# QUESTIONS ON PIPES & CISTERNS (PART-II)

## QUESTIONS ON PIPES & CISTERNS (PART-II)

#### QUERY 11

**Tap A can fill a tank in 10 minutes and tap B can empty it in 12 minutes. If both the taps are opened alternatively in interval of 1 minute, how much time the tank will take to be filled completely.**

A) 1 hour 49 minutes

B) 1 hour 35 minutes

C) 2 hours

D) 1 hour 48 minutes

**MAHA GUPTA**

Both the taps are kept open alternatively for 1 minute each, we’ll find the part of the tank filled in 2 minutes i.e. one interval of time; so it is ^{1}⁄_{10} – ^{1}⁄_{12} = 1/60

It’s obvious that the tap-B which is emptying the tank will not have to be opened in the last interval, otherwise the tank will never fill; so we need to know the work of the tap-A in the last 1 minute.

So, work of tap-A in 1 minute = 1/10

It does mean both the taps work simultaneously to fill = 1 – ^{1}⁄_{10} = ^{9}⁄_{10} of the tank.

Time taken to fill 1/60 of the tank by both the taps working together = 2 minutes

Hence, time taken in all to fill 9/10 of the tank = (2*60) × ^{9}⁄_{10} = 108 minutes

So, total time to fill the full tank = Time taken for ^{9}⁄_{10} of the part + time taken for ^{1}⁄_{10} of the part

= 108 + 1 = 109 minutes = 1 hour 49 minutes (option ‘A’)