RESULTS SUBMITTED FOR PROBLEM #21

Here is the original problem, followed by solutions submitted by Zeno's patrons. Yes, the amounts remaining are equal!! Here's a nice solution submitted by Gregg Boettcher:

comments: I found that both cups would contain exactly the same proportion of their
original contents! This finding very much surprised me, but I can explain the way I
arrived at this answer.

Suppose that each cup contains 16 tablespoons. This may or may not be true, but I hope it
will work as an example to illustrate a general rule. This is how the operation would then
take place.

1. Anonymous customer has 16 tablespoons of tea. Charles has 16 tablespoons of coffee.

2. Exactly 1 tablespoon of coffee is transferred from Charles' cup to the customer's cup.

3. The anonymous customer now has 16 tablespoons of tea and 1 tablespoon of coffee.
Charles now has 15 tablespoons of coffee.

4. Exactly 1 tablespoon of the customer's mixture is transferred to Charles' cup. The
tablespoon will contain the same ratio of tea to coffee that the customer's cup had
contained--namely, a ratio of 16 to 1. In other words, the tablespoon mixture will consist
of exactly 16/17 tablespoon tea and 1/17 tablespoon coffee.

5. Therefore, it can only be that the anonymous customer now has 15 1/17 tablespoons tea
and 16/17 tablespoons coffee. Likewise, Charles must have 15 1/17 tablespoons coffee and
16/17 tablespoons tea. Both cups have exactly the same proportion of their original
contents.

username: marcos@net-k.com.br

state: SP

city: Indaiatuba

country: Brazil

comments: Both cups will have the SAME contain of the original content, and the SAME
contain of the "new added content".

If you transform the problem to an algebraic formulation,

you can easily prove it.

For a given "x" content of coffee ("A") in the cup 1 and the same
"x" content of tea ("B") in the cup 2, if you take "y" (one
tablespoon) from 1 and add it to 2, you will have:

cup 1: (x-y)A

cup 2: xB + yA

And now, you will take "y" from the new content of cup 2,

and add it to cup 1. Considering that cup 2 has a perfect mixture, when you take off
"y", you will have proportional parts of "A" and "B":
(x/(x+y))*yB and (y/(x+y))*y A. Then:

Cup 1:(x-y)A + (x/(x+y))*yB + (y/(x+y))*y A

Cup 2: (xB+yA)- (x/(x+y))*yB - (y/(x+y))*y A

after some "adding and subtracting", you will have:

cup 1: (x**2/(x+y)A and (x*y)/(x+y)B

cup 2: (x**2/(x+y)B and (x*y)/(x+y)A.

username: brentf007@yahoo.com

state: SK

city: Saskatoon

country: Canada

comments: Charles should wager that each cup contains the same percentage

of it's original contents. This is proven if the procedure is worked

out mathematically. Start with assigning a mug a volume of 100 units and a tablespoon

a volume of 10 units, (any designation can be used, providing

it is used consistently through the equation). Then procede step by step.

STEP ONE

100T 100C

+10C -10C

=100T + 10C 90C

or 90.91%T + 9.09%C

STEP TWO

One tbsp taken from the cup of tea will contain 90.91%T and

9.09%C. This works out to 9.09T + 0.91C.

100T + 10C 90C +9.09T + 0.91C

-9.09T - 0.91C

=90.91T + 9.09T =90.91C + 9.09T

Therefore, the amount of coffee in Charles' cup is equal to

the amount of tea in the stranger's cup.

username: mattbrissette@hotmail.com

state: Oh

city: Heath

country: USA

comments: the customer would have more of his original contents because Charles gave up
one tablespoon of his Coffee while the customer gave up a tablespoon of a mixure of coffee
and tea therefore not giving up a full spoo

username: kc5jlv@juno.com

state: TX

city: Welch

country: USA

comments: Should Charles decide to make a wager, the odds are even;

both cups hold the same amount of their original liquid.

At the outset, both cups have one cup of liquid; Charles

has 8 oz. of coffee and the other fellow (call him Dave) has

8 oz. of tea. One tablespoon (0.5 oz.) of Charles' coffee is

transferred to Dave's cup. Now Charles has 7.5 oz. of coffee,

and Dave has 8 oz. of tea plus 0.5 oz. of coffee.

In Dave's cup, the liquid can be thought of as containing

seventeen equal parts (where each part is 0.5 oz.). There are

sixteen parts tea and one part coffee in this mixture. Thus,

Dave's cup (and the liquid which is removed) is 16/17 tea and

1/17 coffee.

