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# Rate problem

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Wilma and Betty ran a 100-meter race at top speed, and Wilma finished when Betty had 10 meters to go. They decided to run again, but this time Wilma gave Betty an advantage. Wilma's starting point was 10 meters behind the original starting point. Given that Wilma and Betty run at the same speeds as the previous race, how many meters will Betty be from the finish line when Wilma crosses it?

Apr 10, 2018

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Call Wilma's speed,  W

So when Wilma runs 100m, Betty runs 100 - 10 = 90m

So...Betty's speed is 9/10 of Wilma's =  (9/10)W

Note that Wilma rums 110 m  and her rate  = W

Call the distance that  Betty runs, D,   and her rate is  (9/10)W

So ...D /R  = Time

so....equating times, we have

110 / W  =  D / [(9/10)W]         multiply both sides by [ (9/10)W ]

110 (9/10)W / W  =  D

110 (9/10)  = D

[(110/ 10) * 9]   =  D

11 * 9     = D

99  = D

So....Betty  runs 99m  in the same time that Betty runs 110m....so....Betty is 1  m   from the finish line when Wilma finishes the 110m   Apr 10, 2018