How To Solve Pipes & Cisterns Questions Quickly

Type 3: Calculate Time Taken When Pipes Are Opened For Different Periods

Question 1: Two taps T1 and T2 can fill a bucket in 10 minutes and 15 minutes. Both the taps are opened together. However, after 2 minutes, tap T1 is closed. What is the total time required to fill the bucket?

Options:

A. 11 min and 20 sec

B. 15 min and 20 sec

C. 13 min and 40 sec

D. 15 min and 40 sec

Solution:    Part filled in 2 min = 2 * ( \frac{1}{10} + \frac{1}{15}) = 2 * \frac{5}{30}= \frac{5}{15}

Remaining part = (1- \frac{1}{3}) = \frac{2}{3}

Now, part filled by tap T2 = \frac{1}{20}

Therefore, \frac{1}{20}: \frac{2}{3}:: 1 : x

On solving we get x = \frac{40}{3} (13 min 20 sec)

Therefore, the bucket will be filled in = 2 minutes + 13 minutes + 20seconds = 15 min and 20 sec

Correct option: B

Question 2. A pond can be filled by two pipes P1 and P2. P1 takes 3 hours and P2 takes 5 hours. There is another pipe P3 attached to the pond which can empty the pond in 7 hours. If all the pipes are opened at the same time then in how much time will the empty pond be fully filled?

Options:

A. \frac{105}{41}hrs

B. 41 hrs

C. \frac{134}{43} hrs

D. \frac{122}{13} hrs

Solution:    Time taken by Pipe P1 to fill the tank = 3 hr

Time taken by pipe P2 to fill the tank = 5 hr

Time taken by pipe P3 to empty the tank = 7 hr

Therefore, pipe P1 and P2 fills \frac{1}{3} th and \frac{1}{5}th part of the pond in 1 hour respectively.

Part of the pond emptied in 1 hour by pipe P3 = \frac{1}{7}

Therefore, in 1 hour part of the pond filled is = ( \frac{1}{3} + \frac{1}{5} – \frac{1}{7}) = \frac{41}{105}

Therefore, tank will be filled completely in \frac{105}{41} hrs

Correct option: A

Question 3.Pipe M takes 12, pipe N takes 16, and Pipe O takes 20 minutes to fill a water container. Pipe M was opened first. After 2 minute, pipe N was opened. After 2 minutes from the start of N, pipe O was also opened. Find the time when water container will be full after opening of pipe O?

Options:

A. \frac{321}{47} min

B. \frac{311}{47} min

C. \frac{1621}{47} min

D. \frac{321}{22} min

Solution:    Let water container get full in x min.

Then part filled by pipe M in x min + part filled by pipe N in (x-2) min + part filled by pipe O in (x-4) min = 1

\frac{x}{12} + \frac{x – 2 }{16} + \frac{x – 4 }{20} = 1

x = \frac{162}{47} = \frac{321}{47} min

Correct option: A