Question

While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hr. Find the original speed/hour of the plane.

Answer 1

Let the original speed of the plane be x km/hr.

Increased speed of the plane = (x + 100) km/hr.
Total Distance = 1500 km.

We know that, $\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}$

Time taken to reach the destination at original speed = t₁ = $\frac{1500}{x}$ hr

Time taken to reach the destination at increasing speed = t₂ = $\frac{1500}{x+100}$ hr

According to the question,

t₁ − t₂ = 30 min
$$\begin{gathered} \Rightarrow \frac{1500}{x}-\frac{1500}{x+100}=\frac{30}{60} \\ \Rightarrow \frac{1500(x+100)-1500x}{x(x+100)}=\frac{1}{2} \\ \Rightarrow \frac{1500x+150000-1500x}{x^2+100x}=\frac{1}{2} \\ \Rightarrow \frac{150000}{x^2+100x}=\frac{1}{2} \\ \Rightarrow 300000=x^2+100x \\ \Rightarrow x^2+100x-300000=0 \\ \Rightarrow x^2+600x-500x-300000=0 \\ \Rightarrow x(x+600)-500(x+600)=0 \\ \Rightarrow (x-500)(x+600)=0 \\ \Rightarrow x-500=0\mathrm{or}x+600=0 \\ \Rightarrow x=500\mathrm{or}x=-600 \end{gathered}$$

Since, speed cannot be negative.

Thus, the original speed/hour of the plane is 500 km/hr.