·901, which subtracted from 224·703, gives 149·802, which, if they be pence, amounts to 12s. 5·802d., in which 1s. 6d. is contained 8 times. Subtract 8 farthings, or 2d., and we have 12s. 3·802d., which differs from the truth only about ¹/₂₀ of a farthing. In the same way, £100 per year is 5s. 5¾d. per day.

234. The following connexion between the measures of length and the measures of surface is the foundation of the application of arithmetic to geometry.

Suppose an oblong figure, a, b, c, d, as here drawn (which is called a rectangle in geometry), with the side a b 6 inches, and the side a c 4 inches. Divide a b and c d (which are equal) each into 6 inches by the points a, b, c, l, m, &c.; and a c and b d (which are also equal) into 4 inches by the points f, g, h, x, y, and z. Join a and l, b and m, &c., and f and x, &c. Then, the figure a b c d is divided into a number of squares; for a square is a rectangle whose sides are equal, and therefore a a f e is square, since a a is of the same length as a f, both being 1 inch. There are also four rows of these squares, with six squares in each row; that is, there are 6 × 4, or 24 squares altogether. Each of these squares has its sides 1 inch in length, and is what was called in (215) a square inch. By the same reasoning, if one side had contained 6 yards, and the other 4 yards, the surface would have contained 6 × 4 square yards; and so on.

235. Let us now suppose that the sides of a b c d, instead of being a whole number of inches, contain some inches and a fraction. For example, let a b be 3½ inches, or (114) ⁷/₂ of an inch, and let a c contain 2½ inches, or ⁹/₄ of an inch. Draw a e twice as long as a b, and a f four times as long as a c, and complete the rectangle a e f g. The rest of the figure needs no description. Then, since a e is twice a b, or twice ⁷/₂ inches, it is 7 inches. And since a f is four times a c, or four times ⁹/₄ inches, it is 9 inches. Therefore, the whole rectangle a e f g contains, by (234), 7 × 9 or 63 square inches. But the rectangle a e f g contains 8 rectangles, all of the same figure as a b c d; and therefore a b c d is one-eighth part of a e f g, and contains ⁶³/₈ square inches. But ⁶³/₈ is made by multiplying ⁹/₄ and ⁷/₂ together (118). From this and the last article it appears, that, whether the sides of a rectangle be a whole or a fractional number of inches, the number of square inches in its surface is the product of the numbers of inches in its sides. The square itself is a rectangle whose sides are all equal, and therefore the number of square inches which a square contains is found by multiplying the number of inches in its side by itself. For example, a square whose side is 13 inches in length contains 13 × 13 or 169 square inches.

236. EXERCISES.

What is the content, in square feet and inches, of a room whose sides are 42 ft. 5 inch. and 31 ft. 9 inch.? and supposing the piece from which its carpet is taken to be three quarters of a yard in breadth, what length of it must be cut off?—Answer, The content is 1346 square feet 105 square inches, and the length of carpet required is 598 feet 6⁵/₉ inches.

The sides of a rectangular field are 253 yards and a quarter of a mile; how many acres does it contain?—Answer, 23.

What is the difference between 18 square miles, and a square of 18 miles long, or 18 miles square?—Answer, 306 square miles.

237. It is by this rule that the measure in (215) is deduced from that in (214); for it is evident that twelve inches being a foot, the square foot is 12 × 12 or 144 square inches, and so on. In a similar way it may be shewn that the content in cubic inches of a cube, or parallelepiped,[48] may be found by multiplying together the number of inches in those three sides which meet in a point. Thus, a cube of 6 inches contains 6 × 6 × 6, or 216 cubic inches; a chest whose sides are 6, 8, and 5 feet, contains 6 × 8 × 5, or 240 cubic feet. By this rule the measure in (216) was deduced from that in (214).

SECTION II.
RULE OF THREE.

