Alternate minutes starting with the faster pipe Two pipes X and Y fill a cistern in 6 min and 7 min. Starting with X, they are opened alternately for 1 minute each. In what time will the cistern be filled (stop immediately when full)?

Introduction / Context:
Alternate-minute problems mirror alternate-hour logic but with smaller time quanta. Compute 2-minute cycle gain, then apply a final fractional minute of the next (faster) inlet if necessary.

Given Data / Assumptions:

• X: 6 min ⇒ 1/6 per min; Y: 7 min ⇒ 1/7 per min.
• Order: X then Y, repeating; stop immediately when full.

Concept / Approach:
Per 2-min cycle, fill = 1/6 + 1/7 = 13/42. After an integer number of cycles, the remainder is filled by the next minute of X.

Step-by-Step Solution:

Verification / Alternative check:
In the last partial minute, we stop the instant the tank becomes full; we do not start Y again.

Why Other Options Are Wrong:
62/7, 63/7, 65/7 assume whole extra minutes or miscount cycles.

Common Pitfalls:
Forcing whole-minute boundaries instead of stopping at the exact completion time.

Final Answer:
45/7 min