## Mixture Problems, Word Problems

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### Mixture Problems, Word Problems

by swerve » Mon Jun 29, 2020 12:31 pm

00:00

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## Global Stats

A solution contains water, milk and liquid chocolate in the ratio of 2:3:5 by volume. If y liters of water and milk each are added to this solution, the resultant solution would contain 25 percent of water by volume. If the volume of this resultant solution is 120 liters, what is the value of y in liters?

A. 8
B. 10
C. 11
D. 12
E. 15

The OA is B

Source: e-GMAT

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### Re: Mixture Problems, Word Problems

by [email protected] » Mon Jun 29, 2020 9:00 pm
swerve wrote:
Mon Jun 29, 2020 12:31 pm
A solution contains water, milk and liquid chocolate in the ratio of 2:3:5 by volume. If y liters of water and milk each are added to this solution, the resultant solution would contain 25 percent of water by volume. If the volume of this resultant solution is 120 liters, what is the value of y in liters?

A. 8
B. 10
C. 11
D. 12
E. 15

The OA is B

Source: e-GMAT
Say initially, the solution had 2x, 3x, and 5x liters of water, milk and liquid chocolate, respectively.

After the addition of y liters of water and milk each, the solution now has (2x + y), (3x + y), and 5x liters of water, milk and liquid chocolate, respectively. And the total content of the solution is (2x + y) + (3x + y) + 5x = (10x + 2y) liters

=> 10x + 2y = 120 => 5x + y = 60 ---(1)

Water in the new solution = 25% of 120 = 30 ltrs

=> 2x + y = 30 ---(2)

Solving (1) and (2), we get x = y = 10.

Hope this helps!

-Jay
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### Re: Mixture Problems, Word Problems

by [email protected] » Mon Apr 19, 2021 6:21 am
swerve wrote:
Mon Jun 29, 2020 12:31 pm
A solution contains water, milk and liquid chocolate in the ratio of 2:3:5 by volume. If y liters of water and milk each are added to this solution, the resultant solution would contain 25 percent of water by volume. If the volume of this resultant solution is 120 liters, what is the value of y in liters?

A. 8
B. 10
C. 11
D. 12
E. 15

The OA is B

Solution:

We can let the amount of water, milk, and liquid chocolate in the original solution be 2x, 3x, and 5x (where x is a positive number). We can create the equations:

2x + 3x + 5x + y + y = 120

and

(2x + y)/(2x + 3x + 5x + y + y) = 1/4

Simplifying the first equation, we have:

10x + 2y = 120

5x + y = 60

y = 60 - 5x

Substituting this in the second equation (and noticing that the denominator is just 120), we have:

(2x + 60 - 5x)/120 = 1/4

60 - 3x = 30

30 = 3x

10 = x

Since y = 60 - 5x, y = 60 - 5(10) = 10.

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