238. Suppose it required to find what 156 yards will cost, if 22 yards cost 17s. 4d. This quantity, reduced to pence, is 208d.; and if 22 yards cost 208d., each yard costs ²⁰⁸/₂₂d. But 156 yards cost 156 times the price of one yard, and therefore cost

208 × 156 pence, or 208 × 156 pence (117).
2222

Again, if 25½ French francs be 20 shillings sterling, how many francs are in £20. 15? Since 25½ francs are 20 shillings, twice the number of francs must be twice the number of shillings; that is, 51 francs are 40 shillings, and one shilling is the fortieth part of 51 francs, or ⁵¹/₄₀ francs. But £20 15s. contain 415 shillings (219); and since 1 shilling is ⁵¹/₄₀ francs, 415 shillings is

⁵¹/₄₀ × 415 francs, or (117) 51 × 415 francs.
40

239. Such questions as the last two belong to the most extensive rule in Commercial Arithmetic, which is called the Rule of Three, because in it three quantities are given, and a fourth is required to be found. From both the preceding examples the following rule may be deduced, which the same reasoning will shew to apply to all similar cases.

It must be observed, that in these questions there are two quantities which are of the same sort, and a third of another sort, of which last the answer must be. Thus, in the first question there are 22 and 156 yards and 208 pence, and the thing required to be found is a number of pence. In the second question there are 20 and 415 shillings and 25½ francs, and what is to be found is a number of francs. Write the three quantities in a line, putting that one last which is the only one of its kind, and that one first which is connected with the last in the question.[49] Put the third quantity in the middle. In the first question the quantities will be placed thus:

22 yds. 156 yds. 17s. 4d.

In the second question they will be placed thus:

20s. £20 15s. 25½ francs.

Reduce the first and second quantities, if necessary, to quantities of the same denomination. Thus, in the second question, £20 15s. must be reduced to shillings (219). The third quantity may also be reduced to any other denomination, if convenient; or the first and third may be multiplied by any quantity we please, as was done in the second question; and, on looking at the answer in (238), and at (108), it will be seen that no change is made by that multiplication. Multiply the second and third quantities together, and divide by the first. The result is a quantity of the same sort as the third in the line, and is the answer required. Thus, to the first question the answer is (238)

208 × 156 pence, or, which is the same thing, 17s. 4d. × 156 .
2222

240. The whole process in the first question is as follows:[50]

The question might have been solved without reducing 17s. 4d. to pence, thus:

The student must learn by practice which is the most convenient method for any particular case, as no rule can be given.

241. It may happen that the three given quantities are all of one denomination; nevertheless it will be found that two of them are of one, and the third of another sort. For example: What must an income of £400 pay towards an income-tax of 4s. 6d. in the pound? Here the three given quantities are, £400, 4s. 6d., and £1, which are all of the same species, viz. money. Nevertheless, the first and third are income; the second is a tax, and the answer is also a tax; and therefore, by (152), the quantities must be placed thus:

£1 : £400 ∷ 4s. 6d.

242. The following exercises either depend directly upon this rule, or can be shewn to do so by a little consideration. There are many questions of the sort, which will require some exercise of ingenuity before the method of applying the rule can be found.

EXERCISES.

If 15 cwt. 2 qrs. cost £198. 15. 4, what does 1 qr. 22 lbs. cost?

Answer, £5 . 14 . 5 ¾ ¹⁸⁵/₂₁₇.

If a horse go 14 m. 3 fur. 27 yds. in 3ʰ 26ᵐ 12ˢ, how long will he be in going 23 miles?

Answer, 5ʰ 29ᵐ 34ˢ(²⁴⁶²/₂₅₃₂₇).

Two persons, A and B, are bankrupts, and owe exactly the same sum; A can pay 15s. 4½d. in the pound, and B only 7s. (6¾)d. At the same time A has in his possession £1304. 17 more than B; what do the debts of each amount to?

Answer, £3340 . 8 . 3 ¾ ⁹/₂₅.

For every (12½) acres which one country contains, a second contains (56¼). The second country contains 17,300 square miles. How much does the first contain? Again, for every 3 people in the first, there are 5 in the second; and there are in the first 27 people on every 20 acres. How many are there in each country?—Answer, The number of square miles in the first is 3844⁴/₉, and its population 3,321,600; and the population of the second is 5,536,000.

