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A hiker walks for two days. On the second day the hiker

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A hiker walks for two days. On the second day the hiker [#permalink] New post 07 Oct 2005, 06:13
A hiker walks for two days. On the second day the hiker walked 2 hours longer and at an average speed 1 mph 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?

(A) 2 mph
(B) 3 mph
(C) 4 mph
(D) 5 mph
(E) 6 mph

Please explain your answer.
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 [#permalink] New post 07 Oct 2005, 06:31
First find the average speed walked:

64 miles / 18 hours = 3.55

next the amount of time walked each day:

X + X + 2 = 18 so X=8

1st day: 8hrs
2nd day: 10 hrs

Now plug back into the average formula:

(8x + 10(x+1)) / 18 = 3.55
18x=54
x=3

So I would choose (B) 3mph

Brendan.
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 [#permalink] New post 07 Oct 2005, 07:49
Overall time on the road: x (first day) + (x + 2) (second day) = 18 -> 2x +2 = 18 -> x = 8 (time on the road for the first day)

Total mileage: 8*v (first day) + 10*(v+1) (second day) = 64
-> 18v = 54 -> v = 3(mph)

-> another B
  [#permalink] 07 Oct 2005, 07:49
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