Question

A plane left 30 minutes later then its sheduval time and in order to reach destination 1500km away in time it had to increase speed by 100km/h from the usaval speed find the usaval speed

Answer 1

let the usual time taken by the aeroplane = x distance to the destination =1500km
Case 1. -
speed = distance /time
=1500/x
Case 2.- time taken by the aeroplane = (x-1/2)
distance =1500km
speed =1500/(x-1/2) {using above formula }
increased speed =100km/hr
[ 1500/(x-1/2)] - [1500/x ]= 100
[1500*2/(2x-1)] -[1500/x] =100
[2/2x-1] - [1/x] =1/15
[2x-2x+1]/(2x^2 -x) =1/15 { by taking lcm }
Equation becomes -2x^2 -x -15 =0
2x^2 -6x+5x -15 =0
2x(x-3) +5(x-3) =0
x =3 or -5/2{neglected}

there fore usual time taken is 3hrs.
so speed, =1500/3 =500km/hr

Answer 2

heres the answer
hope it helps