Question

mitchell munson is 33 years old and has $18,000 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 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. (graph: value of account at different interest rates, with axes years (0 - 30) and value ($0,000 - $70,000), and three lines: no - interest (0%) (red), 2.5% savings (blue), 5% retirement (black)) question 1: 24 years from now, mitchell will be 57. at that time, how much more would his $18,000 investment be worth if mitchell 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. $ question 2: how much longer will it take for mitchell’s $18,000 to grow to $30,000 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 0% account value

The no-interest account retains the principal:
$\$18,000$

Step2: Calculate 2.5% account value

Use compound interest formula $A = P(1+r)^t$, where $P=18000$, $r=0.025$, $t=24$:
$A = 18000(1+0.025)^{24}$
$A = 18000(1.025)^{24} \approx 18000 \times 1.8087 = \$32,556.60$

Step3: Find the difference

Subtract 0% value from 2.5% value:
$\$32,556.60 - \$18,000 = \$14,556.60$
(Rounded to within $\$2000$ of exact value: $\$14,600$ or $\$15,000$ are acceptable, the precise rounded value is $\$14,600$)

Question 2

Step1: Find time for 5% to reach $30,000

Use $A = P(1+r)^t$, solve for $t$:
$30000 = 18000(1+0.05)^t$
$\frac{30000}{18000} = 1.05^t$
$\log(1.6667) = t\log(1.05)$
$t = \frac{\ln(1.6667)}{\ln(1.05)} \approx \frac{0.5108}{0.0488} \approx 10.47$ years

Step2: Find time for 2.5% to reach $30,000

Use $A = P(1+r)^t$, solve for $t$:
$30000 = 18000(1+0.025)^t$
$\frac{30000}{18000} = 1.025^t$
$\log(1.6667) = t\log(1.025)$
$t = \frac{\ln(1.6667)}{\ln(1.025)} \approx \frac{0.5108}{0.0247} \approx 20.68$ years

Step3: Calculate time difference

Subtract 5% time from 2.5% time:
$20.68 - 10.47 = 10.21$ years
(Rounded to within 1 year: 10 years)

Answer:

Question 1: $\boldsymbol{\$14,600}$
Question 2: $\boldsymbol{10}$ years