ChrisLele wrote:
Think of it this way: if we wanted the two to finish at the same time, we would give B would only run 13/20 of the race, or the inverse of 20/13 (20/13 x 13/20 = 1, meaning they finish at the same time).
B finishes 20% ahead of A, so we take the 13/20 and we multiply by 4/5, thereby shortening the amount B has to run by 4/5 or 80%. (4/5 x 13/20) = 52/100.
Then there is the final twist to the problem. Lets think of it this way, B only has to run 52 meters out of one hundred whereas A has to run all 100 meters. However that doesnt mean B gets a head start of 52 (then he would only have to run 48 meters). Because B has to run 52 meters, he only gets a head start of 48 meters.
Therefore the answer is 48/100 or 48% (B).
B finishes 20% ahead of A, so we take the 13/20 and we multiply by 4/5, thereby shortening the amount B has to run by 4/5 or 80%. (4/5 x 13/20) = 52/100.
Then there is the final twist to the problem. Lets think of it this way, B only has to run 52 meters out of one hundred whereas A has to run all 100 meters. However that doesnt mean B gets a head start of 52 (then he would only have to run 48 meters). Because B has to run 52 meters, he only gets a head start of 48 meters.
Therefore the answer is 48/100 or 48% (B).
Ok, so we know that B runs slower than A yet still manages to beat A by a length of 20% of the total distance of the race. As stated in the question, B must start from a distance ahead of A that allows him to win my 20% even though A will be gaining on him the entire duration of the race. Why do we multiply 13/20 by 80%? I get that d=s*t and I see where why 13/20 is used but why is his distance 80% of A's? (I see that it comes from 100-20 but why???) Help!!!