Question 451949: At a high school cross country meet, Jared jogged 8 mph for the first part of the race, then increased his speed to 12 mph for the second part. If the race was 21 mi. long and Jared finished in 2hr., how far did he jog the faster pace
Answer by pedjajov(51)   (Show Source):
You can put this solution on YOUR website! Jared ran a distance d=12mi that consists of two parts d1 and d2 -> d=d1+d2.
 , he ran 8mi/h for t1 hours
 , he ran 12mi/h for t2 hours
8*t1+12*t2 = 21
Since he finished his race in 2 hours it means
t1 + t2 = 2, from here we have t1=2-t2
If we substitute this into first equation we have:
8(2-t2)+12*t2=21, multiply
16-8*t2+12*t2=21, combine
16-4*t2 =21, subtract 16 from both sides
-4*t2 =-5, divide both sides by -4
t2 = 5/4
So he ran faster for 5/4 hours or 75 minutes.
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