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Concept of MIXTURE problem?

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by GMATGuruNY » Thu May 10, 2012 7:21 pm
karthikpandian19 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
Since the resulting percentage (35%) is closer to the original percentage (40%) than to the replacement percentage (25%), less than half of the original solution was replaced with 25% solution.
The correct answer must be A or B.

Answer choice B: 1/3
If 1/3 of the mixture is 25% solution, then in every 3 units, there will be 1 part 25% for every 2 parts 40%.
Average percentage in 3 units = (1*25 + 2*40)/3 = 105/3 = 35.
Success!

The correct answer is B.
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by karthikpandian19 » Thu May 10, 2012 7:33 pm
I could not able to understand your concept / approach.....is there any straight forward approach to these kind of problems???
GMATGuruNY wrote:
karthikpandian19 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
Since the resulting percentage (35%) is closer to the original percentage (40%) than to the replacement percentage (25%), less than half of the original solution was replaced with 25% solution.
The correct answer must be A or B.

Answer choice B: 1/3
If 1/3 of the mixture is 25% solution, then in every 3 units, there will be 1 part 25% for every 2 parts 40%.
Average percentage in 3 units = (1*25 + 2*40)/3 = 105/3 = 35.
Success!

The correct answer is B.
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by Bill@VeritasPrep » Thu May 10, 2012 9:26 pm
karthikpandian19 wrote:I could not able to understand your concept / approach.....is there any straight forward approach to these kind of problems???
If we know the concentrations of the two solutions, and we know the weighted average, then the proportion of the two solutions is given by their distance from the weighted average. If we think about it visually:



25----30----35----40

If we had a 1:1 ratio, the average would be (40+25)/2 = 32.5%. The actual average is 35%, which means we must have more of the 40% solution in order to pull up the average. If we find the distances from the average:

35-25 = 10
40 - 35 = 5

Thus, the ratio of 40% to 25% is 10:5, or 2:1. This means that we have 3 total parts (1 part 25%, 2 parts 40%), so 1/3 of the new mixture must be 25%.

This method is called alligation: https://en.wikipedia.org/wiki/Alligation
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by karthikpandian19 » Thu May 10, 2012 9:57 pm
Bill,


I couldnt follow u here when u say

"Thus, the ratio of 40% to 25% is 10:5, or 2:1. This means that we have 3 total parts (1 part 25%, 2 parts 40%), so 1/3 of the new mixture must be 25%."

40% to 25% ratio is to be 5:10 right?????
or please explain?
Bill@VeritasPrep wrote:
karthikpandian19 wrote:I could not able to understand your concept / approach.....is there any straight forward approach to these kind of problems???
If we know the concentrations of the two solutions, and we know the weighted average, then the proportion of the two solutions is given by their distance from the weighted average. If we think about it visually:



25----30----35----40

If we had a 1:1 ratio, the average would be (40+25)/2 = 32.5%. The actual average is 35%, which means we must have more of the 40% solution in order to pull up the average. If we find the distances from the average:

35-25 = 10
40 - 35 = 5

Thus, the ratio of 40% to 25% is 10:5, or 2:1. This means that we have 3 total parts (1 part 25%, 2 parts 40%), so 1/3 of the new mixture must be 25%.

This method is called alligation: https://en.wikipedia.org/wiki/Alligation
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by Bill@VeritasPrep » Thu May 10, 2012 10:00 pm
If you're familiar with the game "tug of war", you can think of the two solutions (25% and 40%) as the teams on each end of the rope. 40% has pulled the rope towards it, so it controls the bigger portion: the length of 10.

Or you can just remember that each end "controls" the opposite portion of the number line.
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by darontan » Thu May 10, 2012 10:05 pm
A1 as the 40% solution, and A2 as the 25% solution

A1=40, B1=60, Total1=100
A2=(1/4)N, B2=(3/4)N, Total2=N

Sum of A1 + A2 will become 35% of the sum of total solution
40+(1/4)N = (35/100)(100+N)
N=50

Therefore, N/(100+N) = 50/150 = 1/3
Last edited by darontan on Thu May 10, 2012 11:13 pm, edited 2 times in total.
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by karthikpandian19 » Thu May 10, 2012 10:05 pm
Got it...thank you for your detailed explanation....
Bill@VeritasPrep wrote:If you're familiar with the game "tug of war", you can think of the two solutions (25% and 40%) as the teams on each end of the rope. 40% has pulled the rope towards it, so it controls the bigger portion: the length of 10.

Or you can just remember that each end "controls" the opposite portion of the number line.
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by aneesh.kg » Thu May 10, 2012 10:10 pm
Mitch and Bill presented the best way to solve this problem and it will be great if you understand that method well.
However, here is an Alternate Solution:

In the new solution,
Lets say there are x units of 40% solution are y units of 25% solution.

35% of (x + y) = 40% of x + 25% of y
35 = x*40 + 25*y*/(x + y)
35 = 40(1 - y/(x + y)) + 25*(y/x + y)
fraction of 25% solution = fraction of 40% solution removed = y/(x + y) = f (say)
35 = 40(1 - f) + 25f
f = 1/3

[spoiler](B)[/spoiler] is the answer.
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