Sophie deposited money into an account in which interest is compounded semiannually at a rate of 32%. She made no other deposits or withdrawals, and the total amount in her account after 7 years was [tex]$\$13,967.17$[/tex]. How much did she deposit?

A. [tex]$\[tex]$9000$[/tex][/tex]
B. [tex]$\$11,200$[/tex]
C. [tex]$\[tex]$12,125$[/tex][/tex]
D. [tex]$\$13,550$[/tex]



Answer :

To find out how much Sophie initially deposited into the account, we will use the compound interest formula, which is given by:

[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]

where:
- [tex]\( A \)[/tex] is the final amount (the amount in the account after the interest has been applied),
- [tex]\( P \)[/tex] is the principal amount (the initial deposit),
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal),
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year,
- [tex]\( t \)[/tex] is the time the money is invested or borrowed for, in years.

Given the information:
- The final amount [tex]\( A = \$ 13,967.17 \)[/tex],
- The interest rate [tex]\( r = 32\% = 0.32 \)[/tex],
- The interest is compounded semiannually, so [tex]\( n = 2 \)[/tex],
- The time period [tex]\( t = 7 \)[/tex] years.

We need to find the initial deposit [tex]\( P \)[/tex]. Rearranging the formula to solve for [tex]\( P \)[/tex] gives:

[tex]\[ P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}} \][/tex]

Substitute the given values into the formula:

[tex]\[ P = \frac{13,967.17}{\left(1 + \frac{0.32}{2}\right)^{2 \times 7}} \][/tex]

Calculate the term inside the parentheses first:

[tex]\[ 1 + \frac{0.32}{2} = 1 + 0.16 = 1.16 \][/tex]

Next, raise this term to the power of [tex]\( 2 \times 7 \)[/tex] (which is 14):

[tex]\[ 1.16^{14} \approx 7.982874 \][/tex]

Now, divide the final amount [tex]\( A \)[/tex] by this result to find [tex]\( P \)[/tex]:

[tex]\[ P = \frac{13,967.17}{7.982874} \approx 1748.62 \][/tex]

We need to determine which value from the given answer choices is closest to this calculated principal amount:

[tex]\[ \begin{align*} 9000 & \approx 9000 - 1748.62 = 7251.38 \\ 11200 & \approx 11200 - 1748.62 = 9451.38 \\ 12125 & \approx 12125 - 1748.62 = 10376.38 \\ 13550 & \approx 13550 - 1748.62 = 11801.38 \\ \end{align*} \][/tex]

The choice that matches the closest to the initial deposit amount is:

[tex]\[ \boxed{9000} \][/tex]