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?
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.)
Answer:
The initial deposit was $1992.44.
Step-by-step explanation:
Compound interest formula is :
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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.