This problem nearly ripped the Meyer household apart tonight:

Which glass contains more of its original soda?

[WCYDWT] Coke v. Sprite from Dan Meyer on Vimeo.

Justify your answer.

**The Goods**

Download the video.

**2011 Mar 04**: Updated to add the goods.

**2011 Mar 13**: 71 comments as of today means we’ve struck a nerve. Many commenters have put their mark down with an algebraic proof. More interesting to me are those who have included devices for *illustrating* the proof to their students. That’s harder stuff. See MPG’s comment:

Consider a similar problem using discrete objects (e.g., playing cards. Take 10 red cards and 10 black cards face down in separate piles. Take four at random from red pile; mix into black pile. Shuffle. Return four random cards face down to red pile. Ask: more black in the red pile or red in the black pile. Try this several times. If youâ€™re not convinced, do it with the faces showing. Apply principle to soda problem.

## 110 Comments

## Thad Moren

March 3, 2011 - 8:22 pm -You would have to prove to me that Sprite wasn’t thicker than Coke first.

## Mark

March 3, 2011 - 8:23 pm -Assuming you took the same amount of liquid each time, so that each glass both started and ended up with equal amounts of liquid, it has to be the same.

## Mr. K

March 3, 2011 - 8:24 pm ->it has to be the same

And this is why this is a great math lesson.

## Cathy

March 3, 2011 - 8:25 pm -My gut feeling is that the Coke has more of its original soda. Unfortunately I don’t have a strong mathematical way of explaining it.

When you took a dropper full of the Sprite out of its glass you took 100% Sprite. Then you mixed this into the Coke. Then, when you removed the same amount of liquid out of the Coke it wasn’t 100% Coke so less of the Coke was removed. Then when you put this dropper of liquid back into the Sprite you did not return 100% of the Sprite so that is less than the original amount of Sprite which had been removed earlier. So, overall there is less Sprite in its original glass than the Coke.

Is this a trick question?

## Mark

March 3, 2011 - 8:31 pm -Cathy, think of it this way:

If we started with 100 units of Coke and 100 units of Sprite, and now the Sprite glass has 98 units of Sprite and 2 units of Coke, then the Coke glass MUST have what’s left, which would be 98 units of Coke and 2 units of Sprite.

This assumes a LOT, though: that the amounts transferred were equal, that we can ignore any drips that spilled, that the two original amounts were equal, etc. But I don’t think those differences were in the spirit of this problem.

## Geoff

March 3, 2011 - 8:32 pm -Assuming that the two sodas mix perfectly well together, both glasses should end up (on average) with the same amount of their original soda.

To justify, take a simplified example.

100 ml of sprite.

100 ml of coke.

Transfer 10 ml of sprite into the coke glass. Mix.

So now 10/110 = 9.09% is sprite, the remaining 100/110 = 91.81% is coke.

Taking 10 ml from this glass back into the coke. 9.1% of this 10 ml should be sprite, and 90.9% should be coke.

So final totals would be 90 ml sprite + 9.1% of 10 ml = 90.091 ml

and 100 ml coke – 90.9% of 10ml = 90.091 ml

## Jameson Brown

March 3, 2011 - 8:33 pm -Let’s go deeper:

How many times would you have to do that until the contents of each glass are exactly the same?

## Regan

March 3, 2011 - 8:34 pm -I have to agree with Cathy’s logic.

## Todd

March 3, 2011 - 8:36 pm -Each glass has the same amount of its original soda as the other glass. This is true because (1) each glass has the same volume in the end (12 oz) and (2) there is no soda added to the system. Simply put, the volume of Sprite replaced with Coke is the same volume of the Coke replaced with the Sprite.

## Dan Meyer

March 3, 2011 - 8:37 pm -I love that there’s a frequency perceptible only to mathophiles. That extension is just

blaringfrom the problem. My math prof brother-in-law heard it. I heard it. The rest of the clan didn’t.## jcr723

March 3, 2011 - 8:42 pm -100 S 100 C start

-10 S +10 S S -> C xfer (pure)

90 S 100 C + 10 S after 1st xfer

+9C ,+1S -9C ,-1S C -> S xfer (mixture)

91S, 9 C 91C, 9S after 2nd xfer

so they are the same.

## Tom Gaffey

March 3, 2011 - 8:45 pm -Cool concentration problem.

Let’s assume that there are 100 pipettes of soda in each glass and that when you mix them they evenly distribute. Let’s also assume that they each started at 0 parts per hundred of the other soda.

You took a pure concentration of sprite and added it to the coke. Adding 1 pipette of sprite to the coke would mean that there is now about 1 pph sprite molecules in the coke. Then you took a 1 pph concentration of sprite from the coke glass and placed it in the sprite glass.

final sprite glass = original – 100 pph* + 1 pph*

final coke glass = original – 99 pph**

*concentration of sprite

**concentration of coke

They both lost the same amount of liquid.

## jcr723

March 3, 2011 - 8:50 pm -The formatting got corrupted on my input. One more time…

100-S…………100-C………………………start

-10-S………….+10-S………………………spite to coke transfer

90-S……………100-C, 10-S………………after first transfer

+9-C,+1-S……-9-C,-1-S………………….”coke” to sprite transfer

91-S,9-C……….91-C,9-S…………………after second transfer

So, contrary to my orginal guess, the volumes in question are the same

## Cathy

March 3, 2011 - 8:54 pm -Okay…as I was writing my ‘gut reaction’ I was leaning toward the fact that they have the same amount but it just didn’t seem correct. Thanks to everyone for explaining this clearly.

Now I am intrigued about how many times you’d have to do this until both drinks have the same amount of Sprite & Coke. Anyone?

## Mark

March 3, 2011 - 8:56 pm -To Geoff and jcr: With all due respect, I think it’s important to note that the math you’re doing is irrelevent.

