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Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
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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. \(3y + 4.5z\)
_________________________________________________________

\(\frac{x+2y+3z}{x+y+z} = \frac{3}{2}\)

\(x = y+3z\)
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
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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 fat by vol [#permalink]
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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. \(3y + 4.5z\)


We are given that 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. We can use this information to create the following equation:

0.01x + 0.02y + 0.03z = 0.015(x + y + z)

Multiplying the entire equation by 1000, we have:

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

We must now solve for x in terms of y and z:

5y + 15z = 5x

We can divide the entire equation by 5:

y + 3z = x

Answer: A
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
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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. \(3y + 4.5z\)



Let's start with a "word equation" and slowly turn it into an algebraic expression:

Total fat in mixture = 1.5% of (x+y+z)
(1% of x) + (2% of y) + (3% of z) = 0.015(x+y+z)
Rewrite as: 0.01x + 0.02y + 0.03z = 0.015x + 0.015y + 0.015z
Multiply both sides by 100: 1x + 2y + 3z = 1.5x + 1.5y + 1.5z
Rearrange and simplify: 0.5y + 1.5z = 0.5x
Multiply both sides by 2 to get: y + 3z = x

Answer: A

Cheers,
Brent
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
Expert Reply
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

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
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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
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 fat by vol [#permalink]
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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 fat by vol [#permalink]
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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 fat by vol [#permalink]
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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
is the answer
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
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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 fat by vol [#permalink]
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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 fat by vol [#permalink]
Bunuel - Where can I find challenging mixture problems/weighted avg problems to practice? Can you please help?

Thanks.
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Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
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p2bhokie wrote:
Bunuel - Where can I find challenging mixture problems/weighted avg problems to practice? Can you please help?

Thanks.


Check our Questions Bank: https://gmatclub.com/forum/viewforumtags.php

18. Mixture Problems




GMAT PS Question Directory by Topic & Difficulty: https://gmatclub.com/forum/gmat-ps-quest ... 27957.html
GMAT DS Question Directory by Topic & Difficulty: https://gmatclub.com/forum/ds-question-d ... 28728.html

Hope this helps.
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
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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 fat by vol [#permalink]
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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 fat by vol [#permalink]
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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 fat by vol [#permalink]
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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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