Given a cup of coffee and a cup of cream, each contain the same amount of liquid. A spoonful of cream is taken from the cup and put into the coffee cup, then a spoonful of the mixture is put back into the cream cup. Is there now more or less cream in the coffee cup than coffee in the cream cup?

You can check the answer of the people under the question at Quora “and a cup of coffee“

Assume that the cups initially contain n amount and capacity of spoon is k. Also, we need to assume that capacity of the cups is sufficiently large so as to not overflow when after first step, one of the cups contain n+k amont.

Let cup A contain coffee and B contain cream.

Initial configuration:

A B

Coffee n 0

Cream 0 n

A spoonful (k) of cream is taken from cup B and is put into cup A.

So the the configuration becomes:

A B

Coffee n 0

Cream k n-k

Notice that total quantity is still 2n and the sum of rows is still n each (conservation of mass 😉 ).

Now, a spoonful of mixture from A is put into B.

A spoon has capacity k. It takes the amount k from the total quantity of n+k.

Hence the ratio is \dfrac{k}{n+k} . Now assuming the mixture is homogeneous by the time you transfer it back, the same ratio of coffee and cream will be in the spoon.

Hence, the quantity of cream that will be in the spoon is: \dfrac{k^2}{n+k} and that of the coffee that will be in the spoon is: \dfrac{nk}{n+k}

Hence, after transferring, the configuration becomes:

A B

Coffee n – nk/(n+k) nk/(n+k)

Cream k – k^2/(n+k) n – k + k^2/(n+k)

Which, after simplification, yields,

A B

Coffee n^2/(n+k) nk/(n+k)

Cream nk/(n+k) n^2/(n+k)

Hence, cup A has as much coffee as the amount of cream in cup B and cup Bhas as much coffee as the amount of cream in cup A.equal. That happens when the size of spoon is exactly n/2.However, the quantities of cofee and cream in a given cup are not

Qualitative answer:So the total fluid volumes in the end are the same in both cups, and are the same as in the beginning.

Whatever happened, in the end there must be a net transferred cream volume .

Since the volumes at the end are equivalent, the net received coffee volume must be equivalent.

Quantitative answerThe cream to coffee ratio r is determined by the spoon to cup ratio s , and is given by r = s /(1+ s ).

ExampleLet s = 1/10, then r = (1/10)/(1+1/10)=1/11, so taking away a tenth, yields an endratio of one eleventh.

Counterintitive?– “But hey, in the forward spoon there is 100% cream, while in the reverse spoon there is only 90% coffee!”

– “True, but the reverse spoon also contains 10% cream which makes the net cream flux a 90% spoon again!”

This is a gem of a puzzler. I was surprised when I learned how easily it could be answered.

Given: You end up with the same amount of coffee and cream that you started with (but maybe in different places), and each cup still holds the same amount of liquid it started with.

At the end of the transfers, whatever amount of cream there is in the coffee cup is equal to the amount of coffee that’s no longer in that cup, but that’s the coffee that’s now in the cream cup. Therefore, the same amount of cream is in the coffee as there is coffee in the cream.

Note that the size of the cups is irrelevant, and it doesn’t matter if you stir or not, Also, the amount of liquid transferred is irrelevant so long as the same amount of liquid is transferred in each direction.

For some history of this puzzle, see Wine/water mixing problem , or Bill the Lizard’s 2009 article onDodgson (Carroll) .Lewis Carroll, that is, Charles Dodgson (1832–1898).

From Martin Gardner’s The Universe in a Handkerchief , page 80

http://www.logic-books.info/sites/default/files/the_universe_in_a_handkerchief_lewis_carrolls_mathematical_recreations_games_puzzles_and_word_plays.pdf

Lets assume that size of the spoon does not matter. Whats the easiest size to try. My first thought was that take the spoon to be half the size. But we can still make it easier: make the size of the spoon equal to the size of all the liquid in one cup. We start with 1 unit in each cup, then we take one unit from the cream and add it to the coffee. The cream cup not has 0 units and the coffee cup has two units mixed in a 50-50 ratio. Now take one unit from this, and put it in the cream cup. We end up with both cups with 1 unit of a 50-50 ratio.

