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 sections A and D?

Options

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

Answer

To solve the problem step by step, we will denote the number of students in sections A, B, C, and D as \( a, b, c, \) and \( d \) respectively. We will also use the average percentages provided in the question to set up equations.

### Step 1: Set up the equation for the overall average
The overall average marks of the students in sections A, B, C, and D together is 60%. Therefore, we can write the equation:

\[ \frac{45a + 50b + 72c + 80d}{a + b + c + d} = 60 \]

Multiplying both sides by \( a + b + c + d \):

\[ 45a + 50b + 72c + 80d = 60(a + b + c + d) \]

Expanding the right side:

\[ 45a + 50b + 72c + 80d = 60a + 60b + 60c + 60d \]

Rearranging gives us:

\[ 15a + 10b = 12c + 20d \quad \text{(Equation 1)} \]

### Step 2: Set up the equation for sections A and B
The average marks of sections A and B together is 48%. Therefore, we can write:

\[ \frac{45a + 50b}{a + b} = 48 \]

Multiplying both sides by \( a + b \):

\[ 45a + 50b = 48(a + b) \]

Expanding gives us:

\[ 45a + 50b = 48a + 48b \]

Rearranging gives us:

\[ 3a = 2b \quad \text{(Equation 2)} \]

### Step 3: Set up the equation for sections B and C
The average marks of sections B and C together is 60%. Therefore, we can write:

\[ \frac{50b + 72c}{b + c} = 60 \]

Multiplying both sides by \( b + c \):

\[ 50b + 72c = 60(b + c) \]

Expanding gives us:

\[ 50b + 72c = 60b + 60c \]

Rearranging gives us:

\[ 10b = 12c \quad \text{(Equation 3)} \]

### Step 4: Solve the equations
From Equation 2, we have:

\[ b = \frac{3}{2}a \]

Substituting \( b \) into Equation 3:

\[ 10\left(\frac{3}{2}a\right) = 12c \]

This simplifies to:

\[ 15a = 12c \implies c = \frac{5}{4}a \]

Now substituting \( b \) and \( c \) back into Equation 1:

\[ 15a + 10\left(\frac{3}{2}a\right) = 12\left(\frac{5}{4}a\right) + 20d \]

This simplifies to:

\[ 15a + 15a = 15a + 20d \]

Thus:

\[ 30a = 15a + 20d \implies 15a = 20d \implies \frac{a}{d} = \frac{20}{15} = \frac{4}{3} \]

### Final Result
The ratio of the number of students in sections A and D is:

\[ \frac{a}{d} = \frac{4}{3} \]