Milk concentration in mix would be \(1\%x+2\%y+3\%z\) and on the other hand we are told that in (x+y+z) there is 1.5% of milk. Hence:

\(1\%x+2\%y+3\%z=1.5\%(x+y+z)\);

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

\(0.5x=0.5y+1.5z\);

\(x=y+3z\).

Answer: A.

Hope it's clear.
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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.
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We are given that 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. We can use this information to create the following equation:

0.01x + 0.02y + 0.03z = 0.015(x + y + z)

Multiplying the entire equation by 1000, we have:

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

We must now solve for x in terms of y and z:

5y + 15z = 5x

We can divide the entire equation by 5:

y + 3z = x

Answer: A
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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

Cheers,
Brent
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Hi All,

This is an "in terms of" question; these questions are usually built around 4-5 algebra steps and are fairly straight-forward "math" questions. To start, we can translate the given information into the following equation:

[(.01X) + (.02Y) + (.03Z)] / (X + Y + Z) = .015

(.01X + .02Y + .03Z0 = (.015X + .015Y + .015Z)

Let's multiply everything by 1000 to get rid of the decimals....

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

And then combine like terms....

5Y+ 15Z = 5X

And then divide everything by 5...

Y + 3Z = X

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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\(\frac{x+2y+3z}{x+y+z} = \frac{3}{2}\)

\(x = y+3z\)

Its A, just put the words into equation

x+2Y+3Z=1.5(X+2Y+3Z)

X=Y+3Z

Hope it clarifies

+1 Kudos if it helps

0.01x+0.02y+0.03z=0.015(x+y+z)
=> x=y+3z

hence A

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

The mixture divided by the sum of weights?

Yes, it is just the weighted average of the fat concentration.
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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

Using weighted average:

(x) (1) + (y) (2) + (z)(3) = (1.5) (x + y + z)

x + 2y + 3z = 1.5x + 1.5y + 1.5z
0.5y + 1.5z = 0.5x

Removing 0.5 overall:

y +3z = x
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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.
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Bunuel - Where can I find challenging mixture problems/weighted avg problems to practice? Can you please help?

Thanks.

Check our Questions Bank: http://gmatclub.com/forum/viewforumtags.php

18. Mixture Problems

• Theory
  Weighted Average and Mixture Problems on the GMAT
  Tips and Tricks: Mixtures
  
  • Questions
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GMAT PS Question Directory by Topic & Difficulty: http://gmatclub.com/forum/gmat-ps-quest ... 27957.html
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Hope this helps.
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Hi VeritasPrepKarishma

I learnt the method to solve mixture problems from your blog and i applied the same method here. Now not sure if that is correct way or not, but i got the answer. Can you please confirm if what I did was correct?

__________|_________|__________
1% 1.5% 2% 3%
x Y z

==> x(1.5-1) = y(2..0 - 1.5) + z(3.0-1.5)
==> x/2 = y/2 + 3z/2
==> x = y + 3z

Thanks in advance!

Please find the solution as attached

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Yes, great work! In essence, you are using 'the deviation method for mean' which is of course applicable to weighted mean too.
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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
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