Let d be the distance between A and B.

Let x be the planned speed in km/hr and let t be the time it takes in planned speed to reach B in hrs.

 

Therefore, d = xt     (1)

 

Now, speed is increased by 5km/hr (i.e. x+5) and time is reduced by 5 hrs (t-5)

 

Therefore, d = (x+5)(t-5)    (2)

 

Now, speed is increased by 10km/hr (i.e. x+10) and time is reduced by 8 hrs (t-8)

 

Therefore, d = (x+10)(t-8)     (3)

 

From 1, 2 and 3

 

xt = (x+5)(t-5) = xt - 5x + 5t - 25

 

So, x = t-5

 

Now, xt = (x+10)(t-8) = xt - 8x + 10t - 80

 

So, 8x = 10t - 80

 

Substitute x = t-5 in above equation, 

 

8t - 40 = 10t - 80

 

t = 20

 

Therefore, x = t -5 = 15 km/hr

 

Correct option is 15km/hr