Can u plz explain with detail...about d qstn and solve them

let the usual speed be : x
and increased speed be : x+250
let theusual time be : t
and the decreased time be :t-30 =t-1/2
distance will be the same ie:1500
now 
t=1500/x
if we subtract 30 min ie 1/2 hour from the usual time we will get 1500/x+250 as the decreased time 
now put this as a equation 
1500/x-1/2 = 1500/x+250
3000-x / 2x =1500/x+250 
now cross multiply 
3000x - x² + 750000 - 250x = 3000x
​
here 3000x will be canclled  on both the sides 
there fore equation  x² +250x-750000 is left 
x² +1000x-750x -750000 =0
x(x+1000)-750(x+1000) = 0
(x+1000)(x-750)=0
x = -1000,+750
 speed can never be negative therefore usual speed= 750km/hr
hope it helps you