A question to mess with your mind

[ReaperFett]

For your help, I'll switch it from Pounds to Dollars. However, if I say Pound, read Dollar :)

And so the question begins...






Three old ladies go to the local Café. They have some cake and tea. The manager goes to the waiter, and asks how much the meal costs. He tells him $15. The manager, seeing how nice the old ladies look, says to charge them only $10.


The waiter has other plans however. He decides to charge them $15, and keep the excess for himself. He gives them their bill, and each hands him five $1 coins.


However, the waiter begins to feel guilty with himself, and decides to compromise. He decides to keep $2, and refund each of the ladies $1. So, he puts the money in the till, pockets $2, and hands each of the women $1 back.



Now, here is the problem.

Each lady handed in $5, and got $1 back. So they spent $4


$4 x 3 = $12


Waiter kept $2. $2 + $12 = $14



SO WHERE IS THE OTHER $1???????????

[Ken Jiang]

Mwuah hah hah ha haaa!!! >D

[Live Wire]

this is supposed to be hard?


Remind me to post some of my math problems from high school. That stuff will turn your brain to mush.

[ReaperFett]

I notice you dont answer though :)

[Q]

$4 x 3 = $12That's indeed how much they wound up spending, but not all of it covered the meal. $10 were for the meal, $2 went in his pocket, and each old lady got $1 back. To put it another way:

They each handed him $5, for a total of $15.

$10 of that is to cover the bill.

He keeps $2.

He hands back $1 to each of the ladies (X3) for a total of $3 handed back.

Unless I'm missing something, that's $15.

[Taylor Millard]

Originally posted by ReaperFett
Three old ladies go to the local Café. They have some cake and tea. The manager goes to the waiter, and asks how much the meal costs. He tells him $15. The manager, seeing how nice the old ladies look, says to charge them only $10.


The waiter has other plans however. He decides to charge them $15, and keep the excess for himself. He gives them their bill, and each hands him five $1 coins.


However, the waiter begins to feel guilty with himself, and decides to compromise. He decides to keep $2, and refund each of the ladies $1. So, he puts the money in the till, pockets $2, and hands each of the women $1 back.



Now, here is the problem.

Each lady handed in $5, and got $1 back. So they spent $4


$4 x 3 = $12


Waiter kept $2. $2 + $12 = $14


Quiet you ninny ( :mneh ) It's simple.

Total Meal: 15 pounds

Total Charged: 10 pounds

Total in Waiter's Pocket: 2 pounds

Total change: 3 pounds

You ninny there is no other missing pound!

Q: Nice job solving it first

*Slaps Fett.

I'd pan you but I don't think I can pan a mod. :grumble

[Alec Lafeyette]

He decides not to charge them 15$, so you don't need to add it up to that.

They give him $15, he gives back $3. They end up paying $12, so he keeps the extra $2.

[ReaperFett]

You miss the point. How come I can say it more or less the same, and lose $1. Your sums are simple, I think anyone could work that out

[Q]

Because you're counting the $2 the waiter keeps for himself twice.

If you want to look at it your way: They spent $4 dollars each for a total of $12. Fine. But then you tack on the $2 that the waiter keeps, which invalidates it: That's already included in the money they spent.

If you divide it as equally as possibly, rounding to the nearest dollar, two of the ladies paid $3 for the meal, while the third wound up paying $4 (since that's the only way to divide the cost of $10 by 3 people in whole dollars). So one lady spent an extra dollar on the meal, the other two spent an extra dollar that lined the waiter's pockets.

[Alec Lafeyette]

What he said.

[darth_mcbain]

If this type of stuff floats your boat try this one out - one of my personal faves...

A person travels from point A to point B and averages 10 miles an hour. How fast does he need to travel from B back to A to average 20 miles an hour for his entire trip?

[Q]

They'd need to be traveling 30mph on the return trip for an average of 20mph

[Alec Lafeyette]

20 miles an hour.

Yeah I just read your post and found out. Didn't see the "average" part, I thought it was a trick question.

[Q]

20 + 10 / 2 = 15. Sorry.

[ReaperFett]

Are you going the same route? You could have had to use a one way system at a stage :)

[ReaperFett]

If you divide it as equally as possibly, rounding to the nearest dollar, two of the ladies paid $3 for the meal, while the third wound up paying $4 (since that's the only way to divide the cost of $10 by 3 people in whole dollars). So one lady spent an extra dollar on the meal, the other two spent an extra dollar that lined the waiter's pockets.

But each gave $5. Each then got $1 back. $4 each. Leaving $3. $2 of which the waiter keeps. $1 is missing.

This is a great question btw if you're doing something. People will nearly stop what they're doing to think about this :)

[darth_mcbain]

Too bad Q --- 30 isn't correct... Keep trying Mon Capitan... :)

And BTW Reap - you are travelling in a straight line between A and B (it's a two way street)

[Q]

But each gave $5. Each then got $1 back. $4 each. Leaving $3. $2 of which the waiter keeps. $1 is missing.No, it's not. Again, the $2 that the waiter keeps is included in the total $12 that they spent. You're counting it twice.

