Men and children together: 3 men can do a work in 18 days; 6 children can do the same work in 18 days. How many days will 4 men and 4 children together need to finish the work?

Introduction / Context:
This is another mixed-rate problem. First find the relative productivity between a man and a child, then sum the rates for the combined group and invert to get the total time required.

Given Data / Assumptions:

• 3 men * 18 days = 1 job ⇒ man rate m = 1/(3*18) = 1/54 job/day per man.
• 6 children * 18 days = 1 job ⇒ child rate c = 1/(6*18) = 1/108 job/day per child.
• Team = 4 men + 4 children.

Concept / Approach:
Team rate = 4m + 4c. Compute this numeric rate and then compute days = 1 / (team rate). Equivalently, note that 1 man = 2 children since 1/54 is twice 1/108.

Step-by-Step Solution:
m = 1/54; c = 1/108 ⇒ 1 man = 2 children. Team rate = 4*(1/54) + 4*(1/108) = 4/54 + 4/108 = 2/27 + 1/27 = 3/27 = 1/9. Days = 1 / (1/9) = 9.

Verification / Alternative check:
Convert to man-equivalents: 4 men + 4 children = 4 + 2 = 6 man-equivalents; total rate = 6*(1/54) = 1/9 job/day ⇒ 9 days.

Why Other Options Are Wrong:
6, 10, or 12 days contradict the derived combined rate of 1/9 job/day.

Common Pitfalls:
Adding days instead of rates, or assuming equal productivity across men and children without computing it.

Final Answer:
9