4 boys from school A and 6 boysfrom school B together can set up an exhibition in 5 days, which 5 boys from school A and 10 boys from school C together can do in 4 days or 3 boys from school B and 4 boysfrom school C together can do in 10 days. Then how manyboys from schoolA can set up the exhibition in one day?

Options

A. 80
B. 40
C. 60
D. 20

Answer

To solve the problem step by step, we will first define the efficiencies of the boys from each school and then set up equations based on the information given in the question.

### Step 1: Define Variables
Let:
- \( A \) = efficiency of one boy from school A
- \( B \) = efficiency of one boy from school B
- \( C \) = efficiency of one boy from school C

### Step 2: Set Up Equations
From the problem, we have the following scenarios:

1. **4 boys from school A and 6 boys from school B can set up the exhibition in 5 days:**
\[ (4A + 6B) \times 5 = \text{Total work} \]
Thus, the total work can be expressed as:
\[ 20A + 30B \]

2. **5 boys from school A and 10 boys from school C can do it in 4 days:**
\[ (5A + 10C) \times 4 = \text{Total work} \]
Thus, the total work can be expressed as:
\[ 20A + 40C \]

3. **3 boys from school B and 4 boys from school C can do it in 10 days:**
\[ (3B + 4C) \times 10 = \text{Total work} \]
Thus, the total work can be expressed as:
\[ 30B + 40C \]

### Step 3: Equate the Total Work
Since all three scenarios represent the same total work, we can equate them:

From the first and second equations:
\[ 20A + 30B = 20A + 40C \]
Cancelling \( 20A \) from both sides gives:
\[ 30B = 40C \quad \Rightarrow \quad 3B = 4C \quad \Rightarrow \quad B = \frac{4}{3}C \]

From the second and third equations:
\[ 20A + 40C = 30B + 40C \]
Cancelling \( 40C \) from both sides gives:
\[ 20A = 30B \quad \Rightarrow \quad A = \frac{3}{2}B \]

### Step 4: Substitute \( B \) in Terms of \( C \)
Now substitute \( B = \frac{4}{3}C \) into \( A = \frac{3}{2}B \):
\[ A = \frac{3}{2} \left(\frac{4}{3}C\right) = 2C \]

### Step 5: Find Total Work in Terms of \( C \)
Now we can express the total work in terms of \( C \):
Using the first equation:
\[ 20A + 30B = 20(2C) + 30\left(\frac{4}{3}C\right) \]
Calculating this gives:
\[ = 40C + 40C = 80C \]

### Step 6: Determine the Work Done by One Boy from School A
Now, if one boy from school A can complete the work in \( n \) days, then:
\[ A \times n = 80C \]
Substituting \( A = 2C \):
\[ 2C \times n = 80C \]
Dividing both sides by \( C \) (assuming \( C \neq 0 \)):
\[ 2n = 80 \quad \Rightarrow \quad n = 40 \]

### Step 7: Calculate the Number of Boys from School A Required
Since one boy from school A takes 40 days to complete the work, the number of boys required to complete the work in one day is:
\[ \text{Number of boys} = \frac{1}{n} = \frac{1}{40} \text{ (in terms of work done in one day)} \]
Thus, to complete the work in one day:
\[ \text{Number of boys} = 40 \]

### Final Answer
Therefore, **40 boys from school A can set up the exhibition in one day.**

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