Question

money planning
part 3 of 3 1 2 3 trying to earn
nancy ladd is 33 years old and has $8000 in savings. she can either

1. put her money in a no - interest checking account (that is, has 0% return),
2. put it in a savings account and earn 2.5% annual interest, or
3. put it into a retirement account of stocks and bonds earning 5% annual gains

the following graph shows the value of the three different accounts at various times.
value of account at different interest rates
$30,000
$20,000
value
$10,000
$0,000
0 5 10 15 20 25
years
— no - interest (0%) — 2.5% savings — 5% retirement
question 1: 20 years from now, nancy will be 53. at that time, how much more would her $8000 investment be worth if nancy put this money in the savings account (2.5%) instead of keeping this money in a no - interest checking account (0%)?
your answer should be within $2000 of the exact amount. $ 5106
question 2: how much longer will it take for nancy’s $8000 to grow to $10,000 at 2.5% than at 5%?
your answer should be within 1 year of the exact number of years. years

Explanation:

Question 1:

Step1: Calculate value at 2.5% (20yrs)

Use compound interest formula: $A = P(1+r)^t$
$A_{2.5\%} = 8000(1+0.025)^{20}$
$A_{2.5\%} = 8000(1.025)^{20} \approx 8000 \times 1.6386 = 13108.80$

Step2: Calculate value at 0% (20yrs)

No interest, so value stays principal:
$A_{0\%} = 8000$

Step3: Find the difference

$\text{Difference} = 13108.80 - 8000 = 5108.80$

Step1: Time to reach $10k at 5%

Use $A = P(1+r)^t$, solve for $t$:
$10000 = 8000(1+0.05)^t$
$\frac{10000}{8000} = 1.05^t$
$\log(1.25) = t\log(1.05)$
$t_{5\%} = \frac{\log(1.25)}{\log(1.05)} \approx \frac{0.09691}{0.02119} \approx 4.57$ years (≈5 years)

Step2: Time to reach $10k at 2.5%

Same formula:
$10000 = 8000(1+0.025)^t$
$\frac{10000}{8000} = 1.025^t$
$\log(1.25) = t\log(1.025)$
$t_{2.5\%} = \frac{\log(1.25)}{\log(1.025)} \approx \frac{0.09691}{0.01072} \approx 9.04$ years (≈9 years)

Step3: Find time difference

$\text{Difference} = 9.04 - 4.57 \approx 4.47$ years

Answer:

$\$5109$ (matches the approximate $\$5106$ within $\$2000$ range)

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Question 2: