# While boarding an aeroplane, a passenger got hurt.

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/hour.
Find the original speed of the plane.
Do you appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time?

Solution:

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

∴ Actual speed of the plane = (x + 100) km/h

Distance of the journey = 1500 km

Time taken to reach the destination at original speed $=\frac{1500}{x} \mathrm{~h}$       $\left(\right.$ Time $\left.=\frac{\text { Distance }}{\text { Speed }}\right)$

Time taken to reach the destination at actual speed $=\frac{1500}{x+100} \mathrm{~h}$

According to the given condition,

Time taken to reach the destination at original speed = Time taken to reach the destination at actual speed + 30 min

$\therefore \frac{1500}{x}=\frac{1500}{x+100}+\frac{1}{2}$            $\left(30 \mathrm{~min}=\frac{30}{60} \mathrm{~h}=\frac{1}{2} \mathrm{~h}\right)$

$\Rightarrow \frac{1500}{x}-\frac{1500}{x+100}=\frac{1}{2}$

$\Rightarrow \frac{1500 x+150000-1500 x}{x(x+100)}=\frac{1}{2}$

$\Rightarrow \frac{150000}{x^{2}+100 x}=\frac{1}{2}$

$\Rightarrow x^{2}+100 x=300000$

$\Rightarrow x^{2}+100 x-300000=0$

$\Rightarrow x^{2}+600 x-500 x-300000=0$

$\Rightarrow x(x+600)-500(x+600)=0$

$\Rightarrow(x+600)(x-500)=0$

$\Rightarrow x+600=0$ or $x-500=0$

$\Rightarrow x=-600$ or $x=500$

∴ x = 500                 (Speed cannot be negative)

Hence, the original speed of the plane is 500 km/h.

Yes, we appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time. This reflects the caring nature of the pilot and his dedication to the work.