The distance by road between two towns A and B, is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car.

(i) Write down the time taken by the car to reach town B from A, in terms of x.

(ii) Write down the time taken by the train to reach town B from A, in terms of x.

(iii) If the train takes 2 hours less than the car to reach town B, obtain an equation in x and solve it.

(iv) Hence, find the speed of the train.

(i) By formula,

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

Speed of car = x km/hr

Distance between A and B by road = 216 km

Time taken by car = $\dfrac{216}{x}$ hrs

Hence, time taken by the car to reach town B from A is $\dfrac{216}{x}$ hrs.

(ii) By formula,

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

Speed of train = (x + 16) km/hr

Distance between A and B by rail = 208 km

Time = $\dfrac{208}{x + 16}$ hrs

Hence, time taken by the train to reach town B from A is $\dfrac{208}{x + 16}$ hrs.

(iii) Given,

Train takes 2 hours less than the car to reach town B.

$\Rightarrow \dfrac{216}{x} - \dfrac{208}{x + 16} = 2 \\[1em] \Rightarrow \dfrac{216(x + 16) - 208x}{x(x + 16)} = 2 \\[1em] \Rightarrow \dfrac{216x + 3456 - 208x}{x^2 + 16x} = 2 \\[1em] \Rightarrow 8x + 3456 = 2(x^2 + 16x) \\[1em] \Rightarrow 8x + 3456 = 2x^2 + 32x \\[1em] \Rightarrow 2x^2 + 32x - 8x - 3456 = 0 \\[1em] \Rightarrow 2x^2 + 24x - 3456 = 0 \\[1em] \Rightarrow 2(x^2 + 12x - 1728) = 0 \\[1em] \Rightarrow x^2 + 12x - 1728 = 0 \\[1em] \Rightarrow x^2 + 48x - 36x - 1728 = 0 \\[1em] \Rightarrow x(x + 48) - 36(x + 48) = 0 \\[1em] \Rightarrow (x - 36)(x + 48) = 0 \\[1em] \Rightarrow (x - 36) = 0 \text{ or } (x + 48) = 0 \text{….[Using zero-product rule]} \\[1em] \Rightarrow x = 36 \text{ or } x = -48$

Since, speed cannot be negative.

∴ x = 36 km/hr

Hence, obtained equation is x² + 12x - 1728 = 0 and the speed of the train = 36 km/hr.

(iv) Speed of the train = x + 16

= 36 + 16 = 52 km/hr.

Hence, the speed of the train = 52 km/hr.