Two pipes fills a cistern in 15 hrs and 20 hrs respectively. The pipes are opened simultaneously and it is observed that it took 26 min more to fill the cistern because of leakage at the bottom. If the cistern is full, then in what time will the leak empty it?
Explanation:
Consider a pipe fills the tank in 'x' hrs. If there is a leakage in the bottom, the tank is filled in 'y' hrs.
| The time taken by the leak to empty the full tank = |
xy |
hrs |
| y – x |
But, direct values of x & y are not given. So, we need to find the work done by the two pipes in 1 hr = (1/15) + (1/20) = 7/60
Hence, the time taken by these pipes to fill the tank = 60/7 hrs = 8 .57 hrs = 8 hrs 34 min --------( by multiplying '0.57' hrs x 60 = 34 minutes)
Due to leakage, time taken = 8 hrs 34 min + 26 min = 8 hrs 60 min = 9 hrs ----------( because 60 min = 1 hr)
Thus, work done by (two pipes + leak) in 1 hr = 1/9
Hence, work done by leak in 1 hr = work done by two pipes – 1/9
= 7 /60 – 1/9 = 3/ 540 = 1/180
Therefore, leak will empty the full cistern in 180 hours.
Hence, option 4 is the correct answer.
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By: Sandeep Dubey