Let's analyze the problem step-by-step.
Given:
- 8 men and 2 boys can complete the work in 7 days.
- Some number of men and 13 boys can complete the work in 2 days (the number of men is not explicitly given in your statement, but it seems incomplete. Assuming it's the same number of men, 8 men with 13 boys can do it in 2 days).
Goal:
Find the fraction of the work that 11 men and 20 boys can do in 1 day.
Step 1: Define variables
Let:
- MMM = work done by 1 man in 1 day
- BBB = work done by 1 boy in 1 day
Step 2: Write equations from the given data
- 8 men and 2 boys do the work in 7 days
Total work = 1 (whole work) Work done by 8 men and 2 boys in 1 day = 8M+2B8M+2B8M+2B Since they finish the work in 7 days:
7×(8M+2B)=1 ⟹ 8M+2B=177\times (8M+2B)=1\implies 8M+2B=\frac{1}{7}7×(8M+2B)=1⟹8M+2B=71
- 8 men and 13 boys do the work in 2 days
Work done by 8 men and 13 boys in 1 day = 8M+13B8M+13B8M+13B Since they finish the work in 2 days:
2×(8M+13B)=1 ⟹ 8M+13B=122\times (8M+13B)=1\implies 8M+13B=\frac{1}{2}2×(8M+13B)=1⟹8M+13B=21
Step 3: Solve the system of equations
We have:
8M+2B=17(1)8M+2B=\frac{1}{7}\quad (1)8M+2B=71(1)
8M+13B=12(2)8M+13B=\frac{1}{2}\quad (2)8M+13B=21(2)
Subtract (1) from (2):
(8M+13B)−(8M+2B)=12−17(8M+13B)-(8M+2B)=\frac{1}{2}-\frac{1}{7}(8M+13B)−(8M+2B)=21−71
11B=12−17=714−214=51411B=\frac{1}{2}-\frac{1}{7}=\frac{7}{14}-\frac{2}{14}=\frac{5}{14}11B=21−71=147−142=145
B=514×11=5154B=\frac{5}{14\times 11}=\frac{5}{154}B=14×115=1545
Substitute B=5154B=\frac{5}{154}B=1545 into equation (1):
8M+2×5154=178M+2\times \frac{5}{154}=\frac{1}{7}8M+2×1545=71
8M+10154=178M+\frac{10}{154}=\frac{1}{7}8M+15410=71
8M=17−10154=22154−10154=12154=6778M=\frac{1}{7}-\frac{10}{154}=\frac{22}{154}-\frac{10}{154}=\frac{12}{154}=\frac{6}{77}8M=71−15410=15422−15410=15412=776
M=677×8=6616=3308M=\frac{6}{77\times 8}=\frac{6}{616}=\frac{3}{308}M=77×86=6166=3083
Step 4: Calculate work done by 11 men and 20 boys in 1 day
Work per day=11M+20B=11×3308+20×5154\text{Work per day}=11M+20B=11\times \frac{3}{308}+20\times \frac{5}{154}Work per day=11M+20B=11×3083+20×1545
Calculate each term:
11×3308=3330811\times \frac{3}{308}=\frac{33}{308}11×3083=30833
20×5154=100154=507720\times \frac{5}{154}=\frac{100}{154}=\frac{50}{77}20×1545=154100=7750
Find common denominator to add:
- Denominator 308 and 77:
Since 308=4×77308=4\times 77308=4×77, common denominator is 308.
Convert 5077\frac{50}{77}7750 to denominator 308:
5077=50×477×4=200308\frac{50}{77}=\frac{50\times 4}{77\times 4}=\frac{200}{308}7750=77×450×4=308200
Now add:
33308+200308=233308\frac{33}{308}+\frac{200}{308}=\frac{233}{308}30833+308200=308233
Answer:
The fraction of the work that 11 men and 20 boys can do in 1 day is:
233308\boxed{\frac{233}{308}}308233
This means they can complete about 75.65% of the work in one day.
