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Bavarian

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Bavarian - Back to the Logic Puzzles

One glass has 10 cl of tonic water and another 10 cl of fernet. Pour 3 cl of tonic into the glass with fernet and after mixing thoroughly, pour 3 cl of the mixture back into the glass with tonic water.

Is there more tonic in the glass of fernet or more fernet in the glass of tonic?

(Ignore the chemical composition!)

This old topic is locked since it was answered many times. You can check solution in the Spoiler below.

Pls visit New Puzzles section to see always fresh brain teasers.

Bavarian - solution

There is exactly as much tonic in the glass of fernet as there is fernet in the glass of tonic.

Having 2 glasses, in one is 10 cl of tonic and in the other 10 cl of fernet. Pour 3 cl of tonic to the glass with fernet and after thorough mixing pour 3 cl of the mixture back to the glass with the tonic. Is there now more tonic in the glass of fernet or more fernet in the glass of tonic?

(Ignore chemical composition!)

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The math is interesting; when you place the the 3 cl of tonic into the glass of fernet the glass now contains 3/13 parts tonic. Assuming that the tonic is mixed evenly throughout, when you take the 3cl the second time, 10/13 of it is fernent, which is approximatly 2.3 cl. The remaining tonic in the fernent is 10/13 of the 10 cl, which is approximatly 2.3 cl as well.

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whaaaa

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it works for any amount... if x amount of tonic goes into the fernet, obviously x amount of fernet has to go into the tonic (where else it is supposed to go?), so that both still has 10cl.

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This made my brain hurt.

Literally, there are veins popping out of my head.

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Here is another way of looking at it:

When you bring back 3 cl of whatever the name of that drink is you are actually bringing back 2.3 of ferxxx and 0.7 tonic.

So in the end you get 7.7 tonic to the 2.3 fernxxx and in the other glass you are left with 2.3 of the original 3.0 tonic (3.0-0.7) plus the 10-2.3 (frenxxx that went back to the tonic glass) 7.7 fernxxx = 10.0.

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umm... I did it differently... Is this still correct?

After pouring the 3cl of tonic into the fernet, there would be 3/13 tonic and 10/13 fernet in the fernet glass (with 7/7 in the tonic glass).

When you pour the 3 cl back, you would get 1.5 cl of tonic and 1.5 cl of fernet (since they were mixed thoroughly).

That leaves 8.5/10 tonic and 1.5/10 in the tonic glass, and 1.5 tonic and 8.5 fernet in the fernet glass.

That comes with the same answer, but did I do it wrong and was just lucky, or what?

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umm... I did it differently... Is this still correct?

After pouring the 3cl of tonic into the fernet, there would be 3/13 tonic and 10/13 fernet in the fernet glass (with 7/7 in the tonic glass).

When you pour the 3 cl back, you would get 1.5 cl of tonic and 1.5 cl of fernet (since they were mixed thoroughly).

That leaves 8.5/10 tonic and 1.5/10 in the tonic glass, and 1.5 tonic and 8.5 fernet in the fernet glass.

That comes with the same answer, but did I do it wrong and was just lucky, or what?

Neah bro. Here is what I see wrong with the math:

When you pour back 3 cl you are pouring back 3* 3/13 =0.7 tonic + 3* 10/13 =2.3 fernet = 3.0 cl and not 1.5 & 1.5. You see what I mean. So, you end up with 7.7 & 7.7 not 8.5 & 8.5.

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Neah bro. Here is what I see wrong with the math:

When you pour back 3 cl you are pouring back 3* 3/13 =0.7 tonic + 3* 10/13 =2.3 fernet = 3.0 cl and not 1.5 & 1.5. You see what I mean. So, you end up with 7.7 & 7.7 not 8.5 & 8.5.

Ok, I see where I messed up now. I must have been thinking they were equal or something...

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this is getting to complicated. I may be wrong here but it seems to me that there would not be more in either glass. If the mix was even, the amounts of tonic and fernet must be even as well. No calculations need....