If (42½) yds. of cloth, 18 in. wide, cost £59. 14. 2, how much will (118¼) yds. cost, if the width be 1 yd.?

Answer, £332. 5. (2⁴/₁₇).

If £9. 3. 6 last six weeks, how long will £100 last?

Answer, (65¹⁴⁵/₃₆₇) weeks.

How much sugar, worth (9¾d). a pound, must be given for 2 cwt. of tea, worth 10d. an ounce?

Answer, 32 cwt. 3 qrs. 7 lbs. ³⁵/₃₉.

243. Suppose the following question asked: How long will it take 15 men to do that which 45 men can finish in 10 days? It is evident that one man would take 45 × 10, or 450 days, to do the same thing, and that 15 men would do it in one-fifteenth part of the time which it employs one man, that is, in (450 ÷ 15) or 30 days. By this and similar reasoning the following questions can be solved.

EXERCISES.

If 15 oxen eat an acre of grass in 12 days, how long will it take 26 oxen to eat 14 acres?

Answer, (96¹²/₁₃) days.

If 22 masons build a wall 5 feet high in 6 days, how long will it take 43 masons to build 10 feet?

Answer, (6⁶/₄₃) days.

244. The questions in the preceding article form part of a more general class of questions, whose solution is called the Double Rule of Three, but which might, with more correctness, be called the Rule of Five, since five quantities are given, and a sixth is to be found. The following is an example: If 5 men can make 30 yards of cloth in 3 days, how long will it take 4 men to make 68 yards? The first thing to be done is to find out, from the first part of the question, the time it will take one man to make one yard. Now, since one man, in 3 days, will do the fifth part of what 5 men can do, he will in 3 days make ³⁰/₅, or 6 yards. He will, therefore, make one yard in ³/₆6 or (3 × 5)/30 of a day. From this we are to find how long it will take 4 men to make 68 yards. Since one man makes a yard in

3 × 5 of a day, he will make 68 yards in   3 × 5 × 68 days,
3030
or (116) in 3 × 5 × 68 days; and 4 men will do this in one-fourth;
30
of the time, that is (123), in 3 × 5 × 68 days, or in 8½ days. 
30 × 4

Again, suppose the question to be: If 5 men can make 30 yards in 3 days, how much can 6 men do in 12 days? Here we must first find the quantity one man can do in one day, which appears, on reasoning similar to that in the last example, to be 30/(3 × 5) yards. Hence, 6 men, in one day, will make

6 × 30 yards, and in 12 days will make 12 × 6 × 30 or 144 yards.
5 × 35 × 3

From these examples the following rule may be drawn. Write the given quantities in two lines, keeping quantities of the same sort under one another, and those which are connected with each other, in the same line. In the two examples above given, the quantities must be written thus:

SECOND EXAMPLE.

Draw a curve through the middle of each line, and the extremities of the other. There will be three quantities on one curve and two on the other. Divide the product of the three by the product of the two, and the quotient is the answer to the question.

If necessary, the quantities in each line must be reduced to more simple denominations (219), as was done in the common Rule of Three (238).

EXERCISES.

If 6 horses can, in 2 days, plough 17 acres, how many acres will 93 horses plough in 4½ days?

Answer, 592⅞.

If 20 men, in 3¼ days, can dig 7 rectangular fields, the sides of each of which are 40 and 50 yards, how long will 37 men be in digging 53 fields, the sides of each of which are 90 and 125½ yards?

Answer, 75 2451 days.
20720

If the carriage of 60 cwt. through 20 miles cost £14 10s., what weight ought to be carried 30 miles for £5. 8. 9?

Answer, 15 cwt.

If £100 gain £5 in a year, how much will £850 gain in 3 years and 8 months?

Answer, £155. 16. 8.

SECTION III.
INTEREST, ETC.

