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)?

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:

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