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 total time taken to fill the empty cistern is $28$ minutes.

• A. After $10$ minutes
• B. After $14$ minutes
• C. After $15$ minutes
• D. After $12$ minutes

Answer 1

Answer: (D)

After $12$ minutes

Amount filled by A in $1$ min $=\frac{1}{24}$

Amount filled by B in $1$ min $=\frac{1}{32}$

Lets assume that pipe A is turned off after $x$ min

Given total time taken is $28$ min

Now,

$\frac{1}{24}x$ $+$ $\frac{(28-x)}{32}$ $=$ 1

By solving we get $x=12$

Hence pipe A has to be turned off after $12$min