Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol

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. \(3y + 4.5z\)

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.



For Weighted Avg and Mixtures Basics, check:
https://anaprep.com/arithmetic-weighted-averages/
https://anaprep.com/arithmetic-mixtures/
and these videos:
https://www.youtube.com/watch?v=_GOAU7moZ2Q
https://www.youtube.com/watch?v=VdBl9Hw0HBg
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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. \(3y + 4.5z\)
_________________________________________________________

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

\(x = y+3z\)

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

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

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

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: https://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: https://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

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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