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Mathematics
Pipes and Cisterns
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NTPC Graduate Tier 1 2025 Shift-1 📅 06 May, 2025

Pipe A can fill a tank in 18 minutes, while pipe B can empty the completely filled tank in 20 minutes. Initially, pipe A is opened and after 6 minutes pipe B is also opened. In how much time (in minutes) will the remaining tank be filled completely?

A
120
B
137
C
107
D
127
Result Summary
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NTPC Graduate Tier 1
2025 • 06 May, 2025 • Shift-1
Pipe A can fill a tank in 18 minutes, while pipe B can empty the completely filled tank in 20 minutes. Initially, pipe A is opened and after 6 minutes pipe B is also opened. In how much time (in minutes) will the remaining tank be filled completely?
Correct Answer
120
Tank Capacity and Base Efficiency: To establish a workable numerical framework, let the total capacity of the target tank equal the Least Common Multiple (LCM) ......
Ref/Book : NTPC Graduate Tier 1 2025
Ref. Subject Mathematics
Chapter -
Ref. Topic Pipes and Cisterns
Pg. No -
💡 Analysis & Explanation
Tank Capacity and Base Efficiency ▼
To establish a workable numerical framework, let the total capacity of the target tank equal the Least Common Multiple (LCM) of the given timeframes, 18 and 20, which is precisely 180 units. Under this scale, Pipe A operates at a positive filling rate of 10 units/min, while Pipe B operates at a negative draining rate of -9 units/min.
Initial Solitary Operation Phase ▼
Pipe A runs completely solo for the first 6 minutes of the operation, successfully and efficiently filling a subtotal of 6 × 10 = 60 units of the tank's volume.
Calculating the Remaining Task ▼
The leftover capacity required to achieve a completely full tank is naturally 180 - 60 = 120 units.
Combined Operational Dynamics ▼
The moment Pipe B is activated alongside Pipe A, the net effective filling rate plummets to a mere (10 - 9) = 1 unit/min.
Conclusion ▼
Consequently, to fill the residual 120 units at this highly restricted combined rate of 1 unit/min, it will necessitate an additional duration of exactly 120 minutes.
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