* A certain quantity of 40% solution is replaced with 25% s.

[alok.sarsidharan]

A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?

(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) ¾

Ans: Let X be the fraction of solution that is replaced.

Then X*25% + (1-X)*40% = 35%

Solving, you get X = 1/3

Can someone please explain the logic behind this?

[stockish24]

A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?

(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) ¾

Ans: Let X be the fraction of solution that is replaced.

Then X*25% + (1-X)*40% = 35%

Solving, you get X = 1/3

Can someone please explain the logic behind this?

Alok, You have solved it. I am not sure what you wanted. you assumed total amount of solution to be 1 which is correct and then x portion is replaced with 25%
.so now the qty remains the same as per problem. On the other side you have
35% which is the total concentration after making this replacement and the total
qty is retained to 1. So, you are correct.

[Ben Ku]

Please cite the source (author) of this problem. We cannot reply unless a source is cited (and, if no source is cited, we will have to delete the post!). Thanks.

[alok.sarsidharan]

I'm sorry Ben, I understand your concern. This was something I found on the beatthegmat forum. Hope this helps.

[Ben Ku]

A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?

(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) ¾

Ans: Let X be the fraction of solution that is replaced.

Then X*25% + (1-X)*40% = 35%

Solving, you get X = 1/3

Can someone please explain the logic behind this?


The Amount of Solute you have is the Concentration * Volume. (Solute is what you put in to make the solution)

Let's suppose your container is 1 L. If we replace X Liters, then we will have 1 - X liters left.

After replacing, X liters will have a concentration of 25%, so the amount of solute is 0.25 * X. The remaining 1 - X liters will have a concentration of 40%, so the amount of solute is 0.40(1 - X). At the end, when it's all mixed, the 1 L will have a concentration of 35%, so the amount of solute is 0.35 * 1 .

We can set up the equation:
0.25X + 0.40(1 - X) = 0.35(1)

Hope that makes sense.