Compound Interest |

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Notes

Compound Interest Formula

$A=P\left(1+\frac{r}{n} \right)^{nt}$

$A=$Amount at time t
$P=$Principle (amount at time t=0)
$n=$compoundings per year
$r=$annual interest rate ($3\%=.03$ for example)
$t=$time in years

Values of $n$

Monthly $ \rightarrow n=12$
Quarterly $ \rightarrow n=4$
Daily $ \rightarrow n=365$
Weekly $ \rightarrow n=52$
Yearly $ \rightarrow n=1$
Annually $ \rightarrow n=1$
Semi-annually $ \rightarrow n=2$

Continuously Compounded Interest Formula

$A=P e^{rt}$

Interest Earned

$I=A-P$

Problems & Videos

$\textbf{1)}$ You place $4,000 in a bank account with 2.5% interest rate compounded monthly. How much will you have in the account after 4 years?
Show Answer
The answer is $ $4{,}420.22 $
Show Work

$\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}$

$\,\,\,\,\,\,A=4{,}000\left(1+\frac{.025}{12}\right)^{12 \cdot 4}$

$\,\,\,\,\,\,A=4{,}000\left(1+0.00208333\right)^{48}$

$\,\,\,\,\,\,A=4{,}000\left(1.00208333\right)^{48}$

$\,\,\,\,\,\,A=4{,}000\cdot 1.105056$

$\,\,\,\,\,\,$The answer is $ $4{,}420.22 $

$\textbf{2)}$ How much money do you have after 5 years if you invest $6,000 at 3.5% interest compounded quarterly?
Show Answer
The answer is $ $7{,}142.04 $
Show Work

$\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}$

$\,\,\,\,\,\,A=6{,}000\left(1+\frac{.035}{4}\right)^{(4)(5)}$

$\,\,\,\,\,\,A=6{,}000\left(1.00875\right)^{20}$

$\,\,\,\,\,\,A=6{,}000 \cdot 1.19034$

$\,\,\,\,\,\,A=6{,}000 \cdot 1.19034$

$\,\,\,\,\,\,$The answer is $ $7{,}142.04 $

$\textbf{3)}$ Alex puts $800.00 into an account. The account earns 4.1% interest, compounded monthly. How much will be in the account after 7 years?
Show Answer
The answer is $ $1{,}065.42 $
Show Work

$\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}$

$\,\,\,\,\,\,A=800\left(1+\frac{.041}{12}\right)^{12 \cdot 7}$

$\,\,\,\,\,\,A=800\left(1+0.00341667\right)^{84}$

$\,\,\,\,\,\,A=800\left(1.00341667\right)^{84}$

$\,\,\,\,\,\,A=800\cdot 1.331773$

$\,\,\,\,\,\,$The answer is $ $1{,}065.42 $

$\textbf{4)}$ Stevie puts $800.00 into an account. The account earns 4.1% interest compounded continously. How much will be in the account after 7 years?
Show Answer
The answer is $ $1{,}065.94 $
Show Work

$\,\,\,\,\,\,A=Pe^{rt}$

$\,\,\,\,\,\,A=800e^{(0.041)(7)}$

$\,\,\,\,\,\,A=800e^{0.287}$

$\,\,\,\,\,\,A=800(1.332424)$

$\,\,\,\,\,\,A=1065.94$

$\,\,\,\,\,\,$The answer is $ $1{,}065.94 $

$\textbf{5)}$ You need $30,000 in 5 years. How much money should you place in an account with 3% interest compounded continuously?
Show Answer
The answer is $ $25{,}821.24 $
Show Work

$\,\,\,\,\,\,A=Pe^{rt}$

$\,\,\,\,\,\,30{,}000=Pe^{(.03)(5)}$

$\,\,\,\,\,\,30{,}000=Pe^{.15}$

$\,\,\,\,\,\,30{,}000=Pe^{.15}$

$\,\,\,\,\,\,30{,}000=P(1.161834)$

$\,\,\,\,\,\,\displaystyle\frac{30{,}000}{1.161834}=P$

$\,\,\,\,\,\,25821.24468=P$

$\,\,\,\,\,\,$The answer is $ $25{,}821.24 $

$\textbf{6)}$ How long would it take $3,000 to triple if you earned 4% interest compounded continuously?
Show Answer
The answer is $ 27.47$ years
Show Work

$\,\,\,\,\,\,A=Pe^{rt}$

$\,\,\,\,\,\,9{,}000=3{,}000e^{.04t}$

$\,\,\,\,\,\,3=e^{.04t}$

$\,\,\,\,\,\,\ln{3}=\ln{e^{.04t}}$

$\,\,\,\,\,\,\ln{3}=.04t$

$\,\,\,\,\,\,\displaystyle\frac{\ln{3}}{.04}=t$

$\,\,\,\,\,\,27.46530=t$

$\,\,\,\,\,\,$The answer is $ 27.47$ years

$\textbf{7)}$ Kay invests $8,294 in an account that pays an annual interest rate of 4% compounded monthly. What will the account balance be after 6 years?
Show Answer
The answer is $ $10{,}539.53 $
Show Work

