Efficiency comparison: A does 20% less work than B (i.e., A’s rate is 20% lower). If A can complete the job in 7.5 hours, in how many hours can B complete the same job?

Introduction / Context:
When one worker is a certain percentage less efficient than another, their times are inversely related to their efficiencies. A being 20% less efficient than B means B is faster. We use proportional reasoning to compute B’s time from A’s time.

Given Data / Assumptions:

• A's rate = 80% of B's rate (since 20% less).
• A’s time = 7.5 hours for the full job.
• Work amount is the same for both (1 job).

Concept / Approach:
Time is inversely proportional to rate. If r_A = 0.8 * r_B, then T_A = (1 / r_A) and T_B = (1 / r_B) = (1 / (r_A / 0.8)) = 0.8 * T_A. Thus we scale A’s time by 0.8 to get B’s time.

Step-by-Step Solution:
T_B = 0.8 * T_A = 0.8 * 7.5 = 6 hours.

Verification / Alternative check:
Relative rates: If B's rate = 1 unit/hour, A's rate = 0.8 unit/hour. Times are 1 and 1/0.8 = 1.25 units of time; ratio T_A : T_B = 1.25 : 1 = 5 : 4 ⇒ T_B = 4/5 * 7.5 = 6 hours.

Why Other Options Are Wrong:
4 h is too fast; 8 h and 10 h are too slow compared with A’s 7.5 h for a worker who is more efficient.

Common Pitfalls:
Treating “20% less work” as subtracting time rather than scaling rate; the correct relation is via inverse proportionality of time and rate.

Final Answer:
6 h