Question

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

Answer 1

Let the usual speed of plane be x km/h.
Increased speed = (x + 100) km/h.
$\therefore$ Distance to cover = 1500 km.
Time taken by plane with usual speed = $\frac{1500}{x}$ hr.
Time taken by plane with increased speed $=\frac{1500}{(100+x)}$hrs.
According to the question,
$\frac{1500}{x}-\frac{1500}{(100+x)}=\frac{30}{60}=\frac{1}{2}$
$1500[\frac{1}{x}-\frac{1}{x+100}]=\frac{1}{2}$
$1500\left[\frac{x+100-x}{\left(x\right)(x+100)}\right]=\frac{1}{2}$
$\frac{1500\times 100}{x^2+100x}=\frac{1}{2}$
$x^2+100x=300000$
$x^2+100x-300000=0$
$x^2+600x-500x-300000=0$
$x(x+600)-500(x+600)=0$
$(x+600)(x-500)=0$
Ether x + 600 = 0
x = -600 (Rejected)
x - 500 = 0
x = 500
$\therefore$ Usual speed of plane = 500 km/hr.