ARITHMETIC.
REDUCTION.
Reduction is the method of converting numbers from one name, or denomination to another: or the method of finding the value of a quantity in terms of some other higher, or lower quantity.
To reduce from a higher to a lower denomination.
Rule.—Multiply the given number by as many of the lower denomination as make one of the greater;[40] adding to the product as many of the lower denomination as are expressed in the given sum.
Example.—In £6 15s. 5d., how many pence?
| £. | s. | d. |
| 6 | 15 | 5 |
| 20 | ||
| 135 | ||
| 12 | ||
| 1625 | Answer. | |
To convert from a lower to a higher denomination.
Rule.—Divide the given number by as many of the lower denomination as are required to make one of the greater.[41] Should there be any remainder, it will be of the same denomination as the dividend.
Example.—Convert 1625 pence into pounds, shillings, and pence.
| 12 ) | 1625 | pence | |
| 20 ) | 135 | 5 | |
| £6 | 15s. | 5d. Answer. | |
THE RULE OF THREE, OR SIMPLE PROPORTION.
It is called the Rule of Three because three numbers are given to find a fourth. It is also called Simple Proportion, because the 1st term bears the same proportion to the 2nd, as the 3rd does to the 4th. Of the three given numbers, two of them are always of the same kind, or name, and are to be the 1st, and 2nd terms of the question; the 3rd number is always of the same name, or kind as the 4th, or answer sought; and in stating the question it is always to be made the 3rd term. If the answer will be greater than the 3rd term, place the least of the other two given quantities for the 1st term; but if the answer will be less than the 3rd term, put the greater of the two numbers, or quantities, for the 1st term.
Rule.—State the question according to the above directions, and multiply the 2nd and 3rd terms together, and divide this product by the 1st, for the 4th term, or answer sought.
If the 1st and 2nd terms are not of the same denomination, they must be reduced to it; and if the third term is a compound number, it must be reduced to its lowest denomination before the multiplication, or division of the term takes place.
Note 1.—The operation may frequently be considerably abridged, by dividing the 1st and 2nd, or the 1st and 3rd terms, by any number which will exactly divide them, afterwards using the quotients, instead of the numbers themselves.*
Example.—If 2 tons of iron for ordnance cost £40, how many tons may be bought for £360?
As £40 : £360 :: 2 tons : 18 tons.
(Thus 360 × 2) ÷ 40 = 18. The answer.
* Or thus, 9 × 2 = 18. The answer.
Note 2.—A concise method of ascertaining the annual amount of a daily sum of money.
Rule.—Bring the daily sum into pence, and then add together as many pounds, half pounds, groats, and pence, as there are pence in the daily sum, for the amount required. For leap year, add the rate for one day.
Example.—Required the annual amount of 2s. 6d. per diem.
| 2s. 6d. = 30d. | 30 pounds. | |
| 15 = | 30 half pounds. | |
| 10s. = 30 groats. | ||
| 2s. 6d. = 30 pence. | ||
| Annual amount (365 days) ... £45 12s. 6d. | ||
Note 3.—To find the amount of any number of days’ pay, the daily rate (under twenty shillings) being given.
The price of any article being given, the value of any number may be ascertained in a similar manner.
Rule 1. When the rate (or price) is an even number, multiply the given number by half of the rate, doubling the first figure to the right hand for the shillings, the remainder of the product will be pounds.
Example. Required the amount of 243 days’ pay, at 4s. per diem.
| 4/2 = 2 | 243 | |
| 2 | ||
| £48 12s. | Ans. |
Rule 2. When the price is an odd number, find for the greatest number as before, to which add one-twentieth of the given number for the odd shilling.
Example. What is the price of 566 pairs of shoes, at 7s. per pair.
| 566 | 2/0 ) | 56/6 | ||
| 3 | 28 6 | |||
| 169 | 16s. | |||
| 28 | 6 | |||
| £198 | 2s. | Ans. |