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A solution contains water, milk and liquid chocolate in the ratio of \(2:3:5\) by volume.

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A solution contains water, milk and liquid chocolate in the ratio of \(2:3:5\) by volume.

by Gmat_mission » Sun Jan 17, 2021 11:10 am

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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

Answer: B

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Re: A solution contains water, milk and liquid chocolate in the ratio of \(2:3:5\) by volume.

by Scott@TargetTestPrep » Tue Jan 26, 2021 4:38 am
Gmat_mission wrote: ↑
Sun Jan 17, 2021 11:10 am
 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

Answer: B

Solution:

If we let 2x, 3x, and 5x be the number of liters of water, milk, and liquid chocolate in the original solution, and we then add y liters of water and y liters of milk, the total volume of the new solution is::

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

We know that the new amount of water will be (2x + y), and this amount will constitute 25% (or 1/4 ) of the volume of the new solution, so we can create the following proportion:

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

Substituting 120 for the denominator in this equation, we have:

(2x + y) / 120 = 1/4

2x + y = 30 → (Eq. 1)

Simplifying the first equation, we have:

10x + 2y = 120 → (Eq. 2)

Multiplying equation (1) by 5 and subtracting equation (2) from that, we have:

3y = 30

y = 10

Answer: B

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