Quantitative Aptitude Pipes and Cistern Study Material

Quantitative Aptitude Pipes and Cistern Study Material

Quantitative Aptitude Pipes and Cistern ( Study Material)

Pipes and Cistern problems usually consist of a cistern or tank to which one or more pipes are connected to fill or/ and empty it. These are similar to Time and Work problems. Here work done is filling/emptying the cistern or tank. And, this work is done by pipes-connected to the tank in place of persons doing the work. The only difference lies in the fact that here work can be positive (filling the tank) as well as negative (emptying the tank) whereas in Time and Work problems, the work done is always positive.

A pipe connected to the tank which fills it is called an Inlet. If the pipe connected to the tank empties it, it is called an Outlet or exit pipe

General Formulae:
(i) If a pipe’ can fill a tank in .v hours, the part of tank filled in 1 hour = 1/x
(ii) If a pipe can fill a tank in x hours and another pipe can fill the same tank in y hours, the part of the tank filled in 1 hour when both pipes are open simultaneously = (1/x+1/y) = x+y/xy

Time taken to fill completely the tank when both pipes are open simultaneously = xy/x+y

(iii) If a pipe can empty a tank in y hours, the part of the full tank emptied in 1 hour = 1/y
(iv) If a pipe can empty a tank in x hours and another pipe can empty the same tank in y hours, the part of the tank exptied in 1 hour when both pipes are open together, (1/x+1/y) = x+y/xy
Time taken to empty the tank completely when both pipes are open together = xy/x+y/xy

(v) If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours, the net part filled in 1 hour when both pipes are open simultaneously
= (1/x-1/y) = y-x/xy

When both pipes are open simultaneouly, time to fill the tank fully = xy/y-x hours

Note: If the result comes out to be negative, the net effect will be emptying of the tank instead of filling it.

(vi) A pipe can fill a tank in x hours. But due to a leak in its bottom, the tank is now filled in y hours. Then the time taken by the leak to empty the full tank = xy/y-x

(vii) A pipe can fill a tank in x hours and another fills it in y hours. A third pipe can empty the full tank in ; hours. When all the three pipes are open simultaneously, the net part of the tank filled in 1 hour = 1/x+1/y-1/z = yz+xz-xy/xyz

When all three pipes are open, time taken to fill the tank = xyz/yz+xz-xy hours

Note: If the result comes out to be negative, the net effect will be emptying of the tank instead of filling it.

We can write many such formulae for different cases. The sign convention to the remembered is, positive for filling the tank and negative for emptying it.

Ex, 1. A pipe can fill a tank at the rate of 16 litres per hr. In how much time can it fill 1/4 th of a tank of capacity 160 litres?
Sol. 1/4 th of 160 litres = 1/4 x 160 litres = 40 litres
Time taken to fill 40 litres
= 40/16 = 5/2 = 2(1/2) hours

Ex. 2. A pipe can discharge water from a tank at the rate of 10 litres per hour. In how much time can it empty half of the tank which has 40 litres of water?
Sol. 1/2 of 40 litres = 20 litres Time taken to discharge 20 litres
= 20/10 = 2 hours

Ex. 3. Two pipes A and B can fill’a tank in 24 hours and 30 hours respectively. If both pipes are open simultaneously, how much time will they take to fill the tank ?
Sol. Part of the tank filled by A alone in 1 hr. = 1/24
Part of the tank filled by B alone in 1 hr. = 1/30
Part of the tank filled by (A + B) together in 1 hr.
= 1/24+1/30 = 5+4/120 = 9/120 = 3/40
.•. (A + B) together can fill the tank in 40/3
= 13 1/3 hrs

Ex. 4. A pipe can empty a tank in 15 hours. Another pipe can empty the tank in 10 hours. In how much time will the completely filled tank get emptied, if both the pipes are open together?
Sol. This problem is similar to Ex. (3) except that the two pipes are emptying the tank instead of filling it. That’s why the formula in both the cases is same.
Using the Direct Formula (iv),
Time taken to empty the tank when both the pipes are open together xy/x+y = 15×10/15+10 = 150/25 = 6 hours

Ex. 5. A pipe can fill a tank in 20 hours. Due to a leak in the bottom of the tank, it is filled in 25 hours. If the tank is full, how much time will the leak take to empty it completely?
Sol. Part of tank filled, without leakage, in 1 hour = 1/20
Part of the tank filled, with leakage, in 1 hour = 1/25
Part of the tank emptied, only due to leakage, in 1 hour = 1/20-1/25 = 5-4/100 =1/100
The leakage will empty the full tank in = 1/1/100 = 100 hours

Ex. 7. Pipes A and B can fill a tank alone in 12 and 15 hours respectively. Pipe C can empty the full tank in 18 hours. If all the three pipes are open simultaneously, how much time it will take them to fill the tank completely?
Sol. Net part of the tank filled in 1 hour, when all three pipes are open
1/12+1/15-1/18 = 15+12-10/180 = 17/180
Hence, the tank will be full in 180/17 hours
= 10(10/17) hours

Quantitative Aptitude Pipes and Cistern ( Study Material)