Solution X contains 20% of material A and 80% of material B. Solution

Please post the answer choices for a PS question.

For any mixture problem in GMAT, please use the alligation technique:
trigger points:
From any mixture problem, note down the below 3 pointers first.
1) What is solution 1
2) what is solution2
3) what is the resulting solution

For example, i have a solution with 40% milk (==> rest i.e. 60% is water) and i have one more solution which is with 70% milk ==> 30% water
qtn: if we mix these two solutions in a particulr ratio to get a resultant solution with 50% milk, how much % solution1 exists in the resultant solution?

40------------70
------50
20------------10 ==> use the solution 1 and solution2 in the ratio 2:1 to get a resultant solution with 50% milk

means if i use 200liters of sultion 1, then i have to use 100liters of solution 2 resulting 300liters of 50%milk..hope it is clear.

Back to the Original qtn:

solutionX: 20% of A
solutionY: 30% of A
resulting solution - 22% of A
X-------------Y
20------------30
-------22
8--------------2 ==> 4:1

==> 4/5 or 80% of the resultant solution is X and 20% of that is Y

Regards,
Murali.

Kudos?

Had seen the problem somewhere but didnt remember the options. thats y couldnt post them..

Thanks Murali, but i am still not able to figure out how come u got the ratio of 8:2. I followed your steps but didnt get it..
can u pls explain in little more detail?

thnx

I guess the question asks about the share of solution X in the final mixture.

This is a weighted average question: \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\).

\(0.22=\frac{0.2x+0.3y}{x+y}\) --> \(2x=8y\) --> \(\frac{x}{y}=\frac{8}{2}\), so there is 80% of solution X in the final mixture.

Answer: 80%.
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Bunuel. that should be: \(x/y=8/2\). Answer should be 80%

Regards,
Murali.

Sure. Edited a typo.
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We can assume the total weight of the mixture= 100
Conc of A in the final mixture= 22

Let weight of A in the mixture be X. Conc given= 20%=.2
Therefore, weight of B= 100-X. Conc given = 30%= .3

Now, acc to the problem, .2X+ .3(100-X)=22
Solving, we get X=80. Since we assumed the weight of the mixture = 100, Therefore presence of A in the mixture= 80%.

Wow! I'm so glad this answer matches that of all others.. mixture problems is my weakest area!
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