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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 |

Answer: Option C

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.

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