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Two solutions of acid were mixed to obtain 10 litres of new

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by rakeshd347 » Mon Oct 07, 2013 8:08 pm
Two solutions of acid were mixed to obtain 10 litres of new solution.Before they were mixed,the first solution contained 0.8 litres of acid while the second contained 0.6 liters of acid.If the Percentage of acid in the first solution was twice that in the second,what was the volume of the first solution?
a.3 litres
b. 3.2 litres
c.3.6 litres
d.4 litres
e. 4.2 litres

OA is D.

Please explain this mixture problem.
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Source: — Problem Solving |

by theCodeToGMAT » Mon Oct 07, 2013 8:24 pm
A + B = 10

A = 0.8 --> X%
B = 0.6 --> Y%

X = 2B
0.8/A = 2 (0.6/B)
2B = 3A

A + B = 10
2A + 2B = 20
==> 5A = 20 = 4 Litres
Answer [spoiler]{D}[/spoiler]
R A H U L
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by GMATGuruNY » Mon Oct 07, 2013 8:26 pm
rakeshd347 wrote:Two solutions of acid were mixed to obtain 10 litres of new solution.Before they were mixed,the first solution contained 0.8 litres of acid while the second contained 0.6 liters of acid.If the Percentage of acid in the first solution was twice that in the second,what was the volume of the first solution?
a.3 litres
b. 3.2 litres
c.3.6 litres
d.4 litres
e. 4.2 litres
Let F = the first solution and S = the second solution.

In F, there are 0.8 liters of acid, while in S, there are 0.6 liters of acid.
Since the percent of acid in F is twice the percent of acid in S, we get:
(0.8)/F = 2 * (0.6/S)
8/F = 12/S
12F = 8S
F/S = 8/12 = 2/3.

Since F:S = 2:3 = 4:6, F=4 and S=6, for a total of 10 liters.

The correct answer is D.
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by GMATGuruNY » Mon Oct 07, 2013 8:53 pm
rakeshd347 wrote:Two solutions of acid were mixed to obtain 10 litres of new solution.Before they were mixed,the first solution contained 0.8 litres of acid while the second contained 0.6 liters of acid. If the Percentage of acid in the first solution was twice that in the second,what was the volume of the first solution?
a.3 litres
b. 3.2 litres
c.3.6 litres
d.4 litres
e. 4.2 litres
An alternate approach is to PLUG IN THE ANSWERS, which represent the volume of the first solution.
When the correct answer choice is plugged in, the percent of alcohol in the first solution will be twice the percent of alcohol in the second solution.

It is almost certain that the amount of acid in the first solution -- 0.8 liters -- will divide easily into the correct answer choice, so that the percent of alcohol in the first solution is equal to an integer value.
Thus, the correct answer choice is probably B (since .8/3.2 = 8/32 = 1/4 = 25%) or D (since .8/4 = 8/40 = 1/5 = 20%).
Since answer choice D is an integer value, start with D.

Answer choice D: first solution = 4 liters, second solution = 6 liters
In the first solution, alcohol/total = .8/4 = 8/40 = 1/5 = 20%.
In the second solution, alcohol/total = .6/6 = 6/60 = 1/10 = 10%.
Success!

The correct answer is D.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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by Scott@TargetTestPrep » Wed Dec 13, 2017 1:13 pm
rakeshd347 wrote:Two solutions of acid were mixed to obtain 10 litres of new solution.Before they were mixed,the first solution contained 0.8 litres of acid while the second contained 0.6 liters of acid.If the Percentage of acid in the first solution was twice that in the second,what was the volume of the first solution?
a.3 litres
b. 3.2 litres
c.3.6 litres
d.4 litres
e. 4.2 litres
We can let x = the number of litres of the first solution (the one with 0.8 litres of acid), so 10 - x = the number of litres of the second solution (the one with 0.6 litres of acid). Since the percentage of acid in the first solution was twice that of the second solution, we can say:

0.8/x = 2 * 0.6/(10 - x)

0.8/x = 1.2/(10 - x)

1.2x = 8 - 0.8x

2x = 8

x = 4

Answer: D

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