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:
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
Discussion & Comments