Question

Two taps A and B can separately fill a cistern
in 24 minute and 30 minute respectively. Both
the pipes are opened together. Find when the
pipe B must be turned off so that the cistern
may be full in 18 minute.​

Answer 1

Step-by-step explanation:

A can fill in 24 mins

so ,in 18 mins=(1/24)×18=3/4 part will be filled by A

So ,B can fill in (1/4 )×30=7.5=7mins 30 secs

Answer 2

Given :

Time taken by tap A = 24 minutes

Time taken by tap B = 30 minutes

If B is turned off, the cistern may be full in 18 minutes.

To find : the time when the pipe B is turned off.

Solution :

Total units = LCM of 24 and 30 = 120 units

Work done by tap A = $\dfrac{120}{24}=5$

It is opened for 18 minutes,

So, work done in 18 minutes = $5\times 18=90\ units$

Work done by tap B = $\dfrac{120}{30}=4\ units$

According to question, we get that :

Remaining units = $120-90=30\ units$

So, pipe B is turned off in $\dfrac{30}{4}=7\dfrac{1}{2}\ minutes$