Question

linda deposits $90,000 into an account that pays 4% interest per year, compounded annually. bob deposits $90,000 into an account that also pays 4% per year. but it is simple interest. find the interest linda and bob earn during each of the first three years. then decide who earns more interest for each year. assume there are no withdrawals and no additional deposits.

Explanation:

Step1: Calculate Bob's simple - interest formula

The simple - interest formula is $I = Prt$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), and $t$ is the time in years. Given $P=\$90000$, $r = 0.04$, and $t$ varies for each year.

Step2: Calculate Linda's compound - interest formula

The compound - interest formula is $A=P(1 + r)^t$, and the interest earned $I=A - P=P((1 + r)^t-1)$. For $P = 90000$ and $r = 0.04$.

First - year calculations

Bob's interest

$I_{Bob1}=Prt=90000\times0.04\times1 = 3600$.

Linda's interest

$I_{Linda1}=90000\times((1 + 0.04)^1-1)=90000\times(1.04 - 1)=90000\times0.04 = 3600$.
They earn the same amount in the first year.

Second - year calculations

Bob's interest

$I_{Bob2}=Prt=90000\times0.04\times1 = 3600$.

Linda's interest

$A_{Linda2}=90000\times(1 + 0.04)^2=90000\times1.0816$, $I_{Linda2}=90000\times1.0816-90000=90000\times(1.0816 - 1)=90000\times0.0816 = 3744$.
Linda earns more in the second year.

Third - year calculations

Bob's interest

$I_{Bob3}=Prt=90000\times0.04\times1 = 3600$.

Linda's interest

$A_{Linda3}=90000\times(1 + 0.04)^3=90000\times1.124864$, $I_{Linda3}=90000\times1.124864-90000=90000\times(1.124864 - 1)=90000\times0.124864 = 3901.92$.
Linda earns more in the third year.

Answer:

Year	Interest Linda earns (Interest compounded annually)	Interest Bob earns (Simple Interest)	Who earns more interest?
Second	$\$3744$	$\$3600$	Linda earns more.
Third	$\$3901.92$	$\$3600$	Linda earns more.