The average marks of the students in four sections A, B, C and D of a school 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 the number of students in sections A and D?

Let $$a,b,c,d$$ be the number of students in A, B, C and D respectively

Given, average marks of the students in section A, B, C and D of the school = 60

Then $$\frac{45a + 50b + 72c + 80d}{a + b + c + d} = 60$$%

= $$45a + 50b + 72c + 80d = 60a + 60b + 60c + 60d$$

= $$12c + 20d = 15a + 10b$$..........(1)

Average marks of the students of sections A and B together is 48%,

= $$\frac{45a + 50b}{a + b} = 48$$%

= $$45a + 50b = 48a + 48b$$

= $$3a =  2b$$ (or) $$15a = 10b$$...............(2)

Average marks of the students of sections B and C together is 60%

= $$\frac{72c + 80d}{c + d} = 60$$%

= $$72c + 80d = 60c + 60d

= $$12c = 20d$$............(3)

Substitute equations (2) and (3) in equation (1)

$$\Rightarrow 20d + 20d = 15a + 15a$$

$$\Rightarrow 40d = 30a$$

$$\Rightarrow a : d = 4 : 3$$

Hence, option B is the correct answer.