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

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.

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: (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: (c)

Part of the tank filled by both pipes in one minute

= $1/20 + 1/30$

Required time = $1/{1/20 + 1/30}$

= ${20 × 30}/50$ = 12 minutes

Here, x = 20, y = 30

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

= $({20 × 30}/{20 + 30})$ minutes = 12 minutes.

Answer: (a)

Using Rule 1,

Part of tank filled by pipes A and B in 2 hours

= $2(1/6 + 1/8)$

=$2({4 + 3}/24) = 7/12$

Remaining part = $1 - 7/12 = 5/12$

This part is filled by pipe B.

Required time = $5/12$ × 8

= $10/3$ hours = 3$1/3$ hours

Answer: (b)

Using Rule 1,

Part of tank filled by pipes A and B in 1 hour

= $1/2 + 1/3 = {3 + 2}/6 = 5/6$ parts

Required time = $6/5$ hours

= 1 hour $1/5$ × 60

= 1 hour 12 minutes