Question

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

Answer 1

Given:

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

To find:

The usual speed of the plane

Formula used for solving the problem:

$\boxed{\bold{Time = \frac{Distance}{Speed}}}$

Solution:

Total distance to be covered to reach the destination = 1500 km

Let "x" km/hr represent the usual speed of the plane.

So, after increasing, the new speed of the plane will be = "x + 100" km/hr

Also,

Time taken by the plane with the usual speed = $\frac{1500}{x}$ hr

and

Time taken by the plane with the increased speed = $\frac{1500}{x + 100}$ hr

Now, according to the question, we can form an equation as,

$\frac{1500}{x} - \frac{1500}{(x + 100)} = \frac{30}{60}$

$\implies 1500 [\frac{1}{x} - \frac{1}{(x + 100)}] = \frac{1}{2}$

$\implies [\frac{x\: +\: 100 \:-\: x}{x(x + 100)}]= \frac{1}{3000}$

$\implies 300000 = (x^2 + 100x)$

$\implies x^2 + 100x - 300000 = 0$

$\implies x^2 + 600x - 500x - 300000 = 0$

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

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

$\implies x=-600 \:or\: x = 500$

the speed can not have a negative value, therefore, neglecting - 600, we get

$\implies \bold{ x = 500\:km/hr}$

Thus, the usual speed of the plane was 500 km/hr.

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