If both glasses started with the same amount of soda (say, 12 oz) and ended with that same amount (12 oz), they by definition must both have an equal amount of their original soda at the end.

## Colin

March 3, 2011 - 8:59 pm -Jameson, I’d have to go with infinite. Each time the dropper would be more diluted and would asymptotically approach 50% and would do so more quickly proportional to the dropper size. The only way to actually achieve 50% is to mix both glasses together in their entirety (i.e., have a 12 oz dropper).

And reversing your question, how much volume must you swap in one go to achieve X% concentration?

## Mark

March 3, 2011 - 9:00 pm -To answer Jameson’s question, using the method of transfer shown, both glasses would never have exactly the same amount of Sprite and Coke (although the difference could certainly get to the point of being insignificant for all practical purposes).

## josh g.

March 3, 2011 - 9:40 pm -Showing my work.

http://thoughtlost.org/content/random/_MG_7518.jpg

Values are in mL’s, but really they could be (almost) anything and still demonstrate what’s happening.

## Albert

March 3, 2011 - 10:46 pm -It’s a bit of a trick question since it doesn’t matter how much you move from one glass to the next, as long as you return the same amount (as Mark points out).

I had to explain this problem to some co-workers who get confused by percentages (that version of the question had a pitcher of water and one of whine) and to adress the issue of it being counter-intuitive I asked them: “What if I were to pour the entire contents of one of the pitchers into the other? Then pour back half of the mixture in the other pitcher?”

All agreed that logically, both pitchers would have the same concentration of water and whine (or coke and sprite). Having agreed that you could end up with the same concentration in one case, it was much easier to convince them of the general case.

## David

March 3, 2011 - 10:49 pm -For some reason, maybe it’s late at night, I couldn’t wrap my head around the proofs suggested here, so I decided to make a visual proof. The explanation is here because I couldn’t see fitting into a comment easily (especially since I have a picture).

http://davidwees.com/content/do-these-glasses-end-same-amount-each-type-soda.

## Anthony

March 3, 2011 - 11:09 pm -Depends how you define “original soda”. If you assume they were both simply sodas and assumed that Coke and Sprite were actually the same thing (apart from their flavored and colored syrup which is also just the same thing).

A lot of assumptions need to be made in math questions like these, but sometimes these assumptions depend on one’s culture.

Sheena Iyengar’s TED talk and Malcolm Gladwell’s differing TED talk on choice are great in the way they present whether or not Sprite or Coke would be considered a choice.

http://www.ted.com/talks/lang/eng/sheena_iyengar_on_the_art_of_choosing.html

http://www.ted.com/talks/malcolm_gladwell_on_spaghetti_sauce.html

## aupaathletic

March 4, 2011 - 12:06 am -I did the math too:

– 1st glass:

33 ml sprite

– 2nd glass:

33 ml coke

-pick 10 ml from 1st glass and put on the 2nd glass:

you pick 10ml, 100% sprite

1st glass:: 23 ml 100% sprite

2nd glass: 43 ml (23,255814% sprite, 76,744186% coke)

-pick 10 ml from 2nd glass and put on the 1st glass:

you pick 2,3255814ml sprite + 7,6744186ml coke

1st glass: 25.3255814ml sprite + 7.6744186ml coke

2nd glass: 25.3255814 ml coke + 7.6744186 ml sprite

The same…

But after reading Mark’s comment, I have to say: touchÃ©!

If you have only two glasses that start and end with the same amount of liquid, it is impossible to make different mixtures.

## aupaathletic

March 4, 2011 - 12:13 am -PS: I mean… of course they are different, I meant to say that it is impossible to not to have the same amount of original soda on both glasses. No math are needed!

## Thanassis

March 4, 2011 - 2:26 am -I remember looking at this problem several years ago (it was oil and vinegar in my version) and coming up with a few ways of explaining the answer. I started with some algebraic ways but the really interesting one was the visual/geometric way which eventually led to the simpler and elegant explanation I saw in the comments. I created a document about it. I’ll send it through email and Dan feel free to use it, enhance it, share it.

## Brian Wyzlic

March 4, 2011 - 3:24 am -I’m only on 4 hours of sleep here, so this may or may not make sense (I may have to come back later and slap myself for bad math). Whatever. I take that back. No excuses, here’s my thought:

It seems everyone is assuming equal parts were transferred (and equal volumes were transferred). But is the concentration the same? Because the Coke was stirred before transfer, but not the Sprite, it seems logical that more carbon dioxide was transferred from the Sprite to the Coke than from the Coke to the Sprite. This carbon dioxide was then lost to the air in the stirring.

My question, which I can’t quite seem to wrap my head around at this hour, is does that matter? I want to say no. . .but I’m not sure.

## mr.brown

March 4, 2011 - 4:17 am -Same amount. Barring any chemistry funny business.

Consider simpler experiments before doing the calculations that have been done above.

Easier experiment #1: Move no liquids. Each glass has the same amount of the original liquid. 100%.

Easier experiment #2: One drop of each, and you transferred that one drop over then back. each glass would have 50% of the original liquid in it.

From there, the more complex calculations will continue to validate the “easier” observations -hopefully re-enforcing the concept.

## mr.brown

March 4, 2011 - 4:18 am -Hmm… Would I show the guy eating sugar video before or after this?

