Question

a plane left 30 minutes late then its scheduled time and order to reach the destination 1500 km away in time it had to increase its speed by 100km/h from usual speed find its usual speed

Answer 1

Let the usual speed of the plain be x km/hr
According to que we have

1500/x = 1500/x + 100 + 1/2 ( because time = distance/speed)

1500/x - 1500/x+100 = 1/2
1500 (x + 100) -1500x/ x ( x + 100) = 1/2
1500x + 150000 - 1500x / x² + 100 = 1/2
2 ( 150000) = x² + 100
x² + 100 - 300000 = 0

Using quadratic formula,

-100 ± √ (100)² - 4 × (- 300000) / 2
-100 ± √ 10000 + 1200000/2
-100 ± √1210000 / 2
- 100 ± 1100 / 2

on taking - sign we have -1200/2 = -600
or on taking + sign we have 1000 /2 = 500

But speed can never be negative so the usual speed of plain is 500 km /hr ...

Hope it will be contented....

Answer 2

let original speed = x km/hr
Increased speed = x+100 km/hr
Distance = 1500 km

ATQ
1500/x - 1500/(x+100) = 30/60

1500{ 1/x - 1/x+100} = 1/2

1500{ x+100-x/x(x+100) } = 1/2

150000/x^2+100x = 1/2

x^2+100x = 300000

x^2+100x-300000=0

x^2+600x-500x-300000=0
x(x+600)- 500(x+600) =0

x = 500km/hr Answer reject -ve value