Question

Two pipes can fill a tank in 10 hours and 12 hours respectively while a third pipe empties the full tank in 20 hours.

If all the three pipes operate simultaneously, in how much time will the tank be filled?
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Answer 1

Solution ⇒

Formula is = xyz/yz+xz-xy

10*12*20/12*20+10*20-10*12

=2400/240+200-120

= 15/2

=7hrs 30mins  is answer

Answer 2

Time taken by pipe 1 = 10 hours

So, by pipe 1 in one hour, 1/10 th part of the tank is filled

Time taken by pipe 2 to fll the tank = 12 hours

So, by pipe 2 in one hour, 12 th part of the tank is filled.

Time taken by pipe 3 to empty the tank = 20 hours

So, by pipe 3 in one hour, 1/20 th part of the tank is filled.

When three pipes are operated simultaneously

Amount of water present in the tank per hour = water entered into the tank - water left from the tank

Amount of water present in the tank per hour = 1/10 + 1/12 - 1/20

                                                                            = 22/120 - 1/20

                                                                            = (22 - 6)/120

                                                                            = 16/120

                                                                            = 2/15

2/15 th part of the tank is filled in one hour

So,

total time taken to fill the tank by using 3 pipes is 15/2 hours = 7 h 30 min