## Shari

March 4, 2011 - 5:43 am -JCR, I agreed with your “work” until you got to the portion of Sprite and Coke in the second transfer. You posted this:

100-Sâ€¦â€¦â€¦â€¦100-Câ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦start

-10-Sâ€¦â€¦â€¦â€¦.+10-Sâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦spite to coke transfer

90-Sâ€¦â€¦â€¦â€¦â€¦100-C, 10-Sâ€¦â€¦â€¦â€¦â€¦â€¦after first transfer

+9-C,+1-Sâ€¦â€¦-9-C,-1-Sâ€¦â€¦â€¦â€¦â€¦â€¦â€¦.â€cokeâ€ to sprite transfer

91-S,9-Câ€¦â€¦â€¦.91-C,9-Sâ€¦â€¦â€¦â€¦â€¦â€¦â€¦after second transfer

The ratio of Coke to Sprite in the second glass is 10 to 1. You have a 9 to 1 ratio in the second transfer. I think it should instead be this:

+9 1/11 C, +10/11 S…………..- 9 1/11 C, – 10/11 S

90 10/11 S, 9 1/11 C………….90 10/11 C, 9 1/11 S

90.91% Sprite…………………..90.91% Coke

I didn’t believe it until I did this calculation. Both glasses have the same concentration of the original soda.

## Colin

March 4, 2011 - 6:17 am -Left glass:

S1 + C1 = 12 oz

Right glass:

S2 + C2 = 12 oz

Initial amounts of Sprite and Coke

S1 + S2 = 12 oz.

C1 + C2 = 12 oz.

In order for the concentrations to be different then the last two conditions would be violated.

For example:

S1 = 10, C1 = 2 (83% Sprite)

S2 = 3, C2 = 9 (75% Coke)

Clearly C1 + C2 != 12 and S1 + S2 != 12, therefore the concentrations must be the same, otherwise Coke or Sprite was created or destroyed. In this case, there is now 13 oz of Sprite and only 11 oz of Coke.

The only way to get different concentrations is to change the first two equations: meaning unequal volumes in the two glasses.

## Christopher Danielson

March 4, 2011 - 6:35 am -I am going to work the middle school angle. Middle schoolers taught me a lot about mathematical reasoning.

Whatever amount is taken out in the video is too small for my middle-school mind to start thinking about, so I want to think about an easier amount-half, and then argue that the video is a special case of a more general relationship.

If you dumped half the Sprite (glass 1) into the Coke (with a big enough glass), then the Coke (glass 2) is now 1 part Sprite, 2 parts Coke (a 1:2 ratio-all ratios from here on out will be Sprite to Coke).

Stir.

Now we need to take half a can from glass 2 and dump it back into glass 1.

That half can is in the ratio 1:2 and it is added back to the pure Sprite, which is in the ratio 1:0 (forgive me, but this works out).

But the 1 part in each of these ratios is a different part. Recall that the stuff I dump in (at 1:2) measures the same as what was left in the glass (at 1:0). So there are three total parts in the dumped stuff and there must be 3 total parts (all Sprite) in what’s already in the glass.

Therefore I am “adding” 1:2 to 3:0, yielding 4:2, or equivalently 2:1. But that’s the same concentration of Sprite that I have of Coke in the other glass. QED

But maybe 1/2 a can is a special case.

Rerun the computations and a pattern emerges. If we put 1/n of a can into the Coke glass and vice versa, then we have an n:1 and a 1:n concentration in the two glasses.

And delightfully, along the way I get to compute (n+1)(n-1)+1:n, which simplifies in the way I have indicated.

Part to part ratios can take you far!

## Max

March 4, 2011 - 7:54 am -Does it matter whether you stir or not? Some people argued vociferously that we had to assume the stirring was very effective. Other people seem to imply that no matter what happens, the amount of original soda is the same. Dan bothered to show the stirring…

Also, if you thought this problem was fun, check out its friends: http://wordplay.blogs.nytimes.com/2010/10/11/numberplay-the-ultimate-answer/

## boyce

March 4, 2011 - 8:28 am -I just presented this to some “non-math” adults. I just showed the video first and asked what they were curious about. The more math-y questions they came to had to do with the relative amounts of “fizz” in each drink, not the intended problem.

When we got to discussing the actual question, they objected to the idea that they’re the same on a couple grounds:

– That the differing carbonation of each drink might affect their volume so that they are not starting the same.

– Then they argued that the more intuitive way for them to think of “more of its original soda” was NOT volume of soda, but the “number of Coke atoms” v. the “number of Sprite atoms.” Which is obviously flawed since both are already mixtures. But I guess they want to see this measured in mass, rather than volume… Has anyone approached the problem in this way?

## corn walker

March 4, 2011 - 8:41 am -Jameson, that question screamed out to me as well.

Max, the stirring is a red herring, and quite an effective one at that.

## Christopher Danielson

March 4, 2011 - 8:51 am -No the stirring is NOT a red herring.

It is also a place for mathematical inquiry. My part-to-part ratio solution assumed effective stirring, as did most of the other solutions here.

And I found myself objecting instinctively to Thanassis’s oil and vinegar problem because these famously do not mix well (although better than oil and water? Unclear).

So corn walker, I believe the onus is on you to demonstrate that stirring doesn’t matter. I don’t believe it’s obvious that it doesn’t.

And even if stirring doesn’t matter *mathematically*, it sure matters *conceptually*. If it’s oil and vinegar and no stirring, I think of it as equivalent to taking two eyedroppers-one of oil, one of vinegar then transferring each to the other container. But if we stir before making the second transfer, the equivalence to this strategy is lost-neither eyedropper is pure anymore.

And the *amount* of Sprite in glass 1 at the end of the process certainly depends on whether we stir, even if the relationship between this amount and the amount of Coke in glass 2 does not.

## Eric B

March 4, 2011 - 10:02 am -The stirring would only be important if you were trying to figure out the actual amount/percentage of coke and sprite in each glass.

It is not necessary to determine that they have the same amount of original soda.

Consider some examples.

If you did not stir and managed to get a full dropper of coke then each glass would now contain on dropper of the other soda.

If you did not stir and ended up with a dropper that was half sprite and half coke then each glass now contains half a dropper of the other soda.

If you did not stir and through some miracle managed to suck back up all of the sprite again then both glasses would have all of their original liquid.

## Shari

March 4, 2011 - 10:36 am -Eric, that’s a great explanation. I was so convinced that stirring mattered. Now I know why it doesn’t.

