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Difficult Math Question #7

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by 800guy » Tue Sep 05, 2006 7:45 pm
OA coming when some people have answered..

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
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Re: Difficult Math Question #7

by piren » Tue Sep 05, 2006 11:49 pm
800guy wrote:OA coming when some people have answered..

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
Let X be the fraction of 40% solution kept
Then we have

X * 0,4 + (1-X) * 0,25 = 0,35
<=> 0,4*X + 0,25 - 0,25*X = 0,35
<=> 0,15*X = 0,10
<=> X = 2/3

Hence the fraction replaced is 1-X = 1/3 -- answer B
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by Rishabh » Thu Sep 07, 2006 8:26 am
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

Soln:
Solving it using allegations:
(Mean % - % of lower conc) / (% of higher conc - Mean %) =
(ratio of quantity of lower conc [x] to quantity of higher conc [y] in the resultant mixture)
=> 35 - 25 / 40 - 35 = x / y
=> 10 / 5 = x / y
=> x + y = 15
=> quantity of higher conc replaced = x / total quantity = x / x + y = 5 / 15 = 1/3
so, answer 1/3 => B
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by 800guy » Thu Sep 07, 2006 2:24 pm
thanks for the responses!! here's the OA:

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

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

Solving, you get X = 1/3
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by naveen1003 » Thu Jun 09, 2011 3:03 am
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
Let total solution is 100 and x is replaced.
Thus we have 40% of (100-x) solution and 25% of x solution = 35% of 100 solution
40(100-x) + 25(x) = 35*100
500=15x
x=100/3
Thus fraction = x/100 = 1/3
IMO B
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by GMATGuruNY » Thu Jun 09, 2011 4:35 am
800guy wrote:OA coming when some people have answered..

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) �
Using alligation:

The proportion needed of each element in the mixture is equal to the distance between the percent attributed to the other element in the mixture and the percent attributed to the entire mixture.

Proportion of 40% solution = 35-25 = 10.
Proportion of 25% solution = 40-35 = 5.
Total = 10+5 = 15.
Fraction of 25% solution in the mixture = 5/15 = 1/3.

The correct answer is B.
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