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A hiker walked for two days. On the second day the hiker wal

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A hiker walked for two days. On the second day the hiker wal  [#permalink]

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New post 24 Dec 2010, 07:58
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A 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) 2 mph
(B) 3 mph
(C) 4 mph
(D) 5 mph
(E) 6 mph

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Re: A hiker walked for two days  [#permalink]

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New post 24 Dec 2010, 08:22
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ajit257 wrote:
A 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) 2 mph
(B) 3 mph
(C) 4 mph
(D) 5 mph
(E) 6 mph


ALGEBRAIC APPROACH:

As on the second day the hiker walked 2 hours longer than he walked on the first day and spent a total of 18 hours walking then t+(t+2)=18 --> t=8. So the hiker walked 8 hours on the first day and 10 hours on the second day;

Let the rate on the first day be r then: 8r+10(r+1)=64 --> r=3.

Answer: B.

10 SECOND APPROACH:

Average rate of the hiker is (total distance)/(total time)=64/18=~3.6, now r<3.6<r+1 (the weighted average of 2 individual averages, r and r+1, must lie between these individual averages) --> only answer choice B fits.

Answer: B.
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Re: A hiker walked for two days  [#permalink]

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New post 25 Dec 2010, 13:39
ron covered this problem in his workshop on Rate/Time/Distance (April 29, 2010)
http://www.manhattangmat.com/thursdays-with-ron.cfm
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Re: A hiker walked for two days  [#permalink]

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New post 25 Dec 2010, 16:32
10 sec approach is great :)
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Re: A hiker walked for two days  [#permalink]

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New post 25 Dec 2010, 18:34
yep, liked the 10 sec app as well...
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Re: A hiker walked for two days  [#permalink]

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New post 22 Feb 2011, 15:03
Let x be the number of miles in first day ; 64-x = number of miles in second day
given,
t+ t+2 =18 -> t=10
Speed in first day = speed in second day - 1

Therefore, \((64-x)/10\) - \(x/8\) = 1

x = 24

Therefore, speed = \(d/t\) = \(24/8\) = \(3\)
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Re: A hiker walked for two days. On the second day the hiker wal  [#permalink]

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New post 27 Jun 2013, 02:25
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: A hiker walked for two days. On the second day the hiker wal  [#permalink]

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New post 27 Jun 2013, 19:02
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1. Average speed for the first 2 days is 64/18 = 3.6 approx. Since the speed on the first day is lower than that on the second day, the speed on the first day should be less than 3.6 . Only A and B fit this requirement
2. It cannot be choice A because if the speed on the first day is 2 miles/hr then the speed on the second day would be 3 miles/hr and so the average speed cannot be 3.6 miles/hr.

So the answer is choice B.
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Re: A hiker walked for two days. On the second day the hiker wal  [#permalink]

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New post 05 Aug 2013, 14:24
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A 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?

Hiker walked for two days for a total of 18 hours. He spent two more hours walking on the second day than the first: h1+h2=18 ===> h+h+2 = 18 ===> 2h = 16 ===> h=8.
He spent 8 hours the first day and 10 hours the second day.
Day 1 = 8h
Day 2 = 10h

Average speed = total distance/combined rate
Average speed = 64/r1+r2

rate = distance/time
r = d/8
r+1 = d/10 ===> r = (d/10)-1

64/(d/8) + (d/10)-1


Why wouldn't I use that formula? Is it because I would be trying to figure out average speed when I don't know the speed of the hiker on day one or two?

distance=rate*time
d1=r*8
d2=(r+1)*10

d1+d2=total distance
d1+d2=64
(r8)+(10r+10)=64
18r=64
r=3

ANSWER: (B) 3 mph
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A hiker walked for two days. On the second day the hiker wal  [#permalink]

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New post 30 Aug 2015, 14:16
1
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


1-st day: vt=d1(v:velocity, t:time taken, d1:distance)
2-nd day: (v+1)(t+2)=d2 and d1+d2=64

t+t+2=18, 2t=16, t=8 and 8v=d1 <-- equation 1), 10(v+1)=d2, 10v+10=d2 <-- equation 2)

equation 1)+ equation 2), then 8v+10v+10=d1+d2=64, 18v=54, v=3

Therefore the answer is B. thank you!!
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Re: A hiker walked for two days. On the second day the hiker wal  [#permalink]

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New post 14 Feb 2018, 16:27
Hi All,

This question requires a certain degree of note-taking and organization. Once you've got everything on the pad, you can then approach the math in a couple of ways:

Since we're dealing with distance, rate and time, I'm going to make two equations based on the two days of travel:

Day 1: Distance = (R mph)(T hours)
Day 2: Distance#2 = (R+1 mph)(T+2 hours)

Since the TOTAL TIME for both days was 18 hours, we can solve for T...

T + (T+2) = 18
2T = 16
T = 8

Since the TOTAL DISTANCE for both days was 64 miles, we can now set up 1 gigantic equation....

64 miles = (R)(8) + (R+1)(10)

At this point, we have 1 variable and 1 equation, so we can solve for R....

64 = 8R + 10R + 10
54 = 18R
3 = R

The prompt asks for the speed on DAY 1; since the hiker traveled R mph on Day 1, the answer is 3...

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Re: A hiker walked for two days. On the second day the hiker wal &nbs [#permalink] 14 Feb 2018, 16:27
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