It’s the same. The amount of cream in the coffee is the same volume of coffee that’s missing, and so must be in the cream cup. And vice versa.

This works–unless alcohol is involved, since alcohol in water reduces the volume somewhat, throwing the final amounts off.

A related question (once posed by Martin Gardner):

If you want to keep your coffee as warm as possible, do you add the cream immediately on pouring it, or do you wait until you are about to drink the first sip?

Eight O’Clock

The cream is thicker than the coffee and clings to the entire spoon, so the amount of cream added to the coffee is greater than the amount of the coffee-cream mixture (which doesn’t cling to the spoon as much as cream alone) transferred back to the coffee cup.

Notes: I am assuming that everything is mixed thoroughly so that the coffee and milk are evenly distributed when combined. Repeating decimals are indicated by trailing dots (“x.xx…”).

For the sake of simplicity, say we have a big spoon such that one spoonful is equal to 10% of a cup, or 0.1 cups. If we have 1 cup of coffee and 1 cup of cream, and then remove 10% of the cream and add it to the coffee, we have 0.9 cups of cream on one hand, and 1 cup coffee + 0.1 cups cream = 1.1 cups of coffee/cream mixture. The coffee is 1 cup out of 1.1 cups, or 90.9090…%, while the cream is 0.1 cups out of 1.1 cups, or 9.0909…%.

Taking a spoonful of coffee/cream mixture, we get 0.1 cups again which is 90.9090…% coffee (0.0909… cups) and 9.0909…% cream (0.009090… cups). The cream remaining in the coffee cup is 0.1 – 0.0090909… cups, or .0909… cups. The coffee in the cream cup is just the amount we just put in, 0.0909… cups, so the amount of coffee in the cream cup is the same as the amount of cream in the coffee cup.

The numbers are totally arbitrary here, so you could say the spoon is tiny and only holds 0.0001 cups, or enormous and holds 1/2 a cup or even the whole cup. You should always get the same result. In the case of 1 spoon = 1 cup, that’s obvious, since you would just be equally mixing equal quantities of cream and coffee. When you change the size of the spoon, it doesn’t affect the result though since you’re using it to transfer equal volumes between the cups.

Now I really want some coffee (with cream).

Let us assume that each cup has a capacity of x ml and the spoon has a capacity a<

After the first step we have:in Cream Cup :- Cream= x-a

in Coffee Cup :- Cream=a and Coffee=x ; Cream : Coffee= a:x

After second stepIn the second step the teaspoonful contains ‘a’ ml which has cream:coffee =a:x.

Therefore the second teaspoonful has

Cream a.a/(x+a)

Coffee a.x/(x+a)

Total a

Now in Cream cup:- Cream=x-a + a.a/(x+a)= x^2/(x+a) and Coffee=a.x/(x+a)Coffee: Cream= a:xAnd in the coffee cuptaking a teaspoonful as in the second step does not change the ratio and therefore Cream:Coffee = a:xSo both are equal.

Suppose container A has 10u (units) of coffee and container B has 10u of cream.

I’m taking 3u coffee from A and pour it into B

So that now A has 7u coffee(only coffee) and B has 13u (cream + coffee) mixture.

In this mixture \frac{10}{13}u per 1u is cream and \frac{3}{13}u per 1u is coffee.

Now I take 3u from B. So it has 3.\frac{10}{13}u cream and 3.\frac{3}{13}u coffee

I added this to A. Already it has 7u coffee. So now it became 7+3.\frac{3}{13}u coffee in A. Ie, 7+\frac{9}{13}=\frac{100}{13}u coffee

B lost only 3.\frac{10}{13}u cream, which means B now has 10-3.\frac{10}{13}=10-\frac{30}{13}=\frac{100}{13}u cream

So the ratio of coffee to cream if denoted by x:y in A then it is y:x in B

I rarely get into questions related with mixtures. Anyway this was a nice question to solve.

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Answer got collapsed:D I guess this is a punishment for late comers. What should I add to make it lengthy? Should I take cream from B and pour it into A and do the calculation again? I swear the answer will be same:D

Okay I will continue the former one itself.

