Daily puzzles/problems.

Ike The Ice Cream Man

Ike the ice cream man has a problem. A bunch of sorority girls just got off at the shuttle bus stop outside his shop, and they are about to come streaming inside his door. Ike can't resist these girls, they get anything they want from him, and he has foolishly left a 750ml bottle of Schnapps that he recently purchased sitting at the side of the counter, in plain view.

The girls love only one thing more than Schnapps, and that's ice cream, and so Ike starts scooping it into cups at the rate of one cup every 10 seconds (the same rate at which the girls are now entering his ice cream shop).

If it takes a girl 60 seconds to finish one cup of ice cream, and 30 seconds to pour and drink a 50ml shot of Schnapps, How long (from the moment he begins scooping) will it be before the sorority girls finish off Ike's Schnapps?

Can I ask, billl - do the girls pour their own schnapps or does Ike stop scooping ice-cream and pour it for them?

Does he keep them supplied with endless cups of ice cream, or does he stop after one each?

They pour their own Schnapps, and he supplies as much ice cream as he can (he's still scooping when the bottle's finished).

I can't see a reason why it isn't 7.5 mins.

I could have maybe done a better job explaining, the situation is this:

Girls are pouring in at a rate of 1 every 10 seconds, and Ike serves each one of them ice cream at the rate of one cup per 10 seconds (which each of the girls are able to finish in 60 seconds). Those who don't have ice cream will do shots of Schnapps, at a rate of 50ml per 30 seconds, and girls without any ice cream to enjoy will share the bottle. (But any girl with an opportunity to have a cup of ice cream will choose ice cream instead of the Schnapps.)

The girls remain in the shop, eating ice cream as quickly as Ike can serve it to them, and so he is busy with a store filled with sorority girls until the following Friday, when a bunch of mandatory keg parties happen.

So he's okay for the first six girls, but when the seventh girl comes in, he has a problem, because the first girl has finished her ice cream, so he has two girls to give ice cream to, and only ten seconds before another girl's finished and an eighth arrives - and every ten seconds that pass, he has another girl he can't keep occupied with ice cream.

The seventh girl in does a shot of Schnapps in thirty seconds, by which time there are ample girls available to drink further Schnapps.

So from the moment he starts scooping it's 1 minute (while he's holding the first six girls off with ice cream) + 750/50*30secs =

8.5 minutes

The only thing that seems odd to me is that you say it takes thirty seconds "to pour and drink a 50ml shot of Schnapps". I'd'a thought that the next one could be poured as the previous one was being drunk, so what we really care about is how long it takes to pour.

Quote:

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Originally Posted by MarkBastable

The only thing that seems odd to me is that you say it takes thirty seconds "to pour and drink a 50ml shot of Schnapps". I'd'a thought that the next one could be poured as the previous one was being drunk, so what we really care about is how long it takes to pour.

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Yes, I had intended to convey the idea that more than one girl could (would) be working on the Schnapp's at the same time. Sorry, again, I'm afraid the correct answer requires a little more jumping through mathematical hoops. (I'll look over my notes again, but I don't think there ends up being a line of people waiting at the Schnapps bottle.)

EDIT: to clarify, more than one girl might be drinking Schnapps at the same time, and each of them would be separately progressing at a pace of 50ml per 30 seconds. There are, as it happens, no cases of more than one girl finishing their drinks at the same time and arguing over the next pour, however (but such a thing could happen in slightly different scenarios).

Quote:

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Originally Posted by billl

Yes, I had intended to convey the idea that more than one girl could (would) be working on the Schnapp's at the same time. Sorry, again, I'm afraid the correct answer requires a little more jumping through mathematical hoops. (I'll look over my notes again, but I don't think there ends up being a line of people waiting at the Schnapps bottle.)

EDIT: to clarify, more than one girl might be drinking Schnapps at the same time, and each of them would be separately progressing at a pace of 50ml per 30 seconds. There are, as it happens, no cases of more than one girl finishing their drinks at the same time and arguing over the next pour, however (but such a thing could happen in slightly different scenarios).

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So I've obviously gone wrong somewhere, but this was the way it came out for me.

Ike can keep six girls happy with ice cream.

At sixty seconds there'll be seven girls in the store, one of whom'll hit the Schnapps.

At seventy seconds there'll be eight girls in the store - six occupied with ice cream, one with the Schapps and one free. If the Schnapps is free too, she'll hit that - so it depends whether the 'pour' part of 'pour and drink' takes less than thirty seconds. If it takes ten, say, she can hit the Schnapps. If it takes thirty, she has to wait.

By ninety seconds there'll be ten girls in the store - six on ice cream, one starting the second Schnapps (if the thirty-second 'pour and drink' cycles can't overlap) or starting the fourth (if the 'pour' part takes, say, ten seconds), and three hanging about, waiting for some kind of gratification (or two of them could be drinking Schnapps they poured earlier).

