Question

you got it right - just one more to go!
tyrone chandler is 35 years old and has $14,000 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 3% annual interest, or
3. put it into a retirement account of stocks and bonds earning 6% 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 value on y - axis from $0,000 to $70,000 and years on x - axis from 0 to 30, three lines for no - interest (0%), 3% savings, 6% retirement)
question 1: 24 years from now, tyrone will be 59. at that time, how much more would his $14,000 investment be worth if tyrone put this money in the retirement account (6%) instead of keeping this money in a savings account (3%)?
your answer should be within $2000 of the exact amount. $
question 2: how much longer will it take for tyrone’s $14,000 to grow to $30,000 at 3% than at 6%?
your answer should be within 1 year of the exact number of years. years

Explanation:

Question 1

Step1: Calculate 6% account value at 24y

Use compound interest formula: $A = P(1+r)^t$
$A_{6\%} = 14000(1+0.06)^{24}$
$A_{6\%} = 14000(1.06)^{24} \approx 14000 \times 4.04893 \approx 56685$

Step2: Calculate 3% account value at 24y

Use compound interest formula: $A = P(1+r)^t$
$A_{3\%} = 14000(1+0.03)^{24}$
$A_{3\%} = 14000(1.03)^{24} \approx 14000 \times 2.03279 \approx 28459$

Step3: Find difference in values

Subtract the two amounts:
$\text{Difference} = 56685 - 28459 = 28226$

Question 2

Step1: Find time for 6% to hit $30k

Use $A = P(1+r)^t$, solve for $t$:
$t = \frac{\ln(\frac{A}{P})}{\ln(1+r)}$
$t_{6\%} = \frac{\ln(\frac{30000}{14000})}{\ln(1.06)} = \frac{\ln(2.1429)}{\ln(1.06)} \approx \frac{0.7621}{0.0583} \approx 13.1$ years

Step2: Find time for 3% to hit $30k

Use same compound interest time formula:
$t_{3\%} = \frac{\ln(\frac{30000}{14000})}{\ln(1.03)} = \frac{\ln(2.1429)}{\ln(1.03)} \approx \frac{0.7621}{0.0296} \approx 25.7$ years

Step3: Calculate time difference

Subtract the two time values:
$\text{Difference} = 25.7 - 13.1 = 12.6$

Answer:

1. $\$28226$ (within $\$2000$ of exact value)
2. $13$ years (within 1 year of exact value)