An aeroplane takes off 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/h from its usual speed. Find its usual speed.
$$time\ =\ \frac{dis\tan ce}{speed}$$
$$t1=\ \frac{1500}{S}$$
$$t2=\ \frac{1500}{S+250}$$
Given t1 - t2 = 1/2 hours
So, $$\frac{1500}{S}\ -\frac{1500}{S+250}=\frac{1}{2}$$
On solving S=750kmph
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