### Question 11 (Multiple Choice)

An electrician invested [tex]\[tex]$9,000[/tex] in an investment account paying [tex]2.45\%[/tex] interest compounded semiannually. After 30 years, the value of the account will have a balance of [tex]\$[/tex]18,685.71[/tex]. If the electrician's investment was compounded continuously instead of twice per year, what would be the difference in the account balance after 30 years?

A. [tex]\[tex]$42.03[/tex]
B. [tex]\$[/tex]80.38[/tex]
C. [tex]\[tex]$83.63[/tex]
D. [tex]\$[/tex]165.55[/tex]

### Question 12 (Multiple Choice)

The equation [tex]A=2.400\left(1+\frac{0.031}{4}\right)^4[/tex] represents the amount of money earned on a compound interest savings account. What does the value [tex]0.031[/tex] represent?

A. The annual interest rate
B. The quarterly interest rate
C. The monthly interest rate
D. The daily interest rate



Answer :

Sure, let's solve question 11 first.

Given:
- Principal [tex]\( P = \$9000 \)[/tex]
- Annual interest rate [tex]\( r = 2.45\% = 0.0245 \)[/tex]
- Time duration [tex]\( t = 30 \)[/tex] years

We have two different compounding methods for comparison:

1. Semiannual Compounding:

With semiannual compounding, interest is compounded twice per year.

- Compounding frequency [tex]\( n = 2 \)[/tex] times per year.
- The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]

Substituting the given values:
[tex]\[ A_{semiannual} = 9000 \left(1 + \frac{0.0245}{2}\right)^{2 \times 30} \][/tex]
[tex]\[ A_{semiannual} \approx \$18,685.71 \][/tex]

2. Continuous Compounding:

With continuous compounding, interest is compounded at every instant.

- The formula for continuous compounding is:
[tex]\[ A = Pe^{rt} \][/tex]

Substituting the given values:
[tex]\[ A_{continuous} = 9000 \cdot e^{0.0245 \times 30} \][/tex]
[tex]\[ A_{continuous} \approx \$18,769.34 \][/tex]

Now, we calculate the difference between the two account balances:
[tex]\[ \text{Difference} = A_{continuous} - A_{semiannual} \][/tex]
[tex]\[ \text{Difference} \approx 18769.34 - 18685.71 \][/tex]
[tex]\[ \text{Difference} \approx \$83.63 \][/tex]

So, the difference in the account balance after 30 years when compounded continuously instead of semiannually is \[tex]$83.63. Therefore, the correct answer for Question 11 is: \[ \boxed{\$[/tex]83.63}
\]

Let me now proceed to Question 12.

Given:
[tex]\[ A = 2.400 \left( 1 + \frac{0.031}{A} \right)^4 \][/tex]

To interpret this equation, let's break it down:

- [tex]\( A \)[/tex] represents the final amount of money in the savings account.
- [tex]\( 2.400 \)[/tex] likely represents the principal amount or the initial deposit.
- [tex]\( 0.031 \)[/tex] is the interest rate applied per compounding period.
- The term [tex]\( (1 + \frac{0.031}{A}) \)[/tex] inside the expression adjusts the interest rate relative to [tex]\( A \)[/tex], which can be further analyzed based on the context of the savings account.
- The exponent [tex]\( 4 \)[/tex] indicates that the interest is compounded four times over the period being considered.

Therefore, in the context of this problem, the value [tex]\( 0.031 \)[/tex] represents the nominal interest rate that is being applied in each compounding period, adjusted in relation to the amount [tex]\( A \)[/tex].

Thus, for Question 12, the value [tex]\( 0.031 \)[/tex] represents the nominal interest rate applied to the principal.