Solution Y is 40 percent sugar by volume, and solution X...

[swerve]

Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. how many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

(A) 37.5
(B) 75
(C) 150
(D) 240
(E) 450

The OA is E.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.

[Brent@GMATPrepNow]

Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. How many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

A) 37.5
B) 50
C) 62.5
D) 300
E) 450

When solving mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED.

Start with 150 gallons of solution that is 40% sugar:

When we draw this with the ingredients separated, we see we have 60 gallons of sugar in the mixture.

Next, we'll let x = the number of gallons of solution X we need to add.
Since 20% of the solution X is sugar, we know that 0.2x = the volume of sugar in this solution:

At this point, we can ADD the two solutions (PART BY PART) to get the following volumes:

Since the resulting solution is 25% sugar (i.e., 25/100 of the solution is sugar), we can write the following equation:
(60 + 0.2x)/(150 + x) = 25/100
Simplify to get: (60 + 0.2x)/(150 + x) = 1/4
Cross multiply to get: 4(60 + 0.2x) = 1(150 + x)
Expand: 240 + 0.8x = 150 + x
Rearrange: 90 = 0.2x
Solve: x = 450

Answer: E

Cheers,
Brent

Here are some additional mixture questions to practice with:
https://www.beatthegmat.com/liters-of-mi ... 71387.html
https://www.beatthegmat.com/percentage-m ... 68631.html
https://www.beatthegmat.com/rodrick-mixe ... 70387.html
https://www.beatthegmat.com/mixure-probl ... 61767.html
https://www.beatthegmat.com/mixture-rati ... 91643.html

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Hi swerve,

We're told that Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. We're asked for the number of gallons of Solution X that must be added to 150 gallons of Solution Y to create a solution that is 25 percent sugar by volume. This prompt is an example of a Weighted Average question - and while you can certainly solve it Algebraically, you can also TEST THE ANSWERS.

To start, IF we had an EQUAL amount of both Solutions in the mixture, then the average sugar percent would be (40+20)/2 = 30. That's not what we're after though - we want there to be a 25% average, so we clearly need MORE of Solution X than Solution. Y. We're told that there is 150 gallons of Solution Y in the mixture, so there has to be MORE than 150 gallons of Solution X. Thus, we can eliminate Answers A, B and C.

Let's TEST Answer E: 450 gallons

IF we have....
150 gallons of Solution Y and 450 gallons of Solution X, the ratio of Y:X is 1:3 and the average sugar content would be...

[(1)(40) + (3)(20)]/(1+3) =
[40 + 60]/4 =
100/4 = 25 percent

This matches what we were told, so this MUST be the answer.

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

[Matt@VeritasPrep]

If we've got 150 gallons of solution Y, then 40% of 150, or 60 gallons of it are sugar.

From here, we're adding x gallons of Solution X. 20% of that is sugar, so

sugar / total = (60 + 0.2x) / (150 + x)

is our ratio.

We need that to equal 25%, or 1/4, so

(60 + 0.2x) / (150 + x) = 1/4

Then just cross multiply and solve for x:

4 * (60 + 0.2x) = 150 + x

240 + 0.8x = 150 + x

90 = 0.2x

450 = x

[Matt@VeritasPrep]

Another approach would be finding the weighted average.

If X is 20% sugar, Y is 40% sugar, and X + Y = 25% sugar, then

.2x + .4y = .25(x + y)

.2x + .4y = .25x + .25y

.15y = .05x

3y = x

We were told y = 150, so x = 3y = 3*150 = 450.