much of it was already expired, answered, that two thirds of the time past was equal to four fifths of the time to come. Required the time past, and the time to come. X - ; ; 19. It is required to divide the number 50 into two such parts, that three fourths of one part added to five sixths of the other may make 40. X + 2 & 20. Two workmen received equal sums for their work; but if one of them had received 18 dollars more, and the other 3 dollars less, then } of the wages of the latter would have been equal to ; of the wages of the former. How much did each receive? . -- S. 21. A certain man, when he married, found that his age was to that of his wife as 7 to 5; if they had been married 8 years sooner, his age would have been to hers as 3 to 2. What were their ages at the time of their marriage? 2. } { VI. 1. Divide the number 68 into two such paris, that the difference between the greater and 84, may be equal to three times the excess of 40 above the less. Leta = the less. Then 68—a = the greater. 68—a must be subtracted from 84. Observe that 68 —a is not so great as 68 by a. Therefore if I subtract 68 from 84, I shall subtract too much by the quantity r, and I must add a to obtain the true result. Then we have 84–68–H a for the difference between 84 and 68 — ar. The excess of 40 above the less is 40—?, and 3 times this is 120 — 3 a. By the conditions, 84–68 –– a = 120 — 3 a. Transposing and uniting, 4. a = 104 Dividing by 4, a = 26 = less. 68 — 26 = 42 = greater. JNote. In this question 68 — a was subtracted from 84. Instead of a, now put its value, 68–26. Now 68—26=42, that is, the number to be subtracted from 84 is 42, and the answer must be 42. When 68 is subtracted from 84, the result is 16, which is too small by 26, the value of a ; to this it is necessary to add 26, and it makes 42, the true result, 84–68 + 26 = 42. This shows that we did right in adding a after subtracting 68. This will always be found true. Therefore, when any of the quantities to be subtracted have the sign – bejore them, they must be changed to + in subtracting, and those which have + must be changed to —. 2. A gentleman hired a laborer for 20 days on condition that, for every day he worked, he should receive 7s., but for every day he was idle, he should forfeit 3s. At the end of the time agreed on he received 80 shillings. How many days did he work, and how many days was he idle? Let w = the number of days he worked. Then 20–2 = the number of days he was idle. a days, at 7s. a day, would come to 7 a shillings. 20 — w; at 3s. per day, would be 60–3 a shillings. This must be taken out of 7 a. By the above rule 60–3 r, subtracted from 7 a., leaves 7 a. — 60 + 3 a ; for 60 is too much to be subtracted by 3 a. By the conditions, * 7 a. – 60+ 3 + = 80. Transposing and uniting, 10 a = 140. Dividing by 10, a = 14 = days he worked. ". 20 — a = 6 = days he was idle. 3. Two men, A and B, commenced trade ; A had twice as much money as B ; A gained $ 50, and B lost $90, then the difference between A's and B’s money was equal to three times what B then had. How much did each commence with? 4. Two men, A and B, played together; when they commenced they had $20 between them, after a certain number of games, A had won $6, then the excess of A's money above B's was equal to ; of B's money. How much had each when they commenced? 5. Divide the number 54 into two such parts that the less subtracted from the greater, shall be equal to the greater subtracted from three times the less. What are the parts? 6. It is required to divide the number 204 into two such parts that ; of the less being subtracted from the greater, the remainder will be equal to ; of the greater subtracted from four times the less. Let z = greater part. Then 204 — a = the less part. Multiplying by 7, 35 a -2856 – 14 was 28560 — 140 z – 15 a. Transposing and uniting, 204 c = 31416. a = 154 204 —a = 50 Let a denote the less number, and solve the question again. JNote. Observe, that after multiplying by 5 in the above example, the signs of both terms of the numerator were changed, that of 408 to —, and that of 2 a to +; this was done because it was not required to subtract so much as 408 by 2 a. The change of signs could not be made before multiplying by 5, because the sign — before the fraction showed that the whole fraction was to be subtracted. If the signs of the fraction had been changed at first, it would have been necessary to put the sign + before the fraction. This requires particular attention, because it is of great importance, and there is danger of forgetting it. 7. A man bought a horse and chaise for $341. Now if of the price of the horse be subtracted from twice the price of the chaise, the remainder will be the same as if ; of the price of the chaise be subtracted from three times the price of the horse. Required the price of each. - 8. Two men, A and B, were playing at cards; when they began, A had only ; as much money as B. A won of B $23; then I of B's money, subtracted from A's, would leave one half of what A had at first. How much had each when they began? 9. A man has a horse and chaise. The horse is worth $44 -less than the chaise. If ; of the value of the horse be subtracted from the value of the chaise, the remainder will be the same as if from the value of the horse you subtract 3 of the excess of the value of the horse above 84 dollars. What is the value of the horse? VII. The examples in this article are intended to exercise the learner in putting questions into equation. They require no operations which have not already been explained. It was remarked, that no rule could be given for putting questions into equation, but there is a precept which may be very useful. Take the unknown quantity, and perform the same operations on it, that it would be necessary to perform on the answer to see if it was right. When this is done the question is in equation. 1. A and B, being at play, severally cut packs of cards so as to take off more than they left. Now it happened that A cut off twice as many as B left, and B cut off seven times as many as A left. How were the cards cut? Let a = the number B left. Then 2 a = the number A cut off. 52 — a = the number B cut off. 52 — 2 r = the number A left. By the conditions, 7 times 52 — 2 a. are equal to 52—w. 364 — 14 a = 52 — ar. Take the numbers of the answer and endeavor to prove that they are right, and you will see that you take the same course as above. 2. A man, at a card party, betted 3s. to 2 on every deal. After twenty deals he had won 5 shillings. At how many deals did he win? Let a = the number of deals he won. Then 20 — a = the number of deals he lost. Every time he won, he won 2 shillings; that will be 2 a shillings. Every loss was 3 shillings; that will be 3 times 20–2, or 60 — 3 a. The loss must be taken from the gain, and he will have 5 shillings left. 2 z – 60 + 3 r = 5. 3 What two numbers are to each other as 2 to 3; to each of which, if 4 be added, the sums will be as 5 to 7. Let a = the first number. Then o = the second. Adding 4 to each, they become a + 4, and # + 4. The first is now ; of the second, or the second is ; of the first. 7 a -i- 28 3 a. —— =– -H4. 5 2 + 4. A sum of money was divided between two persons, A and B, so that the share of A was to that of B as 5 to 3. Now A's share exceeded ; of the whole sum by $ 50. What was the share of each person? Let a = A's share. 5. The joint stock of two partners, whose particular shares differed by 48 dollars, was to the less as 14 to 5. Required the shares. 6. Four men bought an ox for $43, and agreed that those, who had the hind quarters, should pay 3 cent per pound more than those, who had the fore quarters. A and B had the hind quarters, C and D the fore quarters. A's quarter weighed 158 lb., B's 163 lb., C’s 167 lb., and D's 165 lb. What was each per lb., and what did each man pay? 7. A certain person has two silver cups, and only one cover for both. The first cup weighs 12 oz. If the first cup be covered it weighs twice as much as the other cup, but if the second be covered it weighs three times as much as the first. What is the weight of the cover, and of the second cup? |