Now that A has \frac{100}{13}u coffee, it has 10-\frac{100}{13}u cream.

And B has \frac{100}{13}u cream, now that it has 10-\frac{100}{13}u coffee.

This is a fun question and I understand the intent, but it needs one more point of clarification:

A spoonful of cream is equal in volume to a spoonful of coffee.At first this seems to be obvious, but in actuality it is anything but. Coffee and cream have different levels of surface tension. I would be surprised that one spoonful of each is equal.

Because of this my immediate answer was there is more cream in the coffee – cream is “thicker” than coffee (has more surface tension), so when you remove a spoonful of cream, the total volume of the cream is going to be more than the total volume of the mixture.

I would also make an attempt. Lets assume initial scenario

X = 100x = Coffee

Y = 100y = Cream

Assuming one spoon is 10 units. Take 1 spoon from Y to X. New Scenario.

X = 90x = Coffee

Y = 100y+ 10x = Cream

Now, assuming that the Y is uniform mixture. Take one spoon back fro Y to X.

One spoon is 10 unit comprising

01/11=0.909 Coffee

10/11= 9.091 Cream

Add this spoon to the X.

X = 90x+0.909x + 9.091y = 90.909x+9.091y = Coffee

Y= (10-0.909)x + (100-9.091)y = 90.909y+9.091x = Cream

From the above expression, you can see that both of them have equal concentration. It might be off by few decimal points, but, we got the point. 🙂

The solution becomes clearer when we notice that the spoon-size relative to the cup-size is not specified, and so the answer should hold for any spoon-size. If we choose a spoon the same size as the cup, the result is a complete mixture evenly divided. Similarly a spoon-size of zero points to equality.

Victor Allen’s

We take a spoon from cream cup, put it into coffee cup and then again we take a spoon from coffee cup and put it into cream cup => volume of cream cup is unchanged and volume of coffee cup is also unchanged

During this transition, both the cups have newcomers i.e for coffee cup, cream is the new comer and for cream cup, coffee is the newcomer.

Lets say x amount of cream came into coffee cup, so x amount of cream should be missing in cream cup. But since we know volume of cream cup is intact and coffee is the newcomer. so x amount of coffee is present in cream cup.

Fractions & ratios instead of decimals.

—————————————————————————————

100a = cup1 = 100 units cream

100b = cup2 = 100 units coffee

—————————————————————————————

STEP 1: 10 units from cup1 to cup2, stir.

Cup1: 100a – 10a

Cup2: 100b + 10a

Evaluation 1:

Cup 1: 90a

Cup 2: 100b + 10a

STEP 2: 10 units from cup2 to cup1

cup1: 90a + 1/11(100b + 10a) = 100/11b + 1000/11a

cup2: (100b + 10a) – 1/11(100b + 10a) = 100/11a + 1000/11b

Evaluation 2:

cup1: 1000/11a + 100/11b

cup2: 1000/11b + 100/11a

The ratio of cream/coffee in cup 1 is equal to the ratio of coffee to cream in cup 2.

1000/11 of cream to 100/11 of coffee in cup 1 is an equal ratio/proportion to 1000/11 of coffee to 100/11 of cream in cup 2.

At the end of a valid move, there are three (independent) conservation laws (the three independent initial values) that hold for four unknowns. Whatever the dynamics, if at the end of it you now the value of one variable the other three are determined. Given that the total volume of liquid in each cup is the same as the initial conditions, change in qty of coffee in one cup = change in qty of cream in other cup.

In terms of additive symmetry transformations of matrices:

let i label the liquid, a or b

let I label the cup, A or B. Consider the matrix m_iI = mass of liquid i in cup I

(there are possible volume changes when non-immiscible materials are dissolved in one another, so volume is rarely a conserved quantity, and the above question is better posed as a fixed mass being transferred back and forth)

Mass conservation of the liguids tells you that the only additive symmetry is

( x -x )

( y -y )

Conservation of mass in any one cup gives a further constraint: x + y = 0. So finally the only allowable symmetry is:

( x -x )

( -x x )