SOLUTION: Mr. Smith has $100.00 to buy 100 animals with. Cows cost $10.00 each, pigs cost $3.00 each and chickent cost $.50 each. How many can he buy with exactly $100.00. I have this

Question 53523: Mr. Smith has $100.00 to buy 100 animals with. Cows cost $10.00 each, pigs cost $3.00 each and chickent cost $.50 each. How many can he buy with exactly $100.00.
I have this equation so far, but where do I go from here?
10x + 3y + .5z = 100
Found 2 solutions by stanbon, AnlytcPhil:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I think there is more information involved in this
problem. As you stated it he could buy 10 cows or
200 chickens or many other combinations of cows,
pigs, and chickens. Please check the problem again.
Cheers,
Stan H.
Answer by AnlytcPhil(1806) (Show Source): You can put this solution on YOUR website!

Mr. Smith has $100.00 to buy 100 animals with. Cows cost $10.00 each, pigs cost $3.00 each and chickens cost $.50 each. How many can he buy with exactly $100.00. I have this equation so far, but where do I go from here? 10x + 3y + .5z = 100 Multiply through by 10 to clear decimal: 100x + 30y + 5z = 1000 Divide by 5 20x + 6y + z = 200 x + y + z = 100 19x + 5y = 100 19x = 100 - 5y 19x = 5(20 - y) x = 5(20 - y)/19 Since x is a positive integer, the 19 must divide evenly into 5(20 - y). the 19 will not divide evenly into the 5, so 19 must divide into 20 - y. 20 - y is at most 19, since y must be at least 1. The only integer at least 19 which is divisible by 19 is 19 itself. Therefore 20 - y = 19 -y = -1 y = 1 x = 5(20 - y)/19 x = 5(20 - 1)/19 x = 5(19)/19 x = 5 x + y + z = 100 5 + 1 + z = 100 6 + z = 100 z = 94 Therefore he bought 5 cows, 1 pig, and 94 chickens. Edwin

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