$\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}$

$\,\,\,\,\,\,A=8{,}294\left(1+\frac{.04}{12}\right)^{\left(12\right)\left(6\right)}$

$\textbf{8)}$ Ian invests $3,925 in an account with a fixed annual interest rate of 6.5% compounded twice per year. What will the account balance be after 10 years?
Show Answer
The answer is $ $7{,}441.16 $
Show Work

$\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}$

$\,\,\,\,\,\,A=3925\left(1+\frac{.065}{2}\right)^{\left(2\right)\left(10\right)}$

$\textbf{9)}$ Steph invests $7,498 in an account with an annual interest rate of 4.2% compounded continuously. What will the account balance be after 16 years?
Show Answer
The answer is $ $14{,}682.21 $
Show Work

$\,\,\,\,\,\,A=Pe^{rt}$

$\,\,\,\,\,\,A=7498e^{.042\cdot 16}$

$\textbf{10)}$ Kelsey invests $3,225 in a savings account with an annual interest rate of 7% compounded continuously. What will the account balance be after 12 years?
Show Answer
The answer is $ $7{,}470.28 $
Show Work

$\,\,\,\,\,\,A=Pe^{rt}$

$\,\,\,\,\,\,A=3{,}225\:\:e^{.07\cdot 12}$

$\textbf{11)}$ Ashton invests a sum of money in an account with an annual interest rate of 6.32% compounded monthly. After 6 years, the balance reaches $28,449.75. What was the amount of the initial investment?
Show Answer
The answer is $ $19{,}490.64 $
Show Work

$\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}$

$\,\,\,\,\,\,28449.75=P\left(1+\frac{0.0632}{12}\right)^{\left(12\right)\left(6\right)}$

$\,\,\,\,\,\,P=\displaystyle \frac{28449.75}{\left(1+\frac{0.0632}{12}\right)^{\left(12\right)\left(6\right)}}$

$\textbf{12)}$ Izzy invests a sum of money in an account with an annual interest rate of 3.51% compounded weekly. After 8 years, the balance reaches $10,376.23. What was the amount of the initial investment?
Show Answer
The answer is $ $7{,}836.66 $
Show Work

$\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}$

$\,\,\,\,\,\,10376.24=P\left(1+\frac{.0351}{52}\right)^{\left(52\right)\left(8\right)}$

$\,\,\,\,\,\,P=\displaystyle \frac{10376.24}{\left(1+\frac{.0351}{52}\right)^{\left(52\right)\left(8\right)}}$

$\textbf{13)}$ Jen invests a sum of money in an account with a fixed annual interest rate of 3.45% compounded continuously. After 16 years, the balance reaches $4,567.89. What was the amount of the initial investment?
Show Answer
The answer is $ $2{,}630.18 $
Show Work

$\,\,\,\,\,\,A=Pe^{rt}$

$\,\,\,\,\,\,4567.89=Pe^{.0345\cdot 16}$

$\,\,\,\,\,\,P=\displaystyle \frac{4567.89}{e^{.0345\cdot 16}}$

$\textbf{14)}$ Jason invests a sum of money in an account with an annual interest rate of 7.28% compounded continuously. After 11 years, the balance reaches $18,153.16. What was the amount of the initial investment?
Show Answer
The answer is $ $8{,}150.22 $
Show Work

$\,\,\,\,\,\,A=Pe^{rt}$

$\,\,\,\,\,\,18153.16=Pe^{.0728\cdot 11}$

$\,\,\,\,\,\,P=\displaystyle \frac{18153.16}{e^{.0728\cdot 11}}$

$\textbf{15)}$ David invests $4,495 in an account with a fixed annual interest rate compounded continuously. After 15 years, the balance reaches $9,515.92. What is the interest rate of the account?
Show Answer
The answer is $ 5\% $
Show Work

$\,\,\,\,\,\,A=Pe^{rt}$

$\,\,\,\,\,\,9515.92=4495e^{r\cdot 15}$

$\,\,\,\,\,\,e^{15r}=\frac{9515.92}{4495}$

$\,\,\,\,\,\,15r=\ln{\frac{9515.92}{4495}}$

$\,\,\,\,\,\,r=\displaystyle \frac{\ln{\frac{9515.92}{4495}}}{15}$

$\,\,\,\,\,\,r=.05$

$\textbf{16)}$ Olivia invests $2,854 in an account with an annual interest rate of 6.5% compounded continuously. How long will it take for the account balance to reach $5,122.90?
Show Answer
The answer is $ 9 $ years
Show Work

