Answers Answers
1. CONCERNING A CHECK
The amount must have been $31.63. He received $63.31. After he had spent a nickel there would remain the sum of $63.26, which is twice the amount of the check.
2. DOLLARS AND CENTS
The man must have entered the store with $99.98 in his pocket
3. LOOSE CASH
The largest sum is $1.19, composed of a half dollar, quarter, four dimes, and four pennies.
4. GENEROUS GIFTS
At first there were twenty persons, and each received $6.00. Then fifteen persons (five fewer) would have received $8.00 each. But twenty-four (four more) appeared and only received $5.00 each. The amount distributed weekly was thus $120.00.
5. BUYING BUNS
There must have been three boys and three girls, each of whom received two buns at three for a penny and one bun at two for a penny, the cost of which would be exactly 7¢.
6. UNREWARDED LABOR
Weary Willie must have worked 16â
days and idled 13â
days. Thus the former time, at $8.00 a day, amounts to exactly the same as the latter at $10.00 a day.
7. THE PERPLEXED BANKER
The contents of the ten bags (in dollar bills) should be as follows: $1, 2, 4, 8, 16, 32, 64, 128, 256, 489. The first nine numbers are in geometrical progression, and their sum, deducted from 1,000, gives the contents of the tenth bag.
8. A WEIRD GAME
The seven men, A, B, C, D, E, F, and G, had respectively in their pockets before play the following sums: $4.49, $2.25, $1.13, 57¢, 29¢, 15¢, and 8¢. The answer may be found by laboriously working backwards, but a simpler method is as follows: 7 + 1 = 8; 2 à 7 + 1 = 15; 4 à 7 + 1 = 29; and so on, where the multiplier increases in powers of 2, that is, 2, 4, 8, 16, 32, and 64.
9. DIGGING A DITCH
A. should receive one-third of two dollars, and B. two-thirds. Say B. can dig all in 2 hours and shovel all in 4 hours; then A. can dig all in 4 hours and shovel all in 8 hours. That is, their ratio of digging is as 2 to 4 and their ratio of shovelling as 4 to 8 (the same ratio), and A. can dig in the same time that B. can shovel (4 hours), while B. can dig in a quarter of the time that A. can shovel. Any other figures will do that fill these conditions and give two similar ratios for their working ability. Therefore, A. takes one-third and B. twice as muchâtwo-thirds.
10. NAME THEIR WIVES
As it is evident that Catherine, Jane, and Mary received respectively $122.00, $132.00, and $142.00, making together the $396.00 left to the three wives, if John Smith receives as much as his wife Catherine, $122.00; Henry Snooks half as much again as his wife Jane, $198.00; and Tom Crowe twice as much as his wife Mary, $284.00, we have correctly paired these married couples and exactly accounted for the $1,000.00.
11. MARKET TRANSACTIONS
The man bought 19 cows for $950.00,1 sheep for $10.00, and 80 rabbits for $40.00, making together 100 animals at a cost of $1,000.00.
A purely arithmetical solution is not difficult by a method of averages, the average cost per animal being the same as the cost of a sheep.
By algebra we proceed as follows, working in dollars: Since x + y + z = 100, then ½x + ½y + ½z = 50.
by subtraction, or 99x + 19y = 1900. We have therefore to solve this indeterminate equation. The only answer is x = 19, y = 1. Then, to make up the 100 animals, z must equal 80.
12. THE SEVEN APPLEWOMEN
Each woman sold her apples at seven for 1¢, and 3¢ each for the odd ones over. Thus, each received the same amount, 20¢. Without questioning the ingenuity of the thing, I have always thought the solution unsatisfactory, because really indeterminate, even if we admit that such an eccentric way of selling may be fairly termed a âprice.â It would seem just as fair if they sold them at different rates and afterwards divided the money; or sold at a single rate with different discounts allowed; or sold different kinds of apples at different values; or sold the same rate per basketful; or sold by weight, the apples being of different sizes; or sold by rates diminishing with the age of the apples; and so on. That is why I have never held a high opinion of this old puzzle.
In a general way, we can say that n women, possessing an + (n â 1), n(a + b) + (n â 2), n(a + 2b) + (n â 3), âŚ, n[a + b(n â 1)] apples respectively, can sell at n for a penny and b pennies for each odd one over and each receive a + b(n â 1) pennies. In the case of our puzzle a = 2, b = 3, and n â 7.
13. A LEGACY PUZZLE
The legacy to the first son was $55.00, to the second son $275.00, to the third son $385.00, and to the hospital $605.00, making $1,320.00 in all.
14. PUZZLING LEGACIES
The answer is $1,464.00âa little less than $1,500.00. The legacies, in order, were $1,296.00, $72.00, $38.00, $34.00, and $18.00. The lawyerâs fee would be $6.00.
15. DIVIDING THE LEGACY
The two legacies were $24.00 and $76.00, for if 8 (one-third of 24) be taken from 19 (one-fourth of 76) the remainder will be 11.
16. A NEW PARTNER
We must take it for granted that the sum Rogers paid, $2,500.00, was one-third of the value of the business, which was consequently worth $7,500.00 before he entered. Smuggâs interest in the business had therefore been $4,500.00 (1½ times as much as Williamson), and Williamsonâs $3,000.00. As each is now to have an equal interest, Smugg will receive $2,000.00 of Rogersâs contribution, and Williamson $500.00.
17. POCKET MONEY
When he left home Tompkins must have had $2.10 in his pocket.
18. DISTRIBUTION
The smallest number original...