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

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Manager
Joined: 28 Aug 2010
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A hiker walked for two days. On the second day the hiker wal  [#permalink]

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24 Dec 2010, 06:58
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15% (low)

Question Stats:

81% (02:31) correct 19% (02:50) wrong based on 481 sessions

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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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Joined: 02 Sep 2009
Posts: 52285
Re: A hiker walked for two days  [#permalink]

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24 Dec 2010, 07:22
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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.

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.

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

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25 Dec 2010, 12:39
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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25 Dec 2010, 15:32
10 sec approach is great
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Re: A hiker walked for two days  [#permalink]

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25 Dec 2010, 17:34
yep, liked the 10 sec app as well...
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Re: A hiker walked for two days  [#permalink]

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22 Feb 2011, 14:03
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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27 Jun 2013, 01:25
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=44
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Re: A hiker walked for two days. On the second day the hiker wal  [#permalink]

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27 Jun 2013, 18:02
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2
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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05 Aug 2013, 13:24
1
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

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

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30 Aug 2015, 13:16
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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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13347 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A hiker walked for two days. On the second day the hiker wal [#permalink] ### Show Tags 14 Feb 2018, 15:27 1 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... Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: A hiker walked for two days. On the second day the hiker wal &nbs [#permalink] 14 Feb 2018, 15:27
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