You would have to prove to me that Sprite wasn’t thicker than Coke first.

>it has to be the same

And this is why this is a great math lesson.

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.

Let’s go deeper:
How many times would you have to do that until the contents of each glass are exactly the same?

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.

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.

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

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.

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).

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.

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

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!

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.

Hmm… Would I show the guy eating sugar video before or after this?

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.

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/

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

Jameson, that question screamed out to me as well.

Does it matter whether you stir or not?

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

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.

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!

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.

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: you may. Who am I to decree what anyone “should” do?

:>)

Have we got to go from start to end alone or can we ask questions here? Have a few ideas, nothing solid yet.

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?

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!

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).

Is it possible to get more coke than sprite in the original sprite glass and more sprite than coke in the coke glass?

Classic! :D

Dan Meyer:

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

I love that there’s a frequency perceptible only to mathophiles. […] My math prof brother-in-law heard it. I heard it. The rest of the clan didn’t.

I didn’t hear it! I guess it’s only perceptible to some mathophiles [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 N swaps will be 1/2 + 1/2 b^N, where b is 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 F from the left glass to the right glass, and another pipe carrying soda at the same rate from the right glass to the left glass. Let V be the amount of soda in each glass. Let S be the amount of Sprite in the first glass, and let s = S/V be the proportion of Sprite in the first glass. Let R be the amount of Sprite in the second glass, and let r = R/V be 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 of Fr. Hence, the amount of Sprite in the first glass is changing at a net rate of S‘ = F(-s + r). Remember that S + R = V, so s + r = 1, which means –s + r = 1 – 2s. Hence, the amount of Sprite in the first glass is changing at a net rate of F(1 – 2s). That means the proportion of sprite in the first glass is changing at a net rate of s‘ = (F/V)(1 – 2s). Under the condition that s = 1 at time zero, the solution to this differential equation has the form 1/2 + 1/2 b^t, where t is time.

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?

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?

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

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.

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!

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.

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

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.

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!

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!

No, it fails the simplified case of starting with one dropper full of each soda.

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.

I think the sprite is the one and I dont know why I am writing this

sprite

I suck at fractions

same as oliver

I like cheese

89 91 ARE FAKE NOT MADE BY ME

I WIP MY BROTHERS EVERY DAY

I like xylophones

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?”

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

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…)

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!!

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

James Tanton has this problem along with some variations: