Question

If you invest $3,900 at a 7.83% simple annual interest rate, approximately how long will it take for you to have a total of $10,000? a. 12 years b. 15 years c. 20 years d. 30 years

Answer 1

we know that

The simple interest formula is equal to

$F=P*(1+n*r)$

where

F is the future value

P is the present value

n is the number of years

r is the annual interest rate in decimal

In this problem

we have

$P=$$$3,900$

$r=7.83$% -----> $r=0.0783$

$F=$$$10,000$

Find the value of n

$F=P*(1+n*r)\\ \\ \frac{F}{P} =1+n*r\\ \\ n=\frac{(\frac{F}{P}-1)}{r} \\ \\ n=\frac{(\frac{10,000}{3,900}-1)}{0.0783}\\ \\ n=19.98 years$

therefore

the answer is the option

$c. 20 years$

Answer 2

Apply simple interest formula
future value = present value (1+n*rate)
n=number of years
rate = annual interest rate in decimal

Plug in values,
10000=3900(1+0.0783*t)
1+0.0783t=10000/3900
t=(10000/3900-1)/0.0783
=19.98 years.

With compound interest,  future value=present value (1+rate/n)^(nt)
n=1,rate=0.0783, future value=10000, present value=3900,
t=log(10000/3900)/log(1.0783)=12.49 years.