They did, indeed, spend a total of $12. Two dollars of which he kept. You don't add it to the $12. The remaining $3 is divided amongst the women.

And, I assumed that the problem wasn't quite that easy, McBain. But I don't see any other rational solution from the information you provided.

The person already has completed one way of this two-way trip travelling at 10mph. You want the average to equal 20mph. X is the speed they'd have to travel on the way back to complete the formula.

10 + X / 2 = 20

Working backwards, you'd multiply both sides by two: 10 + X = 40. Subtract 10 from each side, and you get thirty. X = 30.

I realize that there's some trick involved, but that's my train of thought, anyway.

[darth_mcbain]

I don't know if this will help or not, but I'll throw it out there for you to mull over...

Notice I didn't mention the distance between A and B...

Maybe that'll just throw more confusion into the works but what the hey... :D

And yes, while (30+10)/2 does equal 20, 30 is still not the correct answer - no tricks, just straight math - but you're not seeing it yet...

[Q]

Well, assuming that A<->B is equal to B<->A, that shouldn't be relevant. But I'm, of course, missing something today.

[ReaperFett]

60mph

[darth_mcbain]

Originally posted by ReaperFett
60mph

Nope...

[Laran Katern]

To average 20mph from b to a, he needs to.. well 20 mph

[ReaperFett]

I did something like this in Maths IIRC. What we did, I can't remember :)

[Q]

Laran: Although McBain has yet to discount the other person who suggested 20mph, I don't think so. He was asking for the average "for his entire trip", both ways. And the travel speed of 10mph one way would bring down that average.

Unless, of course, he's just being cheap and counting the 'entire trip' as only the return trip. :p

[darth_mcbain]

Laran, Average 20 MPH on the Round trip (A to B and back to A). Otherwise you'd be spot on... :)

Edit: Thank you Q

[ReaperFett]

40? No, cant be.





Clue? :)

[darth_mcbain]

Its a lot faster than you'd think... :)

This might help you to work it out:
I never said the distance between A and B. That's because it is immaterial. Pick some distance, say 100 miles, for that distance. Then plug in some numbers and see what you can see....

Having fun yet :)

[ReaperFett]

Is it impossable? say 100mph each way. To get there and back at 20mph would be a 10 hour journey. To do the first leg takes 10 hours. So basically, you'd need to take 0 time to get there :)

[darth_mcbain]

Go REAP!!!! Yes, you would basically have to teleport from B back to A. Actually that wouldn't work for two reasons - 1) teleporting doesn't exist, and 2) if it did exist, it would probably still be limited to the speed of light, which in this case is still too slow... :)

The correct answer is just that - you'd have to travel infinitely fast...

BTW - Don't feel bad Q. Every time I pose that question I always get bombarded with the obvious answer of 30...

[ReaperFett]

Beat light, sounds fun :)

[HaeLari Draevyn]

Ooooo, I'm not even going to try! Just looking at numbers makes my head hurt.

[darth_mcbain]

Allright - here's another fun one for ya... I'll post the answer tomorrow if you haven't gotten it... (I'd be surprised if someone doesn't get this) :)

There are two trains speeding towards each other on a straight railroad track. Train one is going 60 mph, and train two is going 90 mph. They are exactly 300 miles apart. There is a bird sitting on train one. This bird is unique in that he can only move at one constant speed of 100 mph. Now, at the outset, he will fly straight towards train two. When he meets up with train two, he does an immediate about-face and heads back to train one. He keeps pinging back and forth between the two trains as they are on their collision course, until they collide.

Given that - how much distance has the bird travelled before the trains collide?

[ReaperFett]

200 miles

[darth_mcbain]

Yeah - good Reap... That one is good, because sometimes you'll get someone going into infinite series to solve it. Of course that will work, but if you just think a bit, the answer is right there...

[Alec Lafeyette]

So the answer is 200 miles?

[ReaperFett]

yeah

[darth_mcbain]

Allright - one more and that's it for me...

4 guys are walking at night and come to a bridge. The bridge is rickety and can only support two of them at a time. Further, as it is dark they must use a flashlight to cross, but there is only one flashlight between them, so when two cross, one must cross back to bring back the light.
Each guy walks at a different speed:

Guy 1: can cross the bridge in 1 minute
Guy 2: can cross the bridge in 2 minutes
Guy 3: can cross the bridge in 6 minutes
Guy 4: can cross the bridge in 10 minutes

When two guys cross together, they travel at the slower speed (for example, if guy 1 and guy 3 cross together, it takes them 6 minutes to cross the bridge)

Question: What's the shortest amount of time it will take for all four of the guys to get across the bridge?

[ReaperFett]

10 + 1 + 6 + 1 + 2 = 20minutes

[darth_mcbain]

Oh, you can do better than that Reap... :D

[darth_mcbain]

The best you can do is 17 minutes See if you can figure out how... :)