Converting decimals to fractions and a word problem on interest rate

**Coach** · August 15th 2007, 09:55 AM

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!

**earboth** · August 15th 2007, 10:15 AM

Originally Posted by Coach

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
$x = \frac{1.1}{9}=\frac{11}{90}$

Originally Posted by Coach

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 $p \approx 1.94~\%$

**Coach** · August 15th 2007, 10:19 AM

Thank you, I actually made a typo there.

**Coach** · August 15th 2007, 10:20 AM

Originally Posted by earboth

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?

**earboth** · August 15th 2007, 10:30 AM

Originally Posted by Coach

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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