Question

Sam put $2,000 in a savings account at his bank. After 10 years, his account balance was $4,000. The interest rate on the savings deposit is

Answer 1

$\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \ 4000\\ P=\textit{original amount deposited}\dotfill & \ 2000\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &10 \end{cases} \\\\\\ 4000=2000(1+r\cdot 10)\implies \cfrac{4000}{2000}=10r+1\implies 2=10r+1 \\\\\\ 1=10r\implies \cfrac{1}{10}=r\implies 0.1=r\implies \stackrel{\textit{converting to percentage}}{0.1\cdot 100\implies }~~\stackrel{\%}{10}=r$

Answer 2

Answer:

The answer is approx 7.1%

Step-by-step explanation:

The compound interest formula is :

$A=p(1+r/n)^{nt}$

where, p = 2000

A = 4000

n = 1 (we will assume)

t = 10

r = ?

Now putting the values in formula we get

$4000=2000(1+r/1)^{10}$

or

$2=(1+r)^{10}$

Switching sides:

$(1+r)^{10}=2$

$1+r=\sqrt[10]{2}$

Solving this we get, r=0.07177 and r=-2.07177(neglect the negative)

So, we have r=0.0717 ≈ 0.071 (taking only up to 3 decimal places)

And in percentage, this is 7.1%(approx)