An aeroplane covered a distance of 400 km at an average speed of x km/h. On the return journey, the speed was increased by 40 km/h . Write down an expression for the time taken for :

(i) the onward journey

(ii) the return journey

If the return journey took 30 minutes less than the onward journey , write down an equation in x and find its value.

(i) Distance covered by plane = 400 km

Average speed of plane = x km/h

Time = $\dfrac{\text{ Distance}}{\text{Speed}} = \dfrac{400}{x} \text{ hrs}$

(ii) Distance covered by plane = 400 km

Average speed of plane = (x + 40) km/h

Time = $\dfrac{\text{ Distance}}{\text{Speed}} = \dfrac{400}{x + 40} \text{ hrs}$

According to question,

$\Rightarrow \dfrac{400}{x} - \dfrac{400}{x + 40} = \dfrac{30}{60} \\[1em] \Rightarrow \dfrac{400(x + 40) - 400x}{x(x + 40)} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{400x - 400x + 16000}{x(x + 40)} = \dfrac{1}{2} \\[1em] \Rightarrow 16000 \times 2 = x(x + 40) \\[1em] \Rightarrow 32000 = x^2 + 40x \\[1em] \Rightarrow x^2 + 40x - 32000 = 0 \\[1em] \Rightarrow x^2 + 200x - 160x - 32000 = 0 \\[1em] \Rightarrow x(x + 200) - 160(x + 200) = 0 \\[1em] \Rightarrow (x - 160)(x + 200) = 0 \\[1em] \Rightarrow x - 160 = 0 \text{ or } x + 200 = 0 \\[1em] x = 160 \text{ or } x = -200$

Since speed of aeroplane cannot be negative hence, x ≠ -200

∴ x = 160

Equation in x : $\dfrac{400}{x} - \dfrac{400}{x + 40} = \dfrac{1}{2}$
Hence, the speed of aerolane is 160 km/h.