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# GMATPrep PS Mixture: Three grades of milk

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Manager
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Re: GMAT Prep VIC Problem [#permalink]

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24 Dec 2009, 10:19
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Difficulty:

25% (medium)

Question Stats:

83% (01:59) correct 17% (01:22) wrong based on 47 sessions

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Three grades of milk are 1 percent, 2 percent, and 3 percent by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent are mixed to give x+y+z gallons of the 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

x + 2y + 3z = 1.5(x + y + z)
0.5x = 0,5y + 1,5z
x = y + 3z

[Reveal] Spoiler: OA
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Re: GMAT Prep VIC Problem [#permalink]

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24 Dec 2009, 10:54
Thanks for the replies. I think I am confused with the setup. Originally, I thought this was a weighted average problem and setup (x + 2y + 3z) / 6 = 1.5 which would be the equation for weighted average. Obviously, this is wrong but I think conceptually I am missing something. Why couldn't I have done the weighted average way?
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12 Apr 2010, 01:07
1.5 is a weighted average of the fat content of the various milks.

Thus,$$1.5 = {x/(x+y+x)}*1 + {y/(x+y+z)}*2 + {z/(x+y+z)}*3$$

Solving for X gives y+3z
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27 Jun 2010, 06:54
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The answer is y + 3z.

The total amount of Fat: is 1*x + 2*y + 3*z = 1.5*(x+y+z)
Solving for x we get, x = y+3z
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27 Jun 2010, 21:04
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x+ 2y + 3z = 1.5(x + y + z)
note: we can divide by 100 on both sides to indicate the % amount, but since it would be common to both sides hence the equation will take the above form.
0.5x = 0.5y + 1.5z
x = y + 3z
Hence 'A'
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24 Aug 2010, 00:13
x/100 + 2y/100 + 3z/100 = (x + y + z)*(1.5/100)

Solving the above equation returns:
x = y + 3z

(A)
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30 Sep 2010, 22:22
Michmax3 wrote:
Three grades of milk are 1%, 2%, and 3% fat by volume. If x gallons of the 1% grade, y gallons of the 2% grade, and z gallons of the 3% grade are mixed to give x+y+z gallons a 1.5% 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

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

0.5x= 0.5y + 1.5z

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Senior Manager
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30 Sep 2010, 22:40
wow...so much simpler than I was making it. Thanks
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12 Mar 2011, 23:24
very simple mixture problem
% X variable on each end = total percent X (all variables)
solve
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08 Oct 2011, 10:56
hello all,

Confused as to solving this.please need some input. thx

Three grades of milk are 1 percent, 2 percent, 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
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08 Oct 2011, 11:01
0.01x+0.02y+0.03z = 0.015(+y+z).. then solve the eq to find x.
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08 Oct 2011, 14:18
thnx .. made it so easy
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09 Oct 2011, 18:36
Ans is A.

The weighted avg is 1.5%, so taking the total milk fat and dividing by the total number of gallons of milk:

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

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

.5x = .5y+1.5z

x = y+3z
Re: percents & averages   [#permalink] 09 Oct 2011, 18:36
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