Question

money planning
part 3 of 3 1 2 3 trying to earn
mel hill is 34 years old and has $6000 in savings. he can either:

1. put his money in a no - interest checking account (that is, has 0% return),
2. put it in a savings account and earn 3.5% annual interest, or

3 put it into a retirement account of stocks and bonds earning 7% annual gains
the following graph shows the value of the three different accounts at various times.
value of account at different interest rates
$50,000
$40,000
$30,000
value
$20,000
$10,000
$0,000
0 5 10 15 20 25 30
years
— no - interest (0%) — 3.5% savings — 7% retirement
question 1: 20 years from now, mel will be 54. at that time, how much more would his $6000 investment be worth if mel put this money in the retirement account (7%) instead of keeping this money in a savings account (3.5%)?
your answer should be within $2000 of the exact amount. $
question 2: how much longer will it take for mels $6000 to grow to $10,000 at 3.5% than at 7%?
your answer should be within 1 year of the exact number of years. years

Explanation:

Step1: Calculate 7% retirement value at 20y

We use the compound interest formula $A = P(1+r)^t$, where $P=6000$, $r=0.07$, $t=20$.
$A_{7\%} = 6000(1+0.07)^{20} = 6000(1.07)^{20} \approx 6000 \times 3.8697 = 23218.20$

Step2: Calculate 3.5% savings value at 20y

Use the same formula with $r=0.035$, $t=20$.
$A_{3.5\%} = 6000(1+0.035)^{20} = 6000(1.035)^{20} \approx 6000 \times 1.9898 = 11938.80$

Step3: Find difference for Question1

Subtract the two values.
$\text{Difference} = 23218.20 - 11938.80 = 11279.40$

Step4: Solve for time at 7% to $10k

Rearrange $A = P(1+r)^t$ to $t = \frac{\ln(A/P)}{\ln(1+r)}$, where $A=10000$, $P=6000$, $r=0.07$.
$t_{7\%} = \frac{\ln(10000/6000)}{\ln(1.07)} = \frac{\ln(1.6667)}{\ln(1.07)} \approx \frac{0.5108}{0.0677} \approx 7.55$ years

Step5: Solve for time at 3.5% to $10k

Use the same formula with $r=0.035$.
$t_{3.5\%} = \frac{\ln(10000/6000)}{\ln(1.035)} = \frac{\ln(1.6667)}{\ln(1.035)} \approx \frac{0.5108}{0.0344} \approx 14.85$ years

Step6: Find time difference for Question2

Subtract the two time values.
$\text{Time Difference} = 14.85 - 7.55 = 7.30$ years

Answer:

Question 1: $\$11279.40$ (within $\$2000$ of exact value, acceptable range ~$\$9279$ to $\$13279$)
Question 2: $7$ years (within 1 year of exact value)