## Karl M

March 4, 2011 - 10:37 am -Think this problem is made easier if we look at the transfer, not the glasses or stirring. Pippette is say 10ml, when we transfer it, it’s 10ml of sprite, now the back transfer. Whatever we pickup that isn’t coke, lets say x ml of sprite. So, sprite transfer is 10-x and coke transfer is also 10-x. Therefor sprite and coke transfer are the same, therefore concentration is the same. Ta da!

## Heather

March 4, 2011 - 10:42 am -My theory is that the Sprite glass has more of it’s original soda. The liquid taken from the Sprite initially contaminated the Coke; so even an equal amount of liquid from the Coke glass could have some element of Sprite in it, whereas the initial transfer brought no parts coke with it to the Coke glass.

Sprite glass has more of the initial soda in it.

## Christopher Danielson

March 4, 2011 - 11:07 am -Your idea, Heather, was my initial idea too. But then I got two theories from that: (1) yours, which suggests that more Sprite is in the Sprite glass. And (2) this one: If the liquid taken from the Coke IS contaminated with a little bit of Sprite, then I haven’t removed the same amount of Coke from the Coke glass as I removed Sprite from the Sprite glass. So if less Coke has been removed, it’s more concentrated.

That convinced me I had try a different strategy.

## Karl M

March 4, 2011 - 11:08 am -Your premise is correct about the first pipette Heather however, no matter what we transfer back, we are always putting some sprite back.

## Michael Paul Goldenberg

March 4, 2011 - 12:36 pm -Consider a similar problem using discrete objects (e.g., playing cards. Take 10 red cards and 10 black cards face down in separate piles. Take four at random from red pile; mix into black pile. Shuffle. Return four random cards face down to red pile. Ask: more black in the red pile or red in the black pile. Try this several times. If you’re not convinced, do it with the faces showing. Apply principle to soda problem.

Better problem: 877 identical coins lying heads up on a table and an unknown number of identical coins lying tails up on same table. Coins are all flat on the table and arranged randomly (not grouped). It is impossible to distinguish the coins in any way but by looking at them.

Explain how to arrange the coins into two groups, not necessarily the same size, so the there are an equal number of heads in each pile.

You must do this blindfolded. No one may help you.

If you know a solution (there are at least two that respect the conditions), please don’t post it here. Give others a chance to think. Write me with answers (mikegold at umich dot edu). Post questions here. Consider the soda and card problems as hints.

## Max

March 4, 2011 - 1:30 pm -Michael, may we count the total number of coins? Should we?

## Michael Paul Goldenberg

March 4, 2011 - 1:45 pm -@Max: you may. Who am I to decree what anyone “should” do?

:>)

## Nick

March 4, 2011 - 2:00 pm -Well, after a few rounds of this, there will be a certain amount of sprite in the coke. Because both glasses still have the same volume in them, that must also be the amount of coke in the sprite, or one would have a greater total volume.

## Karl M

March 4, 2011 - 2:53 pm -Have we got to go from start to end alone or can we ask questions here? Have a few ideas, nothing solid yet.

## Karl M

March 4, 2011 - 3:05 pm -Right, couple of questions.

1, we have 877 coins showing heads, we can’t have two piles with an equal amount of heads, so do we have to start turning coins over?

2, this is quite unrealistic to do as an experiment, is there a way of it working with less than 877 heads, so I don’t have to resort to paper?

## Karl M

March 4, 2011 - 3:08 pm -Oh and a third, is it actually possible to get it correct every time or is there some margin for error? I mean if I had the coins and the time, the method would work every one of the infinite times I did it?

## hillby

March 4, 2011 - 3:34 pm -Same, the math I did has been shown several times, I focused on the contents of the pipette.

I engaged with this problem immediately because:

1) I could take a guess, but never know if it was right until I did the math

2) I knew at least what section of my math toolbox to work in: proportions

3) I knew that I could use any volume for the pipette, and using 1 mL would make the math easier

4) I drew a diagram for each step, and checked my math with reasonableness at each step

5) I didn’t get hung up on what he meant by original soda, I assumed he meant the volume of soda in each glass

So if I toss this out to students, those are the supports/scaffolds I need to have in my pocket.

## Nathan Chow

March 4, 2011 - 3:39 pm -love this wcydwt!

my two line proof:

The volume of each soda is conserved.

If x mL of Coke is in Sprite, than that missing x mL of Sprite is in the Coke!

## Nathan Chow

March 4, 2011 - 3:40 pm -(assuming transfer 1 volume = transfer 2 volume)

## Avery

March 4, 2011 - 6:45 pm -Full disclosure…I’ve seen this before and thought about it a bunch.

I’m thinking of all of this in terms of units of pipettes. When you take 1 pipette of liquid from the Coke, you’re removing X amount of Coke and 1-X amount of Sprite. The sprite lost 1 unit of liquid in the first transfer and gets back 1-x units meaning it has X units of Coke in it [original amount – 1 + (1-x)] and [original – X] amount of Sprite in it. The Coke on the other hand has 1 – (1-x) of sprite left (the 1 unit that was added minus the 1-x units that was then transferred back). So the Coke has X units of Sprite in it after the 2nd transfer.

As mentioned above, I really like the thought process of taking this to an extreme and combining everything. A great mathematical habit of mind!

Anyway, I have yet to add anything original so how about…

Do you need to start with the same volume of liquid in both glasses or would the same be true with a 40 and a sprite?

As mentioned above, mixing doesn’t matter (surprisingly).

As mentioned above, this can also be done using discrete units (I’ve done this with M&M’s, but I guess you can pick your favorite product placement).

You can even do this with discrete units and cheat (no need to shuffle cards….feel free to look at the M&Ms you’re transferring).

The coin problem is a fun one (and also surprising to have a solution).

