Five men started a job expected to be done in 15 days. After 5 days, 10 women joined them and together they finished in the next 5 days. If 10 women alone had started the job, in how many days would they have completed it?

Difficulty: Medium

Correct Answer: 15 days

Explanation:


Introduction / Context:
Infer women-to-man efficiency from the mixed phase, then apply it to compute the time for 10 women to complete the full job alone.


Given Data / Assumptions:

  • 5 men for 15 days → job size = 75 man-days.
  • After 5 days (25 man-days done), 10 women join and the remaining is done in 5 more days.


Concept / Approach:
Use remaining work and team rate over the next 5 days to deduce 1 woman in man-equivalents, then compute the women-only time.


Step-by-Step Solution:

Remaining work after 5 days = 75 - 25 = 50 man-daysIn next 5 days: (5 men + 10 women)*5 days = 50 → 25 man-days + 50 woman-days = 50Thus, 50 woman-days = 25 man-days → 1 woman-day = 1/2 man-dayWomen-only rate (10 women) = 10 * (1/2) = 5 man-days/dayTime = 75 / 5 = 15 days


Verification / Alternative check:
Back-substitution to mixed phase confirms consistency with the observed completion time.


Why Other Options Are Wrong:
They reflect incorrect conversion between women and men or wrong total job size.


Common Pitfalls:
Confusing “man-days” with “men-days” arithmetic; always treat efficiency conversion consistently.


Final Answer:
15 days

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