Brain teaser thread - Page 10

Dauphin · October 26, 2002, 19:15

Quote:

Originally posted by Ramo
c₁/c₂ln(5/4) = 2
c₁/c₂ln(6/5) = 1

Solve that (which I can't be arsed to do

).

Those equations are not compatable.

Urban Ranger · October 26, 2002, 23:34

Quote:

Originally posted by Ignorance
12 is XII in roman numerals.

\/ I I
/\ I I

Split in two

\/ I I
------
/\ I I

VII is 7

But /\II don't mean anything.

Urban Ranger · October 27, 2002, 00:18

I agree with SD.

Ramo · October 27, 2002, 01:50

Quote:

Those equations are not compatable.

It should be 5 - T, where T is the starting time, not 5, and so forth.

But ignoring the infinite velocity problem, I worked it out and got crap. I think the problem is contradictory.

Urban Ranger · October 27, 2002, 01:57

Not sure if it is contradictory, but some info seems to be missing.

Ramo · October 27, 2002, 02:00

Well, I got 0 = 1 when I was working out the problem

. I can't see anything wrong with how I set up the problem, so I'm pretty sure that one of the stated proportionalities is screwed up.

Urban Ranger · October 27, 2002, 02:40

Okay, since you are sure, go ahead and post a question.

Ignorance · October 27, 2002, 07:39

r = rate of falling snow (unknown)
ts = when snow has started falling (which we want)
p = plows constant rate of removing snow (unknown)

snow depth at time t: s = r*(t-ts)
speed of plow: v = p/s
which gives us v = p/(r*(t-ts))

Thus speed on first hour is integral from 4pm to 5pm:
v1 = (p/r)*log((5-ts)/(4-ts))
And for the second hour:
v2 = (p/r)*log((6-ts)/(5-ts))
We know that
v1 = 2*v2
So:
(p/r)*log((5-ts)/(4-ts))=(p/r)*log((5-ts)/(4-ts))
Get rid of (p/r):
log((5-ts)/(4-ts))=log((6-ts)/(5-ts))
Get rid of log:
(5-ts)/(4-ts)=((6-ts)/(5-ts))^2
Which simplifies down to:

ts^2-9ts+19=0

Which gives us roots 5,61 and 3,38 and as it started snowing before 4, only latter makes sense.
Turn .38 into minutes and we have the answer: 3:23 pm

Ignorance · October 27, 2002, 09:08

Heres an easier one:
A bookworm eats from the first page of an encyclopedia to the last page.
The bookworm eats in a straight line.
The encyclopedia consists of ten 1000-page volumes and is sitting on a bookshelf in the usual order.
Not counting covers, title pages, etc., how many pages does the bookworm eat? Including the first and the last page.

Zero-Tau · October 27, 2002, 15:57

Quote:

Originally posted by Urban Ranger

But /\II don't mean anything.

No, but it's still two equally big parts.

Ramo · October 27, 2002, 16:35

Quote:

Thus speed on first hour is integral from 4pm to 5pm:
v1 = (p/r)*log((5-ts)/(4-ts))
And for the second hour:
v2 = (p/r)*log((6-ts)/(5-ts))

Distance travelled, not speed.

Quote:

))^2

Ack, of course. Forgot the factor of two.

Jaguar · October 27, 2002, 16:38

Quote:

Originally posted by Ignorance
Heres an easier one:
A bookworm eats from the first page of an encyclopedia to the last page.
The bookworm eats in a straight line.
The encyclopedia consists of ten 1000-page volumes and is sitting on a bookshelf in the usual order.
Not counting covers, title pages, etc., how many pages does the bookworm eat? Including the first and the last page.

5000?

Jaguar · October 27, 2002, 16:46

It could also be 5010, depending on how many cover pages there are (or where the first "real" page is, on the left or on the right.)

Combat Ingrid · October 27, 2002, 16:54

8002
He eats the entire books 2-9 and one page each from the first and the last book.

Jaguar · October 27, 2002, 16:58

Why only 2-9?

Combat Ingrid · October 27, 2002, 17:03

It starts at the first page of the first book and goes directly into the last page of the second book. When it arrives at the last book it only eats the last page.

Combat Ingrid · October 27, 2002, 17:08

Here's another one:

It cannot be seen, cannot be felt.
Cannot be heard, cannot be smelt.
It lies behind stars and under hills,
And empty holes it fills.
It comes first and follows after,
Ends life, kills laughter.

Capt Dizle · October 27, 2002, 17:08

10018

Capt Dizle · October 27, 2002, 17:09

death

Combat Ingrid · October 27, 2002, 17:10

nope

JohnM2433 · October 27, 2002, 17:11

I would think the answer would be ten thousand, but that just seems too obvious. Am I missing something?

Immortal Wombat · October 27, 2002, 17:13

time

Jaguar · October 27, 2002, 17:16

Nothing?

Combat Ingrid · October 27, 2002, 17:18

Quote:

Originally posted by Immortal Wombat
time

no

Combat Ingrid · October 27, 2002, 17:21

Quote:

Originally posted by Jaguar Warrior
Nothing?

If you are immortal. "Nothing ends life"

JohnM2433 · October 27, 2002, 17:32

I mean, "not counting covers, title pages, etc.," the bookworm eats 1000 pages from each of the ten volumes, for a total of 10,000. It sounds like a simple word problem to me...

Combat Ingrid · October 27, 2002, 17:42

Quote:

Originally posted by JohnM2433
I mean, "not counting covers, title pages, etc.," the bookworm eats 1000 pages from each of the ten volumes, for a total of 10,000. It sounds like a simple word problem to me...

Take a look in your bookshelves how the books are positioned

Ignorance · October 27, 2002, 17:43

8004 actually. Every page has 2 sides.
For those of you who thought of 10000, think how the books are in the shelv and where the first and last page actually end up.

Lemmy · October 27, 2002, 17:58

Quote:

8004 actually. Every page has 2 sides.
For those of you who thought of 10000, think how the books are in the shelv and where the first and last page actually end up.

i still don't get it, my first guess would be 5000

Combat Ingrid · October 27, 2002, 18:13

Follow the yellow line, please