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 Twenty litres of the solution is withdrawn from it and is replaced with water.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?
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.
$$\frac{0.3(M-20)}{M} = \frac{24}{100}$$
$$\frac{3(M-20)}{10M} = \frac{24}{100}$$
$$\frac{(M-20)}{M} = \frac{8}{10}$$
$$\frac{(M-20)}{M} = \frac{4}{5}$$
$$5M -100=4M$$
$$M = 100$$