Cool Math/Logic Problem

[Bill Murphy]

Gotta stocking stuffer book of math problems for XMas. Had just glanced it over but then I saw this one. It was composed by Sam Loyd, a very famous 19th century puzzle & chess master.

What I like about it is there's a eureka moment that has nothing to do w/setting up multi-variable equations, or esoteric knowledge. Not real tough, but fun. Here goes.

Two ferrys travel across a river all day. The two piers are directly across from each other, and the ferrys always leave at exactly the same time and keep constant speeds.

On one trip they pass each other 720 yards from Pier A. On the trip back they pass each other 400 yards from Pier B. How wide is the river?

[Alobar]

this answer seems to be easy to be correct, but heres what I got in white below

<font color="white">1120 </font>

[Bill Murphy]

Nope. /images/graemlins/wink.gif

[Alobar]

[ QUOTE ]
Nope. /images/graemlins/wink.gif

[/ QUOTE ]

lol, I knew that couldnt be right...oh well /images/graemlins/frown.gif

[daryn]

do the speeds of each boat ever change?

in other words we have ferry 1 and ferry 2. say ferry 1 travels at speed X, ferry 2 travels at speed Y.

does ferry 1 always travel at X? does ferry 2 always travel at Y? of course they don't have to from the wording, but i'm asking.


if they DO always go at the same speed, then the problem makes no sense and has no solution.

if they DON'T, then the problem has an infinite number of solutions.

[Skjonne]

Solution in white:

<font color="white"> 1440 </font>

[daryn]

</font><blockquote><font class="small">In risposta di:</font><hr />
Solution in white:

<font color="white"> 1440 </font>

[/ QUOTE ]

logic please

[pshreck]

[ QUOTE ]
[ QUOTE ]
Solution in white:

<font color="white"> 1440 </font>

[/ QUOTE ]

logic please

[/ QUOTE ]

Wait to see if that is the right answer first.

[Bill Murphy]

I prolly shoulda just copied the problem word for word from the book, but its not that complicated. Yes, for these two trips Ferry1 is traveling at a constant speed X and Ferry2 at constant speed Y, but that's not the key to it. It's not a per se math problem.

[daryn]

</font><blockquote><font class="small">In risposta di:</font><hr />
</font><blockquote><font class="small">In risposta di:</font><hr />
</font><blockquote><font class="small">In risposta di:</font><hr />
Solution in white:

<font color="white"> 1440 </font>

[/ QUOTE ]

logic please

[/ QUOTE ]

Wait to see if that is the right answer first.

[/ QUOTE ]


does it matter if he is right? he used some sort of logic to arrive at an answer.

[pshreck]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Solution in white:

<font color="white"> 1440 </font>

[/ QUOTE ]

logic please

[/ QUOTE ]

Wait to see if that is the right answer first.

[/ QUOTE ]


does it matter if he is right? he used some sort of logic to arrive at an answer.

[/ QUOTE ]

Yeah I guess, I just want to keep thinking about it. He can tell us.

[Bill Murphy]

Nope!

[thirddan]

[ QUOTE ]
The two piers are directly across from each other

[/ QUOTE ]

wouldnt they just crash into each other? if not then they would have to travel along some kind of arc, and unless they travelled along the same arc each time i don't see how to solve the problem...i suck at math...and logic...

you are all gonna have to help me with my logic class next semester /images/graemlins/smile.gif

[daryn]

</font><blockquote><font class="small">In risposta di:</font><hr />
I prolly shoulda just copied the problem word for word from the book, but its not that complicated. Yes, for these two trips Ferry1 is traveling at a constant speed X and Ferry2 at constant speed Y, but that's not the key to it. It's not a per se math problem.

[/ QUOTE ]

<font color="white"> so is it a trick question like that monopoly joker? if not, it's impossible. the boats would always meet each other 720 yards from ONE of the piers, and it would alternate, so if you say they met 720 yards from A the first time, they would meet 720 yards from B the second time. but you say 400 from B.. breakdown.....</font>

[pshreck]

Yeah Im thinking it can't be a math problem, because thinking purely mathematically, its impossible. Adding other factors that I have yet to figure out Im sure its possible and the answer will make sense.

