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

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

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New post 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

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  [#permalink]

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New post 29 Aug 2016, 10:17
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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New post 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

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  [#permalink]

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New post 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:
https://gmatclub.com/forum/three-grades ... l#p1032710
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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New post 25 Jun 2017, 18:00
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
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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New post 24 Nov 2017, 12:46
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??
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Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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New post 24 Nov 2017, 13:51
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. \(3y + 4.5z\)


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

Hence, the answer is A
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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New post 24 Jun 2018, 20:39
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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Three grades of milk are 1 percent, 2 percent and 3 percent  [#permalink]

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New post 24 Nov 2018, 15:45
A video explanation can be found here:
https://www.youtube.com/watch?v=aXGmN6qg8Hg

In mixture problems we’ll typically see (1) some sort of a substance (in this case, milk) and (2) some characteristic of the substance (in this case, fat) as a fraction or (usually) a percentage of the volume of the substance.

If the volumes of our three types of milk are expressed as x, y, z:

.01x + .02y + .03z = .015(x + y + z)

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

2x + 4y + 6z = 3(x + y + z)

Then isolate x:

x = y + 3z
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Re: 3 grades of milk are 1 %, 2% and 3% by volume. If x gallons  [#permalink]

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Re: 3 grades of milk are 1 %, 2% and 3% by volume. If x gallons   [#permalink] 25 Sep 2019, 05:56

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