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Q.

A cylindrical overhead tank is filled by two pumps $P_1$ and $P_2$. $P_1$ can fill the tank in 8 hr while $P_2$ can fill the tank in 12 hr. There is a pipe $P_3$ which can empty the tank in 8 hr. Both the pumps are opened simultaneously. The supervisor of the tank, before going out on a work, sets a timer to open $P_3$ when the tank is half filled so that the tank is exactly filled up by the time he is back. Due to technical fault $P_3$ opens when the tank is one-third filled. If the supervisor comes back as per the plan what percent of the tank is till empty?

 A.

$15\%$

 B.

$12\%$

 C.

$10\%$

 D.

None of these

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Solution:
Option(C) is correct

$P_1$ and $P_2$ can fill the tank in \(\dfrac{24}{5}\) hr. [In one hour these fill \(\left(\dfrac{1}{8}+\dfrac{1}{12}\right)\)part of tank].

It takes \(\dfrac{12}{5}\) hr in filling half the tank.

For remaining half of the tank $P_3$ will open and this will takes 6 hour.

Supervisor has gone out for \(\left(\dfrac{12}{5}+6\right)\) hr.

Now, \(\dfrac{1}{3}^{rd}\) tank will fill in \(\dfrac{8}{5}^{rd}\)  hr

In remaining \(\dfrac{42}{5}\) hr only \(\dfrac{33}{60}^{th}\) part of the tank will fill.

Empties part of tank

\(=1-\left(\dfrac{1}{33}+\dfrac{33}{60}\right)\)

\(=\dfrac{1}{10}\)

Which is 10% of tank.


(4) Comment(s)


ARUN V
 ()

sorry for typo. it is 34/60 *



ARUN V
 ()

in remaining 34/5 hrs only 32/60 tank is filled. then empty tank is 1-(1/3 + 32/60) .i.e. 6/60 = 1/10. 10%



ARUN V
 ()

its not 1/33 it should be 1/3



RAVI KUMAR
 ()

HOW IS 33/60 CAME IN ABOVE SOLUTION