Compound Interest 1

Question 1 of 5

Timothy opened a savings account and deposited $$3{,}000$ . The account earns

7 %

$7%$ interest compounded yearly. What is the total amount he will have in the bank after

4

$4$ years?

Round your answer to two decimal places

$total amount = $$ $\text(total amount)=$$ (3932.39)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Compound Interest Formula

A = P {(1 + \frac{r}{n})}^{n t}

$A=\color\green{P}\left(1+\frac{\color{blue}{r}}{\color{#a300c1}{n}}\right)^{\large\color{#a300c1}{n}\color{#e85e00}{t}}$

Label Known Values

Principal Value $= P = $3{,}000$

Interest Rate $= r = 7 % per annum = 0.07 (as a decimal)$ $= r = 7% \text(per annum) = 0.07 \text((as a decimal))$

Time Elapsed $= t = 4$ $= t = 4$ years

Number of times compounded per year $= n = 1 (Annually)$ $= n = 1 \text((Annually))$

Solve for the Amount

(A),

$(A),$ using the formula

$A$ $A$	$=$ $=$	$\color\green{P}\left(1+\frac{\color{blue}{r}}{\color{#a300c1}{n}}\right)^{\large\color{#a300c1}{n}\color{#e85e00}{t}}$	Compound Interest Formula

	$=$ $=$	$\color\green{3{,}000}\left(1+\frac{\color{blue}{0.07}}{\color{#a300c1}{1}}\right)^{\large\color{#a300c1}{1}\times\color{#e85e00}{4}}$	Substitute known values

	$=$ $=$	$(3{,}000)(1.07)^4$	Simplify
	$=$ $=$	$(3{,}000)(1.31079601)$
	$=$ $=$	$3{,}932.39$	Round to two decimal places

$ 3,932.39

$$3{,}932.39$ total amount

Question 2 of 5

If you deposit $$9{,}300$ into an account earning

4.5 %

$4.5%$ interest compounded annually, how much money will be in the account after

3

$3$ years?

Round your answer to two decimal places

$total amount = $$ $\text(total amount)=$$ (10612.84)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Compound Interest Formula

A = P {(1 + \frac{r}{n})}^{n t}

$A=\color\green{P}\left(1+\frac{\color{blue}{r}}{\color{#a300c1}{n}}\right)^{\large\color{#a300c1}{n}\color{#e85e00}{t}}$

Label Known Values

Principal Value $= P = $9{,}300$

Interest Rate $= r = 4.5 % per annum = 0.045 (as a decimal)$ $= r = 4.5% \text(per annum) = 0.045 \text((as a decimal))$

Time Elapsed $= t = 3$ $= t = 3$ years

Number of times compounded per year $= n = 1 (Annually)$ $= n = 1 \text((Annually))$

Solve for the Amount

(A),

$(A),$ using the formula

$A$ $A$	$=$ $=$	$\color\green{P}\left(1+\frac{\color{blue}{r}}{\color{#a300c1}{n}}\right)^{\large\color{#a300c1}{n}\color{#e85e00}{t}}$	Compound Interest Formula

	$=$ $=$	$\color\green{$9{,}300}\left(1+\frac{\color{blue}{0.045}}{\color{#a300c1}{1}}\right)^{\large\color{#a300c1}{1}\times\color{#e85e00}{3}}$	Substitute known values

	$=$ $=$	$(9{,}300)(1.045)^3$	Simplify
	$=$ $=$	$(9{,}300)(1.141166125)$
	$=$ $=$	$10{,}612.84$	Round to two decimal places

$ 10,612.84

$$10{,}612.84$ total amount

Question 3 of 5

If you deposit $$3{,}000$ into an account earning

7 %

$7%$ interest compounded quarterly, how much money will be in the account after

4

$4$ years?

Round your answer to two decimal places

$total amount = $$ $\text(total amount)=$$ (3959.79)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Compound Interest Formula

A = P {(1 + \frac{r}{n})}^{n t}

$A=\color\green{P}\left(1+\frac{\color{blue}{r}}{\color{#a300c1}{n}}\right)^{\large\color{#a300c1}{n}\color{#e85e00}{t}}$

Label Known Values

Principal Value $= P = $3{,}000$

Interest Rate $= r = 7 % per annum = 0.07 (as a decimal)$ $= r = 7% \text(per annum) = 0.07 \text((as a decimal))$

Time Elapsed $= t = 4$ $= t = 4$ years

Number of times compounded per year $= n = 4 (Quarterly)$ $= n = 4 \text((Quarterly))$

