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# Three grades of milk are 1 percent, 2 percent and 3 percent

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Joined: 26 Nov 2009
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Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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24 Dec 2009, 09:38
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75% (02:06) correct 25% (02:21) wrong based on 1432 sessions

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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  [#permalink]

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24 Dec 2009, 10:23
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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:

$$1\%x+2\%y+3\%z=1.5\%(x+y+z)$$;

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

$$0.5x=0.5y+1.5z$$;

$$x=y+3z$$.

Hope it's clear.
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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15 Jan 2012, 09:46
22
5
$$\frac{x+2y+3z}{x+y+z} = \frac{3}{2}$$

$$x = y+3z$$
##### General Discussion
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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22 Jan 2012, 18:56
22
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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  [#permalink]

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15 Jan 2012, 17:00
14
2
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  [#permalink]

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22 Mar 2012, 03:16
11
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
E-3y+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  [#permalink]

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29 Jan 2014, 21:22
5
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  [#permalink]

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25 Jul 2013, 21:52
2
3
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  [#permalink]

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12 Jan 2014, 09:51
2
1
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

y + 3z = x --> A
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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17 Sep 2012, 17:06
1
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  [#permalink]

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22 Oct 2012, 20:01
1
(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
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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25 Jul 2013, 14:22
1
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  [#permalink]

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05 Feb 2016, 20:26
1
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

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  [#permalink]

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30 Aug 2016, 01:33
1
neeraj609 wrote:
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

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  [#permalink]

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02 Feb 2017, 23:56
1
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:
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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22 Jan 2012, 08:04
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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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  [#permalink]

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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.
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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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

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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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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   [#permalink] 13 Jan 2014, 00:27

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