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

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

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