Question

4 pipes can fill a reservoir in 15, 20, 30 and 60 hrs respectively. the first was opened at 6 am, second at 7 am, third at 8 am and the fourth at 9 am. when will the reservoir be filled up?

a. 1 am

b. 4 pm

c. 1 pm

d. 4 am

e. none of these

Answer 1

Lets consider situation where all pipes are running together,

They fill = (1/15) + (1/20) + (1/30) + (1/60)

= (4/60) + (3/60) + (2/60) + (1/60)

=10/60 = 1/6 of the tank per hour

From 6 to 7, 1/15 of the tank fills = 4/60

From 7 to 8, another 1/15 + 1/20 fills from the two pipes.

Now there is 4/60 (previously filled) + 4/60 + 3/60 = 11/60

From 8 to 9, another 1/15 + 1/20 + 1/30 fills from 3 pipes.

Now there is 11/60 (previously filled) + 4/60 + 3/60 + 2/60 = 20/60

So 20/60 is full now, now we have all pipes running together,

From 9 on, the rate is 1/6 = 10/60 per hour

Remaining tank left = 40/60

So to fill remaining, time taken will be = 40/60 ÷ 10/60 = 4 hrs

Reservoir takes 4 hrs from 9 am to fill up.

Therefore, Reservoir fills up at 1 pm: Option C

Answer 2

Solution :-

The 1st pipe can fill the reservoir in 15 hours. This means that it fills 1/15 of reservoir in 1 hour.

Similarly, 2nd pipe, 3rd pipe and 4th pipe fill 1/20, 1/30 and 1/60 of the reservoir in 1 hour respectively.

Let 't' be the number of hours taken by 1st pipe. The 2nd pipe opened at 7 AM, i.e. one hour later. Therefore it worked t - 1 hour.

Similarly, 3rd pipe and 4th pipe worked t - 2 hours and t - 3 hours respectively.

Together, all pipes fill the whole reservoir, so we obtain the following equation:

t/15 + (t - 1)/20 + (t - 2)/30 + (t - 3)/60 = 1

Taking LCM of the denominators and then solving it.

⇒ (4t + 3t - 3 + 2t - 4 + t - 3)/60 = 1

⇒ 10t - 10 = 60

⇒ 10t = 60 + 10

⇒ 10t = 70

⇒ t = 70/10

⇒ t = 7 hours

1st pipe was opened at 6 Am

So, 6 AM + 7 hours = 1 PM

The reservoir will be filled at 1 PM.

Answer.