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# The hiker walked for two days. on the second day the hiker

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Manager
Joined: 01 Dec 2007
Posts: 52
The hiker walked for two days. on the second day the hiker [#permalink]

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09 Sep 2008, 04:15
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The hiker walked for two 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?

a)2mph
b)3mph
c)4mph
d)5mph
e)6mph

SVP
Joined: 17 Jun 2008
Posts: 1529
Re: Distance problem paper test code 14 section 5 question #16 [#permalink]

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09 Sep 2008, 04:46
let time taken for day 1 = t and average speed = t

Then, time taken for day 2 = t + 2 and average speed = s + 1.

Now, t + t+2 = 18 => t = 8.

Also, ts + (t+2)(s+1) = 64
Or, s = 3.

Intern
Joined: 02 Sep 2008
Posts: 45
Re: Distance problem paper test code 14 section 5 question #16 [#permalink]

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09 Sep 2008, 05:51
Hi..

I want to explain it very basics..

we have total 18 hours. and on second day hiker walks 2 hours more than yesterday.

let's first day he walked h hours, so second day is h + 2

h + ( h + 2 ) = 18

h = 8 hours.

let's assume he walked at x miles per hour on first day and x + 1 on second day.

And, hiker has walked 64 miles in last 2 days..

so, 8*x + 10*(x + 1) = 64

8x + 10x + 10 = 64

18x = 54

x = 3 mph. It is our answer.
Manager
Joined: 28 Aug 2008
Posts: 100
Re: Distance problem paper test code 14 section 5 question #16 [#permalink]

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09 Sep 2008, 14:51
Good Solution ^^

I agree with 3 mph
Current Student
Joined: 28 Dec 2004
Posts: 3342
Location: New York City
Schools: Wharton'11 HBS'12
Re: Distance problem paper test code 14 section 5 question #16 [#permalink]

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09 Sep 2008, 17:25
here is how I would do it..

suppose speed on 1st day is xmph and time is t hrs.. then on 2nd day speed is (x+1) and time is (t+2)

total distance=x*t + (x+1)(t+2)=64
t+(t+2)=18
t=8hrs

8x+(x+1)(10)=8x+10x+10=64
18x=54

solve for x; x=3

blover wrote:
The hiker walked for two 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?

a)2mph
b)3mph
c)4mph
d)5mph
e)6mph

Re: Distance problem paper test code 14 section 5 question #16   [#permalink] 09 Sep 2008, 17:25
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# The hiker walked for two days. on the second day the hiker

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