But hours is the whole time, therefore, the cistern being 1, General Ans. a + b hours, and the other in 4 hours, and find the answer by the general formula. Ans. 3 hours. Suppose one pipe would fill the cistern in 8 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. Let xA's share of the gain. 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. Since a fraction is divided by dividing its numerator, the ma m + n + p will be found by dividing the numerator a multiplied by m is m a, therefore, m a divided by 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. numbers instead of the letters, A's share is $800, B's share is 3 8 +5 +3 5 8 +5 +3 For, putting in the 8 or of 16 8+5+3 or of it, and C's share is or of it. That is, the sum of all their porportions 16 is 16, and of these A furnished 8; B, 5; and C, 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 p the principal, or sum lent. = r= the rate per annum, which in the above case the time for which it was lent, α= the amount. rp the interest for one year, and Then and trp and p+trp Hence we have = = do. for t years, = the amount. p+trpa 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 2 years. See Arithmetic, page 84. Ans. $249 78875 110123. 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 (or d) cents. How many did he break? For every one unbroken he was to receive 8 or b cents, these will amount to 8 x or b x; and for every one broken he was to pay back 12 or c cents, these will amount to 240 12 x cents, or a c è x; this must be subtracted from the former. 12 x, subtracted from 8 x, is 240 24012 x, or 20 x cx subtracted from bx, is b x quantity ca cx is not so large as ca, by the quantity cx, therefore when we subtract ca from br, 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 20 x a c = d b x + c x = d + ac d+ac x= b+c Particular Ans. 17 unbroken, and 3 broken. d + ac General Ans. Unbroken Putting numbers into the general answer, b+c 100+ 12 X 20 17. 8 +12 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 da c) 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 The above is a good illustration of positive and negative quantities, or quantities affected with the signs and -. sign+ is placed before the quantities, which he is to receive, and the sign before his losses. We observed that the dif - ference between receiving 100 and the difference between + 100 and Also the difference between +d and losing 240 is 340, that it, 240 is 340, or their sum. a c is d+a c. So the |
