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Master | Next Rank: 500 Posts
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by drhomler » Sat Jun 16, 2007 6:59 pm
A hiker walked for 2 days on the second day the hiker walked 2 hours longer and at an average speed 1 mile per hour faster than he walked on the first day if during the two days he walked a total of 64 miles and spent a total of 18 hours walking what was his average speed on the first day

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by gmatfan » Sat Jun 16, 2007 10:07 pm
First, I figured out the times. You know the total time is 18 hours, and that the second day was two hours longer, so if you let t be the time for the first day:

t+(t+2)=18
2t+2=18
2t=16
t=8 for day one and 10 for day two.

Then I set up a dist=Rate * time formula, adding the two rates and times together for the total distance.

Total distance=64 Day one distance=8r Day two distance=10(r+1)

64=8r + 10(r+1)
64=8r + 10r +10
54=18r
r=3

so 3mph
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