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Q. A cask is completely filled with 30% alcohol solution. Twenty litres of the solution is withdrawn from it and is replaced with water. If the concentration now is 24% find the capacity of the cask.

$Ans:100l$

Attempted solution:

Let capacity be M litres.

Amount of alcohol in (M-20) litres $= 0.3(M-20)l $

$Therefore$

$$\frac{0.3(M-20)}{M} = \frac{24}{100}$$ $$ \Rightarrow 0.75M-15=6M$$

The answer seems to match if $6M$ is $0.6M$. Where have I gone wrong?

  • 1
    It's not clear where you went wrong - your equation $\frac{0.3(M-20)}{M} = \frac{24}{100}$ is correct, but the equation you got out of it isn't. Can you show your steps to get from one to the other? – 2017-02-19
  • 0
    Got it sir. It was a calculation mistake. – 2017-02-19

1 Answers 1

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Your mistake is $0.3 \times 25 = 7.5$.

Not 0.75.

So your solution should be -

$7.5M - 150 = 6M$

$1.5M = 150$

$M = 100$.

Also you can simplify using the following way.

$$\frac{0.3(M-20)}{M} = \frac{24}{100}$$

$$\frac{3(M-20)}{10M} = \frac{24}{100}$$

$$\frac{(M-20)}{M} = \frac{8}{10}$$

$$\frac{(M-20)}{M} = \frac{4}{5}$$

$$5M -100=4M$$

$$M = 100$$