SOLUTION: 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 s

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Question 1184018: 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?
Found 3 solutions by ikleyn, MathTherapy, greenestamps:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
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?
~~~~~~~~~~~~~~~~~~ 


Let x be the Susan's rate (in meters per second), and let y be the Jenny's rate.

Susan's time is    seconds;  it is equal to  , according to the condition.


So, we have   = ,  or    =  = .   (1)


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


The problem asks which value is lesser,    or  .


From (1),  we have x = ,  therefore

    Susan's time will be   =  =  =  = .

    Obviously, it is less than the Jenny's time  .


So, the conclusion is:  Susan will win (will be first) in the second scenario run.    ANSWER

Solved and thoroughly explained.


//////////////


There is even more easy and comprehensive explanation.



        From the given part,  Susan runs faster than Jenny.

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

        If we add the same  ANY  arbitrary  distance  D  (like  10 meters in the problem)  to  200 m and 190 m respectively,
        Susan will cover the new distance of  200 + D  meters   QUICKER   than  Jenny will cover her distance of  190 + D  meters.



Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

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?
This TRAVEL problem, just like most of them involves using the travel formula: Distance = Speed * Time

Let Susan’s speed be S
Then time Susan took to complete the race = 
Then, Jenny’s speed, with her time also being  is 

If Susan starts 10 m behind the starting line, she’ll have to travel 200 + 10 = 210 m
Jenny, however, will travel 200 m

Time Susan takes to travel 210 m: 
Time Jenny takes to travel 200 m: 

Susan’s time: 
Comparing Susan’s time:  to Jenny’s time: , we see that Susan’s time is less, so Susan will win the race, again!
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Here is a very different approach to answering the question -- WITHOUT using distance = rate * time.

In the 200m race, Jenny completes 190m while Susan completes 200m. That means Jenny's speed is 19/20 of Susan's speed.

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.

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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