Answer :
Sure, let's go through the detailed steps to solve the problem using the long-hand formula.
First, let's summarize the given information:
- Future value (FV) = [tex]$25,000 - Interest rate per period (r) = 5% = 0.05 - Number of periods (n) = 10 years We are asked to find the annual savings required to reach the future value of $[/tex]25,000 in 10 years with an interest rate of 5%.
To solve for the annuity payment (P), we use the formula for the future value of an annuity:
[tex]\[ FV = P \times \left( \frac{(1 + r)^n - 1}{r} \right) \][/tex]
We can isolate P (the annuity payment) by rearranging the formula:
[tex]\[ P = \frac{FV \times r}{(1 + r)^n - 1} \][/tex]
Now, let's substitute the given values into the formula:
[tex]\[ P = \frac{25000 \times 0.05}{(1 + 0.05)^{10} - 1} \][/tex]
First, calculate [tex]\((1 + r)^n\)[/tex]:
[tex]\[ (1 + 0.05)^{10} = (1.05)^{10} \][/tex]
Next, raise 1.05 to the power of 10:
[tex]\[ (1.05)^{10} \approx 1.62889462677 \][/tex]
Now, subtract 1 from this value:
[tex]\[ 1.62889462677 - 1 = 0.62889462677 \][/tex]
Then, multiply the future value by the interest rate:
[tex]\[ 25000 \times 0.05 = 1250 \][/tex]
Now, divide this result by the value obtained from the previous subtraction:
[tex]\[ P = \frac{1250}{0.62889462677} \][/tex]
[tex]\[ P \approx 1987.60578924 \][/tex]
Finally, we round this amount to 2 decimal places:
[tex]\[ P \approx 1987.61 \][/tex]
So, the annual savings required to buy a boat in 10 years for [tex]$25,000, at an interest rate of 5% annually, is \$[/tex]1987.61.
First, let's summarize the given information:
- Future value (FV) = [tex]$25,000 - Interest rate per period (r) = 5% = 0.05 - Number of periods (n) = 10 years We are asked to find the annual savings required to reach the future value of $[/tex]25,000 in 10 years with an interest rate of 5%.
To solve for the annuity payment (P), we use the formula for the future value of an annuity:
[tex]\[ FV = P \times \left( \frac{(1 + r)^n - 1}{r} \right) \][/tex]
We can isolate P (the annuity payment) by rearranging the formula:
[tex]\[ P = \frac{FV \times r}{(1 + r)^n - 1} \][/tex]
Now, let's substitute the given values into the formula:
[tex]\[ P = \frac{25000 \times 0.05}{(1 + 0.05)^{10} - 1} \][/tex]
First, calculate [tex]\((1 + r)^n\)[/tex]:
[tex]\[ (1 + 0.05)^{10} = (1.05)^{10} \][/tex]
Next, raise 1.05 to the power of 10:
[tex]\[ (1.05)^{10} \approx 1.62889462677 \][/tex]
Now, subtract 1 from this value:
[tex]\[ 1.62889462677 - 1 = 0.62889462677 \][/tex]
Then, multiply the future value by the interest rate:
[tex]\[ 25000 \times 0.05 = 1250 \][/tex]
Now, divide this result by the value obtained from the previous subtraction:
[tex]\[ P = \frac{1250}{0.62889462677} \][/tex]
[tex]\[ P \approx 1987.60578924 \][/tex]
Finally, we round this amount to 2 decimal places:
[tex]\[ P \approx 1987.61 \][/tex]
So, the annual savings required to buy a boat in 10 years for [tex]$25,000, at an interest rate of 5% annually, is \$[/tex]1987.61.
