If Debra invests €40,000 for nine years at a return of 2.5% per annum, what will her investment grow to?

If Debra invests €40,000 for nine years at a return of 2.5% per annum, what will her investment grow to?

• A. €47,800
• B. €49,960
• C. €53,800
• D. €62,000

Answer: (B)

To determine the future value of Debra's investment after nine years at an annual return of 2.5%, we can use the formula for compound interest, which is: \[ A = P \times (1 + r)^n \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial sum of money). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of years the money is invested or borrowed. In this case: - \( P = €40,000 \) - \( r = 2.5\% = 0.025 \) - \( n = 9 \) Plugging these values into the formula gives: \[ A = 40,000 \times (1 + 0.025)^9 \] Calculating the exponent part first: \[ (1 + 0.025)^9 = (1.025)^9 \approx 1.2460 \] Now, multiply this by the principal amount: \[ A \approx 40,000 \times 1.2460 \approx 49,840 \] Rounding

To determine the future value of Debra's investment after nine years at an annual return of 2.5%, we can use the formula for compound interest, which is:

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

Where:

• ( A ) is the amount of money accumulated after n years, including interest.

• ( P ) is the principal amount (the initial sum 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 = €40,000 )

• ( r = 2.5% = 0.025 )

• ( n = 9 )

Plugging these values into the formula gives:

[ A = 40,000 \times (1 + 0.025)^9 ]

Calculating the exponent part first:

[ (1 + 0.025)^9 = (1.025)^9 \approx 1.2460 ]

Now, multiply this by the principal amount:

[ A \approx 40,000 \times 1.2460 \approx 49,840 ]

Rounding