Ten women can complete a piece of work in 6 days, six men can complete the same work in 5 days and eight children can complete it in 10 days. What is the ratio of the efficiencies of a woman, a man and a child respectively?

Introduction / Context:
In this Time and Work problem, we are comparing the efficiencies of three different categories of workers: women, men and children. Each group, working alone, can finish the same job in different times with different numbers of workers. By interpreting total work as constant and using worker-days, we can determine how much work one woman, one man and one child can do per day, and then express these capacities as a ratio.

Given Data / Assumptions:

• 10 women complete the work in 6 days.
• 6 men complete the work in 5 days.
• 8 children complete the work in 10 days.
• All women have equal efficiency, all men have equal efficiency, and all children have equal efficiency within their groups.
• Total work is treated as 1 unit.

Concept / Approach:
If a group of workers completes a job in D days, the total work is equal to (number of workers) * (individual efficiency) * D. For each group, we can write an expression for total work, set each equal to 1 unit, and deduce individual efficiencies. Once we know the individual daily work done by a woman, a man and a child, we express these in a simple ratio form to get the required answer.

Step-by-Step Solution:
Let Ew, Em and Ec be the daily efficiencies of one woman, one man and one child respectively. Total work W = 1 unit. For 10 women in 6 days: 10 * Ew * 6 = 1 ⇒ 60Ew = 1 ⇒ Ew = 1 / 60. For 6 men in 5 days: 6 * Em * 5 = 1 ⇒ 30Em = 1 ⇒ Em = 1 / 30. For 8 children in 10 days: 8 * Ec * 10 = 1 ⇒ 80Ec = 1 ⇒ Ec = 1 / 80. Now we want the ratio Ew : Em : Ec. Substitute values: Ew : Em : Ec = 1 / 60 : 1 / 30 : 1 / 80. To clear denominators, multiply by the LCM of 60, 30 and 80 which is 240. Ew * 240 = 240 / 60 = 4. Em * 240 = 240 / 30 = 8. Ec * 240 = 240 / 80 = 3. Hence, Ew : Em : Ec = 4 : 8 : 3.

Verification / Alternative check:
We can reverse-check by assuming that 1 woman does 4 units of some arbitrary work measure, 1 man does 8 units and 1 child does 3 units each day, with total work scaled accordingly. Then 10 women for 6 days would do 10 * 4 * 6 = 240 units, 6 men for 5 days would do 6 * 8 * 5 = 240 units, and 8 children for 10 days would do 8 * 3 * 10 = 240 units. Since all groups complete the same 240 units of work, the ratio 4 : 8 : 3 is consistent across all three relationships.

Why Other Options Are Wrong:
Option A (4 : 6 : 3) would imply that a man is only 1.5 times as efficient as a woman, which does not satisfy the condition that 6 men finish in 5 days compared to 10 women in 6 days. Option B (4 : 5 : 3) and Option C (2 : 4 : 3) similarly fail when we examine the total work done by each group. Only 4 : 8 : 3 allows all three given scenarios to represent the same total work.

Common Pitfalls:
Students often get confused by the presence of three different groups and attempt to combine all data in a single equation. The correct approach is to treat each group separately, compute individual efficiencies and then form the ratio. Another frequent source of error is failing to take a proper LCM or mishandling the fraction simplification when converting 1 / 60, 1 / 30 and 1 / 80 into a simple ratio.

Final Answer:
The ratio of the efficiencies of a woman, a man and a child is 4 : 8 : 3 respectively.