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#11 unreality

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Posted 12 July 2007 - 07:43 PM

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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#12 PDR

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Posted 04 August 2007 - 05:08 PM

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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#13 erictheread

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Posted 22 August 2007 - 07:31 PM

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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#14 erictheread

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Posted 22 August 2007 - 07:36 PM

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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#15 BoilingOil

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Posted 29 September 2007 - 01:12 AM

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

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#16 slmo

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Posted 10 October 2007 - 08:18 PM

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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#17 kharnath

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Posted 31 October 2007 - 10:47 PM

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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#18 yukino21

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Posted 12 November 2007 - 01:08 PM

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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#19 joeyrz

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Posted 07 December 2007 - 05:37 PM

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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#20 dissatisfied

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Posted 29 December 2007 - 03:21 AM

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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