But a hours is the whole time, therefore, the cistern being 1, Ꮖ Suppose one pipe would fill the cistern in 8 hours, and the other in 4 hours, and find the answer by the general formula. Ans. 3 hours. 12. Suppose it were required to make a rule for Fellowship. First take a particular case. Three men, commencing trade together, furnished money in the following proportions; A $8 as often as B $5, and as often as C $3. They gained $800. What is each man's share of the gain? It is evident that they must receive in the proportion of the capital that they respectively furnished. x= A's share of the gain. Let Now, instead of 8, 5, and 3, suppose they furnished in the proportion of m, n, and p; and let the whole gain be a. Let x = A's share of the gain. m n m part of m a = A's share. m + n + p 1 m Since a fraction is divided by dividing its numerátor, the part of ma by 'm. ma m + n + p will be found by dividing the numerator a multiplied by m is m a, therefore, m a divided by 1 m m + n + p m + n + p Hence to find the share of either, multiply the whole sum to be divided, by the proportion of the stock which he furnished, and divide the product by the sum of their proportions. The propriety of this rule is easily seen. For, putting in the numbers instead of the letters, A's share is $800, B's share is 3 5 8 8+5+3 or of or of it, and C's share is or of it. That is, the sum of all their propor 8+5+3 13. Let it be required to find what sum, put at interest at a given rate, will amount to a given sum in a given time; that is, to find a rule, by which the principal may be found, when the rate, time, and amount are given. First take a particular case. A man lent some money for 3 years, interest at 6 per cent, and received for interest and principal $472. What was the sum lent? It is a custom established among mathematicians to use the first letters of the alphabet for known quantities, and some of the last letters for unknown quantities. It is, however, frequently convenient to choose letters, that are the initials of the words for which they stand, whether the quantities be known or unknown. To generalize the above example, Let is or .06. and p = the principal, or sum lent. r the rate per annum, which in the above case t = the time for which it was lent, That is, multiply the rate by the time, add 1 to the product, and divide the amount by this, and it will give the principal. In the above example the rate is .06, which, multiplied by 3 (the time), gives .18, and one added to this makes 1.18; 472 divided by 1.18 gives 400, as before. Apply this rule to the following example. A man owes $275, due two years and three months hence, without interest." What ought he to pay now, supposing money to be worth 4 per cent. per annum? N. B. 2 years and 3 months is 24 years. See Arithmetic, page 84. 78875 Ans. $249,715. The learner may now make rules for the following purposes: 14. The interest, time, and rate being given, to find the principal. 15. The amount, time, and principal being given, to find the rate. 16. The amount, principal, and rate given, to find the time. 17. A man agreed to carry 20 (or a) earthen vessels to a certain place, on this condition; that for every one delivered safe he should receive 8 (or b) cents, and for every one he broke, he should forfeit 12 (or c) cents; he received 100 (ord) How many did he break? cents. For every one unbroken he was to receive 8 or b cents, these will amount to 8 x orb x; and for every one broken he was to pay back 12 or c cents, these will amount to 240 cx; this must be subtracted from the former. cents, or a c 240 12x, subtracted from 8 X, - 240. 12 x is 8x - 240 + 12 x, or 20 x Also ca cx subtracted from b x, is b x ca+cx; for the quantity ca-cx is not so large as c a, by the quantity c x, therefore when we subtract ca from b x, we subtract too much by cx, and in order to obtain a correct result, it is necessary to add cx. The equation is -240 100 or bx + cx x=17 ac=d bx + cx=d+ac (b + c) x=d+ac d+ac X = b+c Particular Ans. 17 unbroken, and 3 broken. The propriety of this answer may be shown as follows: If he had broken the whole 20 (or a) he must have paid 12 × 20 = 240 (or a c) cents; but instead of paying this, he received 100 (or d) cents. Now the difference to him between paying 240 and receiving 100 is evidently 340, (or d + ac) cents. The difference for each vessel between paying 12 and receiving 8 is 20 (or b+c) cents; 340 divided by 20 gives 17, the an swer. The above is a good illustration of positive and negative quantities, or quantities affected with the signs + and —. The sign+ is placed before the quantities, which he is to receive, and the sign before his losses. We observed that the difference between receiving 100 and losing 240 is 340, that is, the difference between + 100 and 240 is 340, or their sum. Also the difference between + dand-ac is d+ac. So the |