$\,\,\,\,\,\,A=Pe^{rt}$

$\,\,\,\,\,\,5122.90=2854e^{.065\cdot t}$

$\,\,\,\,\,\,e^{.065 t}=\frac{5122.9}{2854}$

$\,\,\,\,\,\,.065 t=\ln{\frac{5122.9}{2854}}$

$\,\,\,\,\,\, t=\displaystyle \frac{\ln{\frac{5122.9}{2854}}}{0.065}$

$\,\,\,\,\,\, t=9$

$\textbf{17)}$ Mike invested P dollars. The value of his invested decreased by 5.5% each year. At the end of the first year, the value of Mike’s investment was $5670. What was P?
Show Answer
The answer is $ $6000 $

$\textbf{18)}$ What if your ancestor put $1 in the bank in the year 0 at 2% interest compounded monthly. What would it be worth today?
Show Answer
The answer is approximately $ $346,394,275,535,600,000. $

See Related Pages

$\bullet\text{ Exponential Functions}$
$\,\,\,\,\,\,\,\,y=a(b)^x$

$\bullet\text{ Half Life Questions}$
$\,\,\,\,\,\,\,\,A_t=A_0e^{kt}$

$\bullet\text{ Graphing Exponential Functions}$
$\,\,\,\,\,\,\,\,f(x)=2^{x-1}-2\,\, $

$\bullet\text{ Inverse of Exponential Functions}$
$\,\,\,\,\,\,\,\,f^{-1}(x)=\log_5(x-1)$

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Compound Interest

Notes

Compound Interest Formula

\(A=P\left(1+\frac{r}{n} \right)^{nt}\)

\(A=\)Amount at time t
\(P=\)Principle (amount at time t=0)
\(n=\)compoundings per year
\(r=\)annual interest rate (\(3\%=.03\) for example)
\(t=\)time in years

Values of \(n\)

Monthly \( \rightarrow n=12\)
Quarterly \( \rightarrow n=4\)
Daily \( \rightarrow n=365\)
Weekly \( \rightarrow n=52\)
Yearly \( \rightarrow n=1\)
Annually \( \rightarrow n=1\)
Semi-annually \( \rightarrow n=2\)

Continuously Compounded Interest Formula

\(A=P e^{rt}\)

Interest Earned

\(I=A-P\)

Problems & Videos

\(\textbf{9)}\) Steph invests $7,498 in an account with an annual interest rate of 4.2% compounded continuously. What will the account balance be after 16 years?
Show Answer
The answer is \( $14{,}682.21 \)
Show Work

\(\,\,\,\,\,\,A=Pe^{rt}\)

\(\,\,\,\,\,\,A=7498e^{.042\cdot 16}\)

\(\textbf{10)}\) Kelsey invests $3,225 in a savings account with an annual interest rate of 7% compounded continuously. What will the account balance be after 12 years?
Show Answer
The answer is \( $7{,}470.28 \)
Show Work

\(\,\,\,\,\,\,A=Pe^{rt}\)

\(\,\,\,\,\,\,A=3{,}225\:\:e^{.07\cdot 12}\)

\(\textbf{17)}\) Mike invested P dollars. The value of his invested decreased by 5.5% each year. At the end of the first year, the value of Mike’s investment was $5670. What was P?
Show Answer
The answer is \( $6000 \)

\(\textbf{18)}\) What if your ancestor put $1 in the bank in the year 0 at 2% interest compounded monthly. What would it be worth today?
Show Answer
The answer is approximately \( $346,394,275,535,600,000. \)

See Related Pages\(\)

\(\bullet\text{ Exponential Functions}\)
\(\,\,\,\,\,\,\,\,y=a(b)^x\)

\(\bullet\text{ Half Life Questions}\)
\(\,\,\,\,\,\,\,\,A_t=A_0e^{kt}\)

\(\bullet\text{ Graphing Exponential Functions}\)
\(\,\,\,\,\,\,\,\,f(x)=2^{x-1}-2\,\, \)

\(\bullet\text{ Inverse of Exponential Functions}\)
\(\,\,\,\,\,\,\,\,f^{-1}(x)=\log_5(x-1)\)

About Andymath.com

Notes

Compound Interest Formula \(A=P\left(1+\frac{r}{n} \right)^{nt}\) \(A=\)Amount at time t \(P=\)Principle (amount at time t=0) \(n=\)compoundings per year \(r=\)annual interest rate (\(3\%=.03\) for example) \(t=\)time in years

