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# Pipes and Cistern #1

 1. Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in: A. $4\dfrac{1}{2}$ hours B. $1\dfrac{13}{17}$ hours C. $3\dfrac{9}{17}$ hours D. $2\dfrac{8}{11}$ hours

Explanation:

Solution 1

Pipes A and B can fill the tank in 5 and 6 hours respectively. Therefore,
part filled by pipe A in 1 hour $=\dfrac{1}{5}$
part filled by pipe B in 1 hour = $\dfrac{1}{6}$

Pipe C can empty the tank in 12 hours. Therefore,
part emptied by pipe C in 1 hour $=\dfrac{1}{12}$

Net part filled by Pipes A,B,C together in 1 hour
$=\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{17}{60}$

i.e., the tank can be filled in $\dfrac{60}{17}=3\dfrac{9}{17}$ hours.

Solution 2

LCM$(5,6,12)=60$

Suppose capacity of the tank is $60$ litre. Then,
Quantity filled by pipe A in 1 hour $=\dfrac{60}{5}=12$ litre.
Quantity filled by pipe B in 1 hour $=\dfrac{60}{6}=10$ litre.
Quantity emptied by pipe C in 1 hour $=\dfrac{60}{12}=5$ litre.

Quantity filled in 1 hour if all the pipes are opened together
$=12+10-5=17$ litre.

Required time $=\dfrac{60}{17}=3\dfrac{9}{17}$ hours.