Uncovering the Mystery of Percentage Labeling on Chocolate Bars

[Semo727]

Today, I was eating a chocolate with 20% of its mass extra.
Then I realized (after some thinking) that the chocolate company colud also write, that the chocolate was with about 18,232155 % extra.
Do you know how that is possible??

 

[Semo727]

I can help you a little. If the chocolate comany wrote "chocolate with 16.6666% extra" it would probably mean that you payed for just
100-16.666 % of chocolate.

 

[HallsofIvy]

Today, I was eating a chocolate with 20% of its mass extra.
Then I realized (after some thinking) that the chocolate company colud also write, that the chocolate was with about 18,232155 % extra.
Do you know how that is possible??

I have no idea what you mean by "20% of its mass extra". I might guess that the candy bar said something like "20% more chocolate" than previous bar that cost the same.

I can help you a little. If the chocolate comany wrote "chocolate with 16.6666% extra" it would probably mean that you payed for just
100-16.666 % of chocolate.

Well, if it meant that, it would be wrong. 16.6666% (is there any reason you use that rather than the 20% you meantioned before?) more would mean 16.6666% of the previous amount: in other words, the chocolate in the bar is 1+ 0.166666= 1.16666 of the previous amount- and therefore, the amount of chocolate in the previous bar (the amount you are "paying" for) is 1/1.16666= 60.0000240001% of the chocolate in the current bar, not 100- 16.6666= 83.3334%. Be careful not confuse the "base" of percents.

Now, how about showing how you use differential equations to show that "the chocolate company colud also write, that the chocolate was with about 18,232155 % extra"

 

[Semo727]

Sorry guys for such an confusing explanation of what I thought.
I will explain it again.
You'll buy that chocolate (for instance it weighs 120g). 16.666% (1/6*100%) of its mass is for free and you paid for the rest (100-16.666% the same as 5/6*100%)
So 20 g is for free. This was the 1st point of view.
The second is, that you bought the chocolate and paid for 100g but they are very nice and gave you 20grams for free, but they want us to see how nice they are so they will write with 20% free. But this 20% is 20% from the original chocolate.
I think, that the chocolate company would be better to write with 18,232155 % free, because now we don't have to know the base of percents. If we just have chocolate of 100grams and we say that it is with 18,232155 % free, and than we will count this way:

100 +
(100*0.18232155/infinite) + (100+(100*0.18232155/infinite))*0.18232155/infinite+
(100+... and we will continue for infinite times we will finally get 120g.
hope I dind't make any mistake..

And number 0.18232155 is ln1.2

 

[mathwonk]

just go to worldwidechocolates.com and don't bother us any more.

 

[Semo727]

just go to worldwidechocolates.com and don't bother us any more.

If you are mathematician, I don't want to become mathematician! (despite I was planing to)

I was just happy that I used diff. eq. to count the thing like this, and I found it interesting.

goodbye!

 

[HallsofIvy]

Sorry, but

1. Your calculation is completely meaningless.

2. Apparently you don't know what a "differential equation" is.

 

[Semo727]

So if it is really meaningless, I apologize. It might be caused by the fact that I didn't write it in math language, what might have been even more confusing because of a fact that I'm not used to it..(no one has ever tought me how to write them)

But I have to say that despite it might be meaningless, it is correct! And then apparently I know what differential equation is.
The prove, that it is correct can be the result of computer program I created just to prove that my calculation made in my head was correct. The program said, that if we use In1.2 (0.18232155) as I said, after 1000000 equations we will get exactly 119.999998055 grams.
But read carefuly what was the base of percentage I took. (I wrote about that in a previous post)
This base was not constant number, it was... i don't know how to write that...sory

Ok, administrator can cancel this whole thread...
but... I was right..

 

[d_leet]

So if it is really meaningless, I apologize. It might be caused by the fact that I didn't write it in math language, what might have been even more confusing because of a fact that I'm not used to it..(no one has ever tought me how to write them)

But I have to say that despite it might be meaningless, it is correct! And then apparently I know what differential equation is.
The prove, that it is correct can be the result of computer program I created just to prove that my calculation made in my head was correct. The program said, that if we use In1.2 (0.18232155) as I said, after 1000000 equations we will get exactly 119.999998055 grams.

I understand what you did, you're finding the sum of an infinite series, but I don't see at all how you used differential equations to come up with this, would you care to explain that?

 

[JasonRox]

If you are mathematician, I don't want to become mathematician! (despite I was planing to)

He is a mathematician, so I guess you won't become one.

 

[mathwonk]

good point, but actually didn't his statement only prove that he doesn't want to become one? he might actually be unable to help himself, if he is like the rest of us in the subject, and thus become one anyway.


lighten up kid, i was sharing a wonderful website with high intensity chocolate. it really exists, and has bars with like 73% cocoa content. but if yo7 really prefer differentiwal equations to chocolate, ok.:shy:

 

[saltydog]

just go to worldwidechocolates.com and don't bother us any more.

