Find the balance after 10 years of a $30,000 savings account that pays 10% interest compounded yearly.

Answer 1

To calculate the balance of a savings account with compounded interest, you can use the compound interest formula:

\[ A = P(1 + r)^n \]

Where:

• \( A \) is the amount of money accumulated after n years, including interest.
• \( P \) is the principal amount (the initial amount of money).
• \( r \) is the annual interest rate (decimal).
• \( n \) is the number of years the money is invested or borrowed.

In this case:

• \( P = 30,000 \)
• \( r = 0.10 \) (10% interest)
• \( n = 10 \)

Now plug these values into the formula:

\[ A = 30,000(1 + 0.10)^{10} \]

This simplifies to:

\[ A = 30,000(1.10)^{10} \]

Calculating \( (1.10)^{10} \):

\[ (1.10)^{10} \approx 2.59374 \]

Now, calculate \( A \):

\[ A \approx 30,000 \times 2.59374 \approx 77,812.20 \]

Therefore, the balance after 10 years will be approximately $77,812.20.

Answer 2

The Answer choice are

$112,070.90
$33,154.81
$77,812.27
$2,245,256.05

Answer 3

Let's double-check the calculation using the compound interest formula to ensure we get the correct balance for the $30,000 savings account after 10 years with 10% interest compounded yearly.

The formula is:

\[ A = P(1 + r)^n \]

Where:

• \( P = 30,000 \) (the principal)
• \( r = 0.10 \) (the interest rate)
• \( n = 10 \) (the number of years)

So computing:

\[ A = 30,000(1 + 0.10)^{10} \]

This becomes:

\[ A = 30,000(1.10)^{10} \]

Calculating \( (1.10)^{10} \):

\[ (1.10)^{10} \approx 2.59374 \]

Now, multiply by the principal:

\[ A \approx 30,000 \times 2.59374 \approx 77,812.20 \]

When we look at the options provided:

• $112,070.90
• $33,154.81
• $77,812.27 (closest option)
• $2,245,256.05

The closest answer to our calculated value ($77,812.20) is $77,812.27.

Thus, the correct answer is $77,812.27.