document.write( "Question 1184018: Susan and Jenny run a 200 m race which Susan wins by 10 m. Jenny suggests that they run
\n" ); document.write( "another race, with Susan starting 10 m behind the starting line. Assuming they run at the same
\n" ); document.write( "speeds as in the first race, what is the outcome of the race?
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Algebra.Com's Answer #814574 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "Here is a very different approach to answering the question -- WITHOUT using distance = rate * time.

\n" ); document.write( "In the 200m race, Jenny completes 190m while Susan completes 200m. That means Jenny's speed is 19/20 of Susan's speed.

\n" ); document.write( "So Jenny and Susan will finish a race together if the distance Susan has to run is 20/19 of the distance Jenny has to run.

\n" ); document.write( "But in the second race, the distance Susan has to run is 210/200 = 21/20 of the distance Jenny has to run. And 21/20 (= 1 1/20) is less than 20/19 (= 1 1/19) -- so Susan will win this race also.

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