Relative Speeds and Combined Duration A takes only 4/5 as many days as B takes to do the same work. Working together, they finish the job in 20/3 days. What is B’s efficiency (as a percentage of the whole job per day)?

Introduction / Context:
“A takes 4/5 as many days as B” means A is faster. We use the relationship between times and rates to find B’s per-day efficiency, expressed as a percent of one full job.

Given Data / Assumptions:

• T_A = 4/5 * T_B.
• Working together time = 20/3 days ⇒ combined rate = 3/20 job/day.

Concept / Approach:
Rates: r_A = 1/T_A and r_B = 1/T_B. Then r_A + r_B = 3/20. Substitute T_A = (4/5)T_B to solve for T_B and r_B, then convert to percent.

Step-by-Step Solution:
1/((4/5)T_B) + 1/T_B = 3/20(5/4)*(1/T_B) + 1/T_B = (9/4)*(1/T_B) = 3/201/T_B = (3/20)*(4/9) = 12/180 = 1/15 job/dayB’s efficiency = (1/15)*100% = 6 2/3%.

Verification / Alternative check:
Then A’s rate is 1/T_A = 1/((4/5)T_B) = 5/(4T_B) = 5/60 = 1/12; sum 1/12 + 1/15 = 3/20, consistent.

Why Other Options Are Wrong:
16%, 10%, 8 1/3%, and 5.55% do not match the exact 1/15 per day.

Common Pitfalls:
Interpreting “4/5 as many days” as 4/5 the rate; it refers to time, not rate.

Final Answer:
6 2/3%