One man, 3 women and 4 boys can do a piece of work in 96 hours, 2 men and 8 boys can do It in 80 hours, 2 men and 3 women can do it in 120 hours. 5 men and 12 boys can do it in

Options

A. `39(1)/(11)` hours
B. `42(7)/(11)` hours
C. `43(7)/(11)` hours
D. 44 hours

Answer

To solve the problem step by step, we will first determine the efficiencies of each worker type (men, women, and boys) based on the information provided.

### Step 1: Determine the efficiency of 1 man, 3 women, and 4 boys.
Given that 1 man, 3 women, and 4 boys can complete the work in 96 hours, we can express their combined efficiency as:
\[ \text{Efficiency} = \frac{1}{\text{Time}} = \frac{1}{96} \]
Thus, the equation is:
\[ M + 3W + 4B = \frac{1}{96} \quad \text{(Equation 1)} \]

### Step 2: Determine the efficiency of 2 men and 8 boys.
For 2 men and 8 boys who can complete the work in 80 hours, their combined efficiency is:
\[ \text{Efficiency} = \frac{1}{80} \]
Thus, the equation is:
\[ 2M + 8B = \frac{1}{80} \quad \text{(Equation 2)} \]

### Step 3: Determine the efficiency of 2 men and 3 women.
For 2 men and 3 women who can complete the work in 120 hours, their combined efficiency is:
\[ \text{Efficiency} = \frac{1}{120} \]
Thus, the equation is:
\[ 2M + 3W = \frac{1}{120} \quad \text{(Equation 3)} \]

### Step 4: Solve for the efficiencies of men, women, and boys.
From Equation 2, we can express the efficiency of 1 man and 4 boys:
\[ M + 4B = \frac{1}{160} \quad \text{(by dividing Equation 2 by 2)} \]

Now, we can substitute this into Equation 1:
\[ \frac{1}{160} + 3W = \frac{1}{96} \]
To solve for W, we need a common denominator (LCM of 160 and 96 is 480):
\[ \frac{3}{480} + 3W = \frac{5}{480} \]
This simplifies to:
\[ 3W = \frac{5}{480} - \frac{3}{480} = \frac{2}{480} = \frac{1}{240} \]
Thus, the efficiency of 3 women is:
\[ W = \frac{1}{720} \]

### Step 5: Substitute W back to find M.
Now substituting W back into Equation 3:
\[ 2M + 3 \left(\frac{1}{720}\right) = \frac{1}{120} \]
This simplifies to:
\[ 2M + \frac{1}{240} = \frac{1}{120} \]
Finding a common denominator (240):
\[ 2M + \frac{1}{240} = \frac{2}{240} \]
Thus:
\[ 2M = \frac{2}{240} - \frac{1}{240} = \frac{1}{240} \]
So:
\[ M = \frac{1}{480} \]

### Step 6: Find the efficiency of boys.
Now we can substitute M back into the equation for boys:
Using the equation \(M + 4B = \frac{1}{160}\):
\[ \frac{1}{480} + 4B = \frac{1}{160} \]
Finding a common denominator (480):
\[ 4B = \frac{3}{480} \]
Thus:
\[ B = \frac{3}{1920} = \frac{1}{640} \]

### Step 7: Calculate the efficiency of 5 men and 12 boys.
Now we can calculate the efficiency of 5 men and 12 boys:
\[ 5M + 12B = 5 \left(\frac{1}{480}\right) + 12 \left(\frac{1}{640}\right) \]
Calculating each term:
\[ 5M = \frac{5}{480} = \frac{1}{96}, \quad 12B = \frac{12}{640} = \frac{3}{160} \]
Finding a common denominator (480):
\[ 5M + 12B = \frac{5}{480} + \frac{9}{480} = \frac{14}{480} = \frac{7}{240} \]

### Step 8: Calculate the time taken by 5 men and 12 boys.
The time taken is the reciprocal of the efficiency:
\[ \text{Time} = \frac{1}{\text{Efficiency}} = \frac{240}{7} \text{ hours} \]

### Final Answer
Thus, 5 men and 12 boys can complete the work in \(\frac{240}{7}\) hours, which is approximately 34.29 hours.

---