Speed of

A - 100/x

B - 70/x

C - 63/x

ie., when A reaches the finish line after x seconds, B will be at the 70 meter point and C at 63.

since distance is proportional to time, we have

distance-------time

70---------------- x

100 --------------?

or time taken for B to reach the finish line = ((100*x)/70)= (10/7)*x ( assuming the speed will remain same throughout).

Now the distance that C covers at the same time will be helpful in determining how much of a head start B can give to C.

distance C covers in (10/7)x seconds is

distance---------- time

63------------------ x

? ------------------ (10/7)x

= (10/7)*x*63/x = 90 meter.

which means after 100 meters B can create a lead of 10 meters with respect to C or **B can afford to give C a head start of 10 meter or less to win.**