Time and Distance Practice Test 2 with solutions

Let original speed of the bus be x km/hr; then
$\frac{72}{\mathit{x}}-\frac{72}{\mathit{x}+10}$ = $\frac{36}{60}$
⇒ $\frac{72\left(\mathit{x}+10-\mathit{x}\right)}{{\mathit{x}}^2+10\mathit{x}}$ = $\frac{3}{5}$
⇒ 1200 = ${\mathit{x}}^2+10\mathit{x}$
⇒ ${\mathit{x}}^2+10\mathit{x}-1200$ = 0
⇒ (x + 40) (x - 30) = 0
∴ x = 30 or x = - 40
Taking positive value only, so speed of the bus = 30 km/hr.