Model 1 Basic Pipes & Cisterns problems Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Part of the tank filled in 1 hour = $1/x$

Part of the tank emptied in 1 hour = $1/y$

Part of the tank filled in 1 hour

when both are opened

= $1/x - 1/y = {y - x}/{xy}$

Tank will be filled in ${xy}/{y - x}$ hours.

Answer: (b)

1 hour = 60 minutes.

Rate of emptying the tank by the two taps are $1/60$ and $1/30$ of the tank per minute respectively.

Rate of emptying the tank when both operate simultaneously

= $1/60 + 1/30 = {1 + 2}/60 = 3/60 = 1/20$

of the tank per minute.

Time taken by the two taps together to empty the tank = 20 minutes

Here, x = 60, y = 30

Required time = $({xy}/{x + y})$ minutes

= $({60 × 30}/{60 + 30})$ minutes = 20 minutes.

Answer: (d)

Part of the tank filled in 1 hour by pipe A = $1/2$

Part of the tank filled by both pipes in1 hour

= $1/2 + 1/6 = {3 + 1}/6 = 2/3$

So, Time taken to fill $2/3$ part = 60 minutes

Time taken to fill $1/2$ part

= ${60 × 3}/2 × 1/2$ = 45 minutes

The tank will be filled at 11:45 A.M.

Answer: (a)

Part of cistern filled by three pipes in an hour

= $1/3 + 1/5 - 1/2 = {10 + 6 - 15}/30 = 1/30$

Hence, the cistern will be filled in 30 hours.