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.