Question

Three pipes a, b, c can fil an empty cistern in 2, 3 and 6 hours respectively. They are opened together. After what time should b be closed, so that the cistern gets filled in exactly 1 hr 15 min?

Answer 1

Three pipes a, b, c can fill an empty cistern in 2, 3 and 6 hours respectively.

Let pipe b is closed after t minutes.

So, pipes a, c are working for entire 1 hour and 15 minutes.

Given

⇒a can fill the cistern in 120 minutes

In one minute, It would fill 1/120th part of cistern.

⇒c can fill the cistern in 360 minutes.

In one minute, It would fill 1/360th part of cistern.

In 1 hour 15 minutes, i.e 75 minutes,

Pipe A fills 75 * 1/120 part of cistern

Pipe B fills 75 * 1/360 part of cistern.

In the given time, Part of cistern filled by a & c is,

$= \frac{75}{120} + \frac{75}{360} \\ \\ = \frac{75}{120} (1 + \frac{1}{3} ) \\ \\ = \frac{5}{8} \times \frac{4}{3} \\ \\ = \frac{5}{6}$

So, a, c fill 5/6th portion of cistern in 75 minutes. So, b must fill the remaining portion of cistern.

⇒Remaining portion of cistern = 1 - 5/6 = 1/6

Given, b can fill an empty cistern in 180 minutes.

It can fill 1/180th portion in one minute.

Let b is stopped after t minutes,

Then portion of cistern filled by b = t * 1/180

⇒Pipe A, C have already filled 5/6th portion of cistern. So b must fill 1/6th portion.

$\frac{1}{6} = \frac{t}{180} \\ \\ \: t = \frac{180}{6} \\ \\ t = 30 \: mins$

Therefore, The pipe b must be closed after 30 minutes.