## Karl M

March 5, 2011 - 12:48 am -If there are different volumes atbthe start, then the concentrations would be different but the molecules of cola and sprite transferred would be the same.

## Ken Ellis

March 5, 2011 - 6:28 am -Is it possible to get more coke than sprite in the original sprite glass and more sprite than coke in the coke glass?

## Ken Ellis

March 5, 2011 - 6:29 am -Is it possible to get more coke than sprite in the original sprite glass vice versa?

## Aaron F.

March 5, 2011 - 10:45 am -Classic! :D

I didn’t hear it! I guess it’s only perceptible to

somemathophiles [jealous].Assuming the drinks are well-mixed, I think proportion of Sprite in the left glass will go to 1/2 exponentially. More precisely, I think the proportion of Sprite in the left glass after

Nswaps will be 1/2 + 1/2b^N, wherebis some constant. I feel like there should be a way to argue this without doing any calculations, but I can’t think of one…In the meantime, here’s an argument with lots of calculations. Assume there’s a pipe carrying soda at a constant rate

Ffrom the left glass to the right glass, and another pipe carrying soda at the same rate from the right glass to the left glass. LetVbe the amount of soda in each glass. LetSbe the amount of Sprite in the first glass, and lets=S/Vbe the proportion of Sprite in the first glass. Let R be the amount of Sprite in the second glass, and letr=R/Vbe the proportion of Sprite in the second glass.Sprite is leaving the left glass at a rate of

Fs, and Sprite is leaving the right glass at a rate ofFr. Hence, the amount of Sprite in the first glass is changing at a net rate ofS‘ =F(-s+r). Remember thatS+R=V, sos+r= 1, which means –s+r= 1 – 2s. Hence, the amount of Sprite in the first glass is changing at a net rate ofF(1 – 2s). That means the proportion of sprite in the first glass is changing at a net rate ofs‘ = (F/V)(1 – 2s). Under the condition thats= 1 at time zero, the solution to this differential equation has the form 1/2 + 1/2b^t, wheretis time.## Aaron F.

March 5, 2011 - 10:54 am -Oooh, I love Ken Ellis’s question! The answer is obviously “no,” but I can’t think of a proof as beautiful and concise as the question deserves. :)

## Michael

March 5, 2011 - 11:06 am -Mark, can you prove that volume conservation implies concentration equilibrium? I’ve been thinking about your comment and I don’t quite see why its true?

## corn walker

March 5, 2011 - 12:44 pm -While the stirring is irrelevant to the question “Which glass contains more of its original soda?” it is relevant to the question of how many transfers are required before the mixtures approach 50% concentration.

## Tedious B

March 5, 2011 - 5:50 pm -Why did people pick 10 or 100 to use in their example. that just leads to ugly decimals.

Imagine 12 parts coke and 12 parts sprite.

take 1/2 the sprite and move it to the coke.

now you have a glass with 6 parts sprite and a glass with 12 parts coke and 6 parts sprite. Take six parts from the “coke” and put it back into the sprite. If it is mixed properly then 4 of those parts will be coke and 2 will be sprite. Both glasses therefore turn out to be the same.

Use the same math to show that if you poured 4 parts, or if you poured all 12 parts into the other glass you get the same answer. It is therefore a linear relation.

It is necessary for the following conditions to be met.

1. There must be perfect mixing.

2. There must be exactly the same amount returned.

Note: the density or thickness of the respective liquids does not matter.

note2: Can people think through the loss of CO2 in stirring?

## Megan Golding

March 6, 2011 - 7:22 am -Hypothesis: The best WCYDWT force me — the student — to change an assumption that was obviously true on first look.

This is one of those problems.

## Luke Hodge

March 6, 2011 - 8:52 am -What a great problem.

Related to Ken’s question

I think you can make an inductive argument that Cup A/B will never be less than 50% sprite/coke. Removing a dropper of fluid does not change the cocentration of the cup it came from. If a cup was mostly sprite/coke before removing a dropper…

Related to the concentrations after several transfers:

I think you can look at this as a recursive formula. Take a weighted average of the current contents in each cup to get the contents in each cup after the next round of transferrs. Unless I’m missing something, the weights do not change – they are determined by the dropper/cup sizes.

For example, if the dropper volume is 10% of a cup, you will use the weights 1/1.1 & .1/1.1 to get the contents after another round of transferrs:

Next Cup A = .909prior A + .0909 prior B

If you start with 1 oz of sprite & the dropper is r oz, the amount of sprite in cup A after each round of transfers creates the interesting values below (I think):

1 , 1/ (1 + r) , (1 + r^2) / (1 + r)^2 , (1 + 3r^2) / (1 + r)^3, (1 + 6r^2 + r^4)/(1 + r)^4

## Jesse

March 6, 2011 - 9:27 am -Imagine the amounts are exaggerated – it’s easier to see.

Pour half of the Sprite into the Coke glass, stir, then pour half of the Coke/Sprite mix into the Sprite glass.

After the first pour, the Coke/Sprite mix will be 1/3 Sprite and 2/3 Coke. After the second pour, the Coke glass will have the same proportions (because of the stirring), but Sprite glass will have 1/2 Sprite (because of the initial 1/2 pouring) and the other half will be 1/3 Sprite and 2/3 Coke.

In other words, the Coke glass has 2/6 Sprite and 4/6 Coke.

The Sprite glass has 4/6 Sprite and 2/6 Coke.

I believe this method is generalizable to any proportion of mixing, but… frankly I can’t be arsed to prove the general case.

## Zeno

March 6, 2011 - 12:42 pm -Let O be the amount of original soda in each glass. Let S be the amount of Sprite transferred from the Sprite glass to the Coke glass in the first transfer. Let C be the amount of Coke transferred from the Coke glass to the Sprite glass in the second transfer. Let X be the amount of Sprite transferred from the Coke glass back to the Sprite glass in the second transfer.