Solve for the Amount

(A),

$(A),$ using the formula

$A$ $A$	$=$ $=$	$\color\green{P}\left(1+\frac{\color{blue}{r}}{\color{#a300c1}{n}}\right)^{\large\color{#a300c1}{n}\color{#e85e00}{t}}$	Compound Interest Formula

	$=$ $=$	$\color\green{3{,}000}\left(1+\frac{\color{blue}{0.07}}{\color{#a300c1}{4}}\right)^{\large\color{#a300c1}{4}\times\color{#e85e00}{4}}$	Substitute known values

	$=$ $=$	$(3{,}000)(1.0175)^{16}$	Simplify
	$=$ $=$	$(3{,}000)(1.3109929351)$
	$=$ $=$	$3{,}959.79$	Round to two decimal places

$ 3,959.79

$$3{,}959.79$ total amount

Question 4 of 5

If you deposit $$5{,}400$ into an account earning

6 %

$6%$ interest compounded quarterly, how much money will be in the account after

5

$5$ years?

Round your answer to two decimal places

$total amount = $$ $\text(total amount)=$$ (7273.02)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Compound Interest Formula

A = P {(1 + \frac{r}{n})}^{n t}

$A=\color\green{P}\left(1+\frac{\color{blue}{r}}{\color{#a300c1}{n}}\right)^{\large\color{#a300c1}{n}\color{#e85e00}{t}}$

Label Known Values

Principal Value $= P = $5{,}400$

Interest Rate $= r = 6 % per annum = 0.06 (as a decimal)$ $= r = 6% \text(per annum) = 0.06 \text((as a decimal))$

Time Elapsed $= t = 5$ $= t = 5$ years

Number of times compounded per year $= n = 4 (Quarterly)$ $= n = 4 \text((Quarterly))$

Solve for the Amount

(A),

$(A),$ using the formula

$A$ $A$	$=$ $=$	$\color\green{P}\left(1+\frac{\color{blue}{r}}{\color{#a300c1}{n}}\right)^{\large\color{#a300c1}{n}\color{#e85e00}{t}}$	Compound Interest Formula

	$=$ $=$	$\color\green{5{,}400}\left(1+\frac{\color{blue}{0.06}}{\color{#a300c1}{4}}\right)^{\large\color{#a300c1}{4}\times\color{#e85e00}{5}}$	Substitute known values

	$=$ $=$	$(5{,}400)(1.015)^{20}$	Simplify
	$=$ $=$	$(5{,}400)(1.4346855007)$
	$=$ $=$	$7{,}273.02$	Round to two decimal places

$ 7,273.02

$$7{,}273.02$ total amount

Question 5 of 5

If you deposit $$3{,}000$ into an account earning

7 %

$7%$ interest compounded monthly, how much money will be in the account after

4

$4$ years?

Round your answer to two decimal places

$total amount = $$ $\text(total amount)=$$ (3966.16)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Compound Interest Formula

A = P {(1 + \frac{r}{n})}^{n t}

$A=\color\green{P}\left(1+\frac{\color{blue}{r}}{\color{#a300c1}{n}}\right)^{\large\color{#a300c1}{n}\color{#e85e00}{t}}$

Identify the given values

Principal Value $= P = $3{,}000$

Interest Rate $= r = 7 % per annum = 0.07 (as a decimal)$ $= r = 7% \text(per annum) = 0.07 \text((as a decimal))$

Time Elapsed $= t = 4$ $= t = 4$ years

Number of times compounded per year $= n = 12 (Monthly)$ $= n = 12 \text((Monthly))$

Solve for the Amount

(A),

$(A),$ using the formula

$A$ $A$	$=$ $=$	$\color\green{P}\left(1+\frac{\color{blue}{r}}{\color{#a300c1}{n}}\right)^{\large\color{#a300c1}{n}\color{#e85e00}{t}}$	Compound Interest Formula

	$=$ $=$	$\color\green{3{,}000}\left(1+\frac{\color{blue}{0.07}}{\color{#a300c1}{12}}\right)^{\large\color{#a300c1}{12}\times\color{#e85e00}{4}}$	Plug in the known values

	$=$ $=$	$(3{,}000)(1.005833)^{48}$
	$=$ $=$	$(3{,}000)(1.327053878)$
	$=$ $=$	$3{,}966.16$	Round to two decimal places

$ 3,966.16

$$3{,}966.16$ total amount

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Nice Job!

Compound Interest Formula

Label Known Values

2. Question

Hint

Excellent!

Compound Interest Formula

Label Known Values

3. Question

Hint

Fantastic!

Compound Interest Formula

Label Known Values

4. Question

Hint

Keep Going!

Compound Interest Formula

Label Known Values

5. Question

Hint

Correct!

Compound Interest Formula

Identify the given values

Quizzes