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JJ2014
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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JJ2014
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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Hi All,

These types of mixture questions can typically be solved with a variety of approaches (the average formula, ratios, TESTing VALUES, TESTing THE ANSWERS, etc.). Here's the standard Weighted Average Formula approach:

Solution X = 20% chemical A
Solution Y = 30% chemical A

Mixture of the two solutions = 22% chemical A

X = number of ounces of Solution X
Y = number of ounces of Solution Y

(.2X + .3Y) / (X + Y) = .22

.2X + .3Y = .22X + .22Y
.08Y = .02X

We can multiply both sides by 100 to get rid of the decimals...

8Y = 2X

The question asks for the percentage of the new mixture that is Solution X....

8/2 = X/Y
4/1 = X/Y

80% X and 20% Y

Final Answer:
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0.20x+(1-x)0.30 = 0.22
0.20x+0.30-0.30x=0.22
-0.10x=-0.08
x=4/5 = 80%
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If the solution is 22% of A we know off the bat that the majority of the solution has to be X. We can eliminate A,B,C.

Solution X is 20% A and Y is 30

So X is 2 units from the combo and Y is 8 units away.

Ration of X:Y for solution A is 8:2
8/10=80%
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Diagram approach

X-----Y
20---30
--22--
8----2
4:1

what percent of the mixture is Solution X = 4/5*100=80%
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JJ2014
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!

Here are two videos that explain the concept behind this problem in detail:
https://www.youtube.com/watch?v=_GOAU7moZ2Q
https://www.youtube.com/watch?v=VdBl9Hw0HBg

or you can check out these posts:
https://anaprep.com/arithmetic-weighted-averages/
https://anaprep.com/arithmetic-mixtures/
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