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# What percent of the mixture is Solution X?

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Intern
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What percent of the mixture is Solution X? [#permalink]

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08 Dec 2012, 11:49
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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!
[Reveal] Spoiler: OA
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Re: What percent of the mixture is Solution X? [#permalink]

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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]

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09 Dec 2012, 01:29
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Re: What percent of the mixture is Solution X? [#permalink]

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09 Dec 2012, 06:31
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Expert's post
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.

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]

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08 Dec 2012, 12:03
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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!

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]

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09 Mar 2013, 00:14
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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!

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]

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02 Feb 2015, 06:27
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Re: What percent of the mixture is Solution X? [#permalink]

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02 Feb 2015, 13:11
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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

[Reveal] Spoiler:
D

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Re: What percent of the mixture is Solution X? [#permalink]

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27 Oct 2016, 16:22
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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Re: What percent of the mixture is Solution X? [#permalink]

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27 Oct 2016, 16:40
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%
Re: What percent of the mixture is Solution X?   [#permalink] 27 Oct 2016, 16:40
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