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

Options

A. 2:3
B. 4:3
C. 5:3
D. 3:5

Answer

To solve the problem, we need to find the ratio of the number of students in sections A and D based on the given average marks. Let's break down the solution step by step.

### Step 1: Define Variables
Let:
- \( A \) = number of students in section A
- \( B \) = number of students in section B
- \( C \) = number of students in section C
- \( D \) = number of students in section D

### Step 2: Set Up the Equations
We know the average marks of the students in sections A, B, C, and D:
- Average of A = 45%
- Average of B = 50%
- Average of C = 72%
- Average of D = 80%

The overall average of all sections together is 60%. Therefore, we can express this as:
\[ \frac{45A + 50B + 72C + 80D}{A + B + C + D} = 60 \]

### Step 3: Multiply Through by Total Students
Multiplying both sides by \( A + B + C + D \):
\[ 45A + 50B + 72C + 80D = 60(A + B + C + D) \]
This simplifies to:
\[ 45A + 50B + 72C + 80D = 60A + 60B + 60C + 60D \]
Rearranging gives:
\[ 15A + 10B - 12C - 20D = 0 \quad \text{(Equation 1)} \]

### Step 4: Use Given Averages of A & B, and B & C
We are also given:
1. The average of A and B together is 48%:
\[ \frac{45A + 50B}{A + B} = 48 \]
Multiplying through by \( A + B \):
\[ 45A + 50B = 48(A + B) \]
This simplifies to:
\[ 45A + 50B = 48A + 48B \]
Rearranging gives:
\[ 3A - 2B = 0 \quad \text{(Equation 2)} \]
Thus, we can express \( A \) in terms of \( B \):
\[ A = \frac{2}{3}B \]

2. The average of B and C together is 60%:
\[ \frac{50B + 72C}{B + C} = 60 \]
Multiplying through by \( B + C \):
\[ 50B + 72C = 60(B + C) \]
This simplifies to:
\[ 50B + 72C = 60B + 60C \]
Rearranging gives:
\[ 10B - 12C = 0 \quad \text{(Equation 3)} \]
Thus, we can express \( C \) in terms of \( B \):
\[ C = \frac{5}{6}B \]

### Step 5: Substitute A and C into Equation 1
Substituting \( A \) and \( C \) in terms of \( B \) into Equation 1:
\[ 15\left(\frac{2}{3}B\right) + 10B - 12\left(\frac{5}{6}B\right) - 20D = 0 \]
This simplifies to:
\[ 10B + 10B - 10B - 20D = 0 \]
Thus:
\[ 10B - 20D = 0 \]
This gives:
\[ B = 2D \quad \text{(Equation 4)} \]

### Step 6: Find the Ratio of A and D
Now we have:
- From Equation 2: \( A = \frac{2}{3}B \)
- From Equation 4: \( B = 2D \)

Substituting \( B \) into the expression for \( A \):
\[ A = \frac{2}{3}(2D) = \frac{4}{3}D \]

### Step 7: Ratio of A to D
The ratio of the number of students in section A to section D is:
\[ \frac{A}{D} = \frac{\frac{4}{3}D}{D} = \frac{4}{3} \]

### Final Answer
Thus, the ratio of the number of students in section A to section D is:
\[ \text{Ratio of A to D} = 4:3 \]