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I bet most of you know this one as it was discussed before. [#permalink]
 21 Jan 2008, 04:52
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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
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Re: a different, faster approach to solve this PS [#permalink]
21 Jan 2008, 05:42
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V_{av}=64/18=3.8...V_1<V_{av}<V_1+1only 3mph works
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Re: a different, faster approach to solve this PS [#permalink]
21 Jan 2008, 05:45
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.
Answer B
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Re: a different, faster approach to solve this PS [#permalink]
21 Jan 2008, 09:49
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
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Manager
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Re: a different, faster approach to solve this PS [#permalink]
22 Jan 2008, 12:56
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?
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CEO
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Schools: Chicago (Booth) - Class of 2011
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Re: a different, faster approach to solve this PS [#permalink]
22 Jan 2008, 14:51
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_2if 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_2Therefore, V_1<V_{av}<V_2
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Re: a different, faster approach to solve this PS
[#permalink]
22 Jan 2008, 14:51
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