# Thread: Converting decimals to fractions and a word problem on interest rate

1. ## 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!

2. 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
$\displaystyle x = \frac{1.1}{9}=\frac{11}{90}$

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 $\displaystyle C_0$ at an interest o p % then the capital grows in n years to

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

Now plug in the values you know:

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

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

I got $\displaystyle p \approx 1.94~\%$

3. Thank you, I actually made a typo there.

4. 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
$\displaystyle x = \frac{1.1}{9}=\frac{11}{90}$

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

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

Now plug in the values you know:

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

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

I got $\displaystyle p \approx 1.94~\%$

Thank you! That helped alot.

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

5. 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 $\displaystyle C_0$ a p % then you have after one year:

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

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

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

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

I hope this helps a little bit further