6 men + 8 boys finish a job in 10 days; 26 men + 48 boys finish the same job in 2 days. In how many days will 15 men + 20 boys finish it?

Problem restatement
Given two group-productivity scenarios, deduce the individual efficiency of a man and a boy, then compute the time for a third team.

Given data

• (6m + 8b) × 10 = 1 job → 6m + 8b = 1/10.
• (26m + 48b) × 2 = 1 job → 26m + 48b = 1/2.

Concept/Approach
Let m, b be daily job-fractions of one man and one boy. Solve the linear system for m and b; then compute (15m + 20b) and invert to get time.

Step-by-step calculation
6m + 8b = 1/10 … (1) 26m + 48b = 1/2 … (2) From (1): m = (1/10 − 8b)/6 Substitute in (2): 26(1/10 − 8b)/6 + 48b = 1/2 (13/3)(1/10 − 8b) + 48b = 1/2 13/30 − (104/3)b + 48b = 1/2 13/30 + (40/3)b = 1/2 → (40/3)b = 1/2 − 13/30 = 1/15 b = (1/15) × (3/40) = 1/200 Then from (1): 6m = 1/10 − 8(1/200) = 1/10 − 1/25 = 3/50 → m = 1/100 For 15 men + 20 boys: rate = 15m + 20b = 15/100 + 20/200 = 0.15 + 0.10 = 0.25 = 1/4 job/day Time = 1 ÷ (1/4) = 4 days

Verification
Plug m = 1/100, b = 1/200 back into (2): 26/100 + 48/200 = 0.26 + 0.24 = 0.50 = 1/2 (correct).

Common pitfalls

• Forgetting to convert group-time statements into per-day equations first.
• Arithmetic errors when handling fractions; keep exact fractions to avoid rounding mistakes.

Final Answer
4 days