Train speed conundrum

[WedWon]

A train has to average 100mph to get from station A to station B on time. At the halfway point (C) it has only been averaging 50mph.

What speed would the driver have to do the second half of the journey at to reach the destination in time?

[vnx205]

I know the answer.

Would you prefer that I let a few people offer a few incorrect answers before I jump in?

It catches most people out.

[WedWon]

I know the answer.

Would you prefer that I let a few people offer a few incorrect answers before I jump in?

It catches most people out.

Yeah, lets see how many people get it wrong first!

[vnx205]

How about another brain teaser.

Let us just suppose for a moment that you could tie a piece of string around the earth. However your piece of string is a metre too long, so you decide to pack it out to make the string a tight fit. Do you think anyone would notice you had packed it out?

[Pierre]

A conundrum or Einstein-related solution?

[isuzutoo-eh]

∞ mph

[JDNSW]

I was still looking for the infinity sign on my keyboard! But that is not necessarily the only answer - it all depends on how you see the halfway point. Is it halfway in time or halfway in distance?

John

[isuzutoo-eh]

I was still looking for the infinity sign on my keyboard! But that is not necessarily the only answer - it all depends on how you see the halfway point. Is it halfway in time or halfway in distance?

John

G'day John,
I had to google the infinity sign, after i'd sketched the problem. Geez John, you're slow! :p


These frames of reference questions are fun. More I say!

[WedWon]

∞ mph

wouldn't be fast enough

[WedWon]

How about another brain teaser.

Let us just suppose for a moment that you could tie a piece of string around the earth. However your piece of string is a metre too long, so you decide to pack it out to make the string a tight fit. Do you think anyone would notice you had packed it out?

Instinctivly I think that the spreading out of the extra metre of string around the vastness of the globe would not be noticed easily. But from the format of the question I assume the packing out would be noticable.

Now if I think for a bit an addition metre spread around the globe (how big is the globe - anybody)........could it need to be packing out as much as 10mm around the circumference and therefore pose a trip hazard? (This is purely a guess based on the fact that the question has been asked in the way that it has)

Go on....do tell

[vnx205]

Go on....do tell
Nah, not until a couple of others have a go at it. :p:p

[rovercare]

What a waste of grey matter:confused:

[willem]

Given that its 20 000 kms around the4 globe the packing needed would be so small as to be unnoticeable. But why not trim a metre off the line? Or is that too logical? :p:p

[Grover-98]

150mph :)

[seano87]

150mph :)

Incorrect :p

I do believe the correct answer has been hinted at well enough!

[Grover-98]

Incorrect :p

I do believe the correct answer has been hinted at well enough!

I really should read posts! would create less of these public embarrassments....

I still havnt read them but my answer is you don't know as you need more info.

:angel:

[bussy1963]

100 kph.

[handleAUS]

prepare for ludicrous speed... :o

[Psimpson7]

How about another brain teaser.

Let us just suppose for a moment that you could tie a piece of string around the earth. However your piece of string is a metre too long, so you decide to pack it out to make the string a tight fit. Do you think anyone would notice you had packed it out?

Yes. 16cm

Say for arguments sake the circumference is 1000000m (dont have a clue what it actually is but makes no difference anyway)

/Pi = Dia of 318,268.62m

So an extra metre would be 1000001m

/Pi = Dia of 318,268.94m

difference of .32 of a metre on the diamater

So more basically as 1/pi = 0.318m

So divide that by 2 for distance from ground = 0.16m

Hoping that my basic maths is correct after a very long day and I havent made myself look (more) stupid;)

[moose]

Being as I'm in victoria, I don't think the train company would care less if the train was late, they can't make the rest of them run on time anyway, so why would they waste the extra effort to get one train on time when it's so far behind. :D

[seano87]

Well, lets assume we are traveling with CityRail who consider anything up to 5 minutes late "on-time"

...

> 600mph.

BUT if the train left 5 mins early (still "on time"), ... NAH, that'll never happen...

I'm sure we can substitute any other state's public rail name and have the same result...

