Math Problem

[ABM]

Three dudes go into a hotel to get a room. The desk clerk says the room is $30. The guys get the room. After the transaction, the clerk realizes he's overcharged the men. So, he calls the bellhop over. He explains to the bellhop that the the room is actually only $25. He hands the bellhop five $1 bills, then asks him to go up to the room and give the men their partial refund.

As the bellhop is going up the elevator, he realizes that $5 can't be split evenly between 3 people, so he stuffs $2 into his pocket. He gets up to the room, explains the situation (in his own words), then proceeds to give each of the men $1 back ($3 total......keeping in mind that the bellhop has the other $2).

So then, now each of the men have paid $9 apiece, right? Right.

Therefore, $9 x 3 = $27....plus the $2 the bellhop pocketed. That equals $29.

What happened to the other dollar?

 

[mgb]

The 2 is part of the 27 they paid, you don't add it afterwards, plus 3 is 30.

 

[ABM]

The 2 is part of the 27 they paid, you don't add it afterwards, plus 3 is 30.

So, you think you're smart, huh.

You'd be correct, sir.

 

[Master Shake]

Lol, more!

 

[ABM]

Lol, more!

Are you a private dick, or a private duck?

Just wonderin'.

 

[Master Shake]

Private Dick. It's in that show a short form of "Detective" but Duckman (Voice my Jason Alexander) is a huge dick and a theif, and it's my fav. show of al time. Google it and try and find some full episodes online, it'll make you laugh.

 

[BalancedMan]

I think I've seen that question before. Those are good. The type of questions Microsoft asks applicants.

 

[BalancedMan]

Here's another I remember:

There are 4 women who want to cross a bridge. They all begin on the same side. You have 17 minutes to get all of them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each woman walks at a different speed. A pair must walk together at the rate of the slower woman's pace.

Woman 1: 1 minute to cross
Woman 2: 2 minutes to cross
Woman 3: 5 minutes to cross
Woman 4: 10 minutes to cross

For example if Woman 1 and Woman 4 walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Woman 4 then returns with the flashlight, a total of 20 minutes have passed and you have failed the mission. What is the order required to get all women across in 17 minutes? Now, what's the other way?

 

[ABM]

Here's another I remember:

That sounds like that fox, chicken, and whatever in a rowboat story problem.

I'll have to think about this one and get back to you.

EDIT: Of course, doing it in 19 minutes is easy. Can't quite get the 17 minute plan figured out....................just yet.

 

[TripTango]

That's a good one -- I'm proud to say I got it without cheating, even though I was a little tempted at one point...

The most obvious choice would be to have Woman 1 ferry the flashlight back and forth, but as ABM pointed out, this results in 19 minutes elapsing. To get down to 17 minutes:

Woman 1 and Woman 2 cross together, Woman 1 returns alone. (3 minutes)
Woman 3 and Woman 4 cross together, Woman 2 (who was already on the other side) returns the flashlight alone. (12 minutes)
Woman 1 and Woman 2 cross together. (2 minutes)

 

[ABM]

That's a good one -- I'm proud to say I got it without cheating, even though I was a little tempted at one point...

The most obvious choice would be to have Woman 1 ferry the flashlight back and forth, but as ABM pointed out, this results in 19 minutes elapsing. To get down to 17 minutes:

Woman 1 and Woman 2 cross together, Woman 1 returns alone. (3 minutes)
Woman 3 and Woman 4 cross together, Woman 2 (who was already on the other side) returns the flashlight alone. (12 minutes)
Woman 1 and Woman 2 cross together. (2 minutes)

Very nice!