Six children and two men complete a job in 6 days. Each child takes twice as long as a man to finish the work alone (i.e., a child is half as efficient as a man). In how many days can five men alone finish the same work?

Introduction / Context:
This problem converts different worker types into a single efficiency scale. Given a child takes twice the time of a man alone, a child’s rate is half of a man’s rate. We use the team’s combined rate to deduce total work and then compute the time for five men.

Given Data / Assumptions:

• 6 children + 2 men finish in 6 days.
• Child is half as efficient as a man.
• Find days for 5 men alone.

Concept / Approach:
Let m = one man’s daily work; then one child = m/2 per day. Compute the team’s daily rate, multiply by days to get total work W. Then divide W by 5m to find the required days.

Step-by-Step Solution:
One child’s rate = m/2. Team daily rate = 6*(m/2) + 2*m = 3m + 2m = 5m. Total work W = (team rate) * days = 5m * 6 = 30m. Five men’s daily rate = 5m. Required days = W / (5m) = 30m / 5m = 6 days.

Verification / Alternative check:
If one man alone needed T days, five men would need T/5; the calculations above are consistent with standard work-rate relations.

Why Other Options Are Wrong:
8, 9, 10, 15 days contradict the direct rate computation of total work and five-men rate.

Common Pitfalls:
Confusing “takes twice the time” with “twice the efficiency,” or adding times instead of adding rates.

Final Answer:
6 days