245. In the questions contained in this Section, almost the only process which will be employed is the taking a fractional part of a sum of money, which has been done before in several cases. Suppose it required to take 7 parts out of 40 from £16, that is, to divide £16 into 40 equal parts, and take 7 of them. Each of these parts is

£16 and 7 of them make 16 × 7, or 16 × 7 pounds (116).
404040

The process may be written as below:

Suppose it required to take 13 parts out of a hundred from £56. 13. 7½.

Let it be required to take 2½ parts out of a hundred from £3 12s. The result, by the same rule is

£3 12s. × 2½ , or 123 £3 12s. × 5 ;
100200

so that taking 2½ out of a hundred is the same as taking 5 parts out of 200.

EXERCISES.

Take 7⅓ parts out of 53 from £1 10s.

Answer, 4s. 1 129 d.
159

Take 5 parts out of 100 from £107 13s. 4¾d.

Answer, £5. 7. 8 and ³/₂₀ of a farthing.

£56 3s. 2d. is equally divided among 32 persons. How much does the share of 23 of them exceed that of the rest?

Answer, £24. 11. 4½ ½.

246. It is usual, in mercantile business, to mention the fraction which one sum is of another, by saying how many parts out of a hundred must be taken from the second in order to make the first. Thus, instead of saying that £16 12s. is the half of £33 4s., it is said that the first is 50 per cent of the second. Thus, £5 is 2½ per cent of £200; because, if £200 be divided into 100 parts, 2½ of those parts are £5. Also, £13 is 150 per cent of £8. 13. 4, since the first is the second and half the second. Suppose it asked, How much per cent is 23 parts out of 56 of any sum? The question amounts to this: If he who has £56 gets £100 for them, how much will he who has 23 receive? This, by 238, is 23 × ¹⁰⁰/₅₆ or ²³⁰⁰/₅₆ or 41¹/₁₄. Hence, 23 out of 56 is 41¹/₁₄ per cent.

Similarly 16 parts out of 18 is 16 × ¹⁰⁰/₁₈, or 88⁸/₉ per cent, and 2 parts out of 5 is 2 × ¹⁰⁰/₅, or 40 per cent.

From which the method of reducing other fractions to the rate per cent is evident.

Suppose it asked, How much per cent is £6. 12. 2 of £12. 3? Since the first contains 1586d., and the second 2916d., the first is 1586 out of 2916 parts of the second; that is, by the last rule, it is ¹⁵⁸⁶⁰⁰/₂₉₁₆, or 54¹¹³⁶/₂₉₁₆, or £54. 7. 9½ per cent, very nearly. The more expeditious way of doing this is to reduce the shillings, &c. to decimals of a pound. Three decimal places will give the rate per cent to the nearest shilling, which is near enough for all practical purposes. For instance, in the last example, which is to find how much £6·608 is of £12·15, 6·608 × 100 is 660·8, which divided by 12·15 gives £54·38, or £54. 7. Greater correctness may be had, if necessary, as in the [Appendix].

EXERCISES.

How much per cent is 198¼ out of 233 parts?—Ans. £85. 1. 8¾.

Goods which are bought for £193. 12, are sold for £216. 13. 4; how much per cent has been gained by them?

Answer, A little less than £11. 18. 6.

A sells goods for B to the amount of £230. 12, and is allowed a commission[51] of 3 per cent; what does that amount to?

Answer, £6 . 18. 4¼ ⁷/₂₅.

A stockbroker buys £1700 stock, brokerage being at £⅛ per cent; what does he receive?—Answer, £2. 2. 6.

A ship whose value is £15,423 is insured at 19⅔ per cent; what does the insurance amount to?—Answer, £3033. 3. 9½ ²/₅.

247. In reckoning how much a bankrupt is able to pay his creditors, as also to how much a tax or rate amounts, it is usual to find how many shillings in the pound is paid. Thus, if a person who owes £100 can only pay £50, he is said to pay 10s. in the pound. The rule is easily derived from the same reasoning as in 246. For example, £50 out of £82 is

£ 50 out of £1, or 50 × 20 shillings,
8282

or 12s. 2½ ¹⁵/₄₁ in the pound.

