A flight, travelling to a destination 11,200 kms away, was supposed to take off at 6:30 AM. Due to bad weather, the departure of the flight got delayed by three hours. The pilot increased the average speed of the airplane by 100 km/hr from the initially planned average speed, to reduce the overall delay to one hour. Had the pilot increased the average speed by 350 km/hr from the initially planned average speed, when would have the flight reached its destination?

Let the original planned speed be V km/hr; because the initial delay of 3 hours was reduced to 1 hour, the flight saved 2 hours of flight time. Using the formula Time equals Distance divided by Speed, the equation becomes 11200 divided by V minus 11200 divided by (V + 100) equals 2. Multiplying by V multiplied by (V + 100) gives 1120000 equals 2 multiplied by V squared plus 200 multiplied by V, which simplifies to V squared plus 100 multiplied by V minus 560000 equals 0. Solving this quadratic equation yields the original speed V as 700 km/hr; with the new speed increase of 350 km/hr, the revised speed becomes 1050 km/hr. The flight time at 1050 km/hr is 11200 divided by 1050, which equals 10 hours and 40 minutes; adding this to the delayed departure time of 9:30 am results in an arrival time of 8:10 PM. The result is 8:10 PM.

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