Now we transfer one tablespoon (0.5 oz.) from Dave's cup

to Charles'. This liquid is 16/17(0.5 oz.) = 8/17 oz. tea and

1/17(0.5 oz.) = 1/34 oz. coffee.

Charles' cup now contains (7.5 + 1/34)oz. coffee and 8/17

oz. tea. Performing the arithmetic, Charles has 7 9/17 oz.

coffee and 8/17 oz. tea (which totals 8 oz., as it should).

Dave's cup, on the other hand, has had liquid removed. He

now has (8 - 8/17) oz. tea and (0.5 - 1/34) oz. coffee. Doing

the arithmetic, this means that Dave has 7 9/17 oz. tea and

8/17 oz. coffee.

Thus, both Charles and his new acquaintance have 7 9/17 oz.

of the original drink in their cups, along with 8/17 oz. of

the other person's drink. Since the mixtures are now exactly

identical, any party who says that one cup holds more of its

original contents than the other will be incorrect.

In this situation, neither person will win by selecting a

cup. But how can you have a wager if there is no winner? In

order to win, Charles has two options.

1) Point out that both of the cups hold identical proportions

of their original liquid, meaning that Dave will have to pay

if Charles is wrong. Since Charles is right, he will win.

(This approach is identical to selecting both cups.)

2) Force Dave to select a cup on which he desires to wager.

After Dave has chosen the cup which he believes to contain

more of its original liquid, Charles can win the bet by proving

Dave to be incorrect.

username: chead@math.umn.edu

state: MN

city: Minneapolis

country: USA

comments: ASSUMING the cups have the same volume, they both have the same amount of their
original contents -- if the cups have volume V and the spoon holds a volume t, they both
have V^2/(V+t) of their original content.

But what if they don't? Let the volume of coffe be C and the volume of tea be T. After the
mixing, there is T^2/(T+t) tea in the tea cup, and C-(Tt)/(T+t) coffee in the coffee cup.
The only way for these to be equal is for T=C, otherwise, the bigger one wins.

In fact, even if we consider proportions, there is 1-(tT/C)/(T+t)% of the coffee in the
coffe, and T/(T+t)% of the tea in the tea, and again, these can only be equal if T=C.

So whoever had the bigger cup wins. If they're the same size, it's a tie.

username: boettcg@ssu.southwest.msus.edu

state: MN

city: Marshall

country: USA

comments: I found that both cups would contain exactly the same proportion of their
original contents! This finding very much surprised me, but I can explain the way I
arrived at this answer.

Suppose that each cup contains 16 tablespoons. This may or may not be true, but I hope it
will work as an example to illustrate a general rule. This is how the operation would then
take place.

1. Anonymous customer has 16 tablespoons of tea. Charles has 16 tablespoons of coffee.

2. Exactly 1 tablespoon of coffee is transferred from Charles' cup to the customer's cup.

3. The anonymous customer now has 16 tablespoons of tea and 1 tablespoon of coffee.
Charles now has 15 tablespoons of coffee.

4. Exactly 1 tablespoon of the customer's mixture is transferred to Charles' cup. The
tablespoon will contain the same ratio of tea to coffee that the customer's cup had
contained--namely, a ratio of 16 to 1. In other words, the tablespoon mixture will consist
of exactly 16/17 tablespoon tea and 1/17 tablespoon coffee.

5. Therefore, it can only be that the anonymous customer now has 15 1/17 tablespoons tea
and 16/17 tablespoons coffee. Likewise, Charles must have 15 1/17 tablespoons coffee and
16/17 tablespoons tea. Both cups have exactly the same proportion of their original
contents.

username: don.keyworth@drake.edu

state: IA

city: Des Moines

country: USA

comments: Excellent Web Site, particularly ZENO's!

Answer: Charles should bet that each cup contains the same

amount of its original contents. Suppose there are exactly

10 teaspoonsfull of liquid in each cup to start. The customer

puts 1/10 of Charles' coffee into his own cup, resulting in

1/11 of his cupful being coffee. Then one teaspoonfull,

containing 10/11 tea and 1/11 coffee is placed back in

Charles' cup. Charles now has a mixture of 9 and 1/11

teaspoonsfull of coffee plus 10/11 spoonfull of tea. The

customer has a mixture of 9 and 1/11 spoonfulls of tea

plus 10/11 spoonfull of coffee.