I feel your comment above is inappropriate. He's not bothering me and I can tolerate his lack of apparent familiarity with differential equations. Your unsympathetic attitude is not conducive to fostering an interest in mathematics and probably does more harm than good. I'm disappointed a mentor here didn't tell you that. I know you're teaching DE's now. I would hope "don't bother me" is not an option for dealing with students.

 

[JasonRox]

I feel your comment above is inappropriate. He's not bothering me and I can tolerate his lack of apparent familiarity with differential equations. Your unsympathetic attitude is not conducive to fostering an interest in mathematics and probably does more harm than good. I'm disappointed a mentor here didn't tell you that. I know you're teaching DE's now. I would hope "don't bother me" is not an option for dealing with students.

I feel your comment above is inappropriate. He's not bothering me and I can tolerate his lack of apparent familiarity with differential equations. Your unsympathetic attitude is not conducive to fostering an interest in mathematics and probably does more harm than good. I'm disappointed a mentor here didn't tell you that. I know you're teaching DE's now. I would hope "don't bother me" is not an option for dealing with students.

I'm guessing his comment was merely geared towards his ignorance towards DE's.

If one of his students went to see him for help and made it obvious he was ignorant about DE's and/or his class, I would hope that "don't bother me" would be his option for dealing with this student. Ignorance should not be tolerated under these circumstances.

Anyways, mathwonk is of great help around here. If you think otherwise, maybe you should spend more time around here.

Note: I may have been offended by mathwonk before, but I certainly do not remember. The bottom line is... you will get offended eventually, usually more than once. No one is on this planet to baby everyone else about their feelings.

 

[mathwonk]

i may be of help some of the time, and i greatly appreciate the defense saltydog has offered for me based on that, but I think i should apologize here. my phrase "don't bother us" was unfortunate, even though I was just joking, I chose my words badly.
basic suggestion, don't take everything so seriously.

i think if the guy had actually looked up the chocolate website, he might possibly have realized it was meant in fun.

 

[saltydog]

i may be of help some of the time, and i greatly appreciate the defense saltydog has offered for me based on that, but I think i should apologize here. my phrase "don't bother us" was unfortunate, even though I was just joking, I chose my words badly.
basic suggestion, don't take everything so seriously.
i think if the guy had actually looked up the chocolate website, he might possibly have realized it was meant in fun.

Jesus, get it straight. This is a math forum after all: Jason defended you. I was "offended" by your use of the term "us" as I believe I'm part of this forum and do not approve of you speaking for me.

 

[JasonRox]

Jesus, get it straight. This is a math forum after all: Jason defended you. I was "offended" by your use of the term "us" as I believe I'm part of this forum and do not approve of you speaking for me.

Alright, let's get back on topic now.

Be more clear about your complaint. I would never have guessed that's what it was.

 

[HallsofIvy]

Semo727, I would still like to know how you used differential equations to determine that.

 

[Semo727]

I understand what you did, you're finding the sum of an infinite series, but I don't see at all how you used differential equations to come up with this, would you care to explain that?

So I'm going to (at last) try to explain everything in math language and I hope I will succeed (at last).

y'/1=x*y, x is the percentage/100 of y what is the original amount of chocolate :) and y' is the free amount of chocolate that is added to chocolate.
I hope that it is clear. If the original chocolate weighs 100g, and x=0.2, y' = 20g and chocolate you buy weighs 120g.
And what happens if we rewrite the whole thing

dy/dt=z*y, What does it mean? In the first case there was 1 instead of dt in denominator on the left side in equation.
In the first case, we could find y' very easily with solving just 1 equation. In second case, we have to solve infinite amount af equations (because dt->0). And also, y will not be the same in all equations.
Or we can solve differential equation, and if we know z, we could easily find out what will y2 be for t=1 (y2 is the final amount of chocolate that we buy in a store), and the integration constant C will be ln100 (becouse the original chocolate weighs 100g) But what we want to know is z, and we know y2 (it is 120g) From the solved diff. equation 120=100*e^(z*t) and t=1 we can easily get z=In1.2 and that is many times mentioned 0,18232. So if I were the owner of chocolate company, and if I wanted to say that 20g of 120g chocolate is for free, I could write on the cover: 20% is free, 16.666% is free or 18.232% is free. Isn't it great?

 

[mathwonk]

i apologize for not properly ,recognizing those who are offended by me as opposed to those are defending me.
I apologize to saltydog for, let's see, not appeciating that he is offended by me.
i also thank again jason for defending me.
i apoologize to the OP for being too subtle for his sensibilities.
i cannot resist however a small suggestion: lighten up.

