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# What percent of the mixture is Solution X?

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Intern
Joined: 06 Sep 2012
Posts: 38
Concentration: Social Entrepreneurship
What percent of the mixture is Solution X?  [#permalink]

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08 Dec 2012, 10:49
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00:00

Difficulty:

25% (medium)

Question Stats:

78% (01:57) correct 22% (02:16) wrong based on 325 sessions

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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!
Manager
Joined: 22 Dec 2011
Posts: 227
Re: What percent of the mixture is Solution X?  [#permalink]

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08 Dec 2012, 11:03
1
2
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 $$X / Total = 4/5 *100 = 80%$$

Cheers
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Joined: 01 Sep 2010
Posts: 3352
Re: What percent of the mixture is Solution X?  [#permalink]

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08 Dec 2012, 11:16
3
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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Joined: 06 Jun 2010
Posts: 153
Re: What percent of the mixture is Solution X?  [#permalink]

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09 Dec 2012, 00:29
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Math Expert
Joined: 02 Sep 2009
Posts: 53069
Re: What percent of the mixture is Solution X?  [#permalink]

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09 Dec 2012, 05:31
2
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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Manager
Joined: 22 Nov 2010
Posts: 221
Location: India
GMAT 1: 670 Q49 V33
WE: Consulting (Telecommunications)
Re: What percent of the mixture is Solution X?  [#permalink]

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08 Mar 2013, 23:14
1
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, 12:11
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

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

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27 Oct 2016, 15:22
1
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, 15:40
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%
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Re: What percent of the mixture is Solution X?  [#permalink]

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05 Sep 2018, 04:45
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Re: What percent of the mixture is Solution X?   [#permalink] 05 Sep 2018, 04:45
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# What percent of the mixture is Solution X?

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