Question

Fernando has a savings account balance of $2,471.23. The interest rate on the account is 2.4% compounded quarterly. If he opened the account nine years ago, what was the value of his initial deposit?

Answer 1

Answer:

The initial deposit was $1992.44.

Step-by-step explanation:

Compound interest formula is :

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

Fernando has a savings account balance of $2,471.23.

Means A = 2471.23

r = 2.4% = 0.024

n = 4

t = 9

Putting the values in formula we get;

$2471.23=p(1+\frac{0.024}{4})^{4*9}$

=> $2471.23=p(1+\frac{0.024}{4})^{36}$

=> $2471.23=p(1.006)^{36}$

=> $2471.23=p(1.24030)$

=> p = 1992.44

Hence, the initial deposit was $1992.44.

Answer 2

You will need a formula for compound interest solved for principal. (See attached).
Principal = $2,471.23 / (1+ (.024/4)^4*9

Principal =  $2,471.23 / 1.2403016056
Principal =  $1,992.44
(I checked this with an online compound interest calculator to check this - it is correct.)