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. \(\frac{y +z}{4}\)

C. \(2y + 3z\)

D. \(3y + z\)

E. \(3y + 4.5z\)

\(\frac{x+2y+3z}{x+y+z} = \frac{3}{2}\)

\(x = y+3z\)

A quick approach:

The question asks you for x in terms of y and z. Whatever values x, y and z can take, this relation should hold.
Since we mix 1%, 2% and 3% milk and get 1.5% milk, one way of mixing them could be that 1% and 2% are mixed in equal quantities (to give 1.5% milk) and 3% milk is not added at all. Which means x = 1, y = 1 and z = 0 should satisfy the relation between x, y and z.
The only relation that satisfies these values is (A).

Note: If multiple options satisfied these values, you could take another set of values e.g. x = 3, y = 0 and z = 1 and check out of the shortlisted options.
_________________

This is essentially viewing this problem as a weighted average correct?

The mixture divided by the sum of weights?

I approached it as a residuals problem

Since 1% fat = -0.5% from average
2% fat = 0.5% from average
3% fat = 1.5% from average

When added together, they have to create a 0 'residual' from 1.5% average:

-0.5x + 0.5y + 1.5z = 0
0.5y + 1.5z = 0.5x

is the answer

If you develop a knack for playing with numbers, you will rarely need to make equations for ratios/percent/mixture/average problems.

What I thought here was that milk of 1% (volume x), 2% (volume y) and 3% (volume z) have to be mixed to give 1.5%. An easy way I can see immediately is that I don't take any 3% milk and mix 1% and 2% in equal quantities to get 1.5%.
i.e. If z = 0, x = y
If we put z = 0, only option (A) gives x = y hence it is the answer.
_________________

Check our Questions Bank: viewforumtags.php

Mixture problems made easy: mixture-problems-made-easy-49897.html

All DS Mixture Problems to practice: search.php?search_id=tag&tag_id=43
All PS Mixture Problems to practice: search.php?search_id=tag&tag_id=114

GMAT PS Question Directory by Topic & Difficulty: gmat-ps-question-directory-by-topic-difficulty-127957.html
GMAT DS Question Directory by Topic & Difficulty: ds-question-directory-by-topic-difficulty-128728.html

Hope this helps.
_________________

Please find the solution as attached

_________________

Responding to a pm:



What is the weighted average concept? It says
Cavg = (C1*w1 + C2*w2 + C3*w3)/(w1 + w2 + w3)

You are given that the average percentage when you mix them all is 1.5 (i.e. 3/2 in fractions)

So,
1.5 = (x + 2y + 3z)/(x + y + z)

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

Now isolate x.

Check another method here:
https://gmatclub.com/forum/three-grades ... l#p1032710
_________________

A video explanation can be found here:
https://www.youtube.com/watch?v=aXGmN6qg8Hg

In mixture problems we’ll typically see (1) some sort of a substance (in this case, milk) and (2) some characteristic of the substance (in this case, fat) as a fraction or (usually) a percentage of the volume of the substance.

If the volumes of our three types of milk are expressed as x, y, z:

.01x + .02y + .03z = .015(x + y + z)

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

2x + 4y + 6z = 3(x + y + z)

Then isolate x:

x = y + 3z
_________________