248. Interest is money paid for the use of other money, and is always a per-centage upon the sum lent. It may be paid either yearly, half-yearly, or quarterly; but when it is said that £100 is lent at 4 per cent, it must be understood to mean 4 per cent per annum; that is, that 4 pounds are paid every year for the use of £100.

The sum lent is called the principal, and the interest upon it is of two kinds. If the borrower pay the interest as soon as, from the agreement, it becomes due, it is evident that he has to pay the same sum every year; and that the whole of the interest which he has to pay in any number of years is one year’s interest multiplied by the number of years. But if he do not pay the interest at once, but keeps it in his hands until he returns the principal, he will then have more of his creditor’s money in his hands every year, and if it were so agreed will have to pay interest upon each year’s interest for the time during which he keeps it after it becomes due. In the first case, the interest is called simple, and in the second compound. The interest and principal together are called the amount.

249. What is the simple interest of £1049. 16. 6 for 6 years and one-third, at 4½ per cent? This interest must be 6⅓ times the interest of the same sum for one year, which (245) is found by multiplying the sum by 4½, and dividing by 100. The process is as follows:

(230)(a) £1049 .16 . 6
a × 44199 .6 . 0
a × ½524 .18 . 3

(82) 100) 47,24 . 4 . 3(£47 . 4 . 10¹¹/₁₀₀

20
(228)  4,84[52]
12
10,11[53]
(b) £47 . 4 . 10¹¹/₁₀₀Int. for one yr.
b × 6283 .9 .0⁶⁶/₁₀₀
b × ⅓15 . 14 . 11³⁷/₁₀₀
£299 . 4 . 0³/₁₀₀Int. for 6⅓ yrs.

EXERCISES.

What is the interest of £105. 6. 2 for 19 years and 7 weeks at 3 per cent?

Answer, £60. 9, very nearly.

What is the difference between the interest of £50. 19 for 7 years at 3 per cent, and for 8 years at 2½ per cent?

Answer, 10s. (2½)d.

What is the interest of £157. 17. 6 for one year at 5 per cent?

Answer, £7. 17. 10½.

Shew that the interest of any sum for 9 years at 4 per cent is the same as that of the same sum for 4 years at 9 per cent.

250. In order to find the interest of any sum at compound interest, it is necessary to find the amount of the principal and interest at the end of every year; because in this case (248) it is the amount of both principal and interest at the end of the first year, upon which interest accumulates during the second year. Suppose, for example, it is required to find the interest, for 3 years, on £100, at 5 per cent, compound interest. The following is the process:

£100 First principal.
5 First year’s interest.
105 Amount at the end of the first year.
(249)    5 . 5 Interest for the second year on £105.
110. 5 Amount at the end of two years.
5 . 10 . 3Interest due for the third year.
115 . 15 . 3Amount at the end of three years.
100 . 0 . 0First principal.
15 . 15 . 3Interest gained in the three years.

When the number of years is great, and the sum considerable, this process is very troublesome; on which account tables[54] are constructed to shew the amount of one pound, for different numbers of years, at different rates of interest. To make use of these tables in the present example, look into the column headed “5 per cent;” and opposite to the number 3, in the column headed “Number of years,” is found 1·157625; meaning that £1 will become £1·157625 in 3 years. Now, £100 must become 100 times as great; and 1·157625 × 100 is 115·7625 (141); but (221) £·7625 is 15s. 3d.; therefore the whole amount of £100 is £115. 15. 3, as before.

251. Suppose that a sum of money has lain at simple interest 4 years, at 5 per cent, and has, with its interest, amounted to £350; it is required to find what the sum was at first. Whatever the sum was, if we suppose it divided into 100 parts, 5 of those parts were added every year for 4 years, as interest; that is, 20 of those parts have been added to the first sum to make £350. If, therefore, £350 be divided into 120 parts, 100 of those parts are the principal which we want to find, and 20 parts are interest upon it; that is, the principal is £(350 × 100)/150, or £291. 13. 4.

