(b) Let the number of students in the sections A, B, C and D be a, b,c and d, respectively. Then, total marks of students of sections A, B, C and D will be 0.45a, 0.50b, 0.72c and 0.80d , respectively.
According to the question,
| 0.45a+0.50b+0.72c+0.80d |
| a+b+c+d |
= 0.60
⇒ 0.45a + 0.50b + 0.72c + 0.80d = 0.60a + 0.60b + 0.60 c + 0.60 d
⇒ 0.12 c + 0.20 d = 0.15a + 0.10 b
⇒ 12c + 20d = 15a + 10 b .........(i)
Also,
= 0.48
⇒ 0.45a + 0.50b = 0.48a+ 0.48b
⇒ 0.03a = 0.02b
⇒ 3a = 2b
⇒ b =
a ......(ii)
And,
= 0.60
⇒ 0.50b + 0.72c = 0.60b + 0.60 c
⇒ 0.10 b = 0.12 c
⇒ 5b = 6c
⇒ c =
b=â‹…a [using Eq. (ii)]
∴ c =
a ............(iii)
On putting the values from Eqs. (ii) and (iii) in Eq. (i),
12â‹…
a + 20d = 15a + 10â‹…
a
⇒ 20d = (15 + 15 -15) a
⇒ 20 d = 15a
⇒
= = 4 : 3