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Rate of increase problem

  1. May 3, 2004 #1
    A gallon of milk was $1.79 two years ago. Today, it's $2.15. Find the rate it increased each year. :smile:
  2. jcsd
  3. May 4, 2004 #2


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    PLEASE do not double post! If this really is homework, then it shouldn't be posted under "general mathematics" and if it is not it shouldn't be posted here!

    The price of milk, according to this problem increased from $1.79 to $2.15, an increase of ($2.15-1.79)= $0.36. Since that occured over two years, the rate of increase is ($0.36)/(2 years)= $0.18 per year.
  4. May 5, 2004 #3


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    I don't think that's what he's looking for. I think he wants a percentage per year. c is initial cost, C is final cost, r is rate, n is how many times (years?)

    [tex]C = (c)(r)^n[/tex]

    [tex]\frac{C}{c} = (r)^n[/tex]

    [tex]^n\sqrt{\frac{C}{c}} = r[/tex]

    [tex]r = 1.095955[/tex]

    After the first year, the price of milk will be 1.96176 which is a change of 0.17176.
    After the second year, the price of milk is 2.15 which is a change of 0.18824
  5. May 6, 2004 #4


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    There is no way to tell. For all you know, the price could have gone up to $2.15 the first day. From the info given, I would of course not assume that. Apparently ShawnD is assuming the rate to be compounded once per year. The answer would be different for the arguably equally valid assumption that it is compounded continuously. I suppose you could declare your assumption, but make sure that you are not required to use a different compounding period.
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