If the dropper is filled with the same amount of soda in the two transfers, then S = C + X.

The amount of original soda in the Coke glass after the two transfers is O – C. The amount of original soda in the Sprite glass after the two transfers is O – S + X.

Using algebra,

S = C + X

C = S – X

O – C = O – (S – X) = O – S + X

Thus the amounts of original soda in the two glasses after the two transfers are equal.

## Chris S

March 7, 2011 - 1:33 pm -My gut was leaning toward the Sprite, because you took out a pure sample and replaced it with a diluted one, but the math looks like they are the same.

I stuck with the idea that that was a teaspoon dropper and that there are 6 teaspoons in an ounce (hence 72 teaspoons in a can of soda).

72 tsps of sprite becomes 71 pure tsps

72 tsps of coke becomes 73 diluted at 72/73

you put one tsp of 72/73 into the 71 giving it 71 + 1 tsp with 1/73 of Sprite

= 71.0136986 of Sprite in 72 teaspoons

You have 72 tsps of Coke at 72/73 dilution which is 72*0.98630137

=71.0136986 of Coke in 72 teaspoons.

## Karl M

March 7, 2011 - 1:46 pm -Did this lesson twice today. Found a couple of things, we went through, volume sense, why, even with the lack of markings on the pipette we can make an educated guess by knowing the volume of the cans.

Kids tried to quantify the amount of Sprite in the Coke glass, one kid made the step to fractions, so I got him to explain this to the class.

They tried to explain the other way. Got lost, so lost. So I explained about using mathematical modelling. How we use maths to help us simplify problems which seem very difficult. So I, probably doing some pseudoteaching, show them the blocks.

We start with 33 blocks. It becomes clear that no matter what we do, even trying to knowingly sabotage it, we always end up with the same dilution. We change the starting amount, the amount we put across and stopping mixing.

My question is, where do I take it from here. All, and I mean ALL of the kids are utterly absorbed.

I’ve got ratio, proportion, volume, volume sense, mathematical modelling, but however it just feels like I can squeeze a bit more out of it. HELP!

## jg

March 9, 2011 - 4:29 am -A gallery of student work on this problem, including algebraic specific-case solutions, graphical general-case solutions, a graph of original drink remaining as a function of dropper size, and my general-case algebraic solution; also some points on checking the units and the special cases to ferret out mistakes:

http://tatnallsbg.blogspot.com/2011/03/coke-v-sprite.html

Thanks for the rich problem!

## Zeno

March 10, 2011 - 9:10 pm -Here’s another solution using algebra:

Let O be the original amount of soda in each glass. Let S be the amount transferred in each direction using the dropper.

After the first transfer, the amount of Sprite remaining in the Sprite glass is O-S. The amount of Coke in the Coke glass is O and the amount of Sprite in the Coke glass is S. The total amount of soda in the Coke glass is O+S, so the fraction of Coke in the mixture is O/(O+S), and the fraction of Sprite in the mixture is S/(O+S).

After the second transfer, the amount of soda remaining in the Coke glass is O. The fraction of Coke in the mixture in the Coke glass is O/(O+S), so the amount of Coke remaining in the Coke glass is O*(O/(O+S)) = (O*O)/(O+S).

The amount of the mixture in the Coke glass that was transferred back to the Sprite glass is S. The fraction of Sprite in the mixture is S/(O+S), so the amount of Sprite transferred back to the Sprite glass is S*(S/(O+S)) = (S*S)/(O+S). This amount was added to the amount of Sprite that remained in the Sprite glass after the first transfer, which was O-S. So the total amount of Sprite in the Sprite glass after the second transfer is (O-S)+(S*S)/(O+S).

Using algebra,

(O-S)+(S*S)/(O+S) = ((O-S)*(O+S))/(O+S)+(S*S)/(O+S) = ((O-S)*(O+S)+(S*S))/(O+S) = (O*O-S*S+S*S)/(O+S) = (O*O)/(O+S)

Thus the amounts of original soda in the two glasses after the two transfers are equal.

## Sean

March 12, 2011 - 6:41 am -Always fun, creative, elegant.

The one nagging question I continue to have about WCYDWT is…what exactly does it accomplish?

Yes, it appeals tremendously to our intuition. Students are looking at the world and designing questions about curiosities. This certainly appears to be the kind of skill we want to teach. But research is at best divided about the kinds of gains similar projects have had in the past.

Dan has pointed out that his students out-performed his entire department, that they showed real and true gains according to every measure we currently use.

But still I wonder. How much of those gains were attributable to the actual WCYDWT lessons? How much of it was attributable to his skill and highly developed craftsmanship as a manager, questioner, evaluator? WCYDWT doesn’t strike me as terribly efficient. And I know that’s not the point, but where is the research showing that these kinds of problems are effective?

I think Dan also mentioned that he would do these kind of lessons bi-weekly (about 1 in 10 days). To which I say, fair enough. He was efficient and skilled enough in those other nine days to experiment with something like this.

I don’t know, though. These problems appear to make students conceptually flexible, which is brilliant. To make them procedurally flexible- arguably more important and more difficult to teach- is probably what Dan was doing the other 90% of the time.

I know the focus of this blog is WCYDWT and debunking conventional textbook wisdom. But when there’s no coke or sprite, escalators, three-pointers, or cheese, how do you do the other stuff?

## Steve Bergen

March 13, 2011 - 6:44 am -This has been on my Math Tricks & Treats website for years and years. I started teaching in 1973 and have used this problem (along with others) to convey to students the difference between intuition, deductive reasoning and inductive reasoning. If you love this problem as much as I do, you may enjoy the BELT AROUND THE EARTH problem posted on the same website as a TREAT on http://www.summercore.com/math/trickstreats.html

Steve Bergen (who heard Dan speak at TEDX in Spring 2010 and thought he was LB-esque)

sbergen33@gmail.com

## Zeno

March 13, 2011 - 11:53 am -Here’s one more solution using algebra:

Let O be the amount of soda in each glass before and after the two transfers. Let S be the amount of Sprite in the Sprite glass after the transfers and let C be the amount of Coke in the Coke glass after the transfers. Let X be the amount of Sprite in the Coke glass after the transfers.

