Pipes A and B can empty a full tank in 5 hours and 12 hours respectively. Pipe C can fill the same empty tank in 2 hours. If all three pipes are opened together, then the tank will be filled in:

Correct option is C

Given:

Pipe A empties tank in 5 hours

Pipe B empties tank in 12 hours 

Pipe C fills tank in 2 hours 

Solution:

Work rate of  A =$$-\frac{1}{5}$$​ tank/hour

Work rate of B = $$-\frac{1}{12}$$​ tank/hour

Work rate of C = $$\frac{1}{2}$$​ tank/hour

Net rate = $$\frac{1}{2} - \frac{1}{5} - \frac{1}{12}$$​​

​$$= \frac{30}{60} - \frac{12}{60} - \frac{5}{60}$$​​

​$$= \frac{30 - 12 - 5}{60} = \frac{13}{60}$$​​

Time to fill tank:

​$$= \frac{1}{\text{Net rate}} = \frac{60}{13} \, \text{hours}$$​​

Alternate Solution: ​

Let total work = LCM of 5, 12, and 2 = 60 units (one full tank).

Pipe A rate =$$\frac{60}{5}$$​ = 12 units/hour (emptying means negative work)

Pipe B rate = $$\frac{60}{12}$$​ = 5 units/hour (emptying)

Pipe C rate = $$\frac{60}{2}$$​ = 30 units/hour (filling)

Net work done per hour when all pipes open:

= 30 - 12 - 5 = 13 units/hour

Time to fill the tank = $$\frac{60}{13} \text{ hours}$$​​