Alternate hours starting from A Two inlets A and B can fill the tank in 16 h and 10 h. They are opened alternately for 1 hour each, starting with A. How long will it take to fill the tank (stop immediately when full)?

Difficulty: Medium

Correct Answer: 62/5 h

Explanation:


Introduction / Context:
Alternate-hour problems are best handled in 2-hour cycles (one hour of A, one hour of B). After a whole number of cycles, a final fractional hour of the next tap may be required to finish the small remainder.



Given Data / Assumptions:

  • A: 16 h ⇒ 1/16 per hour; B: 10 h ⇒ 1/10 per hour.
  • Order: A then B, repeating; stop as soon as the tank reaches full.


Concept / Approach:
Per 2-hour cycle, net fill = 1/16 + 1/10. Accumulate cycles to just below 1 tank, then finish with a partial hour of A.



Step-by-Step Solution:

2-hour gain = 1/16 + 1/10 = (5 + 8)/80 = 13/80After 12 hours (6 cycles): 6 * 13/80 = 78/80 = 39/40Remaining = 1/40, next in order is ATime for A = (1/40) / (1/16) = 16/40 = 2/5 h = 24 minTotal time = 12 + 2/5 = 62/5 h (12 h 24 min)


Verification / Alternative check:
Check contributions: 12 h give 39/40; last 24 min at 1/16 adds 1/40.



Why Other Options Are Wrong:
13 h or 14.5 h ignore the exact fractional finish; 12 h 24 min equals 62/5 h (both are correct renderings).



Common Pitfalls:
Stopping at a whole cycle and forgetting the final fraction with the correct next tap.



Final Answer:
62/5 h

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