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Thread: Increasing by percent

  1. #1
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    Question Increasing by percent

    Number of students at university increased from 3000 to 12000 during 2 years. By how much percent was it increasing annually?
    Answer is 100%

    How can I figure out this without testing possible answers?

    P.S. I have translated this problem from other language and I could translated it a bit wrongly.
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  2. #2
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    Re: Increasing by percent

    Use the "compound interest" formula $\displaystyle A = P (1+i)^n$

    A=12000, P = 3000, n=2 years, i = "interest" rate per year (as a decimal)

    Solve for i. You should get i = 1 = 100%
    Thanks from Jotaro
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  3. #3
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    Re: Increasing by percent

    I didn't know this formula. Thanks! I'm going to search derivation of it.

    But I don't think that I should solve this problem with compound interest formula. There's no word about it in the textbook. Do you know other, more intuitive way solve it?
    Last edited by Jotaro; Apr 3rd 2019 at 01:35 AM.
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  4. #4
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    Re: Increasing by percent

    Well, if you don't want to use cpd. int. formula, then:
    Code:
    YEAR   100%    POPULATION
      0               3000
      1    3000       6000
      2    6000      12000
    2[2(3000)] = 12000
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  5. #5
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    Re: Increasing by percent

    Quote Originally Posted by Jotaro View Post
    I didn't know this formula. Thanks! I'm going to search derivation of it.

    But I don't think that I should solve this problem with compound interest formula. There's no word about it in the textbook. Do you know other, more intuitive way solve it?
    Well assume it is increasing by x% annually.

    After one year the population is $\displaystyle 3000*(1+\frac{x}{100})$

    After 2 years the population is $\displaystyle 3000 * (1 +\frac{x}{100})(1+\frac{x}{100}) = 12000$. Solve for x.
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