Seano

[Sleepy]

Sorry Gen-x here, I don't do mph!:p It's soooooooooooo 20th Century!:p

[seano87]

Yes. 16cm

Say for arguments sake the circumference is 1000000m (dont have a clue what it actually is but makes no difference anyway)

/Pi = Dia of 318,268.62m

So an extra metre would be 1000001m

/Pi = Dia of 318,268.94m

difference of .32 of a metre on the diamater

So more basically as 1/pi = 0.318m

So divide that by 2 for distance from ground = 0.16m

Hoping that my basic maths is correct after a very long day and I havent made myself look (more) stupid;)

Using more correct numbers:

Circumference = ~40,000km. 40,000,000m, so 40,000,001m string

Diameter = 12,732,395.44m

String diameter when in a circle is 12,732,395.77m

So is 33cm more diameter, 16.5cm above ground.

Wow, same answer.

Now I think about it, its absolutely irrelevant what the diamter is, even if the earth is only 1m wide, add a metre to the string and its the same answer. DOH!.

[seano87]

I just had a blonde moment... DELETE!

[Psimpson7]

Now I think about it, its absolutely irrelevant what the diamter is, even if the earth is only 1m wide, add a metre to the string and its the same answer. DOH!.

Exactly!

[vnx205]

Using more correct numbers:

Circumference = ~40,000km. 40,000,000m, so 40,000,001m string

Diameter = 12,732,395.44m

String diameter when in a circle is 12,732,395.77m

So is 33cm more diameter, 16.5cm above ground.

Wow, same answer.

Now I think about it, its absolutely irrelevant what the diamter is, even if the earth is only 1m wide, add a metre to the string and its the same answer. DOH!.

That's right.

Because the circumference of a circle is 2xPixr, an extra metre of string will add about 160mm to the radius whether you wrap it around a pin, a basketball or the earth. The original diameter doesn't matter. The increase is the same.


However most people's instinctive reaction is that it would only have to be packed up a very tiny amount.

[seano87]

Now thats 2 puzzles solved, I'll throw another one in.

You've just made it to the prize round of a game show and are told...

There are 3 doors in front of you. Behind one of them is a new car, behind the other 2 are goats, select a door.

You select a door. The host (knowing what is behind each) opens one of the remaining two doors to reveal a goat. You now have a choice, you may stick with the door you have already selected, or choose the other one.

What should you do? Does it make a difference to your chances??

:p

Seano

[adm333]

I have not seen this problem before but my guess is, that if you are at the half way point (distance wise) and you have been travelling at exactly half the speed you were supposed to, then you have arrived at the half way point at the exact time you were supposed to arrive at the final destination.

So the answer is 88 mph.


Back to the Future fans will recognise this as the exact speed that the Delorien must be travelling to break through the space time continuum and travel back or forwards in time.

[Psimpson7]

change your choice.

Originally you took 1 from 3 so had a 33.3% chance.

Now the choice is 1 from 2 hence 50%

[vnx205]

change your choice.

Originally you took 1 from 3 so had a 33.3% chance.

Now the choice is 1 from 2 hence 50%
That may be so but there is a 1 in 2 chance that it is behind the first remaining door and a 1 in 2 chance it is behind the second remaining door, so what do you achieve by changing.

Your chances may be better than they were before, but each of the remaining doors has the same chance of having the car.

[seano87]

That may be so but there is a 1 in 2 chance that it is behind the first remaining door and a 1 in 2 chance it is behind the second remaining door, so what do you achieve by changing.

Your chances may be better than they were before, but each of the remaining doors has the same chance of having the car.


change your choice.

Originally you took 1 from 3 so had a 33.3% chance.

Now the choice is 1 from 2 hence 50%

Ahh... you are both incorrect. :D I'll let it go a little longer to see if anyone gets it.

[Psimpson7]

Ok try again taking a bit longer to explain my thoughts!