Without fractions: suppose each cup holds 100 units

of liquid. Each spoonful contains 10 units. The customer

moves 10 units of coffee from Charles to himself, leaving

Charles with 90 units of coffee, and the customer with 100

units of tea plus 10 units of coffee. The tea and coffee

are thoroughly mixed. The next transfer moves 9 units of

tea and 1 of coffee, leaving Charles with 91 units of coffee

and 9 of tea, and the customer with 91 units of tea and 9 of

coffee. Hence even. QED.

username: mbaxterp@nas.edu

state: dc

city: Washington

country: USA

comments: Charles should probably hold onto his money. First, there is

probably some reason that Coffee and Tea do not mix uniformly.

So it is difficult to discern what the second spoonful would

actually contain. The game would be more interesting if there

were two cups of the same coffee. In such a case the first spoon

would remove a set amount of the pure liquid in cup one and replace

it with the exact same amount of a diluted liquid. The second cup actually has

a tablespoon removed, but only after it has a tablespoon added. Therefore, the

amount removed is a smaller amount of its "original substance" since it is

bound to have at least some of the cup on liquid removed. This, again,

assumes that the two can mix at all in the second cup. If not

the probability of having more of the same liquid in the

second cup is still higher, but not worth Charles losing any

money.

username: schmidt2@home.com

state: CA

city: Livermore

country: USA

comments: After the transfer, the cups each contain equal amounts of their original
contents, so Charles should bet that the amounts are equal. Say, for ease of calculation
(and because I can't recall off the top of my head how many teaspoons are contained in one
measuring cup), that each cup contains 10 tsp of liquid initially. After one tsp of coffee
(C) is put into the tea cup (which contains 10T of tea), the totals are: coffee cup = 9C,
tea cup = 10T+ 1C. Right. Now, when the customer takes one tsp from the 11 tsp in the tea
cup, he adds one-eleventh of the 10:1 tea:coffee mixture to the coffee, thereby making the
coffee cup contain 9C + 1/11(10T + 1C), or a total of (100C/11 + 10T/11). The tea cup now
contains 10/11(10T + 1C), which = (100T/11 + 10C/11). Therefore (in this example), the
cups each contain about 90.9% (i.e., 10/11) of their original liquid.

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username: kwagner818@aol.com

state: wi

city: sussex

country: USA

comments: They both have the same amount of the original contents because the coffee is
taking up room in the tea and the tea is taking up room in the coffee. If a cup of coffee
is 8 oz and you scoop out 1 oz and put it into the tea, the coffee is at 7 oz and the tea
is at 8 oz tea and 1 oz coffee. When you scoop back into the coffee cup, say you get 3/4
oz tea and 1/4 oz coffee. Then you end up with 7 1/4 of tea and 3/4 oz coffee in the
teaculp and 7 1/4 oz coffee and 3/4 oz tea in the teacup. Any way you measure it, you'll
always have equal amounts of the original contents.

username: Baldmn@AOL.COM

state: NJ

city: Elizabeth

country: USA -Move 2 years ago from P.R.

comments: Both Cups have the most of its original content if the questions refers to the
fact the cup with cofee now has more cofee that the other and vice-versa. But if the
questions is comparing the amounts the cups now have, then both have the same amount.

For lack of better math let's say a cup contains 19 tablespoons(tbs) and a tbs contains 20
drops. After the first exchange of fluids, Charle's cup has 18 tbs (or 360 drops of cofee)
and the Customer's cup has 19 tbs of tea and 1 of cofee (380 drops of tea and 20 of
cofee). The mixed liquid has a proporsion of 19tea to 1cofee. The spoon from this mix
which is mixed back into Charle's cup has 19 drops of tea and 1 drop of cofee. Thus,
Charles now has 361 drps of cofee in his cup. The customer who originally had 380 drops of
tea in his cup now has lost 19 drops of his tea, and he also has 361 drops of the original
liquid in his cup.

username: kiekeben@aol.com

state: OH

city: Columbus

url: http://members.aol.com/kiekeben/home.html

country: USA

comments: He should say both cups contain the same amount of their original contents. If
an x amount of coffee is now in the tea cup, then, given that the tea cup is just as full
as it was initially (for one teaspoon was added and one removed), an x amount of tea must
have been removed from it and placed in the coffee cup. Each cup therefore contains its
initial amount minus x of its original beverage.

username: psyke.11@usa.net

state: NSW

city: Sydney

url: none

country: Australia

comments: obvuiously the Tea cup contains more of its contents.

