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Calculate the value of $200 at 4% interest after 10 years using each of the following methods:

1. Simple Interest
2. Compound Interest (Annually)
3. Compound Interest (Biannually)


Sagot :

To understand what happens to [tex]$200 at 4% interest after 10 years, we will calculate the value using three different methods: Simple Interest, Compound Interest Annually, and Compound Interest Biannually. ### Simple Interest Simple Interest is calculated using the formula: \[ SI = P \times r \times t \] Where: - \( SI \) is the simple interest - \( P \) is the principal amount (initial amount of money) - \( r \) is the interest rate - \( t \) is the time period in years Given: - \( P = 200 \) - \( r = 4\% = 0.04 \) - \( t = 10 \) Plugging in these values: \[ SI = 200 \times 0.04 \times 10 = 80 \] The total amount after 10 years will be: \[ Total (Simple) = Principal + Simple Interest = 200 + 80 = 280 \] ### Compound Interest Annually Compound Interest Annually is calculated using the formula: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] For annual compounding, \( n = 1 \): \[ A = P (1 + r)^{t} \] Given: - \( P = 200 \) - \( r = 4\% = 0.04 \) - \( t = 10 \) - \( n = 1 \) Plugging in these values: \[ A = 200 \left(1 + 0.04\right)^{10} \approx 296.0488569836689 \] ### Compound Interest Biannually Compound Interest Biannually is calculated using a similar formula, but compounded twice a year: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] For biannual compounding, \( n = 2 \): \[ A = P \left(1 + \frac{0.04}{2}\right)^{2 \times 10} \] Given: - \( P = 200 \) - \( r = 4\% = 0.04 \) - \( t = 10 \) - \( n = 2 \) Plugging in these values: \[ A = 200 \left(1 + \frac{0.04}{2}\right)^{2 \times 10} \approx 297.18947919567097 \] ### Summary After 10 years, the values are as follows: - Simple Interest: $[/tex]80, resulting in a total amount of [tex]$280. - Compound Interest Annually: Approximately $[/tex]296.0488569836689.
- Compound Interest Biannually: Approximately $297.18947919567097.

Thus, the methods yield slightly different totals due to the compounding effect.