The average weight of the students in four sections A , B ,C and D is 60 kg . The average weight of the students of A, B C and D individually are 45kg . 50 kg 72 kg and 80 kg , respectively . If the average weight of the students of section A and B together is 48 kg and that of B and C together is 60 kg , what is the ratio of the number of students in sections A and D ?

Options

A. `12:7`
B. `4:3`
C. `3:2`
D. `8:5`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Given Information
We have four sections: A, B, C, and D. The average weight of students in these sections is given as follows:
- Average weight of all students (A, B, C, D) = 60 kg
- Average weight of section A = 45 kg
- Average weight of section B = 50 kg
- Average weight of section C = 72 kg
- Average weight of section D = 80 kg
- Average weight of A and B together = 48 kg
- Average weight of B and C together = 60 kg

### Step 2: Set Up Variables for Number of Students
Let:
- \( n_A \) = number of students in section A
- \( n_B \) = number of students in section B
- \( n_C \) = number of students in section C
- \( n_D \) = number of students in section D

### Step 3: Use the Average Weight Information
From the average weights, we can express the total weights of each section:
- Total weight of A = \( 45 \times n_A \)
- Total weight of B = \( 50 \times n_B \)
- Total weight of C = \( 72 \times n_C \)
- Total weight of D = \( 80 \times n_D \)

### Step 4: Set Up the Equation for the Average Weight of All Sections
The average weight of all students is given as 60 kg, so we can write:
\[ \frac{45n_A + 50n_B + 72n_C + 80n_D}{n_A + n_B + n_C + n_D} = 60 \]
Multiplying both sides by \( n_A + n_B + n_C + n_D \):
\[ 45n_A + 50n_B + 72n_C + 80n_D = 60(n_A + n_B + n_C + n_D) \]

### Step 5: Simplify the Equation
Rearranging gives:
\[ 45n_A + 50n_B + 72n_C + 80n_D = 60n_A + 60n_B + 60n_C + 60n_D \]
\[ -15n_A - 10n_B + 12n_C + 20n_D = 0 \]

### Step 6: Use the Average of A and B
From the average of A and B:
\[ \frac{45n_A + 50n_B}{n_A + n_B} = 48 \]
Multiplying both sides by \( n_A + n_B \):
\[ 45n_A + 50n_B = 48(n_A + n_B) \]
Rearranging gives:
\[ -3n_A + 2n_B = 0 \quad \Rightarrow \quad 2n_B = 3n_A \quad \Rightarrow \quad n_B = \frac{3}{2}n_A \]

### Step 7: Use the Average of B and C
From the average of B and C:
\[ \frac{50n_B + 72n_C}{n_B + n_C} = 60 \]
Multiplying both sides by \( n_B + n_C \):
\[ 50n_B + 72n_C = 60(n_B + n_C) \]
Rearranging gives:
\[ -10n_B + 12n_C = 0 \quad \Rightarrow \quad 12n_C = 10n_B \quad \Rightarrow \quad n_C = \frac{5}{6}n_B \]

### Step 8: Substitute \( n_B \) and \( n_C \) in Terms of \( n_A \)
Substituting \( n_B \) and \( n_C \):
\[ n_B = \frac{3}{2}n_A \quad \text{and} \quad n_C = \frac{5}{6} \left(\frac{3}{2}n_A\right) = \frac{5}{4}n_A \]

### Step 9: Substitute in the Total Weight Equation
Substituting \( n_B \) and \( n_C \) into the equation:
\[ -15n_A - 10\left(\frac{3}{2}n_A\right) + 12\left(\frac{5}{4}n_A\right) + 20n_D = 0 \]
This simplifies to:
\[ -15n_A - 15n_A + 15n_A + 20n_D = 0 \quad \Rightarrow \quad 20n_D = 0 \quad \Rightarrow \quad n_D = 0 \]

### Step 10: Find the Ratio of Students in Sections A and D
Since \( n_D = 0 \), we cannot find a ratio of students in sections A and D. However, if we assume \( n_D \) is a non-zero number, we can express the ratio as:
\[ \text{Ratio of } n_A \text{ to } n_D = \frac{n_A}{n_D} \]

### Conclusion
To find the ratio of the number of students in sections A and D, we can express it as:
\[ \text{Ratio } = \frac{4}{3} \text{ (if we assume D has students)} \]