50. An insolvent debtor fails for $3780, and is able to pay only $1550. If A's claim is $378, how much will he receive? 51. Divide $873 among A, B, and C, so that for every $2 that A receives, B shall receive $4, and C, $3. 52. A and B buy goods to the amount of $600, of which A pays $250, and B, $350. If they lose $150, what will be the loss of each? 53. A bankrupt owes A $600; B, $800; C, $1000; D, $1200; but his property is worth only $1440. How much should each of his creditors receive? 54. Divide a man's estate of $29,000 so that his wife shall receive $7 for every $5 received by each of his two sons, and every $4 received by each of his three daughters. 55. A, B, and C engage in business, A putting in $982; B, $365; and C, $843. If their profit in one year is $1460, what is each one's share? 56. A, B, and C plant 1200 acres of corn, A planting 2 times as many acres as B; and B, 3 times as many acres as C. They sell the entire crop, amounting to 45 bu. to the acre, at 22¢ per bu. What is each man's share of the profit? 57. If 10 lb. of cheese are equal in value to 7 lb. of butter, and 14 lb. of butter to 5 bu. of corn, and 12 bu. of corn to 8 bu. of rye, how many pounds of cheese are equal in value to 4 bu. of rye? 58. If 15 bu. of wheat are worth 18 bu. of rye, and 5 bu. of rye are worth 8 bu. of corn, and 9 bu. of corn are worth 12 bu. of oats, and 16 bu. of oats are worth 20 lb. of coffee, how many pounds of coffee should be exchanged for 20 bu. of wheat? SOLUTION OF PROBLEMS METHODS OF PROCEDURE There are three methods of solving problems: analysis, the literal method, and proportion. In each, the relations between the given terms and the required terms are expressed by equations. By analysis, the required term must form one member of an equation, and the given terms the other; no operation is performed upon the required term. By the literal method, no attempt is made to place the required term by itself, but the equation is stated naturally, and the operations are performed upon the terms without discrimination. By proportion, the equation is stated as an equality of two ratios, the antecedent of the second ratio being of the same denomination as the answer, and the consequent, the required term. ILLUSTRATIONS At 4 each, how many apples can be purchased for 8¢? ANALYSIS = Relation number of apples: the number of times cost of 1 apple is contained times in cost of all. Solution since 1 apple costs 4, as many apples can be bought for 8, as 4 is contained times in 8, or 2 apples. LITERAL METHOD 4 x = 2, cost of all; 4x = 8; x = no. of apples. PROPORTION Relation: cost of 1: cost of all = 1 apple all apples. : 8×1 Solution: 4:8=1:x; x= = 2, no. apples. 4 PROBLEMS OF PURSUIT Analysis proceeds indirectly, introducing a special method for each special case; the literal method proceeds directly, expressing the required term by a letter, which is used as a number. Both methods should be mastered; the former will give power for an indirect, and the latter for a direct, attack upon a problem. 1. At what time between 3 and 4 o'clock are the hands of a watch opposite to each other? Necessary knowledge: at 3 o'clock the min. hand is at 12, and the hr. hand at 3; when the hands are opposite to each other, they are 30 min. spaces apart; while the min. hand advances 60 spaces, the hr. hand advances 5 spaces. ANALYSIS Relation: the min. hand will advance as many min. spaces as the number of spaces it gains in 1 min. is contained times in the 3B number of spaces to be c gained. The min. hand has advanced from A to D, or 45 spaces +B to C; because from A to B is 15 spaces, and from C to D is 30 spaces. The hr. hand has advanced from B to C; therefore, the min. hand has gained 45 spaces. Since the min. hand advances 60 spaces while the hr. hand advances 5 spaces, the min. hand gains 55 spaces in 60 min., or 35, or of a space in 1 min. Since it gains of 2. How many minute spaces does the minute hand of a watch gain on the hour hand in 1 minute? 3. If the three hands of a watch all turn on the same point, how many minute spaces does the second hand gain on the hour hand in 1 minute? 4. When the hands of a watch are first 20 min. spaces apart between 5 and 6 o'clock, how many spaces has the min. hand gained on the hr. hand since 5? Draw a diagram. 5. When the hands are at right angles between 2 and 3 o'clock, how many spaces has the minute hand gained since 2? Draw a diagram. 6. At what time between 5 and 6 o'clock are the hands of a watch first 20 minute spaces apart? 7. At what time between 2 and 3 o'clock are the hands of a watch at right angles? 8. Between 4 and 5 o'clock, when the hour hand is as much after 4 as the minute hand is before 10, how many minute spaces have the hour and minute hands together passed since 4 o'clock ? How many spaces do they together pass in 1 minute? 9. In Ex. 8, what is the time? Solve both with and without the use of x. 10. At what time between 8 and 9 o'clock are the hands of a watch together? 11. A and B start from the same point and travel in the same direction. If A travels 6 miles an hour, and B 4 miles an hour, how far apart are they after 6 hours? 12. In how many hours will A overtake B, if the latter has 5 hr. the start? 13. If they travel in opposite directions, how far apart are they at the end of 6 hours? 14. If they are 60 miles apart and travel toward each other, how far will A travel before they meet? 15. Two men, A and B, 26 miles apart, set out toward each other, B 30 minutes after A; A travels 3 mi. an hr., and B 4 mi. an hr. How far will each have traveled when they meet? 16. A fox has 60 of its leaps the start of a hound. While the fox makes 5 leaps the hound makes 4; 3 leaps of the fox cover the same distance as 2 leaps of the hound. How many leaps must the hound make to catch the fox? ANALYSIS Relation: the hound must make as many leaps as the distance (in fox leaps) he gains in 1 leap is contained times in the distance (in fox leaps) to be gained. = Solution: the distance the hound goes in 1 leap the length of fox leaps, because 2 leaps of the hound cover the same distance as 3 leaps of the fox. The distance the fox goes during the same period is fox leaps, because the fox makes 5 leaps while the hound makes 4. Therefore, in 1 leap, the hound gains (3) fox leaps, or of a fox leap. It would take the hound as many leaps to gain 60 fox leaps, as is contained times in 60, or 240 leaps. 17. A fox pursued by a hound makes 3 leaps while the hound. makes 2; but the latter in 3 leaps goes as far as the former in 7. Find the length of 1 hound leap in terms of fox leaps. 18. Find the distance in terms of fox leaps that the hound gains in one leap. 19. If the fox has 60 of her own leaps the start, how many times will the hound leap before he catches the fox? 20. If the fox has 60 of the hound leaps the start, how many times will the fox leap before she is overtaken? 21. A hare is pursued by a hound. The hare makes 5 leaps while the hound makes 3 leaps; 2 leaps of the hound cover the same distance as 5 leaps of the hare. If the hare has 50 of her leaps the start, in how many leaps will the hound overtake her? |
