Question

A plane left 30 minutes late than the schedule time and in order to reach its destination 1500 km away in time,it has to increase its speed by 100 km/h from its usual speed.find its usual speed.

Answer 1

Let sedule time is T.
Then speed of aeroplane = distance covered/time taken = 1500km/T hour ---(1)

A/C to question,
Time taken by Aeroplane in 2nd case = sedule time - 30 minutes
= (T - 1/2) hour
Now, speed of aeroplane = distance covered/time taken = 1500/(T - 1/2) ----(2)

Now, 1500/(T - 1/2) - 1500/T = 100 [according to question]
1500[ (T - T + 1/2)/(T² - T/2)] = 100
15×1/2 = T² - T/2
15 = 2T² - T
2T² - T - 15 = 0
2T² - 6T + 5T - 15 = 0
T = 3 and -5/2 , T ≠ -5/2 so, T = 3

Hence, usual speed of aeroplane = 1500/T = 1500/3 = 500 I'm/h

Answer 2

Let the usual speed of the plane = x km/h

Time taken to cover 1500 km = (1500/x) hours

New speed = (100 + x) km/h

Time taken = 1500/(100 + x) - This time is 30 minutes (0.5 hours) less than the first time

So;

1500/x - 1500/(100 + x) = 0.5

150000 = 0.5(x² + 100x)

x² + 100x - 300000 = 0

This is a quadratic equation

x = [-200+/-√(100)² - 4(1)(-300000)]/2(1)

x = 500 or - 600

So, the usual speed of the plane = 500 km/h