Hi chaitanya.bhansali,

Mixture questions can also be answered with the Part/Whole formula:

In this case, we're dealing with Sugar/Total

Solution X = 20% sugar
Solution Y = 40% sugar

We're told to mix a certain amount of Solution X with 150 gallons of Solution Y to get a 25% sugar solution:

X = number of gallons of Solution X

[(.2)(X) + (.4)(150)] / (X + 150) = .25

Now, it's just algebra steps:

.2X + 60 = .25X + 37.5

22.5 = .05X

450 = X

Final Answer: E

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

chaitanya.bhansali wrote: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