i am not sure. but acc to me soln wud be like this

suppose after t time B wud catch up A

4(t+2) (1+2+...t)

4(t+2) t(t 1)/2 Ans.

sol: 4(t+2) (1+2+3+...+t); t in hours.

8+4t t(t+1)/2; since sum of the natural number

1+2+3+...+t t(t+1)/2

t^2-7t-16 0;

t 9.81hours

Ans:none

answer is 9.48.A starts at 11.00 am with 4kmph speed that means he covers 8 km in 2 hrs i.e. at 1 pm.Now at 1 pm b starts his speed increases by 1 kmph for every hour.there fore they both can catch up at speeds for 7 hrs 4+4+4+4+4+4+4 28 for A

1+2+3+4+5+6+7 28 for B

But A has a head start of 2 hrs this can be covered in the next hour but again A covers 4hrs in this 1 hr.In this way it can be worked out and the answer is 9.48hrs.