Question

Upon graduation from​ college, Warren Roberge was able to defer payment on his ​$22,000 student loan for 3 months. Since the interest will no longer be paid on his​ behalf, it will be added to the principal until payments begin. If the interest is 6.94 ​% compounded monthly ​, what will the principal amount be when he must begin repaying his​ loan?

Answer 1

Answer:

The principal amount be when he must begin repaying his​ loan is $22383.911.

Step-by-step explanation:

Given : Upon graduation from​ college, Warren Roberge was able to defer payment on his ​$22,000 student loan for 3 months. If the interest is 6.94 ​% compounded monthly.

To find : What will the principal amount be when he must begin repaying his​ loan?

Solution :

Using compound interest formula,

$A=P(1+\frac{r}{n})^{nt}$

Where,

A is the amount

P is the principal P=$22,000

r is the rate r=6.94%=0.0694

t is the time t=3 months

Into years, $t=\frac{3}{12}=\frac{1}{4}$

n is the number of period n=12

Substitute the value in the formula,

$A=22000(1+\frac{0.0694}{12})^{12\times \frac{1}{4}}$

$A=22000(1+0.005783)^{3}$

$A=22000(1.005783)^{3}$

$A=22000(1.01745)$

$A=22383.911$

The principal amount be when he must begin repaying his​ loan is $22383.911.

Answer 2

It sounds to me as I read it, as the principal P, will have the 3 months of interest added, because Warren deferred payments for 3 months, but that doesn't let him off the hook on the interest payments, so what the loaning bank did is say, "ok, start paying it in 3 months, but we'll accumulate the monthly interest to it".

so in short what is a principal of 22,000 accumulating 6.94% monthly compounding interest rate for 3 months?

bearing in mind that 3 months is not even a year, since there are 12 months in a year, 3 months is really just 3/12 year.

$\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\ 22000\\ r=rate\to 6.94\%\to \frac{6.94}{100}\to &0.0694\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{montly, thus twelve} \end{array}\to &12\\ t=years\to \frac{3}{12}\to &\frac{1}{4} \end{cases}$

$\bf A=22000\left(1+\frac{0.0694}{12}\right)^{12\cdot \frac{1}{4}}\implies A=2000\left( 1+\frac{347}{60000} \right)^3 \\\\\\ A=22000\left( \frac{60347}{60000} \right)^3$