Before the transfers, all the Sprite (and no Coke) was in the Sprite glass, so S + X = O. After the transfers, the Coke glass contains the remaining Coke plus the transferred Sprite, so C + X = O

Using algebra,

S + X = O

C + X = O

S + X = C + X

S = C

Thus the amounts of original soda in the two glasses after the two transfers are equal.

## JC

March 14, 2011 - 7:29 am -Okay, here’s a proof (that ratios will be equal/opposite) using logic:

Think of a discrete example (this works perfectly because volumes can be considered as a discrete set of molecules). A simple one is two dugouts in a baseball game. One has a red team, another has a blue team. If players swap between dugouts, then there are only a few set of exchanges that can happen:

1. a blue player from the blue dugout can swap with a blue player from the red dugout.

This will only happen after they are initially mixed, but the result is simple: ratios don’t change, and remain equal/opposite.

2. a red player from the blue dugout can swap with a red player from the red dugout.

This also will only happen after they are initially mixed, but the result is simple: ratios don’t change, and remain equal/opposite.

3. a blue player from the blue dugout can swap with a red player from the red dugout.

In this case, the blue dugout lost one of their players, but so did the red dugout! The ratios remain equal/opposite.

4. a red player from the blue dugout can swap with a blue player from the red dugout.

Again this will only happen after they are initially mixed. In this case, the blue dugout gained one of their players, but so did the red dugout! The ratios remain equal/opposite.

So if players/cards/molecules are just swapping, then the four scenarios above describe ALL the transactions they can have, and the ratios cannot be anything but equal/opposite.

Now if you bring in rigorous chemistry and bring up the possibility that one is a more dilute solution than the other, then you’re not starting with an even number of players, and the ratios will not remain equal/opposite. But that’s probably another level of discussion, and may be question number 2 on the WCYDWT exercise.

## Steve Thomas

March 14, 2011 - 9:54 pm -So the question I have is how can we use thist get kids into the habit of modeling problems? So I decided to try and figure a way to let kids model this problem in Etoys.

My first attempt used kedama to show the “soda molecules” moving around and let the kids move the soda into the eye dropper and then into the coke glass and back.

Then building on Michael Paul Goldenberg comments I build a much simpler model using a set of 10 Sprite boxes in a container and 10 Coke boxes in a container. And asked the kids to move the 2 Sprite boxes into the Coke container, then shuffle the Coke container and take the top four boxes and move them back to the Sprite container. And have them do this a few times and think about how this relates to the Coke v Sprite problem.

My blog post with videos is at: http://mrstevesscience.blogspot.com/2011/03/coke-v-sprite-have-kids-build-model-of.html

The models I created were rendered in Etoys (free open source software for kids and part of OLPC core software from http://www.squeakland.org)

## johniel

March 18, 2011 - 7:13 am -Hi. We are two eighth graders that were able to solve this problem in about 10 minutes. At first we thought that the sprite had more, but when we did the math we realized that they were equal. Great problem though!

## muncher

March 18, 2011 - 7:25 am -we go to school with johniel and we were able to solve it very quickly as well. We basically just drew a picture and it speled its self out from there. I had a lot of fun solving this problem!

## Meh number three

March 18, 2011 - 7:28 am -I go to the same school as johniel and muncher and courtneym, and I agree. It’s a good logic problem because it tricks you when you first watch the video without thinking about it. Its really easy when you plug in numbers and draw. Overall its a good problem!

## Zeno

March 18, 2011 - 10:24 am -Let’s try a slightly different problem.

Suppose we have three kinds of soda, instead of two: Sprite, Coke, and Orange. We pour the sodas into three separate glasses. We transfer a dropper full of soda from the Sprite glass to the Coke glass, then from the Coke glass to the Orange glass, and finally from the Orange glass back to the Sprite glass (mixing the soda thoroughly after each transfer).

Do the three glasses now contain equal amounts of their original sodas?

## Nathan

March 18, 2011 - 2:28 pm -No, it fails the simplified case of starting with one dropper full of each soda.

## Zeno

March 18, 2011 - 6:24 pm -Is there any dropper size for which the three glasses will contain equal amounts of their original liquids?

## R. Wright

March 18, 2011 - 10:20 pm -Zeno: No, since Coke and Orange have some of the original taken away and never returend, while Sprite has the same amount taken away and some returned.

Sadly, I set up and solved a big, messy algebra problem to come to the “dropper size = 0” solution, before realizing that a little simple reasoning would suffice. Maybe there should be a course on answering interesting questions like yours without resorting to algebra. My algebra students are often very good at it, actually.

## kurt

May 3, 2011 - 7:58 pm -i think comment 22 is the most reliable

## oliver

May 3, 2011 - 8:03 pm -I think the sprite is the one and I dont know why I am writing this

## Diederick

May 3, 2011 - 8:05 pm -I think comment 22 is most reliable.

## ????

May 3, 2011 - 8:05 pm -sprite

## Diederick

May 3, 2011 - 8:06 pm -Comment 71 does look like it has a lot of scientific and mathematic proof.

## oliver

May 3, 2011 - 8:20 pm -I suck at fractions

## Celine

May 3, 2011 - 8:22 pm -hard to say , but if the coke is the same liquid after the sprite has been added then its sprite because at first the sprite has been added into the coke which dilutes the coke , making it less coke like ,then the diluted coke is put into the sprite and because the coke is diluted the taste of the sprite over powers the coke .

