LIC AAO 22 Mar 2015 Paper

Let the speed of boat A in still water be x kmph.
Speed of boat B = (x + 2) kmph
Rate downstream of boat A
= $\left(\mathit{x}+\frac{\mathit{x}}{3}\right)$ = $\frac{4\mathit{x}}{3}$ kmph
Rate downstream of boat B
= $\left(\mathit{x}+2+\frac{\mathit{x}}{3}\right)$ kmph
= $\left(\frac{4\mathit{x}}{3}+2\right)$ kmph
According to the question
$\frac{20}{\frac{4\mathit{x}}{3}}$ - $\frac{20}{\frac{4\mathit{x}}{3}+2}$ = $\frac{30}{60}$ = $\frac{1}{2}$
⇒ $\frac{60}{4\mathit{x}}$ - $\frac{60}{4\mathit{x}+6}$ = $\frac{1}{2}$
⇒ 4x (4x + 6) = 360 × 2
⇒ x (2x + 3) = $\frac{360\times 2}{8}$ = 90
⇒ $2{\mathit{x}}^2$ + 3x - 90 = 0
⇒ $2{\mathit{x}}^2$ + 15x - 12x - 90 = 0
x (2x + 15) - 6 (2x + 15) = 0
(x - 6) (2x + 15) = 0
⇒ x = 6 because x ≠ $\frac{-15}{2}$
∴ speed of boat B = 6 + 2 = 8 kmph