Coffee, Cream, and Making a Wish

We have looked at several solutions to the classic Coffee and Cream mixture problem:

Suppose you have a cup of coffee and a cup of cream.  If you take a spoonful of cream, mix it up with the coffee, then take a spoonful of that mixture and add it back to the cream, is there more cream in the coffee, or more coffee in the cream?

I offered a solution that begins by making a wish.  That is, I wished that each cup initially contained 10 spoonfuls of liquid; then I proceed to solve the problem algebraically, by following spoonfuls of liquid back and forth between cups.  To illustrate the real power of the make-a-wish strategy, though, let’s wish that each cup started with one spoonful of liquidIn this case, the act of transferring one spoonful of cream into the coffee amounts to pouring all the cream in the coffee. Now, mix it up.  The mixture on the left is obviously half coffee and half cream. Now, pour a spoonful of the mixture, or half, back into the empty cup.   The two cups now contain exactly the same solution:  half coffee, and half cream!  So the answer to the question is obviously that there are equal amounts of cream in the coffee and coffee in the cream!

Be sure to check out the elegant solution to the problem, as well!

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patrick honner

Math teacher in Brooklyn, New York

Julie Wright · June 18, 2016 at 9:31 pm

Fun problem — I don’t remember seeing it before. I love the elegant solution, and it’s probably true for coffee and cream. The ex-chemist in me wants to point out, though, that you can’t take it absolutely for granted that when you mix two liquids together, the volume of the mixture is the same as the sum of the volumes of the two liquids mixed.

I’d forgotten what this concept is called, but Google says it’s ideal solutions vs. non-ideal. Personally I am not a fan of coffee, so coffee + cream seems highly non-ideal to me, but let’s pretend it’s ideal…

The other two solutions you give are certainly highly suggestive, but by themselves I don’t feel right about trusting them to prove it’s always true. (Maybe I am missing something about the argument.) I feel more comfortable proving it works for any volume greater than 1 spoonful.

My original intuition said there would be less coffee in the cream, because you’re taking a spoonful of “creamy coffee” and adding it instead of adding pure coffee. But when I sat down to work it out algebraically, I realized you’re adding it to less than the full cup of cream, so…

Being a tea drinker I changed it to tea, also so I could use t and c to keep track. Then I kept track of t:c and c:t ratios, assuming you start with x spoonfuls of each (x greater than or equal to 1):

After spoonful of cream mixed in: t:c = x:1 in tea cup ; c:t = x-1:0 in cream cup

Spoonful of “creamy tea” will have x/(x+1) tea and 1/(x+1) cream, so new ratio in tea cup will be
x – x/(x+1) : 1 – 1/(x+1) or, multiplying both sides by x+1 & simplifying & dividing by x, x:1 ratio of t:c.
and new c:t ratio in cream cup will be
x – 1 + 1/(x+1) : x/(x+1) which also ends up being x:1.

Whoa! Not elegant, but convincing, and like I said, not what I expected.

MrHonner · June 19, 2016 at 7:23 pm

Very nice, and thorough, solution, Julie! It’s nice to see the algebra work out.

I understand your reluctance to accept the two “Make a Wish” solutions I’ve offered, which essentially amount to setting your x to a specific numeric value. Since the question is ultimately about a ratio, the actual volume is irrelevant. You basically proved this with your solution.

And fascinating point about the chemistry of the situation! I’m no chemist, but it still seems like whatever impact of the “non-conservation of volume” might have, it would impact both cups proportionally, and thus, cancel out in the end.

Julie Wright · June 19, 2016 at 8:29 pm

For some reason I can’t quite wrap my brain around the ratio argument. It seems equivalent to saying it doesn’t matter how many spoonfuls you take out of each cup, as long as they’re the same. But for me, neither of those lead to the idea that you can just pick your amount and the spoonful:cup ratio is irrelevant. The algebra backs you up, but I can’t quite get there myself without the algebra!

As for the non-ideal solution case, I disagree it would all cancel out. Suppose that when you add a spoonful of cream to coffee, the mixture “shrinks down” to a smaller volume than you’d expect. When you add a spoonful of that back to the cream, basically you’re adding more coffee AND more cream back in than you would if it were an ideal solution (though generally more coffee than cream), which in many cases (I think possibly for all cases except when you take out all the cream in the first spoonful, but I’m not certain) would not preserve the ratio.

As a specific example, if you had 10 spoonfuls of each and took out 1 spoonful, and the non-ideal coffee+cream solution “shrunk down” to 10 spoonfuls (instead of 11)*, then each spoonful of dense solution would have 1 spoonful of [original] coffee and 0.1 spoonful of [original] cream. Add that back to your 9 spoonfuls of cream and you will have a 9.1:1 ratio, not a 10:1 as the coffee-with-cream-added cup still has. Basically, you’ll have added just as much coffee as you would in the ideal-solution case, but the amount of stuff in the cup would have started its pre-mixing life as a larger volume than 10 spoonfuls, unlike the ideal-solution case. This will be true whether or not the cream-with-a-little-coffee “shrinks down” the same amount as the other solution.

* Admittedly this wouldn’t be likely to happen with coffee and cream, chemically. 🙂

Julie Wright · June 19, 2016 at 10:53 pm

Oops. When I said, “Basically, you’ll have added just as much coffee as you would in the ideal-solution case,” I think I was wrong: you will have added MORE coffee (and more cream) than you would with ideal solutions, as I argued elsewhere.

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