By increasing the speed of a car by 10 km/hr, the time of journey of 72 km is reduced by 36 minutes. Find the original speed of the car.

Let the original speed of car be x km/hr.

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Time taken to cover 72 km at original speed = $\dfrac{72}{x}$ hrs

New Speed = x + 10 km/hr

Time taken to cover 72 km at new Speed = $\dfrac{72}{x + 10}$ hrs

Given,

The time taken to cover the distance is reduced by 36 minutes or $\dfrac{36}{60}$ hrs.

$\Rightarrow \dfrac{72}{x} - \dfrac{72}{x + 10} = \dfrac{36}{60} \\[1em] \Rightarrow 72 \times \Big[\dfrac{1}{x} - \dfrac{1}{x + 10}\Big] = \dfrac{3}{5} \\[1em] \Rightarrow \dfrac{1}{x} - \dfrac{1}{x + 10} = \dfrac{3}{5} \times \dfrac{1}{72} \\[1em] \Rightarrow \dfrac{(x + 10) - x}{x(x + 10)} = \dfrac{1}{5} \times \dfrac{1}{24} \\[1em] \Rightarrow \dfrac{10}{x^2 + 10x} = \dfrac{1}{120} \\[1em] \Rightarrow 120 \times 10 = x^2 + 10x \\[1em] \Rightarrow 1200= x^2 + 10x \\[1em] \Rightarrow x^2 + 10x - 1200 = 0 \\[1em] \Rightarrow x^2 + 40x - 30x - 1200 = 0 \\[1em] \Rightarrow x(x + 40) - 30(x + 40) = 0 \\[1em] \Rightarrow (x - 30)(x + 40) = 0 \\[1em] \Rightarrow (x - 30) = 0 \text{ or } (x + 40) = 0 \text{….[Using zero-product rule]} \\[1em] \Rightarrow x = 30 \text{ or } x = -40.$

(Since speed can’t be negative).

Hence, the original speed of the car = 30 km/hr.