Management Aptitude Test 2015 Paper

(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, $\frac{0.45a + 0.50b + 0.72c + 0.80d}{a + b + c + d}$ = 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, $\frac{0.45a + 0.50b}{a + b}$ = 0.48⇒ 0.45a + 0.50b = 0.48a+ 0.48b ⇒ 0.03a = 0.02b ⇒ 3a = 2b ⇒ b = $\frac{3}{2}$ a ......(ii) And, $\frac{0.50b + 0.72c}{b + c}$ = 0.60⇒ 0.50b + 0.72c = 0.60b + 0.60 c ⇒ 0.10 b = 0.12 c ⇒ 5b = 6c ⇒ c = $\frac{5}{6}b = \frac{5}{6} \cdot \frac{3}{2} a$      [using Eq. (ii)] ∴ c = $\frac{5}{4}$ a ............(iii) On putting the values from Eqs. (ii) and (iii) in Eq. (i), 12⋅$\frac{5}{4}$a + 20d = 15a + 10⋅$\frac{3}{2}$a ⇒ 20d = (15 + 15 -15) a ⇒ 20 d = 15a ⇒ $\frac{a}{d} = \frac{20}{15}$ = 4 : 3