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- Two taps P and Q fill a cistern in 2 days and 3 days respectively and another tap R empties the full cistern in 4 days.
Question 1:
Two taps P and Q fill a cistern in 2 days and 3 days respectively and another tap R empties the full cistern in 4 days.If all the taps P,Q and R switched simultaneously then the time taken to fill the cistern is:
A. 1day + 120/7 hours
Feedback The tap P can fill a cistern in 1 hour = 1/48 part
The tap Q can fill a cistern in 1 hour = 1/72 part
The tap R can empty a cistern in 1 hour = 1/96 part
Then the net part filled in 1 hour = 1/48 + 1/72 - 1/96
= 7/288
Tank Filled Time Taken
7/288 1
1 ?
The cistern will be filled in = 288/7
288 / 7 can be expressed as = (24 x 7 + 120) / 7 hours = 24 120/7 hours.
i.e., 1 day and 120 / 7 hours
The tap Q can fill a cistern in 1 hour = 1/72 part
The tap R can empty a cistern in 1 hour = 1/96 part
Then the net part filled in 1 hour = 1/48 + 1/72 - 1/96
= 7/288
Tank Filled Time Taken
7/288 1
1 ?
The cistern will be filled in = 288/7
288 / 7 can be expressed as = (24 x 7 + 120) / 7 hours = 24 120/7 hours.
i.e., 1 day and 120 / 7 hours
B. 2days + 12/7hours
C. 3days + 24/7hours
D. None of these
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