Question

A plane left $$30$$ minutes late than its scheduled time and in order to reach the destination $$1500km$$ away in time, it had to increase its speed by $$100km/h$$ from the usual speed. Find its usual speed.

Answer 1

Hint: Assume the usual speed of the plane to be $$xkm/h$$. Use the fact that speed of plane is the ratio of distance travelled with time taken to travel the given distance, form a linear equation in one variable using the given data and solve the equations to get the usual speed of the plane.


Complete step-by-step answer:
We have to find the usual speed of the plane given the data related to the distance it travels over time.

Let’s assume that the usual speed of the plane is $$xkm/h$$.
We know that the speed of a plane is the ratio of distance travelled with time taken to travel the given distance.

The plane travels at a speed of $$xkm/h$$ to cover the distance of $$1500km$$. So the time taken by plane to cover the distance at a given speed is the ratio of distance covered by the plane to the speed of the plane.

Thus, we have time taken by plane $$={\displaystyle \frac{1500}{x}}h$$.
We know that if the plane left $$30$$ minutes late, it covered the same distance of $$1500km$$ by increasing its speed by $$100km/h$$.
As we know that $$1hr=60min$$, dividing the equation by $$2$$ on both sides, we have $$30min={\displaystyle \frac{1}{2}}hr$$.

Thus, the delayed time of plane $$={\displaystyle \frac{1500}{x}}-{\displaystyle \frac{1}{2}}h$$ and new speed of plane $$=x+100km/h$$.

Thus, we have $$x+100={\displaystyle \frac{1500}{{\displaystyle \frac{1500}{x}}-{\displaystyle \frac{1}{2}}}}$$.
Further simplifying the equation, we have $${\displaystyle \frac{1500}{x}}-{\displaystyle \frac{1}{2}}={\displaystyle \frac{1500}{x+100}}$$.
Rearranging the terms, we get $${\displaystyle \frac{1500}{x}}-{\displaystyle \frac{1500}{x+100}}={\displaystyle \frac{1}{2}}$$.
Thus, we have $${\displaystyle \frac{1500(x+100)-1500x}{x(x+100)}}={\displaystyle \frac{1}{2}}$$.
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{1500\times 100}{x(x+100)}}={\displaystyle \frac{1}{2}}\\ & \Rightarrow x(x+100)=2(1500\times 100)\\ & \Rightarrow x^2+100x-300000=0\\ & \Rightarrow x^2+600x-500x-300000=0\\ & \Rightarrow x(x+600)-500(x+600)=0\\ & \Rightarrow (x+600)(x-500)=0\\ & \Rightarrow x=500,-600\end{array}$$

As the speed can’t be a negative quantity, we have $$x=500km/h$$ as the usual speed of the plane.
Hence, the usual speed of the plane is $$500km/h$$.

Note: We can also solve this question by forming linear equations in two variables taking $$x$$ as the speed of the plane and $$y$$ as the time taken by the plane to cover the distance, and then solve those equations to find the speed of the plane.