Tap P and Q can fill a tank in 7 and 10 hours, respectively. Pipe R can empty the same tank in 15 hours. If all the three pipes are opened simultaneously, then in how much time the tank will be filled completely?
- AOption (a)
\(4\frac{{37}}{{25}}\) hours
- BOption (b)
\(5\frac{{25}}{{37}}\) hours
- COption (c)
\(9\frac{{25}}{{37}}\) hours
- DOption (d)
\(11\frac{{37}}{{25}}\) hours
- EOption (e)
None of the above
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Correct answer
\(5\frac{{25}}{{37}}\) hours
Why this answer is correct
Shortcut Trick
Total Capacity (LCM of 7, 10, 15) = 210 units.
Efficiency of P = 210 ÷ 7 = 30, Efficiency of Q = 210 ÷ 10 = 21, Efficiency of R = −(210 ÷ 15) = −14.
Combined Efficiency (P + Q − R) = 30 + 21 − 14 = 37.
Time Taken = Total Capacity ÷ Combined Efficiency = 210 ÷ 37 = 525⁄37 hours.
∴ The correct answer is 525⁄37 hr.
Alternate Method
| Pipe | Time (hrs) | Total Work (LCM) | Efficiency |
|---|---|---|---|
| P (Inlet) | 7 | 210 | 30 |
| Q (Inlet) | 10 | 21 | |
| R (Outlet) | 15 | −14 |
⇒ Net efficiency when all pipes are open = Efficiency of P + Efficiency of Q + Efficiency of R
⇒ Net efficiency = 30 + 21 − 14 = 37 units/hr
⇒ Total time to fill the tank = Total Work ÷ Net efficiency
⇒ Total time = 210 ÷ 37
⇒ Total time = 525⁄37 hr
∴ The correct answer is 525⁄37 hr.
Additional Information
Inlet and Outlet Pipes
An inlet pipe fills the tank (positive work), while an outlet pipe (leak) empties it (negative work).
Efficiency Rule
Total Work = Time × Efficiency. If a tank is filled in 'x' hours and emptied in 'y' hours, net work in 1 hour = (1/x − 1/y).
Pipes & Cisterns Concept
This topic follows the same principles as Time and Work, treating the volume of the tank as the total work to be completed.