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

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Intern
Joined: 01 Jul 2015
Posts: 14
Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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04 Oct 2015, 04:53
Have anybody tried to solve this problem by Allegation method?
Manager
Joined: 07 May 2015
Posts: 87
Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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05 Feb 2016, 19:26
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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Status: GMATINSIGHT Tutor
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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29 Aug 2016, 09:17
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

Please find the solution as attached
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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, 00:33
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, 22:56
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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Manager
Joined: 23 Dec 2013
Posts: 159
Location: United States (CA)
GMAT 1: 710 Q45 V41
GMAT 2: 760 Q49 V44
GPA: 3.76
Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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25 Jun 2017, 17:00
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

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

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

x + 0.5y + 1.5z = 1.5x

0.5y + 1.5z = 0.5x

y + 3z = x
Intern
Joined: 06 Oct 2017
Posts: 9
Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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24 Nov 2017, 11:46
I tried to solve the problem with numbers.. can someone point out what's wrong with my logic?

400 gallons of X = 4 grams of fat
100 gallons of Y = 2 grams of fat
100 gallons of Z = 3 grams of fat

4+2+3 / 400+100+100 = 1.5% fat

So answer choices A and D work for me??
Intern
Joined: 15 Oct 2017
Posts: 30
Location: Ireland
Concentration: Healthcare, Finance
Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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24 Nov 2017, 12:51
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. $$\frac{y +z}{4}$$

C. $$2y + 3z$$

D. $$3y + z$$

E. $$3y + 4.5z$$

$$\frac{x}{100} + \frac{2y}{100} + \frac{3z}{100} = \frac{15*(x + y + z)}{1000}$$
$$x + 2y + 3z = \frac{15*(x + y + z)}{10}$$
$$0.5y + 1.5z = 0.5x$$
$$y + 3z = x$$

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

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24 Jun 2018, 19:39
Hi All,

This is an "in terms of" question; these questions are usually built around 4-5 algebra steps and are fairly straight-forward "math" questions. To start, we can translate the given information into the following equation:

[(.01X) + (.02Y) + (.03Z)] / (X + Y + Z) = .015

(.01X + .02Y + .03Z0 = (.015X + .015Y + .015Z)

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

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

And then combine like terms....

5Y+ 15Z = 5X

And then divide everything by 5...

Y + 3Z = X

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Re: Three grades of milk are 1 percent, 2 percent and 3 percent &nbs [#permalink] 24 Jun 2018, 19:39

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