Question

An express train takes 30 min less for a journey of 440 km, if its usual speed is increased by 8m/hr. Find its usual speed.

Answer 1

Let the usual speed of the express train be x km/hr.
Then, the time taken by the express train to complete 440 km is $\frac{\mathrm{Distance}}{\mathrm{Speed}}=\frac{440}{x}$ hr.
Now, if the usual speed of the express train is increased by 8 km/hr, then the time taken to complete 440 km will be $\frac{\mathrm{Distance}}{\mathrm{Speed}}=\frac{440}{x+8}$ hr.
By the given condition, we get:
$$\begin{gathered} \frac{440}{x}-\frac{440}{x+8}=30\min =\frac{1}{2}\mathrm{hr} \\ \Rightarrow 440\left[\frac{x+8-x}{x\left(x+8\right)}\right]=\frac{1}{2} \\ \Rightarrow \frac{440\times 8}{x^2+8x}=\frac{1}{2} \end{gathered}$$
$\Rightarrow$ x² + 8x = 7040
$\Rightarrow$ x² + 8x – 7040 = 0
On splitting the middle term 8x as 88x – 80x, we get:
$\Rightarrow$ x² + 88x – 80x – 7040 = 0
$\Rightarrow$ x(x + 88) – 80(x + 88) = 0
$\Rightarrow$ (x + 88)(x – 80) = 0
$\Rightarrow$ x + 88 = 0 or x – 80 = 0
$\Rightarrow$ x = – 88 or x = 80
Since x is the usual speed of the express train, which cannot be negative, x = 80
Thus, the usual speed of the express train is 80 km/hr.