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Chocolate with 20% extra

  1. Jan 14, 2006 #1
    Today, I was eating a chocolate with 20% of its mass extra.
    Then I realised (after some thinking) that the chocolate company colud also write, that the chocolate was with about 18,232155 % extra. :cool:
    Do you know how that is possible??
     
    Last edited: Jan 15, 2006
  2. jcsd
  3. Jan 15, 2006 #2
    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.
     
  4. Jan 15, 2006 #3

    HallsofIvy

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    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.

    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"
     
    Last edited: Jan 15, 2006
  5. Jan 15, 2006 #4
    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 dont 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.. :rolleyes:

    And number 0.18232155 is ln1.2
     
    Last edited: Jan 15, 2006
  6. Jan 15, 2006 #5

    mathwonk

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    just go to worldwidechocolates.com and don't bother us any more.
     
  7. Jan 16, 2006 #6
    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!!
     
  8. Jan 16, 2006 #7

    HallsofIvy

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    Sorry, but

    1. Your calculation is completely meaningless.

    2. Apparently you don't know what a "differential equation" is.
     
  9. Jan 16, 2006 #8
    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 dont know how to write that...sory

    Ok, administrator can cancel this whole thread... :uhh:
    but... I was right..
     
    Last edited: Jan 16, 2006
  10. Jan 16, 2006 #9
    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?
     
  11. Jan 16, 2006 #10

    JasonRox

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    He is a mathematician, so I guess you won't become one. :biggrin:
     
  12. Jan 16, 2006 #11

    mathwonk

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    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:
     
  13. Jan 16, 2006 #12

    saltydog

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    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.
     
  14. Jan 16, 2006 #13

    JasonRox

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    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.
     
  15. Jan 16, 2006 #14

    mathwonk

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    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.:tongue2:
    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.
     
    Last edited: Jan 16, 2006
  16. Jan 17, 2006 #15

    saltydog

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    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.
     
  17. Jan 17, 2006 #16

    JasonRox

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    Alright, let's get back on topic now.

    Be more clear about your complaint. I would never have guessed that's what it was.
     
  18. Jan 17, 2006 #17

    HallsofIvy

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    Semo727, I would still like to know how you used differential equations to determine that.
     
  19. Jan 17, 2006 #18
    So I'm gonna (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?
     
    Last edited: Jan 17, 2006
  20. Jan 17, 2006 #19

    mathwonk

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    i apologize for not properly ,recognizing those who are offended by me as opposed to those are defending me.
    I apologize to saltydog for, lets 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.
     
    Last edited: Jan 18, 2006
  21. Jan 18, 2006 #20

    saltydog

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    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.
     
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