QUESTIONS TO BE SOLVED BY SIMPLE EQUATIONS. PAGE 177. 1. There are two numbers of which the difference is 9, and the sum 43. What are the numbers? Let x represent the smaller of the two numbers, then by the question, the other must be x+9, and their sum x+x+9= 43, that is, 2x = 43 9, .. x = the smaller number, and .. 17 +9=26, the larger number. 34 = 17, 2 2. From two places, 108 miles apart, two persons, A and B, set out at the same time, to meet each other. A travels 17 miles a day, and B travels 18; in how many days will they meet? Suppose they meet in a days: then by the question, the first, A, will have travelled 17x miles, and the second, B, 18 miles; and since the sum of these distances is the distance between the two places, we have the equation 17x18x=108, .. 35x108, .. ≈ = 3 days. 3. Find two numbers of which the difference is 13, and which are such that if 17 be added to their sum, the whole will amount to 62. Let x be the greater number; then, by the question, ≈ - 13 is the less; also x + x 131762, .. 2x = 62 + 13 — 17 — 58, .. x 29, the greater number, and .. 29 13 16, the less number. 4. There are two numbers of which the difference is 15, and which are such that if 7 times the less be subtracted from 5 times the greater, the difference will be 19: what are the numbers ? Let x be the greater number; then, by the question, ≈ — 15 is the less; also (7x-105) = 19, that is, 2x10519, .. — 2x = 5x - 86, 43, the greater number, and .. 43 15 28, the less number. 5. A person starts from a certain place, and travels at the rate of 4 miles an hour. After he has gone 10 hours, a horseman, riding 9 miles an hour, is despatched after him how many hours must the horseman ride to overtake him? Suppose the horseman overtakes him after riding x hours; then, by the question, the pedestrian will have walked 40 + 4x miles, and the horseman will have ridden 9~ miles; and since the distance travelled by each is the same, we have the equation 9x404x, .'. 5x = 40, .. x = 8; hence the horseman has ridden 8 hours, so that each person must have travelled 9 × 8 =72 miles. 6. A person has 264 coins,-sovereigns and florins; he has 4 times as many florins as sovereigns: how many of each coin has he? Suppose he had x sovereigns; then, by the question, he must have had 4 florins; so that x+4x=264, that is 51x 264, or 11a 528, .. x = 48. Consequently he had 48 sovereigns, and .. 48 X 4 = 216 florins. 7. A person spends 4th of his yearly income in board and lodging, 4th in clothes and other expenses, and he lays by £85 a-year: what is his income? To avoid fractions, suppose 28x to be the number of pounds he receives yearly, 28 being chosen because it is divisible by both 4 and 7: then, by the question, he spends in board and lodging 7x pounds, and in clothes, &c., 4x pounds. Consequently, since he lays by £85, we have the equation 7x+4x+85 = 28x, that is, 11x+8528x. Transposing, 85= 17x, x = 5,.. 28x140. Hence his yearly income was £140. 5x 8. What number is that whose third part exceeds its fifth part by 72 ? To avoid fractions, let 15x represent the number; then, by the question, 3x=72, .. 2x = 72, .. x = 36, .. the number required is 36 × 15 = 540. 9. I have a certain number in my thoughts. I multiply it by 7, add 3 to the product, and divide the sum by 2. I then find that if I subtract 4 from the quotient I get 15: what number am I thinking of? Let a represent the number; then, by the question, Multiplying by 2, 7x + 3 = 38, .. 7x = 35, .. x = 5, the number thought of. 10. A man 40 years old has a son 9 years old: the father is therefore more than four times as old as his son. In how many years will the father be only twice as old as his son ? Suppose in x years: the father will then be 40+ years old, and the son 9+ years by the question, the former number is to be double of the latter : hence the equation 40+ 18+ 2x, .. 40 — 18 = x = 22. Therefore the father will be twice as old as his son in 22 years; in which time the father will be 62 and the son 31. 11. Two persons, A and B, 120 miles apart, set out at the same time to meet each other. A goes 3 miles an hour, and B 5 miles: what distance will each have travelled when they meet? Suppose that A has travelled x hours, then B also must have travelled a hours. A must therefore have gone 3x miles, and B 5x miles; and since together they must have travelled 120 miles, we have the equation The time occupied by each is therefore 15 hours; so that A must have travelled 15 × 3 = 45 miles, and B 15 X 5 = 75 miles. Otherwise.