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no math needed. the same amount of liquid is in each glass. so whatever tonic that isnt in the tonic glass is in the other glass, and vice versa

assuming nothing spilled

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Here's a simple way to look at it:

First glass = the glass that starts of with 100% tonic

Second glass = the glass that starts off with 100% fernet

T1 = the portion of tonic left in the first glass after all mixing is done

T2 = the portion of tonic that ends up in the 2nd glass after all mixing is done

F1 = the portion of fernet left in the second glass after all mixing is done

F2 = the portion of fernet that ends up in the first glass after all mixing is done

we need to know if T2 = F2 ("Is there now more tonic in the glass of fernet or more fernet in the glass of tonic?")

we know from the problem statement:

T1 + T2 = 10 (before any mixing - all tonic is in glass 1)

F1 + F2 = 10 (before any mixing - all fernet is in glass 2)

T1 + F2 = 10 (after all mixing - glass 1)

F1 + T2 = 10 (after all mixing - glass 2)

From above:

T1 + T2 = 10

T1 + F2 = 10

therefore

T1+ T2 = T1 + F2

which means

T2 = F2

which is what we were trying to solve for in the first place.

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tonic stays at 100% in step 1 and 2

fernet goes from 100% to 76.92% after step 2 therefore ends up with 7.69 cl in it's glass

step 3 introduces a 3 cl 23.07% solution back into the 7 cl 100% of tonic

the 3 cl of 77% solution contains .6921 cl of tonic

plus the 7 cl of 100% tonic = 7.692 cl of total tonic

I end up with an equal amount of each in their own glass.

cai

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no math needed. the same amount of liquid is in each glass. so whatever tonic that isnt in the tonic glass is in the other glass, and vice versa

assuming nothing spilled

There a great bit of logical thinking, cai

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At first, you pour 3 cl from glass A in glass B, and after mixing you pour 3 cl from glass B back into A.

Ergo, in the end, both glasses still contain 10 cl, just like at the start of the experiment. Therefor the amount of fernet in the glass of tonic must be exactly the same as the amount of tonic in the glass of fernet.

One might be inclined to look for a complicated calculation, although it's as simple as stated above

BoilingOil

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tonic stays at 100% in step 1 and 2

fernet goes from 100% to 76.92% after step 2 therefore ends up with 7.69 cl in it's glass

step 3 introduces a 3 cl 23.07% solution back into the 7 cl 100% of tonic

the 3 cl of 77% solution contains .6921 cl of tonic

plus the 7 cl of 100% tonic = 7.692 cl of total tonic

I end up with an equal amount of each in their own glass.

cai

I don't see ur math here. The fernet never goes below 8.5cl...

Start: Glass 1= 10cl tonic...Glass 2= 10cl fernet

1st pour: Glass 1= 7cl tonic...Glass 2 = 10cl fernet + 3cl tonic

2nd pour: Glass 1 = 7cl tonic+1.5cl tonic+1.5cl fernet... Glass 2 10cl fernet-1.5cl fernet-1.5cl tonic

Totals: Glass 1= 8.5cl tonic + 1.5cl fernet...Glass 2=8.5cl fernet + 1.5cl tonic

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you guys are applying a lot of math that, with proper logical deduction, can have a proper answer. without any complicated math that many would find confusing it would be simpler to explain it in a manner as such.

(note: this will not have anything specific but b general an a way so that it would be applicable to other things similar to this instance as well)if a container with 10 (lets say mL) were to lose 3mL to a container of a second substance also with 10 mL and were to be shaken and have its contents properly distributed. when the extra 3 mL in the second substance are returned to the first substance it would consist of a 50/50 distribution in that 3 mL making half of the 3 mL or 1.5 mL the first substance and the other 1.5 the second substance. therefore when its returned the first substance would now be 8.5 mL in that container and 1.5 mL of the second substance in the same container and visa versa in the second container.(although in actuality this may not necessarily be accurate seeing how if you were to consider it in literal terms the substance would be too spread out for it to work quite this way it explains whats going on in a decent enough manor but the 1.5 split is easier to understand)(also note that even though this explanation is long the actual thought process behind it is really quite short in comparison to the explanation as long as the logical deduction is thought of properly).

