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What percent of the mixture is Solution X?

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What percent of the mixture is Solution X? [#permalink] New post 08 Dec 2012, 11:49
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75% (02:02) correct 25% (00:36) wrong based on 7 sessions
Solution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?

A) 20%
B) 44%
C) 50%
D) 80%
E) 90%

Can someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!
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Re: What percent of the mixture is Solution X? [#permalink] New post 08 Dec 2012, 12:03
JJ2014 wrote:
Solution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?

A) 20%
B) 44%
C) 50%
D) 80%
E) 90%

Can someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!


you can solve this by allegation method.
Given 22% of A is the final mixture; Sol X has 20% chemical A and Sol Y has 30 % of A. (you don't need Chemical B to solve when you r solving the problem using this method)

X / Y = (30-22 ) / (22-20)
X / Y = 8/2 = 4/1

So X and Y are in the ratio 4 : 1
Question asks percent of the mixture is Solution X.

So X / Total = 4/5 *100 = 80%

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Re: What percent of the mixture is Solution X? [#permalink] New post 08 Dec 2012, 12:16
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also you can solve this one with the concept of weighted average or balance point

you care about only of chemical A in both X and Y

20------- 22 ------------------------30

Now the difference between 20 and 22 = 2; between 22 and 30 = 8

So 8y = 2 x

\frac{8}{2} =\frac{x}{y}
Our ratio is 2 and 8 and using the concept of unknown multiplier the sum is 10. Our X is 8 on a total of 10 so 80 %

much more difficult to say that to explain. In that way you can solve a question difficult like this in 30 seconds.

Also algebraic approach works fine, but this is a bit faster
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Re: What percent of the mixture is Solution X? [#permalink] New post 09 Dec 2012, 01:29
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Here you go:

seed-mixture-x-is-40-percent-ryegrass-and-60-more-problems-73381.html

mixtures-problems-102057.html
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Re: What percent of the mixture is Solution X? [#permalink] New post 09 Dec 2012, 06:31
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JJ2014 wrote:
Solution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?

A) 20%
B) 44%
C) 50%
D) 80%
E) 90%

Can someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!


22% of chemical A in X+Y grams of solution comes from 20% chemical A in solution X and 30% chemical A in solution Y, thus:

0.22(X + Y) = 0.2X + 0.3Y --> X = 4Y --> X/(X+Y)=4/5=0.8.

Answer: D.

Check out question banks for similar problems:
DS mixture problems: search.php?search_id=tag&tag_id=43
PS mixture problems: search.php?search_id=tag&tag_id=114

Hope it helps.
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Re: What percent of the mixture is Solution X? [#permalink] New post 09 Mar 2013, 00:14
JJ2014 wrote:
Solution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?

A) 20%
B) 44%
C) 50%
D) 80%
E) 90%

Can someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!


Focus on Chemical A:

X--------------------------------------Y
20%---------------------------------30%

--------------------22%-----------------

30 - 22= 8--------------------22-20 = 2

X:Y = 8:2 = 4:1.

Therefore, %age of X = 4/5 * 100 = 80%
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Re: What percent of the mixture is Solution X?   [#permalink] 09 Mar 2013, 00:14
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