A train covers a distance of 780 km at x km/hr. Had the speed been (x − 5) km/hr, the time taken to cover the same distance would have been increased by 1 hour. Write down an equation in x and solve it to evaluate x.

By formula,

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

Time taken by train at normal speed to cover distance of 780 km = $\dfrac{780}{x}$ hrs

Time taken by train at reduced speed to cover distance of 780 km = $\dfrac{780}{x - 5}$hrs

Given,

On decreasing the speed, the time taken by train increases by 1 hour.

$\Rightarrow \dfrac{780}{x - 5} - \dfrac{780}{x} = 1 \\[1em] \Rightarrow \dfrac{780x - 780(x - 5)}{x(x - 5)} = 1 \\[1em] \Rightarrow \dfrac{780x - 780x + 3900}{x^2 - 5x} = 1 \\[1em] \Rightarrow 3900 = x^2 - 5x \\[1em] \Rightarrow x^2 - 5x - 3900 = 0 \\[1em] \Rightarrow x^2 - 65x + 60x - 3900 = 0 \\[1em] \Rightarrow x(x − 65) + 60(x − 65) = 0 \\[1em] \Rightarrow (x + 60)(x − 65) = 0 \\[1em] \Rightarrow (x + 60) = 0 \text{ or } (x − 65) = 0 \text{….[Using zero-product rule]} \\[1em] \Rightarrow x = -60 \text{ or } x = 65$

Since speed must be positive.

Thus, x = 65 km/hr.

Hence, obtained equation is x² - 5x - 3900 = 0 and the speed of the train = 65 km/hr.