### Brain Teasers

# Soft Drink Dilemma

You have two 8 oz. cups in front of you. One is filled with Root Beer and one is full of Club Soda. You take one tablespoon of Root Beer and pour it into the Club Soda. Stir well. Now, you take a tablespoon of the mixture you just stirred, and pour it into the Root Beer. Now, does the Root Beer have more Club Soda in it or vice versa? Don't account for spilling.

### Answer

They are equal. For every drop of club soda in the root beer, there is a drop of root beer in the club soda.Hide Answer Show Answer

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## Comments

the answer to this teaser is wrong.

If a tablespoon of rootbeer is mixed into the club soda, there would be no way to separate the club soda from the rootbeer. and you would be putting back some of the rootbeer that you initially mixed into the cub soda.

and delema is spelled dilemma

If a tablespoon of rootbeer is mixed into the club soda, there would be no way to separate the club soda from the rootbeer. and you would be putting back some of the rootbeer that you initially mixed into the cub soda.

and delema is spelled dilemma

If they were filled, you woulb not be able to add any more liquid into the soda.

geez picky picky

well bullbutter, what is the answer then? not that i belive you, but, if your right, you should know

I think the root beer has more club soda, but thats only because its a slightly bigger mixture (the other one has had 1 tablespoon spilled out) but percentage wise, they are the same

(I think, I got 100:99.999899

Which is close enough for me taking rounding off errors into account)

I think its a brilliant puzzle nice one!

(I think, I got 100:99.999899

Which is close enough for me taking rounding off errors into account)

I think its a brilliant puzzle nice one!

I must however disagree with my esteem colleague CathalMcCabe, While Mogmatt I agree with you that since neither liquid had anything taken away, you should therefore be left with an equal mixture, one not having any more or less than the other. My only problem I guess, is the difficulty, It was so blatantly obvious that I reread the question 3 times to make sure I wasn't missing anything. Come on man, throw in a tablespoon of sugar into the Club soda or something. Which would increase the volumn of the C.S. and therefore make it seem a little more difficult or something.

it's true. you commenters can't always be right so get used to it.

Ok Mogmatt, here's the scoop. I loved this teaser, you have put up a real good poser here. After initially I tried to let my underpowered intellect try and work this out logically in my head, I almost embarrassed myself and posted complete tripe that was unfounded. Luckily I let my fingers (and pens) do the walking and discovered the following to mathematically prove that you are in fact correct - heh, when the title of my original post was to say it was incorrect... PHEW!

To show my workings I am going to have to use different numbers I'm sorry, but it's just to make the calc's easier. However, since we are talking equations here the numbers shouldn't really matter.

So here we go:

I am going to say that your two containers can hold 100ml each. The table spoon holds 20ml (big sucker!). The first exchange sees the following happen:

1.)

20ml RB -> CS container

NEW AMOUNTS:

In RB container - 80ml RB

In CS container - 120ml = 100ml of CS + 20ml RB (MIXED EVENLY)

OR, SHOWN AS FRACTIONS:

In CS container - 100/120 of CS + 20/120 of RB in that container (5/6 + 1/6)

2.)

To show my workings I am going to have to use different numbers I'm sorry, but it's just to make the calc's easier. However, since we are talking equations here the numbers shouldn't really matter.

So here we go:

I am going to say that your two containers can hold 100ml each. The table spoon holds 20ml (big sucker!). The first exchange sees the following happen:

1.)

20ml RB -> CS container

NEW AMOUNTS:

In RB container - 80ml RB

In CS container - 120ml = 100ml of CS + 20ml RB (MIXED EVENLY)

OR, SHOWN AS FRACTIONS:

In CS container - 100/120 of CS + 20/120 of RB in that container (5/6 + 1/6)

2.)

2.)

2.)

20ml of liquid moved to RB container

This 20ml is comprised of 5/6 CS + 1/6 RB. ie. 16.67 CS + 3.33 RB

NEW AMOUNTS:

In RB container 100ml = (80 + 3.33 RB) + 16.67 CS.

83.33RB + 16.67CS

In CS container 100ml = (100 - 16.67 CS) + (20 - 3.33RB)

83.33CS + 16.67RB

VOILA!!! The ratios of liquid within each container are the same and my head hurts and all this talk of root BEER is making me terribly thirsty.

Please if there are any holes in my calculations anyone point them out - or come up with some formula that supports it.. or go have a beer.

cheers.

20ml of liquid moved to RB container

This 20ml is comprised of 5/6 CS + 1/6 RB. ie. 16.67 CS + 3.33 RB

NEW AMOUNTS:

In RB container 100ml = (80 + 3.33 RB) + 16.67 CS.

83.33RB + 16.67CS

In CS container 100ml = (100 - 16.67 CS) + (20 - 3.33RB)

83.33CS + 16.67RB

VOILA!!! The ratios of liquid within each container are the same and my head hurts and all this talk of root BEER is making me terribly thirsty.

Please if there are any holes in my calculations anyone point them out - or come up with some formula that supports it.. or go have a beer.

cheers.

Grrrrr, sorry for that previous rubbish in my posts, I didn't realise you could use HTML (at least I'm assuming that's why it stuffed up!), as that is what caused the weirdness.

I can prove this myself of course, and if I'm right, this will be a new paragraph, if not, oh well, could someone let me know the story

I can prove this myself of course, and if I'm right, this will be a new paragraph, if not, oh well, could someone let me know the story

The solution is, in fact, correct. Initially both cups have equal amount of liquid. In the end also they have the same amounts of fluid. So the amount of root beer lost by cup 1 = the amount of soda gained by cup 1 at the end of the whole transaction. Hence the solution.

