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Math Help - Converting decimals to fractions and a word problem on interest rate

  1. #1
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    Converting decimals to fractions and a word problem on interest rate

    Hello!


    I have two questions.


    1) Convert 1.2(with the 2 as a recurring number) into a fraction.
    My book instructs to multiply the decimal number by ten if there is only one recurring number so


    x= 0.12

    10x: 1.2

    Then the book instructs me to subtract 0.12 from 1.2 .

    Then 9X= 1.08

    and the fraction for this decimal is 1.08/9.


    Am I correct?


    2)

    Can someone please help me with this word problem?


    Jessica invests 10 000 . In four years her investment has grown to 10 800.



    What is the interest rate?




    Thank you!
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  2. #2
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    Quote Originally Posted by Coach View Post
    Hello!


    I have two questions.


    1) Convert 1.2(with the 2 as a recurring number) into a fraction.
    My book instructs to multiply the decimal number by ten if there is only one recurring number so
    x= 0.12
    10x: 1.2
    Then the book instructs me to subtract 0.12 from 1.2 .

    Then 9X= 1.08
    and the fraction for this decimal is 1.08/9.

    Am I correct?
    Hello,

    if I understand your problem correctly you have
    x = 0.122222222.... then
    10x = 1.22222222.... thus
    9x = 1.1 Therefore
    x = \frac{1.1}{9}=\frac{11}{90}

    Quote Originally Posted by Coach View Post
    2)

    Can someone please help me with this word problem?

    Jessica invests 10 000 . In four years her investment has grown to 10 800.

    What is the interest rate?
    If you invest a capital of C_0 at an interest o p % then the capital grows in n years to

    C_n=C_0 \cdot (1+\frac{p}{100})^n

    Now plug in the values you know:

    10800=10000 \cdot (1+\frac{p}{100})^4. Divide by 10000 and calculate the 4th root:

    \sqrt[4]{\frac{10800}{10000}}=1+\frac{p}{100}. Solve for p.

    I got p \approx 1.94~\%
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  3. #3
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    Thank you, I actually made a typo there.
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  4. #4
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    Quote Originally Posted by earboth View Post
    Hello,

    if I understand your problem correctly you have
    x = 0.122222222.... then
    10x = 1.22222222.... thus
    9x = 1.1 Therefore
    x = \frac{1.1}{9}=\frac{11}{90}



    If you invest a capital of C_0 at an interest o p % then the capital grows in n years to

    C_n=C_0 \cdot (1+\frac{p}{100})^n

    Now plug in the values you know:

    10800=10000 \cdot (1+\frac{p}{100})^4. Divide by 10000 and calculate the 4th root:

    \sqrt[4]{\frac{10800}{10000}}=1+\frac{p}{100}. Solve for p.

    I got p \approx 1.94~\%


    Thank you! That helped alot.


    Is there any way one could logically come to that formula?
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  5. #5
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    Quote Originally Posted by Coach View Post
    Thank you! That helped alot.


    Is there any way one could logically come to that formula?
    Hello,

    if you start with a capital C_0 a p % then you have after one year:

    C_1=C_0 + \frac{p}{100} \cdot C_0 = C_0 \cdot (1+\frac{p}{100}) . After 2 years:

    C_2=C_0 \cdot (1+\frac{p}{100}) + \frac{p}{100} \cdot C_0 \cdot (1+\frac{p}{100}) =  C_0 \cdot (1+\frac{p}{100}) \cdot (1+\frac{p}{100}) = C_0 \cdot (1+\frac{p}{100})^2. After 3 years:

    C_3=C_0 \cdot (1+\frac{p}{100})^2 + \frac{p}{100} \cdot C_0 \cdot (1+\frac{p}{100})^2 =  C_0 \cdot (1+\frac{p}{100})^2 \cdot (1+\frac{p}{100}) = C_0 \cdot (1+\frac{p}{100})^3. After n years:

    C_n=C_0 \cdot (1+\frac{p}{100})^n

    I hope this helps a little bit further
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