A farmer goes to market with $100 to buy a total of 100 animals. He has to buy cows, which are $10 each; sheep, which are $3 each; and chickens, which are 50 cents each. How many of each animal did he buy for $100? contributed by Fred Raby, Fenelon Falls, Ontario
A potato and a tomato cost 40 cents. A tomato and an onion cost 50 cents. An onion and a potato cost 60 cents. How much does each single vegetable cost? contributed by Sidney Kravitz, Dover, New Jersey
If a hen-and-a-half lays an egg-and-a-half in a day-and-a-half, how many eggs can six hens lay in six days?
Two men travel by day in the same direction around an island twenty-four miles in circumference, and camp at night. Mr. A starts one mile ahead of Mr. B and goes one mile in the first day, three miles in the second day, and so on, increasing his rate by two miles each day. Mr. B goes five miles every day. When do they camp together?
No answers here. These are elementary and are great warm-ups. The answers are in the 2008 Old Farmer's Almanac by Robert B. Thomas. Leave your solutions in the comments!
Technorati tags: Technorati Tags: Old Farmer's Almanac, math, puzzles
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ReplyDeleteCows, Sheep & Chickens:
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He has to average $1 per animal. Since the only animal under $1 is a chicken, he needs to use Chickens to balance out the other animals.
For each Cow he buys, he has to buy 18 chickens. For each Sheep he buys, he has to buy 4 chickens.
So he can buy animals in multiples of 19 or 5. We need to find natural numbers A and B such that 19A + 5B = 100. Since this can only be solved if 19A is a multiple of 5, and 19 is prime, A must be 5, and then B must be 1
So he buys:
5 Cows for $50
1 Sheep for $3
94 Chickens for $47
Total 100 Animals for $100
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Potatoes, Tomatoes and Onions
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Given:
(1) P + T = 40
(2) T + O = 50
(3) O + P = 60
Add the 3 equations to get:
2P + 2T + 2O = 150
Divide both sides by 2 to get:
P + T + O = 75
Subtract equation (1):
O = 35
Sub for O in (2):
T = 15
Sub for T in (1):
P = 25
So, a tomato costs 15 cents, a potato costs 25 cents, an onion costs 35 cents.
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Hen-and-a-Half
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Ignoring the basic concerns about meaningfulness of half-chickens and half-eggs...
If a hen-and-a-half lays an egg-and-a-half in a day-in-a half, then a single hen can lay a single egg in a day-and-a-half, or 4 eggs in 4 times a day-and-a-half which is 6 days. Six hens can lay 6x4=24 eggs in six days.
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2 Men on an Island
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A's location on the nth night is
1+ (1+3+...2n-1) = 1 + n^2 (mod 24)
B's location on the nth night is
5n (mod 24)
So n^2 + 1 + 24A = 5n + 24B or
n^2 -5n + 1 = 24(B-A)
But... I'm not sure where to go from there.
I can solve it experimentally using an excel worksheet :)
Hi Mom! I love the wording of the hen-and-a-half puzzle. I know my students will make tables to solve the $100 animal one. Hopefully they will use proportions to solve the hen-and-a-half puzzle. And the vegetable one is of a type they love to solve with pictures. These type have been touted as preparing the kids for variables and algebraic notation.
ReplyDeleteYour solution of the men on the island is fascinating. Now I'm thinking it is a great way to introduce "clock arithmetic." Thank you so much.
The $100 animal one is harder than the usual problems of that type to solve using tables, since there are three kinds of animals. (Usually there are two, which means if you guess a number for one, you can compute the number for the other.) It will be interesting to see if they come up with any good observations to help them hone in on an answer. If I were just doing guess and check, I think it would have been a long time before I guessed 5 cows. My first intuition was that you'd have to avoid cows to keep the average low enough.
ReplyDeleteWhat I discovered in my experimental solution to the men on the island problem is that they will never camp together! (Unless I messed something up, which is entirely possible.) I found that A's pattern of camping spots repeats after 12 days, and B's repeats after 24. (I wasn't sure A's pattern would repeat at all!) They never line up in the first 24 days, so they are going to keep repeating that 24-day pattern of never lining up. I'm guessing that this is why the quadratic in my algebraic solution doesn't factor cleanly.
Nice ones, and nice solutions.
ReplyDeleteThe problem sort of implies that there needs to be one of each animal, but I might make it explicit (otherwise we have one additional solution).
Mathmom is right about the campers. Look at it this way: pretend the mile markers are alternately colored red and blue (and there are 24 of them, so they alternate, and after 24blue you get back to 1red).
So one guy starts on blue, and goes an odd number each day (5 is always odd). So start on blue, next day red, next day blue, etc.
The other guy starts on red, and goes an odd number each day (1, 3, 5, all odd, right). So he starts red, then blue, then red, and so forth.
Put them together. One starts on red, the other on blue, and they switch colors each night...
JD, thanks for the parity hint on the men on the island problem. (You don't really need red and blue in this case -- you could just as easily stick with even and odd.) I "should have" thought of that, since I like to do parity problems with my kids!
ReplyDeleteAndrée, have you had a chance to try these with your kids yet? I'd be interested to hear how that went.
ReplyDelete