Values of \(n\) Monthly \( \rightarrow n=12\) Quarterly \( \rightarrow n=4\) Daily \( \rightarrow n=365\) Weekly \( \rightarrow n=52\) Yearly \( \rightarrow n=1\) Annually \( \rightarrow n=1\) Semi-annually \( \rightarrow n=2\)

Continuously Compounded Interest Formula \(A=P e^{rt}\)

Interest Earned \(I=A-P\)

Problems & Videos

\(\textbf{9)}\) Steph invests $7,498 in an account with an annual interest rate of 4.2% compounded continuously. What will the account balance be after 16 years? Show AnswerThe answer is \( $14{,}682.21 \)Show Work \(\,\,\,\,\,\,A=Pe^{rt}\) \(\,\,\,\,\,\,A=7498e^{.042\cdot 16}\)

\(\textbf{10)}\) Kelsey invests $3,225 in a savings account with an annual interest rate of 7% compounded continuously. What will the account balance be after 12 years? Show AnswerThe answer is \( $7{,}470.28 \)Show Work \(\,\,\,\,\,\,A=Pe^{rt}\) \(\,\,\,\,\,\,A=3{,}225\:\:e^{.07\cdot 12}\)

\(\textbf{17)}\) Mike invested P dollars. The value of his invested decreased by 5.5% each year. At the end of the first year, the value of Mike’s investment was $5670. What was P? Show AnswerThe answer is \( $6000 \)

\(\textbf{18)}\) What if your ancestor put $1 in the bank in the year 0 at 2% interest compounded monthly. What would it be worth today?Show AnswerThe answer is approximately \( $346,394,275,535,600,000. \)

See Related Pages\(\)

\(\bullet\text{ Exponential Functions}\) \(\,\,\,\,\,\,\,\,y=a(b)^x\)

\(\bullet\text{ Half Life Questions}\) \(\,\,\,\,\,\,\,\,A_t=A_0e^{kt}\)

\(\bullet\text{ Graphing Exponential Functions}\) \(\,\,\,\,\,\,\,\,f(x)=2^{x-1}-2\,\, \)

\(\bullet\text{ Inverse of Exponential Functions}\) \(\,\,\,\,\,\,\,\,f^{-1}(x)=\log_5(x-1)\)

About Andymath.com

Compound Interest Formula

\(A=P\left(1+\frac{r}{n} \right)^{nt}\)

\(A=\)Amount at time t
\(P=\)Principle (amount at time t=0)
\(n=\)compoundings per year
\(r=\)annual interest rate (\(3\%=.03\) for example)
\(t=\)time in years

Values of \(n\)

Monthly \( \rightarrow n=12\)
Quarterly \( \rightarrow n=4\)
Daily \( \rightarrow n=365\)
Weekly \( \rightarrow n=52\)
Yearly \( \rightarrow n=1\)
Annually \( \rightarrow n=1\)
Semi-annually \( \rightarrow n=2\)

Continuously Compounded Interest Formula

\(A=P e^{rt}\)

Interest Earned

\(I=A-P\)

\(\textbf{9)}\) Steph invests $7,498 in an account with an annual interest rate of 4.2% compounded continuously. What will the account balance be after 16 years?
Show Answer
The answer is \( $14{,}682.21 \)
Show Work

\(\,\,\,\,\,\,A=Pe^{rt}\)

\(\,\,\,\,\,\,A=7498e^{.042\cdot 16}\)

\(\textbf{10)}\) Kelsey invests $3,225 in a savings account with an annual interest rate of 7% compounded continuously. What will the account balance be after 12 years?
Show Answer
The answer is \( $7{,}470.28 \)
Show Work

\(\,\,\,\,\,\,A=Pe^{rt}\)

\(\,\,\,\,\,\,A=3{,}225\:\:e^{.07\cdot 12}\)

\(\textbf{17)}\) Mike invested P dollars. The value of his invested decreased by 5.5% each year. At the end of the first year, the value of Mike’s investment was $5670. What was P?
Show Answer
The answer is \( $6000 \)

\(\textbf{18)}\) What if your ancestor put $1 in the bank in the year 0 at 2% interest compounded monthly. What would it be worth today?
Show Answer
The answer is approximately \( $346,394,275,535,600,000. \)

\(\bullet\text{ Exponential Functions}\)
\(\,\,\,\,\,\,\,\,y=a(b)^x\)

\(\bullet\text{ Half Life Questions}\)
\(\,\,\,\,\,\,\,\,A_t=A_0e^{kt}\)

\(\bullet\text{ Graphing Exponential Functions}\)
\(\,\,\,\,\,\,\,\,f(x)=2^{x-1}-2\,\, \)

\(\bullet\text{ Inverse of Exponential Functions}\)
\(\,\,\,\,\,\,\,\,f^{-1}(x)=\log_5(x-1)\)