Question
Download Solution PDFTwo pipes S1 and S2 alone can fill an empty tank in 15 hours and 20 hours respectively. Pipe S3 alone can empty that completely filled tank in 40 hours. Firstly both pipes S1 and S2 are opened and after 2 hour pipe S3 is also opened. In how much time tank will be completely filled after S3 is opened?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
wo pipes S1 and S2 alone can fill an empty tank in 15 hours and 20 hours respectively.
Pipe S3 alone can empty that completely filled tank in 40 hours.
Firstly both pipes S1 and S2 are opened and after 2 hour pipe S3 is also opened.
Concept used:
Efficiency = (Total work / Total time taken)
Efficiency = work done in a single day
Calculation:
Let total work is 120 units ( LCM of 15 , 20 and 40)
The efficiency of pipe s1 is 120 /15 = 8 units
The efficiency of pipe s2 is 120 / 20 = 6 units
The efficiency of pipe S1 + S2 is 8 + 6= 14 units
In 2 hrs they will fill 14 × 2 = 28 units
Remaining amount 120 - 28 = 92 units
The efficiency of pipe s3 is 120 / 40 = - 3 units
Now combined efficiency of all three pipes is 14 - 3 = 11 units
Time to fill remaining 92 units is
⇒ 92 / 11 hr
∴ The correct option is 4
Last updated on Jul 21, 2025
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