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Having 2 glasses, in one is 10 cl of tonic and in the other 10 cl of fernet. Pour 3 cl of tonic to the glass with fernet and after thorough mixing pour 3 cl of the mixture back to the glass with the tonic. Is there now more tonic in the glass of fernet or more fernet in the glass of tonic?

(Ignore chemical composition!)

ahm..there is more fernet in the glass of fernet and more tonic in the glass of tonic...

well,..im not so sure with my answer but,..there's no harm in trying is it???

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Glass 1 - 10 cl of Fernet - 10/10

Glass 2 - 10 cl of Tonic - 10/10

3cl of tonic in fernet

Glass 1 - 13 cl, 3/13 (23%) of tonic, 10/13 (77%) of fernet.

Glass 2 - 7 cl, 7/10 (70%) of tonic

3cl of fernet with tonic back in tonic.

.69/3 (23%) tonic and 2.31/3 (77%) of fernet to be added to Glass 2 and substracted from Glass 1.

Glass 1 - 7.69/10 of fernet, 2.31/10 of tonic

Glass 2 - 7.69/10 of tonic, 2.31/10 of fernet

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no math needed. the same amount of liquid is in each glass. so whatever tonic that isnt in the tonic glass is in the other glass, and vice versa

assuming nothing spilled

the math will validate what you have only claimed to be true

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One's examination of this problem is based on one of six parts involved; and it is from these options that the confusion stems, because it is easy to lose track of which part is important. The following ratios are relevant:

# Liquid T / Total

# Liquid F / Total

# Liquid T / Liquid F

# Liquid F / Liquid T

# Liquid T / 3 cups

# Liquid F / 3 cups

At no point would the mixtures have a 50-50 distribution for the same reason that a drop of red dye in a glass of water is not distributed 50-50. Problems such as these do assume an "uniform distribution" but this refers to the idea that a spoonful of the mixture from the top, bottom, or middle of the mixture would all have the same ratio of Liquid T / Liquid F.

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At first, you pour 3 cl from glass A in glass B, and after mixing you pour 3 cl from glass B back into A.

Ergo, in the end, both glasses still contain 10 cl, just like at the start of the experiment. Therefor the amount of fernet in the glass of tonic must be exactly the same as the amount of tonic in the glass of fernet.

One might be inclined to look for a complicated calculation, although it's as simple as stated above

BoilingOil

Why "must" it be exactly the same as the amount of tonic in the glass of fernet. What proof do you have? The mathematics will validate what you have only claimed to be true.

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Why "must" it be exactly the same as the amount of tonic in the glass of fernet. What proof do you have? The mathematics will validate what you have only claimed to be true.
Don't over think this one. It was eloquently explained earlier:

no math needed. the same amount of liquid is in each glass. so whatever tonic that isnt in the tonic glass is in the other glass, and vice versa

assuming nothing spilled

It's a closed system. There's a total of 20cl of liquid, 10 of each type, so no matter how it's mixed, whatever isn't in one must be in the other. Since per instructions, after pouring there is again 10 cl in each container, whatever tonic is now in the fernet flask must be equal to the amount of fernet that is now in the tonic flask.

ipso facto no matho needed...

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Don't over think this one.

Isn't that what we do here? While the proof offered by unreality, yourself, and others is not incorrect, it is not rigorous enough. An extremely beautiful proof could be written using your idea, a better proof than the one I was imagining, but unless your thinking is expressed in a more rigorous language than the common sense approach you've adopted, it will not possess the beauty it deserves.

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I guess the question is asking the balance between the amount of tonic water and that of fernet in respective mixed liquid.

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