Question

How much money will you have if you started with $\$1200$ and put it in an account that earned 7.3% every year for 10 years?

Answer 1

Hint: Now we are given with the principal amount which is $\$1200$. We know the rate of interest is 7.3% and the time span is 10 years. Now we want to calculate the final amount. We know that the final amount after n years compounded annually is given by $A=P{{\left( 1+r \right)}^{n}}$ where A is the amount P is principal amount, r is rate of interest and n is the number of years or time span. Hence using this formula we will find the amount after 10 years.

Complete step by step solution:
Now $\$1200$ has been put in an account. Hence $\$1200$ is the principal amount. The rate of interest is 7.3 % per annum. Now since the amount is been put for years the time period is years.
Now since the amount is invested in the Bank we will calculate the interest with the help of Compound interest.
Now compound interest is interest calculated on Principal amount and the interest which is accumulated.
Now the amount after n years on principal amount P at the rate of interest r percent is given by
$A=P{{\left( 1+r \right)}^{n}}$
Hence substituting P = 1200, $r=\dfrac{7.3}{100}$ and n = 10 we get,
$\begin{align}
  & \Rightarrow A=1200{{\left( 1+\dfrac{7.3}{100} \right)}^{10}} \\
 & \Rightarrow A=1200{{\left( 1+0.073 \right)}^{10}} \\
 & \Rightarrow A=1200{{\left( 1.073 \right)}^{10}} \\
 & \Rightarrow A=1200\times \left( 2.023 \right) \\
 & \Rightarrow A=2427.60 \\
\end{align}$
Hence the amount received after 10 years is $\$1200$.

Note: Now remember there are two types of interest. Simple interest is simply interest calculated over principal amount. The amount is given by $A=P\left( 1+rt \right)$. Similarly compound interest is calculated over principal amount as well as interest accumulated and is given by the formula $A=P{{\left( 1+r \right)}^{t}}$ . Note that here r is the rate of interest and is in percentage. Hence the factor of 100 will be divided when substituted in the formula.