Efficiency comparison: A alone can do a job in 24 days. Worker B is 60% more efficient than A. How many days will B alone take to finish the same job?

Introduction / Context:
This tests efficiency-to-time conversion. If B is more efficient than A, B’s time is proportionally lower because time is inversely proportional to efficiency for the same amount of work.

Given Data / Assumptions:

• A’s time = 24 days ⇒ rate_A = 1/24 job/day.
• B is 60% more efficient than A ⇒ rate_B = 1.60 * rate_A.

Concept / Approach:
If efficiency increases by k%, multiply the base rate by (1 + k/100). Time = 1 / rate. So B’s time = A’s time / 1.60.

Step-by-Step Solution:
rate_A = 1/24rate_B = 1.6 * (1/24) = 1/15Time_B = 1 / (1/15) = 15 days

Verification / Alternative check:
Compare: 24 vs 15. A 60% boost in rate is a 37.5% reduction in time (since 9/24 = 37.5%), consistent with 24 → 15.

Why Other Options Are Wrong:

• 12 and 10: too optimistic; imply even higher efficiency.
• 9.6: corresponds to 150% higher efficiency, not 60%.

Common Pitfalls:

• Subtracting 60% from 24 (i.e., 24 − 60% of 24) which is wrong; time is not reduced linearly by the efficiency percentage.

Final Answer:
15