[daryn]

it can't really deal with currents either because they would be irrelevant, since the boats maintain their constant speeds.

[Bill Murphy]

No, no, the pier placements and speeds of the ferrys(other than each being a constant) don't matter. Maybe one ferry loads &amp; unloads on the north side of each pier and the other on the south side! /images/graemlins/laugh.gif

[Skjonne]

[ QUOTE ]
Nope!

[/ QUOTE ]

Huh????

[pshreck]

[ QUOTE ]
No, no, the pier placements and speeds of the ferrys(other than each being a constant) don't matter. Maybe one ferry loads &amp; unloads on the north side of each pier and the other on the south side! /images/graemlins/laugh.gif

[/ QUOTE ]

is that a clue?

[daryn]

</font><blockquote><font class="small">In risposta di:</font><hr />
No, no, the pier placements and speeds of the ferrys(other than each being a constant) don't matter. Maybe one ferry loads &amp; unloads on the north side of each pier and the other on the south side! /images/graemlins/laugh.gif

[/ QUOTE ]

very stupid problem if so.

maybe one ferry just unloads 15 ft before it gets to the pier.. just throws all cargo into the ocean. maybe the other ferry runs aground before it unloads.

there are an infinite number of sulutions then, if anything can be done.

[Bill Murphy]

No, its not a trick. I don't think I made a mistake in the initial post but I can check. Don't have the book here tho.

[pshreck]

yup, if thats a clue then Im feeling Ive been wasting my time being so interested in this.

[daryn]

</font><blockquote><font class="small">In risposta di:</font><hr />
yup, if thats a clue then Im feeling Ive been wasting my time being so interested in this.

[/ QUOTE ]


yup, but what did i expect?

[pshreck]

Bill, can you confirm that it's the right question. Can you answer it from all the info you gave? It seems like 1 small detail would be very important.

[Skjonne]

[ QUOTE ]
Nope!

[/ QUOTE ]

How about this then?

<font color="white"> 1760 </font>

[DBowling]

i got the same answer, damn, i guess ill have to think harder before reading any other's posts.

[Bill Murphy]

Alright, here it is(I knew I should've just copied it word for word).

The key is, when the ferries first pass each other, what's the total distance traveled by each? The width of the river, of course. Now, when they pass again, what's the total distance traveled? Three times the width, right? So, assuming constant speeds, the ferry that left Pier A has now traveled 720 yds x 3 = 2160 yds. And as its now 400 yds from Pier B the width of the river is 1760 yds or a mile.

[pshreck]

[ QUOTE ]
[ QUOTE ]
Nope!

[/ QUOTE ]

How about this then?

<font color="white"> 1760 </font>

[/ QUOTE ]


I got the same, answer, Im pretty positive thats it now.

[fatmongo]

** spoiler **
if anybody wants to check it out for themselves i googled and found it here (http://home.att.net/~numericana/answer/recreational.htm) . half way down the page

[Bill Murphy]

No, that had nothing to do with it. No tricks at all. *sigh* I knew it was a bad idea posting this, but I thought it was a neat lil' puzzle. Live &amp; learn.

Sam Loyd came up w/some insane chess problems, BTW.

[thirddan]

booo, i like my original answer better...nothing even crashed in your answer /images/graemlins/frown.gif

[Bill Murphy]

Wish I'd've known about that site... /images/graemlins/shocked.gif /images/graemlins/laugh.gif

[Bill Murphy]

/images/graemlins/cool.gif

[Bill Murphy]

Yep.

[daryn]

and the ferrys always leave at exactly the same time


this is what threw me off i guess. taking what you said literally, one deduces that if the faster ferry gets to the other side first (which it will), it waits for the other one to get to the other side and unload before they both start up again.


if this were the case, which i assumed it was (and rightly so) then the problem is impossible.


in closing (and this should be obvious), when you post these, make sure the wording is absolutely correct.

