Question 97502This question is from textbook
: Heres the question:
A farmer wants to buy 100 animals for $100.00 (must buy at least one of them)
Horses-$5.00
Cows-$3.00
Chicks-.50
How many horses,cows and chicks does the farmer need to have 100 and 100.00?
This question is from textbook
Found 3 solutions by Earlsdon, Soxxie95, ikleyn:
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! This problem involves diophantine equations, where you have more unknowns than equations.
The solutions, however, are limited by the fact that the unknowns must be integers, i.e. whole cows, whole horses, etc.
We can set up two equations as follows: H = number of horses, C = number of cows, and ch = number of chicks.
From the problem description, we can write:
H+C+ch = 100 The total number of animals is 100.
($5)H+($3)C+($0.50)ch = $100 The total amount the farmer spends.
The two equations are:
1) H+C+ch = 100
2) 5H+3C+0.5ch = 100 Multiply this equation by 2 and subtract equation 1) from the result.
1) H+C+ch = 100
2a) 10H+6C+ch = 200
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3) 9H+5C = 100 Subtract 9H from both sides.
3a) 5C = 100-9H Now divide both sides by 5.
3b) C = 20-(9/5)H Now C can only be an integer if H is a multiple of 5. Let's try H = 5.
3c) C = 20-(9/5)(5)
C = 20 - 9
C = 11 This is one possibility. Now let's try H = 10
3d) C = 20 - (9/5)(10)
C = 20 - 18
C = 2 This is another possibility.
H cannot be greater than 10 otherwise you would end up with C being a non-integer or a negative number.
So, if C = 11 when H = 5, then ch = 100-11-5 = 84 - the number of chicks.
Let's check this solution.
H+C+ch = 100
5+11+84 = 100 and...
5($5) + 11($3) + 84($0.50) = $25 + $33 + $42 = $100 This is a valid solution!
Now if C = 2 when H = 10, then ch = 100-2-10 = 88 - the number of chicks.
Let's check this solution.
H+C+ch = 100
10+2+88 = 100 and...
10($5)+2($3)+88($0.50) = $50+$6+$44 = $100 This solution is also valid!
The two solutions are:
1) 5 horses, 11 cows, and 84 chicks.
2) 10 horses, 2 cows, and 88 chicks.
Answer by Soxxie95(1) (Show Source):
Answer by ikleyn(52794)  (Show Source):
You can put this solution on YOUR website! .
A farmer wants to buy 100 animals for $100.00 (must buy at least one of them)
Horses-$5.00
Cows-$3.00
Chicks-.50
~~~~~~~~~~~~~~
Let x = # of horses
y = # of cows
z = # of chicks
Then
x + y + z = 100 (1)
5x + 3y + 0.5z = 100 (2)
Multiply equation (2) by 2 (both sides). Keep equation (1) as is.
x + y + z = 100 (3)
10x + 6y + z = 100 (4)
Subtract equation (3) from equation (4). You will get
9x + 5y = 100 (5)
All equations (1) - (5) are in integer numbers x, y, z. In particular, equation (5) is in integer numbers x and y.
In equation (5), right side and the term 5y are multiples of 5. Hence, the term 9x in equation (5) is a multiple of 5.
It gives for x only these two possibilities: x= 5 or x= 10.
So, consider both cases separately.
a) If x= 5, then from (5) 5y = 100 - 9*5 = 55; hence, y= 11.
Then from (1) z = 100 - 5 - 11 = 84.
b) If x= 10, then from (5) 5y = 100 - 9*10 = 10; hence, y= 2.
Then from (1) z = 100 - 10 - 2 = 88.
Thus the problem has two solutions in integer numbers:
1) 5 horses, 11 cows and 84 chicks,
and
2) 10 horses, 2 cows and 88 chicks.
Solved.
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