Question

Two pipes P and Q would fill an empty cistern in 24 minutes and 32 minutes respectively. Both the pipes being opened together, find when the first pipe must be turned off so that the empty cistern may be just filled in 16 minutes.

Answer 1

P's one minutes's work $=\frac{1}{24}$
Q's one minutes's work $=\frac{1}{32}$
Let the first pipe must turn off after x minutes.
The cistern filled in 16 minutes
Then P's x minutes work + Q's 16 minutes work = 1
$\Rightarrow \frac{1}{24}\times x+\frac{1}{32}\times 16=1$
$\frac{x}{24}+\frac{1}{2}=1$
$\Rightarrow \frac{x}{24}=1-\frac{1}{2}=\frac{1}{2}$
$\Rightarrow \frac{x}{24}=\frac{1}{2}$
$\Rightarrow x=\frac{24}{2}=12$
$\therefore$ After 12 minutes pipe P would turned off.