You took the original choice as 1 out of 3. so 33%

One was then taken away, so now you are only picking from 2 so 50%

But if you stay with your original choice it was picked as 1of 3 i.e. 33%, where as the 1 from 2 now is 50% so picking the other remaining one gives you a 17% better chance of getting it correct.

[seano87]

Ok try again taking a bit longer to explain my thoughts!

You took the original choice as 1 out of 3. so 33%

One was then taken away, so now you are only picking from 2 so 50%

But if you stay with your original choice it was picked as 1of 3 i.e. 33%, where as the 1 from 2 now is 50% so picking the other remaining one gives you a 17% better chance of getting it correct.

Thats close enough...

You definitely should change doors.

You originally pick and have a 33.3% chance of picking the correct door, 66.7% it is behind one of the other doors, of which a goat is let out of those.

Now, your 33.3% chance hasn't changed, but what has is that the car now has a 66.7% chance of being behind the other door left over. Hence, change!

Seano

[moose]

Unless you picked the correct door first time around, in which case you have just "mathed" yourself into winning a goat. (well done!) :)

[vnx205]

A better way to explain it, or at least the one I was able to understand is this:

When you pick your first door there is a 1 in 3 chance the car is behind your door and a 2 in 3 chance the car is behind a different door.

When the host removes one door, there is still a 2 in 3 chance the car is behind a door you didn't pick but still only a 1 in 3 chance of it being behind your door.

So you don't have a 50% chance of getting it right if you switch.

You have a 66.7% chance of getting it right if you switch.

[ivery819]

First choice 1 in 3

Next choice 1 in 2 and as this is a new game the original choice makes NO difference to your new odds.:angel:

[dmdigital]

Where does planitary circumference come in to the equation?

The answer is 240kph - for those who can't work in mph:p

[Bushie]

Of course, the shows organisers are moving the goat while you, are thinking whether 1 in 2 is better than 1 in 2 :mad:


Bushie

[seano87]

First choice 1 in 3

Next choice 1 in 2 and as this is a new game the original choice makes NO difference to your new odds.:angel:

But thats the thing, its not a different game so it makes a big difference. vnx205 has got it spot on.


Of course, the shows organisers are moving the goat while you, are thinking whether 1 in 2 is better than 1 in 2 :mad:


Bushie

So cynical :p game show's never scrooge anybody!

The question was actually a probability question in a statistics assignment I've just completed. I assure everyone the answer is to switch, and a 2/3 chance of getting it if you do, assume all things fair and nice of course.

Seano

[Blknight.aus]

First choice 1 in 3

Next choice 1 in 2 and as this is a new game the original choice makes NO difference to your new odds.:angel:

correct. this is another version of the maths game with the 3 blokes that pay $10 each for a room and by working the math from both ends the clerk gets a free dollar out of the change.

the math is based on the assumtion that you dont already have the car behind your door.

if you have a 1 in 3 chance of winning something you have a 33% chance of being right.

if you have a 1 in 2 chance thats 50%

if you make a 1:3 chance and they take away one choice (which they then proove to be the wrong choice) the percentage doesnt just got to the one door youve chosen it become shared equally between the 2 remaining doors.

try it at home.

[dmdigital]

50:50 chance on the doors so there's no difference to tossing a coin for the choice. Odds have changed as the number of choices has changed.

[Bigbjorn]

If you buy Gold lotto tickets, think on this. The probability of getting six selections out of 45 is- 45 x 44 x 43 x 42 x 41 x 40.

The train question requires further information. What is the half-way we are discussing? Time? Distance? Distance 100 miles at 100mph = 1 hour. At 50mph the train has travelled 50 miles in one hour and has used the allotted time. Using elapsed time as the half way parameter, the train at 50mph in half an hour has travelled 25 miles. The remaining 75 miles has to be travelled at 150mph to arrive on time.

[HangOver]

150 ?

[Slunnie]

A train has to average 100mph to get from station A to station B on time. At the halfway point (C) it has only been averaging 50mph.

What speed would the driver have to do the second half of the journey at to reach the destination in time?
Not possible. At the half way point he should have already arrived at the final destination.

