Question

kelly puts $350 in a savings account. The savings account accrues interest at a flat rate of 1.05% a month. How much will it count be worth in 7 months?

Answer 1

This is a problem concerning exponential growth/decay. This formula, y = ab^x, will come in handy.

The "a" will represent your initial value, the value that you are already given in the problem, in this case, it is $350. The "b" will be your growth/decay factor, in this case is 1.05%. However, it cannot stay a percent, but must be changed into a decimal. For this factor in particular, since there's a decimal already included, you'd move the decimal two places to the left, giving you 0.0105.

Since the factor is more than 1, you'd add the decimal by one, giving you 1.0105. Now, the "x" represents the amount of time. In this case, it would be 7 months. Let's set up the equation:

y = 350(1.0105)^7 = 376.5 (377 if rounded). In 7 months, Kelly will have $376.5.

Answer 2

the simple interest equation uses "t" as years, but is just cycles, using an APR rate.

now, if we nevermind "t" as years and just use it as an interest cycle, then we can say the rate is 1.05% and the period is 7 cycles.


`\bf \qquad \textit{Simple Interest Earned Amount}\\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to& \$350\\ r=rate\to 1.05\%\to (1.05)/(100)\to &0.0105\\ t=cycles \to 7 \end{cases} \\\\\\ A=350(1+0.0105\cdot 7)`