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’)