Two men and seven boys can do a piece of work in 14 days, while three men and eight boys can do the same work in 11 days. In how many days can eight men and six boys together complete three times this work?

Introduction / Context:
This is a multi concept time and work problem involving two different types of workers, men and boys, with different daily work rates. The question first presents two conditions that allow us to find the relative efficiencies of a man and a boy. It then asks how long a new combination of men and boys will take to complete three times the original work. Such questions test algebraic translation, proportional reasoning and careful simplification.

Given Data / Assumptions:

• Two men and seven boys can finish one unit of work in 14 days.
• Three men and eight boys can finish the same unit of work in 11 days.
• We must find the time taken when eight men and six boys work together to do three times this unit of work.
• Each man has a constant daily rate of work, and each boy also has a constant daily rate, possibly different from that of a man.

Concept / Approach:
We introduce variables for daily work rates. Let the daily work done by one man be M units and by one boy be B units. Then the total work in the first case is (2M + 7B) * 14 and in the second case is (3M + 8B) * 11. Since both equal the same amount of work, we can equate them to find the ratio between M and B. After obtaining M and B up to a common multiple, we compute the total amount of work and the daily work of the group of eight men and six boys. Finally, we find the time required for three times the original work.

Step-by-Step Solution:
Step 1: Express total work W using the first condition: W = (2M + 7B) * 14.Step 2: Express the same work W using the second condition: W = (3M + 8B) * 11.Step 3: Set these equal: (2M + 7B) * 14 = (3M + 8B) * 11.Step 4: Expand both sides: 28M + 98B = 33M + 88B. Rearranging gives 98B - 88B = 33M - 28M, so 10B = 5M and M = 2B.Step 5: Substitute M = 2B into W = (2M + 7B) * 14. Then 2M + 7B = 2 * 2B + 7B = 11B, so W = 11B * 14 = 154B. Three times the work is 3W = 462B. The daily work rate of eight men and six boys is 8M + 6B = 8 * 2B + 6B = 22B. The required time is T = 3W / (8M + 6B) = 462B / 22B = 21 days.
Verification / Alternative check:
As a check, we can see that if 2 men and 7 boys do the work in 14 days, their combined daily work is W / 14. If 3 men and 8 boys do it in 11 days, their combined daily work is W / 11. The second combination does more work per day, consistent with one extra man and one extra boy. Our algebra identified that a man is equivalent to twice a boy in daily work. When eight men and six boys work together, their rate is 22B per day, and 462B units of work divided by 22B again gives 21 days, reinforcing the earlier result.

Why Other Options Are Wrong:
Eighteen days would require a higher daily rate than 22B, implying either more workers or more efficient workers than given in the problem.Twenty four days would imply a smaller combined daily rate and would not be consistent with the efficiencies derived from the original conditions.Thirty days would be even slower and clearly contradicts the computed work rate of the group of eight men and six boys.
Common Pitfalls:
Students sometimes try to apply chain rule directly without setting up variables for the work rates. This often leads to confusion when multiple types of workers are involved. Another common mistake is algebraic: errors when expanding and simplifying the equation that relates M and B. Writing each step carefully, especially when equating total work from two scenarios, prevents these mistakes and leads to a clean solution.

Final Answer:
Eight men and six boys will take 21 days to complete three times the given work.