Problem Solving

Three grades of milk: 1%, 2%, and 3% fat by volume. X gallons of 1%, Y gallons
of 2%, and Z gallons of 3% are mixed to give X + Y + Z gallons of 1.5%. What is X
in terms of Y and Z?
a. Y + 3Z
b. (Y + Z)/4
c. 2Y + 3Z
d. 3Y + Z
e. 3Y + 4Z


According to me 4 gallons of X + 1 gallon of Y + 1 gallon of Z will give:
4(1%) + 1(2%) +1(3%) = 9% Fat
% of fat / Total Number of gallons = Average % of Fat.
Therefore, 9%/6 = 1.5 % Fat

Let X = 20 , So Y = 5 and Z=5.

Substituting in answer 1 we get X = 20.
Now, Substituting in answer 4 we again get X = 20.

Any help, please!! Where did I go wrong?

0.01x+0.02y+0.03z=0.015x+0.015y+0.015z
0.005x-0.005y-0.015z=0
5x-5y-15z=0, x-y-3z=0, x=y+3z
a

winniethepooh wrote:Three grades of milk: 1%, 2%, and 3% fat by volume. X gallons of 1%, Y gallons
of 2%, and Z gallons of 3% are mixed to give X + Y + Z gallons of 1.5%. What is X
in terms of Y and Z?
a. Y + 3Z
b. (Y + Z)/4
c. 2Y + 3Z
d. 3Y + Z
e. 3Y + 4Z


According to me 4 gallons of X + 1 gallon of Y + 1 gallon of Z will give:
4(1%) + 1(2%) +1(3%) = 9% Fat
% of fat / Total Number of gallons = Average % of Fat.
Therefore, 9%/6 = 1.5 % Fat

Let X = 20 , So Y = 5 and Z=5.

Substituting in answer 1 we get X = 20.
Now, Substituting in answer 4 we again get X = 20.

Any help, please!! Where did I go wrong?

x+2y+3z=1.5(x+y+z)

On Solving - x=y+3z

Thus a

x + 2y + 3z = 1.5(x+y+z)
0.5y+ 1.5z = 0.5 x
y+ 3z= x is clear.