91Ó°ÊÓ

Given, Jim places $1000 in a bank account that pays 5.6% compounded continuously. It is to be checked if after 1 year, he will have enough money to buy a computer system that costs $1060. Also another bank will pay Jim 5.9% compounded monthly. It is to be checked if this is a better deal.

According to the compound interest formula, the amount A after t years for the principal P with an annual rate of interest r compounded continuously is $A=P{e}^{rt}$

Here the amount invested is $1,000 i.e. $P=1000$

The rate of interest is 5.6% i.e. $r=0.056$

The time for investment is 1 year i.e. $t=1$

Plugging the values:

$$\begin{gathered} A=P{e}^{rt} \\ A=\left(1000\right)e^{\left(0.056\right)\left(1\right)} \\ Aâ‰\dots 1000\left(1.057597\right) \\ Aâ‰\dots 1057.60 \end{gathered}$$

Since he needs $1060 after 1 year, he would not have enough money to buy the computer.

According to the compound interest formula, the amount A after t years for the principal P with an annual rate of interest r compounded n times per year is $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

Here the amount invested is $1,000 i.e. role="math" localid="1646729321993" $P=1000$

The rate of interest is 5.9% i.e. role="math" localid="1646729309789" $r=0.059$

The time for investment is 1 year i.e. $t=1$

Since the amount is compounded monthly, $n=12$

Plugging the values:

$$\begin{gathered} A=P{\left(1+\frac{r}{n}\right)}^{nt} \\ A=1000{\left(1+\frac{0.059}{12}\right)}^{\left(12\right)\left(1\right)} \\ A=1000{\left(1.00491\right)}^{12} \\ Aâ‰\dots 1060.62 \end{gathered}$$

Since he needs $1060 after 1 year, he would have enough money to buy the computer.

Since the monthly compounding gives better final amount, that is a better deal.