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.

Answer:

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.