Question

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

Answer 1

Let the usual time taken by the aeroplane = x km/hr
Distance to the destination = 1500 km
Case (i)
Speed = Distance / Time = (1500 / x) Hrs
 
Case (iI)
Time taken by the aeroplane = (x - 1/2) Hrs
Distance to the destination = 1500 km
Speed = Distance / Time = 1500 / (x - 1/2) Hrs
 
Increased speed = 100 km/hr
 
⇒ [1500 / (x - 1/2)] - [1500 / x] = 100
⇒ 1/(2x2 - x) = 1/2
⇒ 2x2 - x = 2
=2x^2-x-2=0
=2x^2-2x+x-2=0
⇒ 2x(x - 1)+2(x-1) = 0
⇒ x = 1,-1

Since, the time can not be negative,
The usual time taken by the aeroplane = 1 hrs
and the usual speed = (1500 / 1) = 1500 km/hr.

Answer 2

1500/x-1500/x+100=1/2
1500x+150000-1500x/x(x+100)=1/2
150000*2=x(x+100)
300000=x2+100x
b2-4ac= 10000-4*1*-300000
= 1210000
√1210000=1100
using quadratic eqn we can find x