QuestionPS-58 Number of years needed to acccumulate a future amount For each of the following cases, determine the number of years it will take for the initial deposit to grow to equal the furure amount at the given interest rate.
CHAPTER 5 Time Value of Money 295 \begin{tabular}{cccc} Case & Initial deposit & Future amount & Interest rate \\ \hline A & $\$ 300$ & $\$ 1,000$ & $7 \%$ \\ B & 12,000 & 15,000 & 5 \\ C & 9,000 & 20,000 & 10 \\ D & 100 & 500 & 9 \\ E & 7,500 & 30,000 & 15 \end{tabular}

Studdy Solution

STEP 1

1. The interest is compounded annually.
2. We will use the formula for compound interest to solve for the number of years.
3. The formula for future value with compound interest is $FV = PV \times (1 + r)^n$ , where $FV$ is the future value, $PV$ is the present value or initial deposit, $r$ is the interest rate, and $n$ is the number of years.

STEP 2

1. Set up the compound interest formula for each case.
2. Solve the equation for the number of years $n$ .
3. Calculate $n$ for each case.

STEP 3

For each case, set up the equation using the compound interest formula:
Case A: $1000 = 300 \times (1 + 0.07)^n$
Case B: $15000 = 12000 \times (1 + 0.05)^n$
Case C: $20000 = 9000 \times (1 + 0.10)^n$
Case D: $500 = 100 \times (1 + 0.09)^n$
Case E: $30000 = 7500 \times (1 + 0.15)^n$

STEP 4

Rearrange each equation to solve for $n$ by isolating the exponential term:
Case A: $\frac{1000}{300} = (1.07)^n$
Case B: $\frac{15000}{12000} = (1.05)^n$
Case C: $\frac{20000}{9000} = (1.10)^n$
Case D: $\frac{500}{100} = (1.09)^n$
Case E: $\frac{30000}{7500} = (1.15)^n$

STEP 5

Take the natural logarithm of both sides to solve for $n$ :
Case A: $n = \frac{\ln\left(\frac{1000}{300}\right)}{\ln(1.07)}$
Case B: $n = \frac{\ln\left(\frac{15000}{12000}\right)}{\ln(1.05)}$
Case C: $n = \frac{\ln\left(\frac{20000}{9000}\right)}{\ln(1.10)}$
Case D: $n = \frac{\ln\left(\frac{500}{100}\right)}{\ln(1.09)}$
Case E: $n = \frac{\ln\left(\frac{30000}{7500}\right)}{\ln(1.15)}$

STEP 6

Calculate $n$ for each case:
Case A: $n \approx \frac{\ln(3.3333)}{\ln(1.07)} \approx 18.05$
Case B: $n \approx \frac{\ln(1.25)}{\ln(1.05)} \approx 4.56$
Case C: $n \approx \frac{\ln(2.2222)}{\ln(1.10)} \approx 8.41$
Case D: $n \approx \frac{\ln(5)}{\ln(1.09)} \approx 17.67$
Case E: $n \approx \frac{\ln(4)}{\ln(1.15)} \approx 9.67$
The number of years needed for each case is approximately:
Case A: $18.05$ years \\ Case B: $4.56$ years \\ Case C: $8.41$ years \\ Case D: $17.67$ years \\ Case E: $9.67$ years

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