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Milk %

This topic has 3 expert replies and 2 member replies
750+ Junior | Next Rank: 30 Posts Default Avatar
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Milk %

Post Sun Jun 26, 2016 9:19 pm
Can someone please provide a solution
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Marty Murray Legendary Member
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Post Sun Jun 26, 2016 10:23 pm
750+ wrote:
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
The first thing I noticed was that if z = 0, then you can just have even amounts of 1 percent and 2 percent grade, to get a 1.5 percent mixture. For instance, if x = 2 and y = 2, you get 4 gallons of 1.5 percent grade.

Similarly, you can set y = 0. Then to get a 1.5 percent mixture you need enough 1 percent with every gallon of 3 percent to get a weighted average of 1.5 percent. Since 1.5 is closer to 1 than to 3, x has to be greater than z.

If x > z then choices B and D are out.

If x and y are equal, actually the only answer that works is A.

So that's one hacking way to quickly get to the answer.

Alternatively, you could use the mixture formula, just using it with three components rather than the usual two.

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

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

.5y + 1.5z = .5x

y + 3z = x

The correct answer is A.

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Post Mon Jun 27, 2016 5:35 am
Quote:
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

Cheers,
Brent

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Post Mon Jun 27, 2016 5:50 am

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800_or_bust Master | Next Rank: 500 Posts Default Avatar
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Post Mon Jun 27, 2016 8:42 am
Marty Murray wrote:
750+ wrote:
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
The first thing I noticed was that if z = 0, then you can just have even amounts of 1 percent and 2 percent grade, to get a 1.5 percent mixture. For instance, if x = 2 and y = 2, you get 4 gallons of 1.5 percent grade.

Similarly, you can set y = 0. Then to get a 1.5 percent mixture you need enough 1 percent with every gallon of 3 percent to get a weighted average of 1.5 percent. Since 1.5 is closer to 1 than to 3, x has to be greater than z.

If x > z then choices B and D are out.

If x and y are equal, actually the only answer that works is A.

So that's one hacking way to quickly get to the answer.

Alternatively, you could use the mixture formula, just using it with three components rather than the usual two.

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

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

.5y + 1.5z = .5x

y + 3z = x

The correct answer is A.
Nice. I like that setting z to 0 approach. Not too often you can knock out four answer choices with a simple trick like that!

_________________
800 or bust!

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GMAT/MBA Expert

Post Mon Jun 27, 2016 9:06 am
Hi 750+,

This is an "in terms of" question; these questions are usually built around 4-5 algebra steps and are fairly straight-forward "math" questions.

First, translate the equation:

[(.01x) + (.02y) + (.03z)] / {x + y + z] = .015

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

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

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

5y + 15z = 5x

y + 3z = x

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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