SOLUTION: Justin can run 10 kilometers in the same amount of time that Leo can run 12 kilometers. If Leo can run 1 kilometer per hour faster than Justin, how fast can Leo run?

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Question 1205756: Justin can run 10 kilometers in the same amount of time that Leo can run 12 kilometers. If Leo can
run 1 kilometer per hour faster than Justin, how fast can Leo run?

Found 3 solutions by ikleyn, greenestamps, josgarithmetic:
Answer by ikleyn(52887)   (Show Source): You can put this solution on YOUR website!
.
Justin can run 10 kilometers in the same amount of time that Leo can run 12 kilometers. If Leo can
run 1 kilometer per hour faster than Justin, how fast can Leo run?
~~~~~~~~~~~~~~~~~

Let x be Leo' speed, in kilometers per hour.

Then Justin' speed is (x-1) km/h.


From the problem, the time equation is

     =   (Justin's time for 10 km is the same as Leo's time for 12 km).


So

    10x = 12(x-1)

    10x = 12x - 12

    12 = 12x - 10x

    12 = 2x

    x  = 12/2 = 6.


ANSWER.  Leo' speed is 6 km/h.

Solved.



Answer by greenestamps(13209)   (Show Source): You can put this solution on YOUR website!


Since the times are the same, the ratio of their speeds is the same as the ratio of the distances, which is 12:10 or 6:5.

Then, since Leo's speed is 1 km/hr faster than Justin's, simple reasoning or formal algebra shows that Leo's speed is 6 km/hr and Justin's is 5 km/hr.

ANSWER: 6 km/hr


Answer by josgarithmetic(39630)   (Show Source): You can put this solution on YOUR website!
-----------------------------------------------------------------------------------------------
Justin can run 10 kilometers in the same amount of time that Leo can run 12 kilometers. If Leo can
run 1 kilometer per hour faster than Justin, how fast can Leo run?
----------------------------------------------------------------------------------------------
              SPEED           TIME           DISTANCE

JUSTIN        r-1             10/(r-1)         10

LEO                          12/r             12

"... in the same amount of time."






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