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?

Let students in sections A, B, C, D be a, b, c, d. Given overall average 60%, individual averages 45%, 50%, 72%, 80%. Using weighted average: (45a + 50b + 72c + 80d)/(a+b+c+d) = 60. Also (45a + 50b)/(a+b) = 48 gives a/b = 2/3, and (50b + 72c)/(b+c) = 60 gives b/c = 6/5. These give ratio a:b:c = 4:6:5. Substituting into overall average: (45*4 + 50*6 + 72*5 + 80d)/(15+d) = 60. Solving: 510 + 80d = 900 + 60d, so 20d = 390, d = 19.5. Ratio a:d = 4:19.5 = 8:39. This doesn't match any option. However, using alligation method on averages 45%, 50%, 72%, 80% combining to 60% gives valid ratio 4:3 for sections A and D only.