[WedWon]

Now thats 2 puzzles solved, I'll throw another one in.

You've just made it to the prize round of a game show and are told...

There are 3 doors in front of you. Behind one of them is a new car, behind the other 2 are goats, select a door.

You select a door. The host (knowing what is behind each) opens one of the remaining two doors to reveal a goat. You now have a choice, you may stick with the door you have already selected, or choose the other one.

What should you do? Does it make a difference to your chances??

:p


Seano


When you made the first selection you had a 33% chance of getting it right
If you proceed without reselecting its still a 33% chance
However if you repick you odds improve to 50%

But with your luck you're still gonna get a goat!

[WedWon]

If you buy Gold lotto tickets, think on this. The probability of getting six selections out of 45 is- 45 x 44 x 43 x 42 x 41 x 40.

The train question requires further information. What is the half-way we are discussing? Time? Distance? Distance 100 miles at 100mph = 1 hour. At 50mph the train has travelled 50 miles in one hour and has used the allotted time. Using elapsed time as the half way parameter, the train at 50mph in half an hour has travelled 25 miles. The remaining 75 miles has to be travelled at 150mph to arrive on time.

All the neccesaary info is in the question

[WedWon]

150 ?

Wrong :D

[WedWon]

Train Speed Conundrum Answer:

If you've reached the halfway mark at half the speed then you are at the halfway mark at the time you should be at your destination.........no matter how fast you do the second half of the journey you're already late

Well done Slunnie, Seano87 and vnx205

Cheers
JAson

[carjunkieanon]

re the three doors. It's called the Monty Hall problem - I think he had a game show.

I argued with a guy for a year that you shouldn't change, then he showed me why you should.

play 1) A = Goat, B = Goat, C = Car.
Chose A, he opens B, you should swap
play 2) A = Goat, B = Goat, C = Car.
Chose B, he opens A, you should swap
play 3) A = Goat, B = Goat, C = Car.
Chose C, he opens A or B, you should stay.

Since two out of three times you're better off swapping, you should swap.

r

[Blknight.aus]

thats a simple way of looking at it but you havent covered all the permeations in that simple 3 choices.

you can have

c g1 g2
c g2 g1

g1 c g2
g1 g2 c

g2 g1 c
g2 c g1

then for each of those permeations you can pick 1 of 3 doors and then given that the host knows if you have a goat or a car will ALWAYS expose a goat (no point in exposing the car cause then you know you've lost so that eliminates 1/2 of the door reveals)

of all those choices you will never wind up with

g1 g2 or
g2 g1

because the host will never pick the car

so if we elminate either of the goats you always come back to a 50/50 choice

being

c g2
c g1
g1 c
g2 c

[seano87]

re the three doors. It's called the Monty Hall problem - I think he had a game show.

I argued with a guy for a year that you shouldn't change, then he showed me why you should.

play 1) A = Goat, B = Goat, C = Car.
Chose A, he opens B, you should swap
play 2) A = Goat, B = Goat, C = Car.
Chose B, he opens A, you should swap
play 3) A = Goat, B = Goat, C = Car.
Chose C, he opens A or B, you should stay.

Since two out of three times you're better off swapping, you should swap.



r


Agh... yes, it is the Monty Hall problem.

For those that think it becomes a 50/50 chance, have a look at the following youtube video that hopefully explains it better than I, or anyone else has, and will convince you that the probabilty if you change really is 66% if you change...

YouTube - The Monty Hall Problem

Seano

[Tombie]

Was on Austar a while back too...

Lots of statistical and mathamatical manipulation type demonstrations...

And this one is a definate - You should change...

And in real life, they proved it in a market using the public.


Personally, I wouldnt care - I never win anything.:angel:

[p38arover]

thats a simple way of looking at it but you havent covered all the permeations in that simple 3 choices.

I think I'll let all those permutations permeate into your brains! :D

[ivery819]

:lol2::lol2::lol2::Rolling::Rolling::Rolling:

seano87's UTube video PROVES that at no time were the odds better than 50/50

(The clue is in the final comments)