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# I bet most of you know this one as it was discussed before.

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Manager
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I bet most of you know this one as it was discussed before. [#permalink]

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21 Jan 2008, 03:52
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I bet most of you know this one as it was discussed before. I'm less interested in the algebraic way to solve this (though any post is much appreciated) but more in a different, faster approach
So how would you solve it?

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
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Re: a different, faster approach to solve this PS [#permalink]

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21 Jan 2008, 04:42
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Expert's post
$$V_{av}=64/18=3.8...$$

$$V_1<V_{av}<V_1+1$$

only 3mph works
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Director
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Re: a different, faster approach to solve this PS [#permalink]

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21 Jan 2008, 04:45
x+x+2=18
2x = 16
x = 8

he walked for 8 hours during the first day and 10 hours during the second day

Here is where I would do very quick and easy mental math without writing anything down.

Take each of the answer choices, add one and multiply by 10 to see how far he walked the second day. Then just take the original answer choice, multiply it by 8 and add it to that answer to get total distance.

3*10=30 + 2(8) = 46
4*10=40 + 3(8) = 64 DONE

and that's all you have to do.

Average speed of the first day was 3mph, average speed of the second day was 4mph.

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Re: a different, faster approach to solve this PS [#permalink]

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21 Jan 2008, 08:49
GGUY wrote:
I bet most of you know this one as it was discussed before. I'm less interested in the algebraic way to solve this (though any post is much appreciated) but more in a different, faster approach
So how would you solve it?

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

R x x+1
T t t+2
D d 64-d

2t+2=18 --> t=8

x8=d

(x+1)(10)=64-d --> 10x+10=64-8x --> 18x=54

x=3
Manager
Joined: 26 Jun 2007
Posts: 104
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Kudos [?]: 59 [0], given: 0

Re: a different, faster approach to solve this PS [#permalink]

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22 Jan 2008, 11:56
walker wrote:
$$V_{av}=64/18=3.8...$$

$$V_1<V_{av}<V_1+1$$

only 3mph works

Nice +1 from me too.

How did you arrive to this formula? why does $$V_{av}$$ is between those 2 rates?
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 544

Kudos [?]: 3550 [0], given: 360

Re: a different, faster approach to solve this PS [#permalink]

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22 Jan 2008, 13:51
GGUY wrote:
How did you arrive to this formula? why does $$V_{av}$$ is between those 2 rates?

It is logic and common sense.
if we have to parts with rates $$V_1$$ and $$V_2$$ we always will have $$V_{av}$$ between this two rates.

$$V_1<V_2$$

if we had had $$V_2$$=$$V_1$$ the average rate would have been $$V_1$$. But $$V_1<V_2$$ and so $$V_1<V_{av}$$.

the same logic for $$V_{av}<V_2$$

Therefore, $$V_1<V_{av}<V_2$$
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Re: a different, faster approach to solve this PS   [#permalink] 22 Jan 2008, 13:51
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