Question

Two pipes P and Q would fill an empty cistern in 24 minute 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

Solution:

Let volume of cistern be 'V'.

Flow rate through:

$\implies$ Pipe P = V/24

$\implies$ Pipe Q = V/32

Flow rate through both pipes:

$\implies$ $\sf{\frac{V}{32} + \frac{V}{24}}$

$\implies$ $\sf{\frac{7V}{96}}$

Let the first pipe must turned off after x minutes.

Note: Rest of the volume, be filled by pipe Q for (16-x) minutes.

Volume filled by Q, for (16-x) minute:

$\implies$ $\boxed{\sf{\frac{V\times(16-x)}{32}}}$

So we have:

$\boxed{\sf{\frac{7Vx}{96} + \frac{V(16-x)}{32} = V}}$

$\implies$ $\sf{7x + 48 - 3x = 96}$

$\implies$ $\sf{4x = 48}$

$\implies$ $\sf{x = 12}$

Thus:

First pipe must be turned off after 12 minutes.

Answer 2

Answer:

I don't know please delete my answer