In a factory there are equal numbers of women and children. Women work 6 hours per day and children work 4 hours per day. During festival time, workload increases by 50%. By rule, children cannot work more than 6 hours per day. If women and children are equally efficient per hour and the extra work must be absorbed by women, how many extra hours per day must each woman work?

Difficulty: Medium

Correct Answer: 3

Explanation:


Introduction / Context:
When workload increases proportionally, compute new total hours required and distribute under constraints. Since children are capped at 6 hours, women must pick up the remaining hours.


Given Data / Assumptions:

  • Equal counts: n women, n children; equal hourly efficiency.
  • Normal: women 6 h/day, children 4 h/day → total = n*(6 + 4) = 10n.
  • Festival: +50% workload → need 15n hours/day.
  • Children capped at 6 h/day → children contribute 6n.


Concept / Approach:
Compute required women-hours as the difference between total needed and capped children-hours, then compare with normal women-hours to find the extra per woman.


Step-by-Step Solution:

Required total hours = 15nChildren hours (capped) = 6nWomen required hours = 15n - 6n = 9nNormal women hours = 6n → Extra women hours = 9n - 6n = 3nExtra per woman = (3n)/n = 3 hours


Verification / Alternative check:
New total = 9n + 6n = 15n which equals 150% of original 10n → consistent.


Why Other Options Are Wrong:
They miscount either the cap or the proportional increase.


Common Pitfalls:
Assuming children also bear extra unlimited hours; here the legal cap binds.


Final Answer:
3

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