Mixture problem

[Mani_mba]

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) 3/4

Plz explain.

[dmateer25]

https://www.beatthegmat.com/fraction-t18180.html

[Mani_mba]

let x be the fraction replaced. if x is the fraction then the total is 1.

(1-x)*40 + x*25 = 35
40 - 40x + 25x = 35
5 = 15x
x = 1/3

Still i am not able to get it.

Could someone elaborate this ?

Thanks

[nitin86]

let x be the fraction replaced. if x is the fraction then the total is 1.

(1-x)*40 + x*25 = 35
40 - 40x + 25x = 35
5 = 15x
x = 1/3

Still i am not able to get it.

Could someone elaborate this ?

Thanks

Say, the actual solution is X liters and Y liters of the solution was replaced with another solution.

Now, the first solution is 40% solution, this means that 40% is some substance other than water (say alchol
=> this means that out of X liters of solution 40% or .4X is alchol

Similarily, another solution is 25% solution, in Y liters of another solution
=> %age of alchol is 25% and hence
=> amount of alchol in Y liters of the solution is .25Y

Now, if Y liters of 1 solution was replaced with Y liters of second solution, the amount of alchol that will change in the first solution will be given by

=> .4X - .4Y + .25Y = .35X

where
.4X - .4Y => Amount of alchol left after Y amount of first solution was removed from the X amount of the solution. So, since %age of alchol is 40 in first solution, on removing Y amount of the solution, amount of alchol decreased by .4Y from .4X

Now, Y liters of second solution was added, so the amount of alchol that got added with Y liters of second solution is .25Y ( as second solution is 25% solution)

Now, after this change, the solution becomes 35 %, that means now the amount of alchol in the solution is .35Y

And we have to calculate Y/X.
Hope, this helps you.