Practice | Hashtable

Home > Practice > Arithmetic Aptitude > Time and Work > Miscellaneous

1. Three pipes A,B and C attached to a cistern. A can fill it in 10 min, B in 15 min, C is a waste pipe for emptying it. After opening both the pipes A and B, a man leaves the cistern and returns when the cistern should have been just full. Finding, however, that the waste pipe had left open, he closes it and the cistern now gets filled in 2 min. In how much time the pipe C, if opened alone, empty the full cistern?

A. 12 min

B. 16 min

C. 18 min

D. 15 min

Solution:

Let pipe C alone can empty the cistern in x min.
A fills cistern in 1 min =

\frac{1}{10}

B fills cistern in 1 min =

\frac{1}{15}

A and B together fill in 1 min

= \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6 min

Since, waste pipe was left open for 6 min then,
6 min,

\frac{6}{x}

part of cistern will be emptied by C
Now,

\frac{6}{x}

part of the cistern would be filled by A and B in 2 min.
Hence,
cistern will be filled in

\frac{3}{x}

min. And

\frac{x}{3}

= 6
x = 18 min.

You must login to add comments. Login now.