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36% (01:41) correct
63% (00:47) wrong based on 22 sessions

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?

Re: Mixture problem [#permalink]
22 Jul 2011, 02:30

3

This post received KUDOS

Expert's post

gmatprep09 wrote:

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

Or use our standard mixtures formula for replacements too. In a certain quantity of 40% solution, 25% solution is added to give 35% solution. w1/w2 = (A2 - Aavg)/(Aavg - A1) = (40 - 35)/(35 - 25) = 1/2 So quantity of 40% sol:25% solution = 2:1 This means the initial total solution was 3 and the fraction of 25% now in the mixture is 1. Therefore, 1/3 of the 40% solution was removed and replaced with 25% solution.

Re: Mixture problem [#permalink]
21 Jul 2011, 20:09

2

This post received KUDOS

gmatprep09 wrote:

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

Let actual solution be "T" Replaced solution be "R"

Re: Mixture problem [#permalink]
08 May 2009, 10:41

gmatprep09 wrote:

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

oringal quantity of solution=1 solution replaced = x

0.4 *(1-x)+0.25x= 0.35 *1

0.05= 0.15 x--> x= 1/3
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

Re: Mixture problem [#permalink]
19 May 2009, 23:23

I back solved it: Since it deals with percent, lets take 100 as the base.

When 1/4 - total mixrure is 40 + 25*4 / 500 = 140/500 ; this is not eq 35. When 1/3 - total mixture is 40 + 25*3/ 400 = 115/400 = 35% - this is the ans. When 1/2 - total mixture is 40 + 20*2/ 300 = 80/300 ; not eq 35% Similarly for D & E.

Re: Mixture problem [#permalink]
20 May 2009, 02:13

tkarthi4u wrote:

Can u please the below eqn ?

i belive here 1-x is the remaining soln. Now can you why are we taking 40 % of 1-x. I do not get it. Appreciate ur help.

0.4 *(1-x)+0.25x= 0.35 *1

Let me try out, the logic here is

You removed x quantity of 40% concentration solution from 1 and added same x quantity of 25% concentration solution, which total to original quantity 1 of solution with 35% concentration. Hence

Remaining quantity of 40% solution + added quantity of 25% solution = Total solution with 35% concentration. 0.4(1-x) + 0.25x = .35

Re: Mixture problem [#permalink]
23 Jul 2011, 13:23

These types of quesitons can also be solved using the MGMAT Table method..have you looked into their guide #2 which explains the table method?
_________________

Re: Mixture problem [#permalink]
23 Jul 2011, 22:14

fluke wrote:

gmatprep09 wrote:

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

Let actual solution be "T" Replaced solution be "R"