A plane left 30 min later than its scheduled time to reach its destination 1500 km away. In order to reach in time it increases its speed by 250 km/h. What is its original speed?
Let the usual speed of the plane be v kmph and the time taken by the plane to cover 1500m = t hrs
To reach the destination on time, the increased speed of the plane = v+250 kmph
Since he started 30 min later, the time taken should be t-$$\ \frac{\ 1}{2}$$ hr
v*t =Â (v+250)(t-$$\ \frac{\ 1}{2}$$)Â
v = (250t-125)2
$$\ \frac{\ 1500}{v+250}=t-\ \frac{\ 1}{2}$$
$$\ \frac{\ 1500}{500t-250+250}=t-\ \frac{\ 1}{2}$$
$$\ \frac{\ 3}{t}=t-\ \frac{\ 1}{2}$$
On solving for t, we get
t= 2 hrs
His original speed =Â $$\ \frac{\ 1500}{2}$$ = 750 kmph
B is the correct answer.
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