SOLUTION: Maureen can run at a rate that is 2 miles per hour faster than her friend Hector's rate. While training for a mini marathon, Maureen gives Hector a half-hour head start and then be

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Question 942720: Maureen can run at a rate that is 2 miles per hour faster than her friend Hector's rate. While training for a mini marathon, Maureen gives Hector a half-hour head start and then begins chasing Hector on the same route. If Maureen passes Hector 12 miles from the starting point, how fast is each running?
Found 2 solutions by josmiceli, MathTherapy:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = Hector's speed in mi/hr
= Maureen's speed in mi/hr
----------------
Hector's head start is:


-------------------
Start a stopwatch when Maureen leaves
Let = time in hrs on stopwatch
when she catches Hector
----------------------
Hector's equation:
(1)
Maureen's equation:
(2)
---------------------
(1)
(1)
(1)
-------------------------
(2)
(2)
(2)
------------------------
By substitution:




( just a wild guess )

( can't have negative time )
(2)
(2)
(2)
(2)
(2)
and

-----------------
Hector runs at 6 mi/hr
Maureen runs at 8 mi/hr
-----------------------
check:
(1)
(1)
(1)
(1)
OK
Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

Maureen can run at a rate that is 2 miles per hour faster than her friend Hector's rate. While training for a mini marathon, Maureen gives Hector a half-hour head start and then begins chasing Hector on the same route. If Maureen passes Hector 12 miles from the starting point, how fast is each running?


Let Maureen’s speed be S

Then Hector’s is: S - 2

Since she passed Hector 12 miles from the starting point and with Hector getting a ˝ hr head start,

we get: 

12(2)(S – 2) – S(S – 2) = 12(2S) ---- Multiplying by LCD, 2S(S – 2)









(S - 8)(S + 6) = 0

S, or Maureen’s speed =  mph                OR                S = - 6 (ignore)

Hector’s speed: 8 – 2, or  mph 

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