 

[saltydog]

Well, alright me too then:

First and foremost: Semo . . . your presentation is awkward and caused some . . . conflict. Jesus, I should have just stayed away and kept my mouth shut. But I like differential equations and so am drawn to this quagmire for whatever reason I'm not sure. Now they're beating me up down there about messy posts of differential equations . . . frankly I'd make them turn in ALL their homework in LaTex . . . but I digress.

Mathwonk, I appreciate you recognizing my concerns and respect your wisdom sir.

 

[HallsofIvy]

So I'm going to (at last) try to explain everything in math language and I hope I will succeed (at last).

y'/1=x*y, x is the percentage/100 of y what is the original amount of chocolate :) and y' is the free amount of chocolate that is added to chocolate.
I hope that it is clear. If the original chocolate weighs 100g, and x=0.2, y' = 20g and chocolate you buy weighs 120g.

Okay, so y' means the amount of chocolate added, is not the derivative of y and this is not a differential equation.

And what happens if we rewrite the whole thing
dy/dt=z*y,

What justification do you have for writing it like that? What does t represent and what is z?

What does it mean? In the first case there was 1 instead of dt in denominator on the left side in equation.

Okay, and what does it mean to "have dt in the denominator"?

In the first case, we could find y' very easily with solving just 1 equation. In second case, we have to solve infinite amount af equations (because dt->0).

I don't follow this. By "second case" are you referring to dy/dt= z*y? In what sense is this an "infinite amount of equations"?

And also, y will not be the same in all equations.
Or we can solve differential equation, and if we know z, we could easily find out what will y2 be for t=1 (y2 is the final amount of chocolate that we buy in a store), and the integration constant C will be ln100 (becouse the original chocolate weighs 100g)

Okay, I am starting to grasp this: you are imagining increasing by some "infinitesmal" amount over and over. The differential equation is the limit of that, t represents the "progression" through the infinite number of "increases", you are taking t= 1 to mean the end of the progression. And you are assuming (without any justification) that the "rate of increase" r stays constant.

But what we want to know is z, and we know y2 (it is 120g) From the solved diff. equation 120=100*e^(z*t) and t=1 we can easily get z=In1.2 and that is many times mentioned 0,18232. So if I were the owner of chocolate company, and if I wanted to say that 20g of 120g chocolate is for free, I could write on the cover: 20% is free, 16.666% is free or 18.232% is free. Isn't it great?

Okay, thanks, I see what you are doing and, yes, you did use a differential equation! It's a bit whacky- but then that was how it was intended wasn't it?

But just because I'm feeling grumpy: Why do you sometimes use the period, "." to indicate a decimal fraction and sometimes the comma, ","?
Ohh, and natural logarithm is "ln", "In"!

 

[mathwonk]

my apologies also to jason.now let me try to make my original point as clearly and politely as possible, since it seems to have been missed totally by some.

This whole thread is completely inappropriate to this forum. It seems to have begun by a question about arithmetic and percentages, which revealed very clearly that the OP did not understand the meaning of the words "x percent more" much less differential equations.

this topic is far below even the level of a college homework question, and certainly has no place here.

hence i blew it off with a remark about actual chocolates. I suggest however that my sentiments were justified, if not my words.

of course you are free to differ.

 

[Semo727]

Oh Got! Probably this thread has taken much af your time, and as I can see, it was huge mistake to try to write it here, on this forum. I apologize to everyone!
As I said, I'm not very used to writing this kind of things in math language (as I'm just high school student) . That might have been the main reason of every misunderstanding.
My explanation was not good, I know, but I was doing my best.

To HallsofIvy: I could now answer some of your questions, unfortunately you didn't get most of I wanted to say (but that is probably my fault)
But, I think you sholud not have been looking for every mistake a did. I hoped just that you will be able to get the point of what I was saying... I think that this math language, symbols and stuff are just human made tool how to understand some problems and if I used anything, and you could understand that, I thing there would be no problem..
the last thing:

Why do you sometimes use the period, "." to indicate a decimal fraction and sometimes the comma, ","?

In Slovakia (that's where I come from and live) we use "," for
that purpose, but as this forum is in English, i wanted to use "."
I feel really sorry for that, and also for "In" instead od "ln". Of course it is ln (as it is logarithm)

Once again, sorry to everyone! And I hope I will never create thread like this, (that no-one will understand me... and everyone will be angry with me)
I agree with Mathwonk, lighten up!

 

[HallsofIvy]

I feel really sorry for that, and also for "In" instead od "ln". Of course it is ln (as it is logarithm)

The reason I mention it is because I see it all the time. Personally I blame it on calculators: where the "l" looks like "I".