I have an algebra brainteaser, 5.00x+2.50y+.10z=100?

Question

I have a brainteaser that I need an answer for.
You have exactly 100 dollars to spend and you must buy exactly 100 head of cattle.
Bull are 5.00 each
cows are 2.50 each
calves are .10 each.
I need to know how many of each to buy to make eactly 100 dollars and have 100 animals.

Thank.

Answer 1

5x + 2.5y + 0.1z = 100
Multiply by 10 to remove decimals
50x + 25y + z = 1000 ...(1)

Also x+y+z = 100 .. (2)

That's 2 equations with 3 unknowns. But we also have a constraint that x, y and z must be integers.
Looking at equation 1,
a) z must be a multiple of 5 because everything else in that equation is. In fact z must be a multiple of 25.
b) x < 20
c) y < 40

Subtract equ 1 and 2 to eliminate z
49x + 24y = 900 ... (3)
d) x must be a multiple of 3

Let's try z = 75
x+y = 25
50x + 25y = 925
We get y = 13, x = 12
So this works.

If we try z = 50, x turns out to be -ve.

Unique solution: 12 bulls, 13 cows, 75 calves

Answer 2

We have two equations: 50x + 25y + z = 1000 and x+y+z=100. Using z = 100-x-y into the first eqn, we get 49x+24y=900. Using the standard techniques for solving a linear diophantine equation, we find there are infinintely many solutions, namely x = 900 + 24t , y = -1800 - 49t , and z = 1000 + 25t , where t is any integer. But all three variables must be non-negative and less than or equal to 100. Thus,
0 <= 900 + 24 t <= 100 tells us that

(1) -37 <= t <= -33.

Using the same constraints on y tells us that

(2) -38 <= t <= -36.

Finally, from 0 <= Z <= 100 we find

(3) -40 <= t <= -36.

Since all of (1), (2), and (3) must be satisfied, we find that t =
-37. This gives x = 12, y = 13, and z = 75.

Answer 3

Ha, seems like you have 3 unknowns and 2 simultaneous equations in this question......

Let the number of bulls, cows and claves be x, y and z respectively.
NOTE: x, y and z are positive integers which are between 0 and 100.

5x + 2.5y + 0.1z = 100
50x + 25y + z = 1000 --- (1)
x + y + z = 100 --- (2)
z = 100 - x - y

(1) - (2)
49x + 24y = 900
x = (900 - 24y)/49
NOTE: (900 - 24y) must be a multiple of 49 for y to be an integer and y must be less or equals to 37 for x to be positive.
(because 24 x 38 = 912 which exceeds 900)

By trail and error, y = 13 is the only possible solution.
So, when y = 13,
x = [900 - 24(13)]/49 = 12
z = 100 - 12 - 13 = 75

Therefore, you need to buy 12 bulls, 13 cows and 75 calves.

Answer 4

three variables requires three distinct equations.

5x+2.5y+.1z=100
x+y+z=100 z=100-x-y
50x+25y+z=1000
50x+25y+100-x-y=1000
49x+24y=900

since you don't have a third equation you have to use the knowledge that x and y are integers greater than or equal to 0 and less than or equal to 100, and just use trial and error. and x must be even since 24 and 900 are even.

try multiples of 5 and 10 first
49x+24y=900
490 + 410 = 900 but 410/24 isn't an integer
588 + 312 = 900 hey , how do you like that?
49 *12 =588, 24*13=312
x=12, y=13, z must equal 75.

12*5=60, 13*2.5=32.5, 75*.1=7.5
60+32.5+7.5=100

Answer 5

You'd need a system

You got the first part right, you just need a second part

5x + 2.5y +.1z = 100
x+y+z=100

And then solve the system.