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
Difficult Math Question #7
This topic has expert replies
Let X be the fraction of 40% solution kept800guy 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
Then we have
X * 0,4 + (1X) * 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 1X = 1/3  answer B
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
(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

 Junior  Next Rank: 30 Posts
 Posts: 10
 Joined: Thu Jun 09, 2011 12:54 am
 Location: India
Let total solution is 100 and x is replaced.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
Thus we have 40% of (100x) solution and 25% of x solution = 35% of 100 solution
40(100x) + 25(x) = 35*100
500=15x
x=100/3
Thus fraction = x/100 = 1/3
IMO B
 GMATGuruNY
 GMAT Instructor
 Posts: 15539
 Joined: Tue May 25, 2010 12:04 pm
 Location: New York, NY
 Thanked: 13060 times
 Followed by:1906 members
 GMAT Score:790
Using alligation: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) ï¿½
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 = 3525 = 10.
Proportion of 25% solution = 4035 = 5.
Total = 10+5 = 15.
Fraction of 25% solution in the mixture = 5/15 = 1/3.
The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 testtakers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia  a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua  a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 testtakers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia  a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua  a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3