Time and Distance Practice Test 4 with solutions

Let x km/hr be the usual speed and d km be the distance of destination, then
$\frac{\mathit{d}}{\mathit{x}}-\frac{\mathit{d}}{\mathit{x}+3}$ = $\frac{40}{60}$
⇒ 3d = $\frac{2}{3}\mathit{x}\left(\mathit{x}+3\right)$
∴ d = $\frac{2}{9}\mathit{x}\left(\mathit{x}+3\right)$ ... (i)
Again, $\frac{\mathit{d}}{\mathit{x}-2}-\frac{\mathit{d}}{\mathit{x}}$ = $\frac{40}{60}$
⇒ 2d = $\frac{2}{3}\mathit{x}\left(\mathit{x}-2\right)$
∴ d = $\frac{1}{3}\mathit{x}\left(\mathit{x}-2\right)$ ... (ii)
From (i) & (ii), we get,
$\frac{2}{9}\mathit{x}\left(\mathit{x}+3\right)$ = $\frac{1}{3}\mathit{x}\left(\mathit{x}-2\right)$
⇒ 2x + 6 = 3x – 6
∴ x = 12 km/hr
Now, from (ii), d = $\frac{1}{3}\times 12\left(12-2\right)$ = 4 × 10 = 40 km