If $10,000 is locked in a savings account offering 6.5% per annum, compounded monthly. How fast is the balance growing three years later? Round to the nearest dollar. Ensure that your answer intuitively makes sense to you.
First, we need to calculate the balance in the account after 3 years. We can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where A is the final amount, P is the principal amount ($10,000 in this case), r is the annual interest rate (6.5% or 0.065 as a decimal), n is the number of times interest is compounded per year (monthly means 12 times), and t is the number of years (3 years in this case).
A = 10000 * (1 + 0.065/12)^(12*3)
A = 10000 * (1.005417)^(36)
A ≈ 12394.84
The balance after 3 years is approximately $12,394.84.
Now we need to find out how fast it's growing at this point. We can do this by finding the interest earned in one month and annualizing it (since the interest rate is given per annum).
First, let's calculate the balance in the account 1 month before the three-year mark:
A_prev_month = P(1 + r/n)^(nt - 1)
A_prev_month = 10000 * (1 + 0.065/12)^(12*3 - 1)
A_prev_month = 10000 * (1.005417)^(35)
A_prev_month ≈ 12355.18
Now we subtract the balance at the month before from the balance at the three-year mark.
Interest earned in 1 month = A - A_prev_month
Interest_earned_1_month ≈ 12394.84 - 12355.18
Interest_earned_1_month ≈ 39.66
Now, we can calculate the annualized growth rate:
Annualized_growth_rate = Interest_earned_1_month * 12
Annualized_growth_rate ≈ 39.66 * 12
Annualized_growth_rate ≈ 475.92
So, the balance is growing at a rate of approximately $476 per year 3 years after the $10,000 was first placed into the account.
To find the balance three years later, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final balance
P = the initial principal amount ($10,000)
r = annual interest rate (6.5% or 0.065)
n = number of times interest is compounded per year (12, since it's compounded monthly)
t = time in years (3 years)
Plugging in the values:
A = 10000(1 + 0.065/12)^(12*3)
Now, let's calculate this expression step by step:
1 + 0.065/12 = 1.005416667
12 * 3 = 36
(1.005416667)^36 ≈ 1.231017
Now, we can find the final balance by multiplying the initial principal by this value:
A = 10000 * 1.231017 ≈ $12,310.17
Therefore, the balance after three years would be approximately $12,310.17.
To determine how fast the balance is growing, we can calculate the difference between the final balance and the initial principal. In this case, it would be:
Balance growth = $12,310.17 - $10,000 = $2,310.17
So, the balance is growing by approximately $2,310.17 over the course of three years.
Looking at the numbers, this intuitively makes sense since compound interest allows for exponential growth over time. In this case, the balance is increasing by over $2,000, which is a significant growth rate for a $10,000 investment.
To find out how fast the balance is growing three years later, we need to calculate the compound interest earned over three years.
Starting balance: $10,000
Interest rate: 6.5% per annum (or 0.065 in decimal form)
Compounding: Monthly
First, we need to calculate the monthly interest rate:
Monthly interest rate = Annual interest rate / Compounding periods per year
Monthly interest rate = 0.065 / 12 = 0.00541667 (rounded to 8 decimal places)
Next, we can use the compound interest formula to calculate the future value:
Future value = Principal amount × (1 + Monthly interest rate)^(Number of compounding periods)
Future value = $10,000 × (1 + 0.00541667)^(3 * 12)
Calculating this in steps:
1.00541667 = 1 + 0.00541667
(1.00541667)^36 ≈ 1.222099079
Future value ≈ $10,000 × 1.222099079 ≈ $12,220.99
The balance three years later is approximately $12,220.99.
To find out how fast the balance is growing, we subtract the initial balance from the future balance:
Growth = Future value - Initial balance
Growth = $12,220.99 - $10,000
Growth ≈ $2,220.99
So, the balance is growing at a rate of approximately $2,220.99 over three years.