Question

While boarding an aeroplane a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize 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. What values are depicted here?

Answer 1

Step 1: Given information in the question.

Let the original speed of the plane be $x$.

The time is taken to cover 1500 km with the usual speed $x$ km/hrs =$\frac{1500}{x}\text{hrs}$.

The time is taken to cover 1500 km with increased speed $\left(x+100\right)$ km/hrs =$\frac{1500}{x+100}\text{hrs}$.

Step 2: Apply the given condition.

$\begin{array}{rcl}\frac{1500}{x}-\frac{1500}{x+100}& =& \frac{1}{2}\\ \frac{1500(x+100-x)}{x\left(x+100\right)}& =& \frac{1}{2}\\ 150000\times 2& =& x^2+100x\\ x^2+100x-300000& =& 0\\ x^2+600x-500x-300000& =& 0\\ x\left(x+600\right)-500\left(x+600\right)& =& 0\\ \left(x+600\right)\left(x-500\right)& =& 0\\ x& =& -600\text{or}500\end{array}$

Neglect the negative values because speed cannot be negative.

Hence, the original speed of the plane is $500\text{km/hrs}$.