Pipes A and B together can fill a cistern in 10 hours. Pipes B and C together can fill the same cistern in 20 hours, and pipes C and A together can fill it in 12 hours. Find the time taken by pipe B alone to fill the cistern.
- AOption (a)
24 hours
- BOption (b)
25 hours
- COption (c)
28 hours
- DOption (d)
30 hours
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Correct answer
30 hours
Why this answer is correct
Shortcut Trick
Let the total capacity of the cistern be the LCM of (10, 20, 12) = 60 units.
Combined Efficiency: (A+B) = 6, (B+C) = 3, (C+A) = 5.
Summing them: 2(A+B+C) = 6 + 3 + 5 = 14 ⇒ A+B+C = 7 units/hr.
Efficiency of Pipe B = (A+B+C) − (C+A) = 7 − 5 = 2 units/hr.
Time taken by B = 60 ÷ 2 = 30 hr.
∴ The correct answer is 30 hr.
Alternate Method
Given:
Time taken by (Pipe A + Pipe B) = 10 hr
Time taken by (Pipe B + Pipe C) = 20 hr
Time taken by (Pipe C + Pipe A) = 12 hr
Formula Used:
Efficiency = Total Work ÷ Time taken
| Pipes | Time (hr) | Total Work (LCM) | Efficiency (units/hr) |
|---|---|---|---|
| A + B | 10 | 60 | 60 ÷ 10 = 6 |
| B + C | 20 | 60 ÷ 20 = 3 | |
| C + A | 12 | 60 ÷ 12 = 5 |
Calculations:
⇒ Adding efficiencies: (A + B) + (B + C) + (C + A) = 6 + 3 + 5
⇒ 2(A + B + C) = 14
⇒ A + B + C = 14 ÷ 2 = 7 units/hr
⇒ We know efficiency of (C + A) = 5 units/hr
⇒ Efficiency of B = (A + B + C) − (C + A) = 7 − 5 = 2 units/hr
⇒ Time taken by Pipe B alone = Total Work ÷ Efficiency of B
⇒ Time = 60 ÷ 2 = 30 hr
∴ The correct answer is 30 hr.
Additional Information
Combined Efficiency Rule
When pipes work in pairs, the sum of their individual efficiencies is half the sum of their pairwise efficiencies: (A+B+C) = [(A+B) + (B+C) + (C+A)] ÷ 2.
Pipes and Cisterns Basics
Inlet pipes perform positive work (filling), while outlet pipes perform negative work (emptying). Total work is the volume of the tank/cistern.
Time and Work Relation
Time taken and efficiency are inversely proportional to each other. Time = 1 ÷ Efficiency (for constant work).