An airplane was supposed to cover the distance of 2900 km.
However, after covering 1700 km, it had to land and wait on the ground for 1 hours and 30 minutes.
After it took off again, its average speed was 50 km/h less than before.
Find the original average speed of the plane if it completed the flight 5 hours after departure.
Let x = original speed of the plane (km/hr).
It flew 1700 km at x km/hr.
. . This took: .1700/x hours.
It flew the remaining 1200 km at (x - 50) km/hr.
. . This took: .1200/(x - 50) hours.
The plane flew a total of 3½ hours.
. . And there is our equation: .1700/x + 1200/(x - 50) .= .7/2
Multiply through by 2x(x - 50): .3400(x - 50) + 2400x .= .7x(x - 50)
This simplifes to the quadratic: .7x² - 6150x + 170,000 .= .0
. . which factors: .(x - 850)(7x - 200) .= .0
. . and has roots: .x .= /850, 200/7
We reject the smaller root: x = 200/7
At about 28.6 km/hr, it would take over 100 hours to fly 2900 km.
Therefore, the original speed of the plane was: 850 km/hr.