[pshreck]

I figured out the answer in a weird way... but I still don't understand how this is mathematically possible.

[pshreck]

[ QUOTE ]
and the ferrys always leave at exactly the same time


[/ QUOTE ]

Yeah... this was the clear error in your original post. I don't think it is so tough if we knew they actually left at different times.

[Bill Murphy]

http://home.att.net/~numericana/answer/recreational.htm#ferry

Thanks to fatmongo for finding the site.

[daryn]

exactly.

[Reef]

Two ferry boats start at the same instant on opposite sides of the river. One is faster than the other. They cross at a point 720 yards from the left shore on their way to their respective destinations, where each one spends 10 minutes to change passengers before the return trip. They meet again at a point 400 yards from the right shore.
How wide is the river?

[pshreck]

Yeah, that is significantly different than the original posted question, which implied something mathematically impossible without some sort of trick.

[daryn]

</font><blockquote><font class="small">In risposta di:</font><hr />
Yeah, that is significantly different than the original posted question, which implied something mathematically impossible without some sort of trick.

[/ QUOTE ]

in conclusion, daryn again was first with the correct answer!

[Bill Murphy]

.

[pshreck]

[ QUOTE ]
[ QUOTE ]
Yeah, that is significantly different than the original posted question, which implied something mathematically impossible without some sort of trick.

[/ QUOTE ]

in conclusion, daryn again was first with the correct answer!

[/ QUOTE ]

Daryn, some kid like 10 years ago figured out an error on a math problem on the SAT's, and actually refused to answer it and wrote down the error on the form, on his way to scoring a perfect 800 on math.

That was you right?

[daryn]

wow that is almost so correct it's scary /images/graemlins/shocked.gif

[Skjonne]

[ QUOTE ]
Alright, here it is(I knew I should've just copied it word for word).

The key is, when the ferries first pass each other, what's the total distance traveled by each? The width of the river, of course. Now, when they pass again, what's the total distance traveled? Three times the width, right? So, assuming constant speeds, the ferry that left Pier A has now traveled 720 yds x 3 = 2160 yds. And as its now 400 yds from Pier B the width of the river is 1760 yds or a mile.

[/ QUOTE ]

/images/graemlins/smile.gif That was simple. I did it the math'ish way:

Ferry A travels with a speed of A. Ferry B with a speed of B. After T1 hours they meet for the first time, therefore

A*T1 = 720 and
B*T1 = L-720, L beeing the lenght of the river.

After their fist meeting they travel for T2 hours and meet again. Therfore

A*T2 = (L-720) + 400 and
B*T2 = 720 + (L-400)

Elimínating T1 from the first set of equations gives
A/B = (720)/(L-720)

Elimínating T2 from the second set of equations gives
A/B = (L-320)/(L+320)

Now we can eliminate A/B:
(720)/(L-720)=(L-320)/(L+320)

Solving this for L gives

L=1760

[jakethebake]

[ QUOTE ]
Cool Math/Logic Problem

[/ QUOTE ]

What kind of geek thinks math problems are "cool"?

[pudley4]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Yeah, that is significantly different than the original posted question, which implied something mathematically impossible without some sort of trick.

[/ QUOTE ]

in conclusion, daryn again was first with the correct answer!

[/ QUOTE ]

Daryn, some kid like 10 years ago figured out an error on a math problem on the SAT's, and actually refused to answer it and wrote down the error on the form, on his way to scoring a perfect 800 on math.

That was you right?

[/ QUOTE ]

Here's the question:

Circle A has a circumference of 4pi. Circle B has a circumference of 1pi.

Circle B sits on top of Circle A, and rolls clockwise one time around the entire circumference of Circle A, stopping when it gets back to the top of Circle A.

How many times does Circle B make a complete revolution of itself?

(If anyone needs one, I'll try to put up a picture)