But if it's not necessary to know how long the 'pour' bit is, this may be where I've gone wrong - because in my way of working it out, you only need seven girls in the store - six on ice cream (on the 10 and 60 cycle for serve and consume) and one on Schnapps (on the 30 cycle for drink and pour). Any more girls than that are just hanging about. It seems unlikely to me that bill would have set up the problem in this way if he only needed seven girls - so my logic might have gone awry somewhere around there.

Quote:

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Originally Posted by MarkBastable

So I've obviously gone wrong somewhere, but this was the way it came out for me.

Ike can keep six girls happy with ice cream.

At sixty seconds there'll be seven girls in the store, one of whom'll hit the Schnapps.

At seventy seconds there'll be eight girls in the store - six occupied with ice cream, one with the Schapps and one free. If the Schnapps is free too, she'll hit that - so it depends whether the 'pour' part of 'pour and drink' takes less than thirty seconds. If it takes ten, say, she can hit the Schnapps. If it takes thirty, she has to wait.

By ninety seconds there'll be ten girls in the store - six on ice cream, one starting the second Schnapps (if the thirty-second 'pour and drink' cycles can't overlap) or starting the fourth (if the 'pour' part takes, say, ten seconds), and three hanging about, waiting for some kind of gratification (or two of them could be drinking Schnapps they poured earlier).

But if it's not necessary to know how long the 'pour' bit is, this may be where I've gone wrong - because in my way of working it out, you only need seven girls in the store - six on ice cream (on the 10 and 60 cycle for serve and consume) and one on Schnapps (on the 30 cycle for drink and pour). Any more girls than that are just hanging about. It seems unlikely to me that bill would have set up the problem in this way if he only needed seven girls - so my logic might have gone awry somewhere around there.

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Yes, you're right--the problem assumes (but doesn't state) that the pouring time is inconsequential. Unfortunately for me, it turns out that my solution involves more than one girl pouring herself Schnapps at the same time on two occasions (during the 50 second period before the bottle is finally emptied).

Anyhow, a bit of a screw-up there, sorry--but you've really saved the day on this one, Mark. You've nailed the issues down perfectly well, and since the replies haven't been exactly pouring in like sorority girls at a late-night ice cream parlour, I'm going to post the solution here in an attachment.

Too much ice cream and schnapps for me.

Got a good one, Mark?

This one was lifted straight from the 'Net, so it's easy to cheat. I'm going to have only intermittent access to LitNet over the next couple of weeks, so if everyone gets sick of the question, you might all agree to Google it.

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Three men were standing in a row, all facing the same direction, so that there was one in back who could see the two in front of him, one in the middle who could see the guy at the front, and the one in front who could not see either of the other two.

They were shown five hats - three blue and two red.

One hat was placed on each man, without them knowing which, or knowing which two were left over.

First the man in the back was asked if he could deduce what color hat he had on.

"No, I can't," he said.

The man in the middle was asked the same question.

"Nope," he said.

Then man in the front was asked - and he knew what color hat he was wearing.

What was his answer, and why?

I had a beautiful lucid moment just then and worked out it was blue. Trouble is, I can't remember why.

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Originally Posted by prendrelemick

I had a beautiful lucid moment just then and worked out it was blue. Trouble is, I can't remember why.

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How about this:

The back guy says no, so the guys in front of him can't BOTH be wearing red hats (that would mean only blue ones were left over, so he could deduce "blue").

So the second guy knows that at least one blue must be on either his head or the front guy's head (and maybe both are wearing blue...). If the person in front is wearing blue, then the middle guy can't be sure if he is wearing a blue hat himself--it could be blue OR red. However, if the guy in front of him is wearing a red hat, then he (the guy in the middle) CAN'T be wearing a red hat, because of the situation with the guy in the back (noted above, i.e. he couldn't have seen two red hats without knowing he was wearing blue).

Therefore, since the middle guy can't deduce anything, he must be seeing a blue hat on the front guy. So the front guy deduces that he is wearing a blue hat.

Yep, good. Me and the family are off to NY and the Virgin Islands. Carry on.

Quote:

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Originally Posted by billl

How about this:

The back guy says no, so the guys in front of him can't BOTH be wearing red hats (that would mean only blue ones were left over, so he could deduce "blue").

So the second guy knows that at least one blue must be on either his head or the front guy's head (and maybe both are wearing blue...). If the person in front is wearing blue, then the middle guy can't be sure if he is wearing a blue hat himself--it could be blue OR red. However, if the guy in front of him is wearing a red hat, then he (the guy in the middle) CAN'T be wearing a red hat, because of the situation with the guy in the back (noted above, i.e. he couldn't have seen two red hats without knowing he was wearing blue).

Therefore, since the middle guy can't deduce anything, he must be seeing a blue hat on the front guy. So the front guy deduces that he is wearing a blue hat.

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