## kurt

May 3, 2011 - 8:23 pm -same as oliver

## kurt

May 3, 2011 - 8:23 pm -what most of the people are trying to say is

fill up a glass of coke & a glass of sprite (equal amounts)

take 1/2 of the sprite & put it in the coke

mix the coke/sprite

take 1/2 of the coke/sprite & put it in the pure sprite

there is half of the mixed soda stuff

half of the pure soft drink in each glass PEOPLE

## Kurt

May 3, 2011 - 8:25 pm -I like cheese

## Izzy

May 3, 2011 - 8:27 pm -I TOTALLY AGREE WITH CELINE BECAUSE I HE DILUTED THE COKE FIRST SO THAT MEANS HE PUT COKE/SPRITE IN THE SPRITE!

YO

## kurt

May 3, 2011 - 8:27 pm -89 91 ARE FAKE NOT MADE BY ME

## Celine

May 3, 2011 - 8:29 pm -lolz kurt “i like cheese”

## OLIVER

May 3, 2011 - 8:30 pm -I WIP MY BROTHERS EVERY DAY

## babra the cow

May 3, 2011 - 8:30 pm -i MAKE cheese

## xylophone headed cat

May 3, 2011 - 8:32 pm -I like xylophones

## anonymous

May 3, 2011 - 8:32 pm -anonymous

lol

## Paul Wolf

October 18, 2011 - 12:49 pm -Used this in class today. Totally awesome.

I went in with the discrete version of the problem all wrong in my head, though. It actually didn’t seem to matter whether it was “mixed” i.e. the ratio of Sprite to Coke put back in the Sprite didn’t have to be consistent with the ratio of Sprite to Coke in the mixture, so I got to challenge them to figure out “glass sizes” where it would be. (As someone mentioned, if you start with 12 blocks and move four “Sprites” over…)

The other best part was the two kids who demanded to use algebra–one of whom never shows his work, oddly enough.

I actually showed this yesterday as a teaser and a student was like “they’re mixing them slowly,” leading to Dan’s “when will they be the 50/50 in both?”

## Dan Meyer

October 18, 2011 - 12:54 pm -Awesome. What were the demographics on the class?

## GM

October 31, 2011 - 1:02 pm -Coke side after addition of fraction of S:

C+0.1S

Coke side after removal of same volume just added:

C+0.1S – 0.1/1.1 (C+0.1S)

Sprite side after adding back above removal:

S-0.1S + 0.1/1.1 (C+0.1S)

Multiply both formulae by 1.1:

1.1C+0.11S -0.1 (C+0.1S) = 1.1C-0.1C +0.11S-0.01S

1.1S-0.11S +0.1(C+0.1S) = 1.1S-0.11S+0.01S+0.1C

Simplify and divide by 1.1

(1C+0.1S)/1.1

(1S+0.1C)/1.1

## GM from MooseJaw

October 31, 2011 - 1:31 pm -Generalizing:

x is any fractional value 0

Coke side after addition of fraction of S:

C+xS

Coke side after removal of same volume just added:

C+xS â€“ x/(1+x) (C+xS)

Sprite side after adding back above removal:

S-xS + x/(1+x) (C+xS)

Multiply both formulae by (1+x):

(1+x)C+(x+xx)S -x(C+xS) = C+xC+xS+xxS-xC-xxS

(1+x)S-(x+xx)S +x(C+xS) = S+xS-xS-xxS+xC+xxS

Simplify and divide by 1.1

(1C+xS)/(1+x)

(1S+xC)/(1+x)

## GM from MooseJaw

October 31, 2011 - 1:34 pm -slight edit to above:

x is any fractional value less than 1 greater than 0

(my use of the less than and greater than sign in my post removed the contents between…)

## Alexis Piazza

November 24, 2011 - 8:56 pm -Leave it to my sister to develop an elegant, visual way to solve this problem. Using the smallest units

The link is here: http://mrpiazzamath.tumblr.com/

Frame 1 – Two equals parts of Sprite and Coke

Frame 2 – Move 1 part of Sprite to Coke, which is now 2 parts Coke and 1 part Sprite.

Frame 3 – The mixture dissolves. 1/3 of each unit is now Sprite, 2/3 is now coke.

Frame 4 – Move 1 unit of SpriteCoke (1/3 Sprite, 2/3 Coke) to Sprite. You can see there are 4/3 Sprite on the left and 4/3 Coke on the right.

## Erin

January 22, 2012 - 7:40 am -Kids (7&9) figured this out using legos and then starbursts. Their first instinct was that the “coke” side would end up with more, then they tried to figure it our using simpler numbers and my son changed his mind to “same”. We wanted him to prove it but the numbers were more difficult to use for him so he grabbed 20 of each of two colors of legos and demonstrated his thinking. Then, for fun, did it with starbursts too. Each time they worked it out with the manipulatives it ended up the same. My daughter (7) needed to see it over and over again to believe it but was finally convinced. Great way to spend a Sunday morning!!

## zaq

March 9, 2012 - 8:59 am -sprite for sure

## AstridWang

May 24, 2012 - 3:01 pm -THE SAME

Assue Y is how much original liquid you have for each cup (which is both Y)

X is how much liquid you take out each time (and in this case you take out the same amount of liquid in two transfer).

how much original liquid in sprite: Y-X+X(X/X+Y), which =Y^2/X+Y

how much original liquid in:Y-X(Y/X+Y), which =Y^2/X+Y

if you simplify your expressions nicely LOL

am a 9th grader, my teacher showed us this today and she told us she didn’t no how to do this, WHAT?!

I think this is the easiest way to do this, I did it with simple number too(you should’ve have ratio thing comes in!), think this is more convincing and abstract, and more like Mathematics

Be friends w/me ppl :p

## AstridWang

May 24, 2012 - 3:01 pm -assume* sorry

## mr bombastic

August 12, 2012 - 5:22 am -James Tanton has this problem along with some variations: