Question

Sam initially invested $4,500 into a savings account that offers an interest rate of 3% each year. He wants to determine the number of years, t, for which the account will have less than or equal to $7,020.

Answer 1

Every year, Sam will have 103% of the amount from the previous year.

$p\%=\dfrac{p}{100}\\\\103\%=\dfrac{103}{100}=1.03$

After the first year:

$1.03\cdot\ 4,500$

After the second year:

$1.03\cdot1.03\cdot\ 4,500=\ 4,500(1.03)^2$

After the t-th year:

$\ 4,500(1.03)^t$

Therefore we have the inequality:

$\ 4,500(1.03)^t\leq\ 7,020\ \ \ \ |\text{divide both sides by \ 4,500}\\\\(1.03)^t\leq1.56\\\\(1.03)^{15}\approx15.56\\\\\text{therefore}\ t\leq15$

Answer: t ≤ 15.