BOOK II.
COMMERCIAL ARITHMETIC.
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 expressedeither 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 ispossibleto 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 ofastronomicalstandards, 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]
214. Thesecondhaving 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]
A geographical mile is¹/₆₀th of a degree, and three such miles are one nautical league.
In the measurement of cloth or linen the following are also used:
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.
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.
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 ANDOF ALL DRY GOODS.
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 poundaverdupois, and 5760 a poundtroy. The first pound is always used, except in weighing precious metals and stones, and also medicines. It is divided as follows:
AVERDUPOIS WEIGHT.
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.
The pound troy contains 5760 grains. A cubic foot of water weighs 75·7374 pounds troy, or 908·8488 ounces.
APOTHECARIES’ WEIGHT.
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.
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 acompound 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 calledReduction, 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.
220. The same may be done where the number first expressed is fractional. For example, how many shillings and pence are there in⁴/₁₅of a pound? Now,⁴/₁₅of a pound is⁴/₁₅of 20 shillings;⁴/₁₅of 20 is
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), but⁸⁰/₂₄₀ = ⅓(108); therefore, 6s.8d.=£⅓.
EXERCISES.
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 AppendixOn Decimal Money). I should strongly recommend the reader to make himself perfectly familiar with the modesgiven 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
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.Subtractionis 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:
In this example, and almost every other, the process may be a littleshortened 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 ofMultiplication, 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
which is therefore £7. 13. 6¼ × 13.
This process is usually written as follows:
228.Divisionis 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.
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 calledPractice. 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 02. Since 10s. is £½, 153 tons, at 10s.each, will cost £15³/₂, which is76 10 03. 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 is38 5 04. 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 is3 16 65. 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 is0 19 1½The sum of all these quantities is425 10 7½
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 is425 10 7½
which is, therefore, £2 . 15 . 7½ × 153.
The whole process may be written down as follows:
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:
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 give360 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.