Question 1184018
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Susan and Jenny run a 200 m race which Susan wins by 10 m. Jenny suggests that they run
another race, with Susan starting 10 m behind the starting line. Assuming they run at the same
speeds as in the first race, what is the outcome of the race?
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<pre>
Let x be the Susan's rate (in meters per second), and let y be the Jenny's rate.

Susan's time is  {{{200/x}}}  seconds;  it is equal to  {{{(200-10)/y}}}, according to the condition.


So, we have  {{{200/x}}} = {{{190/y}}},  or   {{{x/y}}} = {{{200/190}}} = {{{20/19}}}.   (1)


Next, in the other scenario, Susan's time to complete 210 m run will be  {{{210/x}}}  seconds,
while the Jenny's time to complete her run of 200 m  will be  {{{200/y}}}.


The problem asks which value is lesser,  {{{210/x}}}  or  {{{200/y}}}.


From (1),  we have x = {{{(20/19)y}}},  therefore

    Susan's time will be  {{{210/x}}} = {{{210/((20/19)y)}}} = {{{(210*19)/(20y)}}} = {{{(21*19)/(2y)}}} = {{{199.5/y}}}.

    Obviously, it is less than the Jenny's time  {{{200/y}}}.


So, the conclusion is:  Susan will win (will be first) in the second scenario run.    <U>ANSWER</U>
</pre>

Solved and thoroughly explained.



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<H3>There is even more easy and comprehensive explanation.</H3>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;From the given part, &nbsp;Susan runs faster than Jenny.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;They run &nbsp;200 &nbsp;meters &nbsp;(Susan) &nbsp;and &nbsp;190 &nbsp;meters &nbsp;(Jenny) &nbsp;in the same time.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;If we add the same &nbsp;ANY &nbsp;arbitrary &nbsp;distance  &nbsp;D  &nbsp;(like &nbsp;10 meters in the problem) &nbsp;to &nbsp;200 m  and  190 m respectively,

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Susan will cover the new distance of &nbsp;200 + D &nbsp;meters &nbsp;&nbsp;QUICKER &nbsp;&nbsp;than &nbsp;Jenny will cover her distance of &nbsp;190 + D &nbsp;meters.