An aircraft was to take off from a certain airport at 8 am but it was delayed by 30 min To make up for the lost time it was to increase its speed by 250 km/hr from the normal speed to reach its destination 1500 km on time What was the normal speed of the aircraft?

Let the normal speed be v km/hr; using the formula time equals distance divided by speed, the normal time is 1500 / v and the delayed time is 1500 / (v + 250). The difference in time is 30 minutes or 0.5 hours, so 1500 / v - 1500 / (v + 250) = 0.5. Multiplying by v(v + 250) gives 1500v + 375000 - 1500v = 0.5v^2 + 125v, which simplifies to v^2 + 250v - 750000 = 0. Factoring this quadratic equation results in (v - 750)(v + 1000) = 0, so the normal speed is 750 km/hr.

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