House distance puzzle

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DrMatt
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Re: House distance puzzle

Post by DrMatt » Thu Oct 31, 2013 9:41 pm

Cool Hand wrote:It's the same problem we all saw in 7th or 8th grade algebra in disguise.
The part where they worked in the monkey's mother was a red herring.
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ceptimus
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Re: House distance puzzle

Post by ceptimus » Thu Oct 31, 2013 10:07 pm

no one in particular wrote:
Spoiler:
Image
Yes, you have the correct answer. Well done!
Spoiler:
You don't need to make the assumption that Alice travels at 150% the speed of Bob - in fact you can work out the answer without explicitly calculating the two friends' relative speeds.

The first thing to note is that as both Alice and Bob stop at the others house for the same time, the length of time they stop for makes no difference: instead of two minutes it could be three days or zero seconds and the answer would still be the same.

We're told that Alice travels faster than Bob, but she can't travel twice as fast (or more) or she would get back to her own house before Bob had left it or before he even arrived there.

Because Alice travels faster than Bob, she will be closer to Bob's house when they first pass, and closer to her own house when they meet again - also the coffee shop must be closer to her house than the first passing place is to Bob's house.

It helps to make a diagram. Label Alice's house A, Bob's house B, the place where they passed first P, and the coffee house meeting place M.

A -------- M -------------- P -------------- B

So 400 yards must be the distance AM, and 800 yards the distance PB.

Now notice that the first time they pass at P, if you add the distance traveled by Alice to the distance traveled by Bob the answer must be the distance between their houses.

Similarly, when they meet at M, the total distance traveled is now three times the distance between their houses (when they both eventually get back home they'll each have traveled twice the distance between their houses).

As each person (ignoring stopping time) has traveled at a steady speed, then each of them has traveled three times as far to get to M as they had when they reached P.

Now we see that Bob traveled 800 yards to get to P, so he's traveled 3 x 800 = 2400 yards to reach M. And he's traveled 400 yards back from Alice's house, so the distance between their houses must be 2400 - 400 = 2000 yards.

The puzzle is a variant of Sam Loyd's Ferry Boat Problem first published about a hundred years ago.

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Re: House distance puzzle

Post by Abdul Alhazred » Thu Oct 31, 2013 11:18 pm

DrMatt wrote:The part where they worked in the monkey's mother was a red herring.
Green. You could paint it green. :notsure:
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Re: House distance puzzle

Post by DrMatt » Fri Nov 01, 2013 12:57 am

Abdul Alhazred wrote:
DrMatt wrote:The part where they worked in the monkey's mother was a red herring.
Green. You could paint it green. :notsure:
This is approximately how it read back in 8th grade. It might even be exactly the same.

Green monkey.
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Re: House distance puzzle

Post by Witness » Fri Nov 01, 2013 1:55 am

Reminds me of the puzzles of Lewis Carroll.