Suppose you deposit $2,000 in a savings account that pays interest at an annual rate of 4%. If no money is added or withdrawn from the account, answer the following questions. a. How much will be in the account after 4 years? b. How much will be in the account after 18 years? c. How many years will it take for the account to contain $2,500? d. How many years will it take for the account to contain $3,000?
we know that
The simple interest formula is equal to
[tex]A=P(1+rt)[/tex]
[tex]A=P\mleft(1+rt\mright)[/tex]where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
in this problem we have
Part a) How much will be in the account after 4 years?
we have
P=$2,000
r=4%=0.04
t=4 years
substitute the given values
[tex]\begin{gathered} A=2,000(1+0.04\cdot4) \\ A=\$2,320 \end{gathered}[/tex]Part b How much will be in the account after 18 years?
we have
P=$2,000
r=4%=0.04
t=18 years
substitute the given values
[tex]\begin{gathered} A=2,000(1+0.04\cdot18) \\ A=\$3,440 \end{gathered}[/tex]Part c How many years will it take for the account to contain $2,500?
we have
P=$2,000
r=4%=0.04
t=? years
A=$2,500
substitute
[tex]\begin{gathered} 2,500=2,000(1+0.04\cdot t) \\ \frac{2,500}{2,000}=1+0.04t \\ \\ 0.04t=\frac{2,500}{2,000}-1 \\ t=6.25\text{ years} \end{gathered}[/tex]Part d How many years will it take for the account to contain $3,000?
we have
P=$2,000
r=4%=0.04
t=? years
A=$3,000
substitute
[tex]\begin{gathered} 3,000=2,000(1+0.04\cdot t) \\ \frac{3,000}{2,000}=1+0.04t \\ \\ 0.04t=\frac{3,000}{2,000}-1 \\ t=12.5\text{ years} \end{gathered}[/tex]