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# Mixture problem

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Manager
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Mixture problem [#permalink]  07 May 2009, 14:26
00:00

Difficulty:

5% (low)

Question Stats:

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?

(A) 1/4

(B) 1/3

(C) 1/2

(D) 2/3

(E) 3/4
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Re: Mixture problem [#permalink]  22 Jul 2011, 02:30
3
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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.

For explanation of the formula, see:
http://www.veritasprep.com/blog/2011/03 ... -averages/
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Re: Mixture problem [#permalink]  21 Jul 2011, 20:09
2
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"

(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}=3

Invertendo:
\frac{R}{T}=\frac{1}{3}
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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
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Re: Mixture problem [#permalink]  10 May 2009, 03:29
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]  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.

Thus B is correct. IMO
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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
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Re: Mixture problem [#permalink]  20 May 2009, 04:47
Thanks a lot for the detailed explanation in the last post by humans.
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Re: Mixture problem [#permalink]  05 Jul 2011, 06:20
Correct answer should be (D) 2/3. Some math-genius confirm this.
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Re: Mixture problem [#permalink]  21 Jul 2011, 13:46
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, 22:02
Thanks fluke, kuddos given!
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Knewton Knerd

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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?
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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"

(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}=3

Invertendo:
\frac{R}{T}=\frac{1}{3}

Nice explanation. Kudos
Re: Mixture problem   [#permalink] 23 Jul 2011, 22:14
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