Pipe A can fill a tank in 17 hours, pipe B can fill the same tank in 21 hours and pipe C can fill the same tank in 19 hours. The time taken by them to fill the same tank if they operate together is::
- AOption (a)
\(13\frac{309}{1079}\) hours
- BOption (b)
\(5\frac{309}{1079}\) hours
- COption (c)
\(6\frac{309}{1079}\) hours
- DOption (d)
\(11\frac{309}{1079}\) hours
View answer & explanationHide answer & explanation
Correct answer
\(6\frac{309}{1079}\) hours
Why this answer is correct
Shortcut Trick
Let total work be the LCM of 17, 21, and 19 = 6783 units.
Combined efficiency = (6783 ÷ 17) + (6783 ÷ 21) + (6783 ÷ 19) = 399 + 323 + 357 = 1079 units/hour.
Time required when working together = Total Work ÷ Combined Efficiency.
⇒ Time = 6783 ÷ 1079 = 6 309/1079 hours.
∴ The correct answer is 6 309/1079 hours.
Alternate Method
Given:
Time taken by pipe A = 17 hours
Time taken by pipe B = 21 hours
Time taken by pipe C = 19 hours
Formula Used:
Total Work = LCM of times taken by all individual pipes.
Efficiency = Total Work ÷ Time Taken
| Pipe | Hours | Total Work | Efficiency |
|---|---|---|---|
| A | 17 | 6783 | 399 |
| B | 21 | 323 | |
| C | 19 | 357 |
Calculations:
⇒ Total Efficiency of pipes A, B, and C together = 399 + 323 + 357 = 1079 units/hour
⇒ Total Time taken together = Total Work ÷ Total Efficiency
⇒ Time = 6783 ÷ 1079
⇒ Time = 6 309/1079 hours
∴ The correct answer is 6 309/1079 hours.
Additional Information
Pipes and Cisterns
Inlet pipes fill the tank (assigned positive efficiency), while outlet pipes empty it (assigned negative efficiency). Net efficiency is the algebraic sum of all actively working pipes.
Concept of Efficiency
Efficiency is inversely proportional to the time taken. For any given setup, Work = Efficiency × Time.
Work Equivalence
If multiple independent units contribute to a shared objective, their per unit outputs (1/TA + 1/TB + 1/TC) directly sum up to equal the net output rate (1/TTotal).