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.

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.

state: SK

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.

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

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.

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.

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.

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.

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.

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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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.

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.

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.

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.

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.

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.

state: oz

city: thereville

country: USA

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

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.

country: Australia

comments: Niether cups have more of there original content.
They now contain a mixture of the original content of both cups.

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.

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.

state: usa

city: HERE

country: USA

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

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.