Question

Two planes left simultaneously from the same airport and headed in the same direction towards another airport 1800 km away. The speed of one of the planes was 100 km/hour slower than the speed of the other plane, and so it arrived at its destination 36 minutes after the faster plane. Find the speeds of both planes.

Answer 1

speed of faster plane --- x km/h
speed of slower plane ---- x - 100 km/h

time for faster plane = 1800/x
time for slower plane = 180/(x-100)

1800/(x-100) - 1800/x = 36/60 = 3/5
multiply each term by 5x(x-100) , the LCD

9000x - 9000(x-100) = 3x(x-100)
9000x - 9000x + 900000 = 3x^2 - 300x
3x^2 - 300x - 900000 = 0
x^2 - 100x - 300000 = 0
(x - 1000)(x +
x^2 - 100x = 300000
x^2 - 100x + 2500 = 300000+2500
(I completed the square)
(x - 50)^2 = 302500
x - 50 = ±√302500 = ± 550

x = 600 or a negative

the faster plane went 600 km/h
the slower plane went 500 km/h

check:
time of faster = 1800/600 = 3 hrs = 180 minutes
time of slower plane = 1800/500 = 3.6 hrs = 216 minutes

difference = 216-180= 36 minutes
My answer is correct

Answer 2

your answer is not correct