Problem Solving

Hello,

can you give me an explanation. Thanks!

Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x + y + z gallons of a 1.5 percent grade, what is x in terms of y and z ?

a) y + 3z
b) (y+z)/4
c) 2y + 3z
d) 3y + z
e) 3y + 4.5z

OA A

szDave wrote:Hello,

can you give me an explanation. Thanks!

Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x + y + z gallons of a 1.5 percent grade, what is x in terms of y and z ?

a) y + 3z
b) (y+z)/4
c) 2y + 3z
d) 3y + z
e) 3y + 4.5z

OA A

If EQUAL AMOUNTS of 1% grade and 2% grade are used, the grade of the mixture will be the AVERAGE of 1% and 2%:
(1+2)/2 = 1.5%.

Let x=100, y=100, and z=0.
Since the mixture will be composed only of equal amounts of 1% grade (the value of x) and 2% grade (the value of y), the grade of the mixture will be 1.5%.

To illustrate:
Percent in x = .01(100) = 1.
Percent in y = .02(100) = 2.
Percent in z = .03(0) = 0.
Percent in x+y+z = (1+2+0) / (100+100+0) = 3/200 = 1.5/100 = 1.5%.

The question asks for the value of x=100. This is our target.
Now we plug y=100 and z=0 into the answers to see which yields our target of 100.
A quick scan of the answer choices reveals that only A works:
y + 3z = 100 + 3(0) = 100.

The correct answer is A.

Algebraically:
Percent in x = 1x.
Percent in y = 2y.
Percent in z = 3z.
Percent in x+y+z = 1.5(x+y+z).

Since the sum of the amounts in the 3 ingredients is equal to the amount in the mixture, we get:
1x + 2y + 3z = 1.5(x+y+z)

10x + 20y + 30z = 15x + 15y + 15z

5y + 15z = 5x

y + 3z = x.