How much pure alcohol should be added to 400 ml of a 15% solution to

Asked: How much pure alcohol should be added to 400 ml of a 15% solution to make the strength of the solution 32%?

In 400 ml solution:
Alcohol = 60 ml

Let x ml of Pure alcohol be added to the solution to make the strength of the solution 32%
(60 + x)/(400+x) = .32
60 + x = 128 + .32x
.68x = 68
x = 100 ml

IMO B

Testing some of the answers was the fastest way in my opinion.
I always start with either B or D and then move up-/downwards based on the result (if necessary).

Say I start with (D) 150 ml:

First of all, how much pure alc contains the solution at the beginning?
400 * 15% = 60 ml

In other terms, you can say that the concentration = volume of the solute / total volume

Now lets add 150 ml to the solution and check its concentration:
volume of solute / total volume = \(\frac{(60 + 150)}{(400 + 150)} = \frac{210}{550} = \frac{21}{55}\) which is slightly less than \(\frac{2}{5}\), therefore slightly less than \(\frac{40}{100} = \) 40%

This concentration is too high, therefore I need to add less alcohol.

Let's check answer (B) 100 ml:

\(\frac{(60 + 100)}{(400 + 100)} = \frac{160}{500} = \frac{16}{50} = \frac{32}{100} = \) 32%

That's the correct solution, Answer B
Notice that this was the slower option: if you had picked B right away, you were done after 1 calculation.

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