252. Suppose that A was engaged to pay B £350 at the end of four years from this time, and that it is agreed between them that the debt shall be paid immediately; suppose, also, that money can be employed at 5 per cent, simple interest; it is plain that A ought not to pay the whole sum, £350, because, if he did, he would lose 4 years’ interest of the money, and B would gain it. It is fair, therefore, that he should only pay to B as much as will, with interest, amount in four years to £350, that is (251), £291. 13. 4. Therefore, £58. 6. 8 must be struck off the debt in consideration of its being paid before the time. This is called Discount;[55] and £291. 13. 4 is called the present value of £350 due four years hence, discount being at 5 per cent. The rule for finding the present value of a sum of money (251) is: Multiply the sum by 100, and divide the product by 100 increased by the product of the rate per cent and number of years. If the time that the debt has yet to run be expressed in years and months, or months only, the months must be reduced to the equivalent fraction of a year.

EXERCISES.

What is the discount on a bill of £138. 14. 4, due 2 years hence, discount being at 4½ per cent?

Answer, £11. 9. 1.

What is the present value of £1031. 17, due 6 months hence, interest being at 3 per cent?

Answer, £1016. 12.

253. If we multiply by a + b, or by a-b, when we should multiply by a, the result is wrong by the fraction

b + b, or b ,
aa - b

of itself: being too great in the first case, and too small in the second. Again, if we divide by a + b, where we should have divided by a, the result is too small by the fraction b/a of itself; while, if we divide by a-b instead of a, the result is too great by the same fraction of itself. Thus, if we divide by 20 instead of 17, the result is ³/₁₇ of itself too small; and if we divide by 360 instead of 365, the result is too great by ⁵/₃₆₅, or ¹/₇₃ of itself.

If, then, we wish to find the interest of a sum of money for a portion of a year, and have not the assistance of tables, it will be found convenient to suppose the year to contain only 360 days, in which case its 73d part (the 72d part will generally do) must be subtracted from the result, to make the alteration of 360 into 365. The number 360 has so large a number of divisors, that the rule of Practice (230) may always be readily applied. Thus, it is required to find the portion which belongs to 274 days, the yearly interest being £18. 9. 10, or 18·491.

274 18·491
180is ½ of 360 9·246
94
90is ½ of 180 4·623
4 is ¹/₉₀ of 360  ·205
9)14·074
8)1·564
·196

13·878 = £13 . 17 . 7 Answer.

But if the nearest farthing be wanted, the best way is to take 2-tenths of the number of days as a multiplier, and 73 as a divisor; since m ÷ 365 is 2m ÷ 730, or (²/₁₀)m ÷ 73. Thus, in the preceding instance, we multiply by 54·8 and divide by 73; and 54·8 × 18·491 = 1013·3068, which divided by 73 gives 13·881, very nearly agreeing with the former, and giving £13. 17. 7½, which is certainly within a farthing of the truth.

254. Suppose it required to divide £100 among three persons in such a way that their shares may be as 6, 5, and 9; that is, so that for every £6 which the first has, the second may have £5, and the third £9. It is plain that if we divide the £100 into 6 + 5 + 9, or 20 parts, the first must have 6 of those parts, the second 5, and the third 9. Therefore (245) their shares are respectively,

£100 × 6 , £ 100 × 5 and £ 100 × 9
202020

or £30, £25, and £45.

EXERCISES.

Divide £394. 12 among four persons, so that their shares may be as 1, 6, 7, and 18.—Answer, £12. 6. 7½; £73. 19. 9; £86. 6. 4½; £221. 19. 3.

Divide £20 among 6 persons, so that the share of each may be as much as those of all who come before put together.—Answer, The first two have 12s. 6d.; the third £1. 5; the fourth £2. 10; the fifth £5; and the sixth £10.