-Suppose A goes x miles, then B goes 120 x miles: the time occupied hours; and the time occupied by B, at 5 hours; but the times are equal. Hence the equation 8/00 x = 120 72 360 by 5, 5x - 3x, .. 8x, = 360, .. x 45, A's distance. 4575, B's distance. 12. Divide £250 among A, B, and C, so that B may have £23 more than A, and C £105 more than B. Let A's share be x pounds, then B's is x + 23; and C's ≈ + 23 + 105; and the sum of the shares is 250: hence the equation x + x + 23 + x + 23 + 105 = 250. Collecting, and transposing, 3x : 99, .. x = 33: hence the shares are as follow:A's share £33; B's £56; C's £161; and their sum is £250. 13. A can execute a piece of work in 3 days, which takes B 7 days to perform: in how many days can it be done if A and B work together? Suppose they can do it in æ days; then since A can do one-third of it in 1 day, he can do in a days in like manner B can do of the whole in x days: hence, when x 3 7 working together, they do + in x days; but these parts make up the whole work 3 7 by 7, 7x+3x= 21, .. 10x = 21, Hence they complete the work in one day and one-tenth. 14. A cistern can be filled by three pipes; by the first in 2 hours, by the second in 3, and by the third in 4; in what time can it be filled by all the pipes running together? Suppose it can be filled in x hours; then since the first can supply one-half in 1 hour, it can supply in x hours; in like manner the second can supply in x hours, x and the third ; and by the question the sum of these parts is the whole. X 3 Multiplying by 4, in order that the first and third fractions may be removed at the same time, we have 12 Multiplying by 3, 9x+4x=12, .. 13x = 12, .. ∞ = ; hence, when they 13 12 all run together, the pipes will fill the cistern in h. 55 min. 23 sec. 13 15. Solve the preceding question, when the first pipe fills the cistern in 1 hour 20 minutes; the second in 3 hours 20 minutes, and the third in 5 hours. Imitating the foregoing solution, a being the required number of hours as before, the part of the whole supplied by the first pipe is 3x Or, multiplying the terms of the first pair of fractions by 3, + 4 6.x 4x Multiplying by 4, 3x + + = 4, or 3x + 2x = 4, 5 16. After A has been working 4 days at a job, which he can finish in 10 days, B is sent to help him; they finish it together in 2 days: in what time could B alone have done the whole ? 1 Suppose B can finish it in x days; then he can do of it in 1 day, so that in the 2 days x 10 1 of it. Now, as A does of it in a day, in the 4 days, working alone, he of it; hence, when B commences there is only 3 of it to be done, so that + or + = = 5 5 Multiplying by x, and then by 5, we have 5=x; hence, B can finish the work alone in 5 days. 17. Divide £143 among A, B, and C, so that A may receive twice as much as B, and B three times as much as C. £14 6s.; B's, £42 18s.; A's, £85 16s.; and the sum of these is the 18. A person has 40 quarts of superior wine worth 7s. a quart; he wishes, however, so to reduce its quality as that he may sell it at 4s. 6d. a quart; how much water must he add? Suppose the water to be a quarts, then the entire number of quarts in the mixture will be 40+x, and by the question the worth of the pure wine is 280s., and that of the reduced wine is 41(40+x) shillings. As the worth is to remain the same, we have the equation hence, the quantity of water to be added is 22 quarts; so that the mixture will make 62 quarts. The value of this at 44s. a quart is found, as in the margin, to be 280s., which is the value of the unreduced 40 quarts. 19. Divide 90 into four parts, such, that if the first be increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, the results may all be equal. Multiplying by 2, 8x + x = 180, .. 9x = 180, .. x = 20; hence, the required parts are 18, 22, 10, and 40, which together make 90. 20. Divide 39 into four parts, such, that if the first be increased by 1, the second diminished by 2, the third multiplied by 3, and the fourth divided by 4, the results may all be equal. Multiplying by 3, 18x + x = 114, .'. 19x — 114, .'. ≈ = 6; hence, the required parts are 5, 8, 2, and 24, the sum of which is 39. From the last two examples you will perceive that although in general the unknown quantity sought is best represented by a single symbol a, yet the conditions of the question may be such as to suggest a more convenient form for the unknown: a judicious form of representation at the outset will often save several steps of work in the solution. When fractions are foreseen to enter the equation, when the symbol for the unknown is x, it will always be better to use instead of x, such a multiple of x as will preclude their entrance, as in examples 7 and 8 above. |