A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less…

(a) Given,

Distance covered = 400 km

Let the original speed of the car be $x$ km/h.

Since the speed is 12 km/h more in the second case,

Increased speed = $(x + 12)$ km/h.

By formula,

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

∴ Time taken with the original speed = $\frac{400}{x}$ hours

∴ Time taken with the increased speed = $\frac{400}{x + 12}$ hours

Converting the given difference of time into hours,

1 hour 40 minutes = $1\frac{40}{60}$ hours = $\frac{5}{3}$ hours.

The question says that with the increased speed the journey takes 1 hour 40 minutes less. So the time taken with the increased speed is smaller, and subtracting it from the time taken with the original speed must give $\frac{5}{3}$ hours.

∴ Time with original speed − Time with increased speed = $\frac{5}{3}$

According to the question,

$$\begin{aligned}\Rightarrow \frac{400}{x} - \frac{400}{x + 12} = \frac{5}{3} \\[1em] \Rightarrow 400\Big[\frac{(x + 12) - x}{x(x + 12)}\Big] = \frac{5}{3} \\[1em] \Rightarrow 400 \times \frac{12}{x^2 + 12x} = \frac{5}{3} \\[1em] \Rightarrow \frac{4800}{x^2 + 12x} = \frac{5}{3} \\[1em] \Rightarrow 5(x^2 + 12x) = 14400 \\[1em] \Rightarrow x^2 + 12x = 2880 \\[1em] \Rightarrow x^2 + 12x - 2880 = 0 \\[1em] \Rightarrow x^2 + 60x - 48x - 2880 = 0 \\[1em] \Rightarrow x(x + 60) - 48(x + 60) = 0 \\[1em] \Rightarrow (x - 48)(x + 60) = 0 \\[1em] \Rightarrow x = 48 \text{ or } x = -60.\end{aligned}$$

Since speed cannot be negative, $x = 48$.

Hence, the original speed of the car = 48 km/h.

(b) Increased speed = $x + 12 = 48 + 12 = 60$ km/h.

By formula,

Time taken with increased speed = $\frac{\text{Distance}}{\text{Speed}}$

$$\begin{aligned}= \frac{400}{60} \\[1em] = \frac{20}{3} \\[1em] = 6\frac{2}{3} \text{ hours}.\end{aligned}$$

= 6 hours 40 minutes.

Hence, the time taken with the increased speed = 6 hours 40 minutes.