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3 grades of milk

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3 grades of milk

by DevB » Mon Jun 16, 2014 10:14 am
Hello Everyone,

Please help me in answering this question from GMATPrep:

3 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?

1. y+3z
2. (y+z)/4
3. 2y+3z
4. 3y+z
5. 3y+4.5z
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by GMATGuruNY » Mon Jun 16, 2014 10:44 am
3 grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of 1% grade, y gallons of 2% grade and z gallons of 3 % grade are mixed to give x+y+z gallons of 1.5% grade, what is x in terms of y & z?
1) y+3z
2) (y+z)/4
3) 2y+3z
4) 3y+2
5) 3y+4.5z
The desired grade -- 1.5% -- is equal to the AVERAGE of x=1% and y=2%:
(1% + 2%)/2 = 1.5%.
Thus, a mixture composed of equal amounts of x and y will be 1.5% grade.

Let x=2, y=2, and z=0, implying that the mixture will composed of equal amounts of x and y (2 units each).
The question stem asks for the value of x=2. This is our target.
Now plug y=2 and z=0 into the answers to see which yields our target of 2.
Only A works:
y + 3z = 2 + 3(0) = 2.

The correct answer is A.

Algebraically:

(1% of X) + (2% of Y) + (3% of Z) must be equal to (1.5% of X+Y+Z).
Thus:
x + 2y + 3x = 1.5(x + y + z)
10x + 20y + 30z = 15x + 15y + 15z.

Since the question stem asks for the value of x, solve for x:
20y + 30z = 5x + 15y + 15z
5y + 15z = 5x
y + 3z = x.

The correct answer is A.
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by DevB » Mon Jun 16, 2014 10:56 am
Thanks a lot Mitch for the detailed explanation...the answer is absolutely correct. :)
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by tass@ » Mon Jun 16, 2014 12:53 pm
Here mixed
x*1 + y*2 + 3*z = 3/2 (x+y+z)
2x +4y + 6z = 3x + 3y + 3z
so x = y+3z
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by Brent@GMATPrepNow » Mon Jun 16, 2014 2:29 pm
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
Let's start with a "word equation" and slowly turn it into an algebraic expression:

Total fat in mixture = 1.5% of (x+y+z)
(1% of x) + (2% of y) + (3% of z) = 0.015(x+y+z)
Rewrite as: 0.01x + 0.02y + 0.03z = 0.015x + 0.015y + 0.015z
Multiply both sides by 100: 1x + 2y + 3z = 1.5x + 1.5y + 1.5z
Rearrange and simplify: 0.5y + 1.5z = 0.5x
Multiply both sides by 2 to get: y + 3z = x

Answer = A

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