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Three grades of milk are 1 percent, 2 percent and 3 percent
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24 Dec 2009, 09:38
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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. \(3y + 4.5z\)
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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24 Dec 2009, 10:23
JimmyWorld wrote: I got this problem wrong on the GMAT Prep and don't really understand it. 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 OA is 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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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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15 Jan 2012, 09:46
\(\frac{x+2y+3z}{x+y+z} = \frac{3}{2}\)
\(x = y+3z\)




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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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22 Jan 2012, 18:56
MSoS wrote: Hi, would someone please so kind and explain the question:
Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the q 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
Thanks a lot... 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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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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15 Jan 2012, 17:00
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



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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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22 Mar 2012, 03:16
imadkho 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 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 E3y+4.5z 0.01x+0.02y+0.03z=0.015(x+y+z) => x=y+3z hence A



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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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29 Jan 2014, 21:22
Impenetrable wrote: Three grades of milk are 1%, 2% and 3% fat by volume. If x gallons of 1%, y gallons of 2% and z gallons of 3% are mixed together to give x+y+z gallons of a 1.5%, what is x in terms of y and z?
y+3z (y+z)/4 2y+3z 3y+z 3y+4.5z
My idea was:
(x+2y+3z)/(x+y+z) = 1.5 from here on I have no idea how to get x to one side...
Cheers, Lars 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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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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25 Jul 2013, 21:52
NvrEvrGvUp wrote: MBAhereIcome wrote: \(\frac{x+2y+3z}{x+y+z} = \frac{3}{2}\)
\(x = y+3z\) 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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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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12 Jan 2014, 09:51
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



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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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17 Sep 2012, 17:06
three grades of milk are 1 percent 2 percent and 3 percent fat by volume. if x gallons of the 1 percent y gallons of the 2 percent and z gallons of the 3 percent are mixed to give x +y+z gallons of a 1.5 percent grade, what is x in terms of y and z? Thanks in advance!!!



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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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22 Oct 2012, 20:01
(x/100)+ (2y/100)+(3z/100) = 1.5 (x+y+z)/100  > cancel out 100 on each side. x+2y+3z = 1.5x+1.5y+1.5z > bring x to one side of = sign .5x=.5y+1.5x > multiply by 2 on both side x=y+3z ______________ is the answer



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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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25 Jul 2013, 14:22
MBAhereIcome wrote: \(\frac{x+2y+3z}{x+y+z} = \frac{3}{2}\)
\(x = y+3z\) This is essentially viewing this problem as a weighted average correct? The mixture divided by the sum of weights?



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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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05 Feb 2016, 20:26
Hi VeritasPrepKarishmaI 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(1.51) = y(2..0  1.5) + z(3.01.5) ==> x/2 = y/2 + 3z/2 ==> x = y + 3z Thanks in advance! VeritasPrepKarishma wrote: NvrEvrGvUp wrote: MBAhereIcome wrote: \(\frac{x+2y+3z}{x+y+z} = \frac{3}{2}\)
\(x = y+3z\) 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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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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30 Aug 2016, 01:33
neeraj609 wrote: Hi VeritasPrepKarishmaI 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(1.51) = y(2..0  1.5) + z(3.01.5) ==> x/2 = y/2 + 3z/2 ==> x = y + 3z Thanks in advance! 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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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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02 Feb 2017, 23:56
JimmyWorld 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 Responding to a pm: Quote: I solved it by : x+2y++3z/x+y+z = but i am not getting how did we get 3/2 on the left hand side . could you please elaborate the same
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/threegrades ... l#p1032710
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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22 Jan 2012, 08:04
Thanks! Sounds easy after i read your reply! Did not think through the whole question..



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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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10 May 2012, 11:33
Hi all, I think I have seen a problem like this somewhere, but in DS form
If (1) is x= y + 3z and (2) gives you the y:z ratio Is the second one sufficient? I somehow feel that it should be, but can't find the reasoning for that.What can we do here? pick numbers? Or is it exessive info and thus is sufficient? sorry, I can't quote the exact second choice.



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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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12 May 2012, 23:29
Yep this one would have seemed more obtuse to me until I realized that the percentages were meant to be for the fat content in milk. Combining the various milk types we got a 1.5% of fat content in the resulting mixture. Got the same answer y+3z.



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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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28 Sep 2012, 03:10
1%x + 2%y + 3%z = 1.5% (x+y+z) 2%y  1.5%y + 3%z  1.5%z = 1.5%x  1%x 2(.5y + 1.5z = .5x)
y + 3z = x
Answer: A



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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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13 Jan 2014, 00:27
JimmyWorld 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 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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Re: Three grades of milk are 1 percent, 2 percent and 3 percent
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