A Sherando boy and a Handley girl ran a 100
meter race. The Handley girl crossed the
finish line when the Sherando boy had gone 95
meters, so she won the race by 5 meters.
When they raced a second time, the Handley
girl wanted to make the contest more even so
she handicapped herself by starting 5 meters
behind the start line.
If the two ran at the same constant speed as
before, who won the second race? Explain (or
give their times relative to each other).
Solution to Problem: The Handley girl runs 100 meters in the same time that the Sherando boy runs 95 meters; therefore, the rate of the girl is 100 meters per some unit of time and the rate of the boy is 95 meters per the same unit of time.
Below is the Rate-Time-Distance table for the second race:
| Runner | Rate | Time | Distance |
| Boy | 95 m/unit | 100/95 units | 100 m |
| Girl | 100 m/unit | 105/100 units | 105 m |
So the Boy's time is 100/95 or 1.05266316 units of time, and the Girl's time is 105/100 or 1.05 units of time. Therefore, the Handley Girl won the second race as well! She passed the Sherando Boy at 95 meters.
Correctly solved by:
| 1. Michael Leatherman | Norfolk, VA |
| 2. Elizabeth Pleacher Cotter | Centreville, VA |
| 3. Kaveh Sadeghzadeh | Williamsburg, VA |
| 4. Andrew Crosby | Winchester, VA |
| 5. Barrett Waybright | Winchester, VA |