WHen the customer takes a tablespoon of coffee to mix in with the tea, he takes pure
coffee, but when he then takes tea to put in the coffee, the tea has been tainted, the
tablespoon of tea is in fact a fraction of coffee as well.

of course we are assuming that the cups were at the same level, that you are posing the
question as to the cups containing "more of its original contents" in proportion
to its own previous volume, and not in comparison to the other cup.

username: magi336@aol.com

state: Il

city: Berwyn

url: members.aol.com/magi336

country: USA

comments: If Charles is to win, he should bet that both cups contain the same amount of
their original

contents, leaving both other (incorrect) options for the

"customer". Since this "customer" was the one to

initiate the bet, he will probably laugh out loud at

Charles, and gladly accept. Charles should bet a lot of money.

username: kaparoo@vassar.edu

state: NY

city: Poughkeepsie

country: USA

comments: They each have the same amount of original and the same amount of foreign liquid

in their cups. To demonstrate this, I will assume quantities.

If each had 100ml of liquid to begin, after the first exchange, Charles could

be said to have 90ml of coffee, and customer had 100mlT+10mlC (91%tea). The liquid

transferred for the second exchange would contain this 10:1 ratio of tea/coffee as

well, so instead of transferring 10ml of liquid, the customer would lose 9.1ml, bringing

his total liquid down to 90.9ml of tea and 9.1ml of coffee. Charles would gain .9ml

of coffee back (9% of the original 10ml lost), making his total coffee content 90.9ml.

username: ha357@hotmail.com

state: oz

city: thereville

country: USA

comments: the coffee cup.

because one tablespoonful of his coffee was sent to the tea and the assumption is that it
will mix with the tea so that the volume of it's content, coffee, more or less equally.

That means that when he tries to get it back to make the volumes equal, via the returning
tablespoon, it will not be 100% tea comming back, but a mixture of tea and coffee.

soo, although the volumes will be the same, the coffee cup will have more of its original
contents

username: okapal.2@osu.edu

state: OH

city: Columbus

country: USA

comments: It mostly depends on if Charles drank any of his coffee during the conversation.

If in fact neither he nor the new aquantance drank either of

their beverages, Charles should bet on his own.

username: norlando@comcen.com.au

country: Australia

comments: Niether cups have more of there original content.

They now contain a mixture of the original content of both cups.

username: tmh@nanoteq.com

city: Centurion

country: RSA

comments: Charles should bet that they both contain equal proportions of the original
contents. Since charles lost a full tsp and gained a dilluted proprtion of coffee back,
the cutomer lost a tsp of tea minus an equally dilluted proportion of coffee. The equation
balances if both cups contained the same amount of fluid.

username: georgev@ma.ultranet.com

state: MA

country: USA

comments: The amount of coffee in "the man's" tea is the same as

the amount of tea in Charles' coffee. This is the case

even if the cups are different sizes. Note that if they

are different sizes, the amounts will be the same, but

the percentages will not. In this case, Charles should

wager so that he has the equality (e.g. "my cup has less

than or the same amount...").

The other cases are less interesting. If you interpret

"amount" to be actual size, then bet on the larger cup.

If they are the same size, but the amount transfered is

not the same, bet on the biggest transfer. Finally, if

you don't assume mixing before transfer, bet on the cup

that has the lowest density if it transfered first, or

equality if it didn't. Viscosity, etc. deleted for brevity.

Interpreting the "Explain" in the answer to mean "explain

why there is the same amount in each cup under the usual,

customary and reasonable assumptions," the following

example is submitted:

Two cups, each 10 volume units (v). Cup A contains coffee

and cup B contains tea. Take 1v of A and put it in B.

A now contains 9v coffee and B contains 1v coffee and 10v

tea. Cup B is 1/11 coffee. Take another v of this

mixture and put it back in cup A. The mixture is 10/11

tea - and it's being put into a 9v cup to make 10v total.

Cup B is still 1/11 coffee and cup A is now (10/11)*(1/10)

or 1/11 tea.

I hope the spoon was clean.

username: PUBLIUS345@AOL.COM

state: usa

city: HERE

country: USA

comments: CHARLES'S CUP HAS MORE BECAUSE THE OTHER CUSTOMER IS CHARLES, HE ORDERED TWO
DRINKS

username: swtbean@azstanet.com

state: AZ

city: Tucson

country: USA

comments: 1. Charles' cup lost 1 tablespoon of undiluted coffee

formula.

2. Charles cup regained 1 tablespoon of tea/coffee mix.

3. The tablespoon of coffee/tea mix contains less coffee

than the tablespoon it is replacing.

4. Charles cup will contain less of it's original

contents.

username: scott_barbour@tekno.com

state: KY

city: Glasgow

country: USA

comments: He should bet that they are both the same. Charles losses one tablespoon of
coffe and then gets a small percentage back when the customer puts the mixture in the
coffee. The customer gains coffee but gives Charles that same amount of tea that he
retains of the coffee