255. When two or more persons employ their money together, and gain or lose a certain sum, it is evidently not fair that the gain or loss should be equally divided among them all, unless each contributed the same sum. Suppose, for example, A contributes twice as much as B, and they gain £15, A ought to gain twice as much as B; that is, if the whole gain be divided into 3 parts, A ought to have two of them and B one, or A should gain £10 and B £5. Suppose that A, B, and C engage in an adventure, in which A embarks £250, B £130, and C £45. They gain £1000. How much of it ought each to have? Each one ought to gain as much for £1 as the others. Now, since there are 250 + 130 + 45, or 425 pounds embarked, which gain £1000, for each pound there is a gain of £¹⁰⁰⁰/₄₂₄. Therefore A should gain 1000 × ²⁵⁰/₄₂₅ pounds, B should gain 1000 × ¹³⁰/₄₂₅ pounds, and C 1000 × ⁴⁵/₄₂₅ pounds. On these principles, by the process in (245), the following questions may be answered.

A ship is to be insured, in which A has ventured £1928, and B £4963. The expense of insurance is £474. 10. 2. How much ought each to pay of it?

Answer, A must pay £132. 15. (2½).

A loss of £149 is to be made good by three persons, A, B, and C. Had there been a gain, A would have gained 4 times as much as B, and C as much as A and B together. How much of the loss must each bear?

Answer, A pays £59. 12, B £14. 18, and C £74. 10.

256. It may happen that several individuals employ several sums of money together for different times. In such a case, unless there be a special agreement to the contrary, it is right that the more time a sum is employed, the more profit should be made upon it. If, for example, A and B employ the same sum for the same purpose, but A’s money is employed twice as long as B’s, A ought to gain twice as much as B. The principle is, that one pound employed for one month, or one year, ought to give the same return to each. Suppose, for example, that A employs £3 for 6 months, B £4 for 7 months, and C £12 for 2 months, and the gain is £100; how much ought each to have of it? Now, since A employs £3 for six months, he must gain 6 times as much as if he employed it one month only; that is, as much as if he employed £6 × 3, or £18, for one month; also, B gains as much as if he had employed £4 × 7 for one month; and C as if he had employed £12 × 2 for one month. If, then, we divide £100 into 6 × 3 + 4 × 7 + 12 × 2, or 70 parts, A must have 6 × 3, or 18, B must have 4 × 7, or 28, and C 12 × 2, or 24 of those parts. The shares of the three are, therefore,

£6 × 3 × 100 , £ 4 × 7 × 100 , and £ 12 × 2 × 100
6 × 3 + 4 × 7 + 12 × 6 × 3 + 4 × 7 + 12 × 26 × 3 + 4 × 7 + 12 × 2

EXERCISES.

A, B, and C embark in an undertaking; A placing £3. 6 for 2 years, B £100 for 1 year, and C £12 for 1½ years. They gain £4276. 7 How much must each receive of the gain?

Answer, A £226. 10. 4; B £3432. 1. 3; C £617. 15. 5.

A, B, and C rent a house together for 2 years, at £150 per annum. A remains in it the whole time, B 16 months, and C 4½ months, during the occupancy of B. How much must each pay of the rent?[56]

Answer, A should pay £190. 12. 6; B £90. 12. 6; C £18. 15.

257. These are the principal rules employed in the application of arithmetic to commerce. There are others, which, as no one who understands the principles here laid down can fail to see, are virtually contained in those which have been given. Such is what is commonly called the Rule of Exchange, for such questions as the following: If 20 shillings be worth 25½ francs, in France, what is £160 worth? This may evidently be done by the Rule of Three. The rules here given are those which are most useful in common life; and the student who understands them need not fear that any ordinary question will be above his reach. But no student must imagine that from this or any other book of arithmetic he will learn precisely the modes of operation which are best adapted to the wants of the particular kind of business in which his future life may be passed. There is no such thing as a set of rules which are at once most convenient for a butcher and a banker’s clerk, a grocer and an actuary, a farmer and a bill-broker; but a person with a good knowledge of the principles laid down in this work, will be able to examine and meet his own future wants, or, at worst, to catch with readiness the manner in which those who have gone before him have done so for themselves.