How to Solve Pipes Opened Alternately Problems
Solve pipes-opened-alternately aptitude problems using cycle-based work accounting, with a worked example and practice questions.
Expected Interview Answer
When pipes fill a cistern on alternating turns rather than simultaneously, find the total work done in one complete cycle of turns, then determine how many full cycles are needed and add the fractional time from the final partial cycle separately.
Unlike simultaneous filling, alternate opening means only one pipe runs per turn (commonly one hour or one specified unit each), so the one-hour-work fractions do not simply add continuously — they add per completed cycle. Compute the work done by each pipe in its own turn, sum them for one full cycle, and divide the total job (1 unit) by that cycle sum to see how many whole cycles complete before the job is nearly done. After the last whole cycle, track the remaining fractional work turn-by-turn (not cycle-by-cycle) since the job may finish partway through the next pipe’s turn, which requires solving for the exact fraction of that final turn needed.
- Cycle-based accounting avoids the error of treating alternation as simultaneous filling
- Separates whole-cycle progress from the final partial-turn calculation
- Generalizes to any number of pipes and any turn-length pattern
AI Mentor Explanation
Two bowlers alternating overs, each contributing a fixed number of wicket-equivalents per over rather than bowling simultaneously, only add up their combined progress once every full two-over cycle — you cannot simply add both bowlers' rates as if they bowled together. Sum each bowler's contribution across one full cycle of alternating overs, find how many cycles clear the required total, then handle the final partial over separately since the target may be reached mid-over rather than at a clean cycle boundary — exactly the accounting alternate-pipe problems require.
Worked example
One cycle (A then B)
- 1/6 + 1/4 = 5/12
After 2 cycles (4 hrs)
- 5/6 done, 1/6 left
Final turn (A, hour 5)
- 1/6 remaining = A's full hour
- Total = 5 hours
Step-by-Step Explanation
Step 1
Find each pipe's per-turn work
One-turn-work = 1 ÷ (hours that pipe takes alone), assuming equal turn lengths.
Step 2
Sum one full cycle
Add every pipe's per-turn work once, in turn order, to get total progress per cycle.
Step 3
Count whole cycles
Divide 1 (the whole job) by the cycle sum; take the integer part as complete cycles.
Step 4
Resolve the final partial turn
Track remaining work turn-by-turn after the last whole cycle until it reaches zero.
What Interviewer Expects
- Correct recognition that alternation is not simultaneous filling
- Accurate per-turn work computed for each pipe individually
- Correct cycle-sum accounting for whole cycles
- Correct turn-by-turn resolution of the final partial cycle
Common Mistakes
- Adding all pipes' rates as if they worked simultaneously
- Losing track of whose turn it is when resolving the final partial cycle
- Assuming the job finishes exactly at a cycle boundary without checking
- Using unequal turn lengths without adjusting each pipe's per-turn work accordingly
Best Answer (HR Friendly)
“The key is that alternating turns is not the same as running pipes together — I only add up progress once per complete cycle of turns, not continuously. I find how much each pipe does in its own turn, sum one full cycle, and see how many whole cycles get close to finishing the job. Then I switch to tracking turn-by-turn for whatever is left, because the job usually finishes partway through someone's turn, not neatly at the end of a cycle.”
Follow-up Questions
- How does the method change if the pipes alternate every half hour instead of every hour?
- How would you solve this with three pipes alternating in a fixed rotation?
- What if a leak is also present during the alternation?
- How do you verify which pipe's turn corresponds to the exact moment the job finishes?
MCQ Practice
1. Pipe A fills a tank in 4 hours, Pipe B in 12 hours. Opened alternately for 1 hour each, starting with A, how much is filled after one full cycle (2 hours)?
One cycle = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3.
2. Using the same pipes as above (A: 4 hrs, B: 12 hrs, alternating 1-hour turns starting with A), how many full cycles are needed before the tank is completely full?
Each cycle fills 1/3, so 3 cycles fill exactly 3 × 1/3 = 1, completing the tank in exactly 6 hours with no partial turn.
3. In alternate-opening problems, why can't you just add each pipe's hourly rate directly as in simultaneous-pipe problems?
Alternating pipes are never running at the same time, so the combined rate only applies once a full cycle of turns is summed.
Flash Cards
Key difference from simultaneous-pipe problems? — Only one pipe runs per turn, so rates add per completed cycle, not continuously.
How to find progress in one cycle? — Sum each pipe's per-turn work once, in the given turn order.
How to find whole cycles needed? — Divide 1 (whole job) by the cycle sum; take the integer part.
How to handle the leftover work? — Track it turn-by-turn after the last whole cycle until the job completes.