Management Aptitude Test 2016 Paper

(d) Let the side of first square = a m
and the side of second square = b m
According to the question,
29${\mathit{a}}^2$ = 6${\mathit{b}}^2$ - 1 ...........(i)
Also, 9a = 4b + 1
⇒ a = $\frac{4\mathit{b}+1}{9}$ ............(ii)
On putting the value of a in Eq. (i), we get
29${\left(\frac{4\mathit{b}+1}{9}\right)}^2=6{\mathit{b}}^2$ - 1
⇒ 29$\left(\frac{16{\mathit{b}}^2+1+8\mathit{b}}{81}\right)=6{\mathit{b}}^2$ - 1
⇒ $\left(\frac{464{\mathit{b}}^2+29+232\mathit{b}}{81}\right)=6{\mathit{b}}^2$ - 1
⇒ 464${\mathit{b}}^2$ + 29 + 232b = 486${\mathit{b}}^2$ - 81
⇒ 22${\mathit{b}}^2$ - 232b - 110 = 0
∴ b = $232\pm \frac{\sqrt{{\left(232\right)}^2+4\times 22\times 110}}{2\times 22}$
= $\frac{232\pm \sqrt{53824+9680}}{44}$
= $\frac{232\pm 252}{44}=\frac{484}{44}$ = 11
Now, putting the value of b in Eq. (ii), we get
a = $\frac{4\mathit{b}+1}{9}$ = $\frac{4\times 11+1}{9}=\frac{45}{9}$ = 5
∴ Difference in sides of squares = b - a = 11 - 5 = 6 m