• question_answer A plane left 30 minutes 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.

 Let the usual speed of plane be x km/h. Increased speed = (x+100) km/h. $\therefore$ Distance to cover $=1500\text{ }km$ 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, $\frac{1500}{x}-\frac{1500}{(100+x)}=\frac{30}{60}=\frac{1}{2}$ $1500\left[ \frac{1}{x}-\frac{1}{x+100} \right]=\frac{1}{2}$ $1500\left[ \frac{x+100-x}{(x)(x+100)} \right]=\frac{1}{2}$ $\frac{1500\times 100}{{{x}^{2}}+100x}=\frac{1}{2}$ ${{x}^{2}}+100x=300000$ ${{x}^{2}}+100x-300000=0$ ${{x}^{2}}+600x-500x-300000=0$ $x(x+600)-500(x+600)=0$ $(x+600)(x-500)=0$ Either                       $x+600=0$ $x=-600$ (Rejected) Or                            $x-500=0$ $x=500$ $\therefore$ Usual speed of plane = 500 km/hr. You will be redirected in 3 sec 