If 12 men and 16 boys can finish a work in 5 days, while 13 men and 24 boys can finish the same work in 4 days. Compare the one day work of 1 man and 1 boy.

Answer

To solve the problem, we need to find the one-day work of 1 man and 1 boy based on the information given. Let's break it down step by step.

### Step 1: Understand the Given Information
We have two scenarios:
1. 12 men and 16 boys can finish the work in 5 days.
2. 13 men and 24 boys can finish the same work in 4 days.

### Step 2: Calculate the Total Work Done
We can express the total work done in terms of "man-days" and "boy-days".

- For the first scenario:
- Total work = (Number of workers) × (Number of days)
- Total work = (12 men + 16 boys) × 5 days

- For the second scenario:
- Total work = (13 men + 24 boys) × 4 days

### Step 3: Express Work in Terms of One Day Work
Let's denote the one-day work of 1 man as \( M \) and the one-day work of 1 boy as \( B \).

From the first scenario:
\[ 12M + 16B = \frac{1}{5} \quad \text{(since they finish the work in 5 days)} \]

From the second scenario:
\[ 13M + 24B = \frac{1}{4} \quad \text{(since they finish the work in 4 days)} \]

### Step 4: Set Up the Equations
We can rewrite the equations:
1. \( 12M + 16B = \frac{1}{5} \) (Equation 1)
2. \( 13M + 24B = \frac{1}{4} \) (Equation 2)

### Step 5: Eliminate One Variable
To eliminate one variable, we can multiply Equation 1 by 4 and Equation 2 by 5 to make the right-hand sides equal:
\[ 4(12M + 16B) = 4 \times \frac{1}{5} \implies 48M + 64B = \frac{4}{5} \quad \text{(Equation 3)} \]
\[ 5(13M + 24B) = 5 \times \frac{1}{4} \implies 65M + 120B = \frac{5}{4} \quad \text{(Equation 4)} \]

### Step 6: Solve the System of Equations
Now we can solve Equations 3 and 4:
1. \( 48M + 64B = \frac{4}{5} \)
2. \( 65M + 120B = \frac{5}{4} \)

We can multiply Equation 3 by 5 and Equation 4 by 4 to eliminate the fractions:
\[ 5(48M + 64B) = 5 \times \frac{4}{5} \implies 240M + 320B = 4 \quad \text{(Equation 5)} \]
\[ 4(65M + 120B) = 4 \times \frac{5}{4} \implies 260M + 480B = 5 \quad \text{(Equation 6)} \]

### Step 7: Subtract to Find Relationship Between M and B
Now we subtract Equation 5 from Equation 6:
\[ (260M + 480B) - (240M + 320B) = 5 - 4 \]
\[ 20M + 160B = 1 \]
\[ M + 8B = \frac{1}{20} \quad \text{(Equation 7)} \]

### Step 8: Substitute Back to Find Values
Now we can use Equation 1 to find the values of \( M \) and \( B \). Substitute \( M \) from Equation 7 into Equation 1:
\[ 12\left(\frac{1}{20} - 8B\right) + 16B = \frac{1}{5} \]
\[ \frac{12}{20} - 96B + 16B = \frac{1}{5} \]
\[ \frac{3}{5} - 80B = \frac{1}{5} \]
\[ -80B = \frac{1}{5} - \frac{3}{5} = -\frac{2}{5} \]
\[ B = \frac{2}{400} = \frac{1}{200} \]

### Step 9: Find M
Substituting \( B \) back into Equation 7:
\[ M + 8\left(\frac{1}{200}\right) = \frac{1}{20} \]
\[ M + \frac{8}{200} = \frac{1}{20} \]
\[ M + \frac{1}{25} = \frac{1}{20} \]
\[ M = \frac{1}{20} - \frac{1}{25} \]
Finding a common denominator (100):
\[ M = \frac{5}{100} - \frac{4}{100} = \frac{1}{100} \]

### Conclusion
Thus, the one-day work of 1 man \( M = \frac{1}{100} \) and the one-day work of 1 boy \( B = \frac{1}{200} \).

### Final Comparison
To compare their work:
\[ M = 2B \]
This means that the work done by 1 man in a day is twice the work done by 1 boy in a day.