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Weighted Avg/Mixture

Expert replies
by jswesth » Mon Apr 18, 2016 7:49 am
Is it best to substitute simple values here or continue with variables? Please help in solving this.

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Source: — Problem Solving |

by Brent@GMATPrepNow » Mon Apr 18, 2016 8:23 am
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
Here's one approach:

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
Brent Hanneson - Creator of GMATPrepNow.com
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by [email protected] » Mon Apr 18, 2016 9:20 am
Hi jswesth,

This is an "in terms of" question; these questions are usually built around 4-5 algebra steps and are fairly straight-forward "math" questions. In most cases, it's actually rather difficult to TEST VALUES on these types of prompts.

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
Contact Rich at [email protected]
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by ceilidh.erickson » Mon Apr 18, 2016 1:37 pm
I would actually pick simple values here.

If the total weighted average is a 1.5% grade, the easy thing to think of is equal parts 1% and 2%, but no 3%. So, say:
x = 1
y = 1
z = 0

If the question is asking for x in terms of y and z, then we want the answer choice that equals 1 when we plug in y = 1 and z = 0.

A. y + 3z --> 1 + 3(0) = 1 --> Success!

B. (y + z)/4 --> (1 + 0)/4 = 1/4 --> No.

C. 2y + 3z --> 2(1) + 3(0) = 2 --> No.

D. 3y + z --> 3(1) + 0 = 3 --> No.

E. 3y + 4z --> 3(1) + 4(0) = 3 --> No.

The answer is A.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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by GMATGuruNY » Mon Apr 18, 2016 1:40 pm
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by chetan.sharma » Tue Apr 19, 2016 8:37 am
jswesth wrote:Is it best to substitute simple values here or continue with variables? Please help in solving this.

Image
Hi,
finally this Q is all about weighted mixture so lets work on this concept for the answer...
Now the average 1.5% is much closer to 1 than 3 and at same distance from 2..

so If there were only 1% and 2% , both would be same so x and y will be same..(i)
if there were only 1% and 3%, 1.5% average means x = (3-1.5)/(1.5-1)*z = 1.5/0.5= 3z..(ii)..
we have all three so x=y+3z, SINCE the average lies between 1% on one side and 2 and 3 on other side of average and therefore the qty of 1% balances qty of BOTH 2% and 3%..

Best would be DRAW a line with 1,2 and 3 marked on it...
mark 1.5% and you can also get the above values from it..
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