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Question Stats:
22% (01:53) correct
77% (00:47) wrong based on 2 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?
(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) 3/4
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Re: Mixture problem [#permalink]
 08 May 2009, 11:41
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
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Re: Mixture problem [#permalink]
10 May 2009, 04:29
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
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Re: Mixture problem [#permalink]
20 May 2009, 00:23
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.
Thus B is correct. IMO
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Re: Mixture problem [#permalink]
20 May 2009, 03:13
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
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Re: Mixture problem [#permalink]
20 May 2009, 05:47
Thanks a lot for the detailed explanation in the last post by humans.
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Re: Mixture problem [#permalink]
05 Jul 2011, 07:20
Correct answer should be (D) 2/3. Some math-genius confirm this.
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Re: Mixture problem [#permalink]
21 Jul 2011, 14:46
Prabhakar wrote: Correct answer should be (D) 2/3. Some math-genius confirm this. Well, i'm certainly no math genius, but I'm getting the answer (d) 2/3 as well.
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Re: Mixture problem [#permalink]
21 Jul 2011, 21:09
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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" (T-R)-> 40% R->25% Average->35% (T-R)0.4+R*0.25=0.35T 0.4T-0.4R+0.25R=0.35T 0.05T=0.15R R/T=0.05/0.15=1/3 Ans: "B" OR using other form of Weighted Average: \frac{T-R}{R}=\frac{35-25}{40-35}\frac{T-R}{R}=\frac{10}{5}=2\frac{T}{R}-1=\frac{10}{5}=2\frac{T}{R}=3Invertendo: \frac{R}{T}=\frac{1}{3}
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Re: Mixture problem [#permalink]
21 Jul 2011, 23:02
Thanks fluke, kuddos given!
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Re: Mixture problem [#permalink]
22 Jul 2011, 03:30
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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. For explanation of the formula, see: http://www.veritasprep.com/blog/2011/03 ... -averages/
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Re: Mixture problem [#permalink]
23 Jul 2011, 14:23
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?
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Re: Mixture problem [#permalink]
23 Jul 2011, 23:14
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" (T-R)-> 40% R->25% Average->35% (T-R)0.4+R*0.25=0.35T 0.4T-0.4R+0.25R=0.35T 0.05T=0.15R R/T=0.05/0.15=1/3 Ans: "B" OR using other form of Weighted Average: \frac{T-R}{R}=\frac{35-25}{40-35}\frac{T-R}{R}=\frac{10}{5}=2\frac{T}{R}-1=\frac{10}{5}=2\frac{T}{R}=3Invertendo: \frac{R}{T}=\frac{1}{3}Nice explanation. Kudos
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Re: Mixture problem
[#permalink]
23 Jul 2011, 23:14
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