Management Aptitude Test Dec 2012 Paper

(b) Let the four numbers be a, b, c and d.
Then, a : b = c : d =1 : 3
⇒ $\frac{\mathit{a}}{\mathit{b}}=\frac{\mathit{c}}{\mathit{d}}=\frac{1}{3}$
⇒ b = 3a, d = 3c ..........(i)
Also, ${\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2+{\mathit{d}}^2$ = 50
⇒ ${\mathit{a}}^2+9{\mathit{a}}^2+{\mathit{c}}^2+9{\mathit{c}}^2$ = 50
⇒ $10{\mathit{a}}^2+10{\mathit{c}}^2$ = 50
⇒ ${\mathit{a}}^2+{\mathit{c}}^2$ = 5 ..............(ii)
Also, sum of means = 5
b + c = 5
⇒b = 5 - c
⇒ a = $\frac{5-\mathit{c}}{3}$ [using Eq. (i)
Putting this value of a in Eq (ii), we get
${\left(\frac{5-\mathit{c}}{3}\right)}^2+{\mathit{c}}^2$ = 5
⇒ $\frac{25+{\mathit{c}}^2-10\mathit{c}}{9}+{\mathit{c}}^2$ = 5
⇒ $10{\mathit{c}}^2$ - 10c - 20 = 0
⇒ ${\mathit{c}}^2$ - c - 2 = 0
⇒ c = 2
Therefore, a = $\frac{5-\mathit{c}}{3}=\frac{5-2}{3}$ = 1
b = 3a = 3
c = 2
d = 3c = 6
Hence, required average
= $\frac{\mathit{a}+\mathit{b}+\mathit{c}+\mathit{d}}{4}$ = $\frac{1+3+2+6}{4}=\frac{12}{4}$ = 3