# CAT 2018 Quant Questions

Question:
A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is

 52% 50% 55% 48%

Let the volume of the first and the second Solution be 100 and 300.

When they are mixed, quantity of ethanol in the mixture

= (20 + 300S)

Let this Solution be mixed with equal volume i.e. 400 of third Solution in which the strength of ethanol is 20%.

So, the quantity of ethanol in the final Solution

= (20 + 300S + 80) = (300S + 100)

It is given that, 31.25% of 800 = (300S + 100)

or, 300S + 100 = 250

or S = $\frac{1}{2}$ = 50%

Hence, 50 is the correct answer.

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