Management Aptitude Test 2015 Paper

(d) In a rectangle

Here, a be the length and b be the width and c be the diagonal,
According to the question,
a + b - c = $\frac{1}{2}$ a .......(i)
⇒ 2 (a + b - c) = a
From Eq. (i), a - $\frac{\mathit{a}}{2}$ + b = c
$\frac{\mathit{a}}{2}$ + b = c ...........(ii)
We have, 2(a + b - c) = a
On squaring both sides, we get
4${\left(\mathit{a}+\mathit{b}-\mathit{c}\right)}^2={\mathit{a}}^2$
4 [${\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2$+ 2ab - 2be - 2ca] = ${\mathit{a}}^2$
By Pythagoras theorem,
${\mathit{c}}^2$ = ${\mathit{a}}^2+{\mathit{b}}^2$
4${\left(\mathit{a}+\mathit{b}-\mathit{c}\right)}^2={\mathit{a}}^2$
4 [${\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2$+ 2ab - 2be - 2ca] = ${\mathit{a}}^2$
${\mathit{a}}^2$ = [4${\mathit{a}}^2$ + 4${\mathit{b}}^2$ + 4${\mathit{a}}^2$ + 4${\mathit{b}}^2$ + 8ab - 8bc- 8ca]
From Eq. (ii),
${\mathit{a}}^2$ = 8${\mathit{a}}^2$ + 8${\mathit{b}}^2$ + 8 ab - 8 × b × $\left(\frac{\mathit{a}}{2}+\mathit{b}\right)$ - 8$\left(\frac{\mathit{a}}{2}+\mathit{b}\right)$
⇒ ${\mathit{a}}^2$ = 8${\mathit{a}}^2$ + 8${\mathit{b}}^2$ + 8ab - $\frac{8\mathit{a}\mathit{b}}{2}$ - 8${\mathit{b}}^2$ - $\frac{8{\mathit{a}}^2}{2}$ - 8ab
⇒ ${\mathit{a}}^2$ = 8${\mathit{a}}^2$ - $\frac{8\mathit{a}\mathit{b}}{2}$ - $\frac{8{\mathit{a}}^2}{2}$
⇒ ${\mathit{a}}^2$ = 8${\mathit{a}}^2$ - 4ab - 4${\mathit{a}}^2$
⇒ ${\mathit{a}}^2$ = 4${\mathit{a}}^2$ - 4ab
⇒ 4ab = 3${\mathit{a}}^2$
⇒ 4b = 3a
∴ $\frac{\mathit{b}}{\mathit{a}}=\frac{3}{4}$
Ratio of shorter to longer side = 3 : 4
Hence, option (d) is correct.