Time and Distance Practice Test 3 with solutions

Let usual speed be x miles/hr and distance of the destination be d miles
Now, $\frac{\mathit{d}}{\mathit{x}}-\frac{\mathit{d}}{\mathit{x}+1}$ = $\frac{5}{2}$
⇒ d = $\frac{5}{2}\mathit{x}\left(\mathit{x}+1\right)$ ... (i)
Again, $\frac{\mathit{d}}{\mathit{x}-1}-\frac{\mathit{d}}{\mathit{x}}$ = 5
⇒ d = 5x (x - 1) ... (ii)
From equations (i) and (ii)
$\frac{5}{2}\mathit{x}\left(\mathit{x}+1\right)$ = 5x (x - 1)
⇒ x + 1 = 2x - 2
∴ x = 3 mile/hr