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.
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