Time and work comparison: A does half as much work as B in three-fourths of B's time. If together A and B complete the whole job in 18 days, how many days would B alone require to finish the entire work?

Introduction / Context:
This time-and-work problem compares efficiencies. The statement “A does half as much work as B in three-fourths of the time” encodes a relationship between their daily work rates. Using that, we combine their rates to match the given combined time and then isolate B’s solo time.

Given Data / Assumptions:

• Together time to finish 1 job = 18 days.
• In 3/4 unit of time, A does half as much work as B does in 1 unit of time.
• Work is uniform; rates are constant; 1 job is the total work.

Concept / Approach:
Translate the wording into a rate ratio. If B’s rate is r_B (job/day), the comparison gives A’s rate r_A as a multiple of r_B. Then r_A + r_B equals the together rate. Finally, invert B’s rate to get B’s time alone.

Step-by-Step Solution:
Interpretation: In 3/4 time, A does 1/2 of what B does in 1 time ⇒ r_A = (1/2) / (3/4) * r_B = (2/3) * r_B. Together rate r_A + r_B = (2/3)r_B + r_B = (5/3)r_B. Given together time = 18 ⇒ (5/3)r_B = 1/18 ⇒ r_B = 3/(5*18) = 1/30 job/day. Therefore, B alone takes 30 days.

Verification / Alternative check:
Together rate computed from B’s rate: (5/3)*(1/30) = 1/18 job/day, matching the given 18 days for the whole work.

Why Other Options Are Wrong:
40 days and 35 days contradict the derived rate relation and the 18-day team time. “None of these” is invalid because 30 days is consistent with all conditions.

Common Pitfalls:
Misreading “half as much work in three-fourths of the time” and setting r_A = 1/2 r_B (ignoring the time scaling). Always convert comparative statements to rates carefully.

Final Answer:
30 days