I don't get this one, If you took a spoon of pure root beer, say 10mls, and add it to the other cup, then stir. Then you take a 10ml spoon of club soda and root beer mix and place it into the root beer. So the first spoon was 100% root beer at 10mls and the second spoon was a mixture of both also at 10mls, there fore the club soda recieves a full 10mls of root beer, where as the root beer does not recieve the full 10ml of club soda in return as it is not 100% club soda, but a mixture of which a small percentage, probably less than 1%, is root beer. so the root beer has recieved minutely less club soda than the Club soda recieved root beer.

I'm totally with mad-ade on this one. It's too bad then that the math supports the solution. Boiling down the proof above and using the actual values of one cup(US) equals 16 tablespoons, the club soda container ends up with 1/17th of the contents being root beer. The tablespoon of the mixture that is brought back to the root beer represents 1/16th of the final volume. That tablespoon contains 16/17ths club soda. And 16/17ths of 1/16th is 1/17th. Easy, huh?

Actually Krishnan is probably correct. I hear that indians love to do math, so they are really good at it.

You are right Mad-ade, the root beer gets slightly less club soda than than the club soda got. EXCEPT you forgot that when you took that teaspon out of the club soda, you were taking a bit of the root beer out of the club soda back into the root beer. Try like Dsquared did, I used 100 ml and a 50 ml "teaspoon" (makes it easier to picture and if there was a difference in the amount of liquid it would be even larger if you traded a larger amount).

Great puzzle. I tried to be lazy and just picture it in my head and I got it wrong. I had to actually use my brain to figure this one out!

Great puzzle. I tried to be lazy and just picture it in my head and I got it wrong. I had to actually use my brain to figure this one out!

Great puzzle. I tried to be lazy and just picture it in my head and I got it wrong. I had to actually use my brain to figure this one out!

while mathematical proof may be correct, the wording of the question is certainly wrong. As it says that the 8 oz. club soda cup is filled with club soda,so there is no space to add any root beer

krishnan is completely correct, though it took em a mminute to agree. regardless of how much root beer you take and put back, it will equal the club soda taken. in mad-ades example, say, 10mls of root beer taken first. now, whatever concoction-mix is taken from CS and RB, say, 8 club soda and 2 rootbeer, means that of the original 10mls of rootbeer added, only 8 are left (taking 2 there), and you are adding 8 mls of club soda to the restof the root beer. very simple once grasped. same amount of liquid yields exactly equal ratios.

very good teaser.

the only problem with Mogmatt16 is that he didn't show his solution. but by simple ratio and proportion, it can be proven indeed!

a short-cut solution: for as long as you start with the same amount of liquid in the 2 cups (say 8oz. each), and end up with the same amount (still 8oz. each but each cup a mixture of the 2 liquids), the 2 mixtures will have have the same percentages of mixture.

but if lets say that you pour 6oz. of the 1st liquid into the 2nd liquid, mix them, then return 4oz. of the mixture into the 1st liquid, and end up with a 6oz. mixture and a 10oz. mixture, then it would have to be solved using ratio and proportion.

my very 1st favorite teaser!

the only problem with Mogmatt16 is that he didn't show his solution. but by simple ratio and proportion, it can be proven indeed!

a short-cut solution: for as long as you start with the same amount of liquid in the 2 cups (say 8oz. each), and end up with the same amount (still 8oz. each but each cup a mixture of the 2 liquids), the 2 mixtures will have have the same percentages of mixture.

but if lets say that you pour 6oz. of the 1st liquid into the 2nd liquid, mix them, then return 4oz. of the mixture into the 1st liquid, and end up with a 6oz. mixture and a 10oz. mixture, then it would have to be solved using ratio and proportion.

my very 1st favorite teaser!

Wow, a lot of discussion for what would seem to be a quite obvious and easily understood solution.

My only issue with this would be that is belongs in the logic category since there's no math or calculations involved.

My only issue with this would be that is belongs in the logic category since there's no math or calculations involved.

Of course there is Mathematics involved in the solution, otherwise there would not be all this arguing about which answer is logically correct. As for comments that the Mathematics supports the answer but maybe its still wrong ? Wha?

Two very powerful strategies to use in problem solving are [1] Solve a simpler problem and [2] Invent your own numbers. Lets do some simple arithmetic by inventing simpler numbers. Say cups hold 550 ml and spoon holds 55 ml. The actual quantities are irrelevant.

CUP A CUPB

550 ml root 550 ml soda

Now take out 55 ml from cup A

CUP A CUPB

495 ml root 550 ml soda + 55 ml root

Now take 55/605 = 1/11 of cup B = 50 ml soda + 5 ml root Add to CUP A

CUP A CUP B

500 ml root + 50 ml soda 500 ml soda + 50 ml root

So the soda contains just as much root beer as the root beer contains soda.

Good Teaser.

Two very powerful strategies to use in problem solving are [1] Solve a simpler problem and [2] Invent your own numbers. Lets do some simple arithmetic by inventing simpler numbers. Say cups hold 550 ml and spoon holds 55 ml. The actual quantities are irrelevant.

CUP A CUPB

550 ml root 550 ml soda

Now take out 55 ml from cup A

CUP A CUPB

495 ml root 550 ml soda + 55 ml root

Now take 55/605 = 1/11 of cup B = 50 ml soda + 5 ml root Add to CUP A

CUP A CUP B

500 ml root + 50 ml soda 500 ml soda + 50 ml root

So the soda contains just as much root beer as the root beer contains soda.

Good Teaser.

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