Compound Interest 2

Question 1 of 5

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

7 %

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

4

$4$ years?

Round your answer to two decimal places

$total amount = $$ $\text(total amount)=$$ (3968.64)

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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 $= n = 52 (Weekly)$ $= n = 52 \text((Weekly))$

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

	$=$ $=$	$\color\green{3{,}000}\left(1+\frac{\color{blue}{0.07}}{\color{#a300c1}{52}}\right)^{\large\color{#a300c1}{52}\times\color{#e85e00}{4}}$	Substitute known values

	$=$ $=$	$(3{,}000)(1+0.00134615384)^{208}$
	$=$ $=$	$(3{,}000)(1.3228807)$	Simplify
	$=$ $=$	$3{,}968.64$	Round to two decimal places

$ 3,968.64

$$3{,}968.64$ total amount

Question 2 of 5

An amount of

$ 375

$$375$ invested for

2

$2$ years grows to

$ 431.35

$$431.35$ . The amount is compounded annually. What is the interest rate?

Round your answer to two decimal places

$Interest Rate =$ $\text(Interest Rate )=$ (7.25) $%$ $%$

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Compound Interest Formula

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

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

Label Known Values

Amount Earned

= A = $ 431.35

$= A = $431.35$

Principal $= P = $ 375$ $= P = $375$

Number of Years $= n = 2$ $= n = 2$ years

Solve for the Interest Rate $(r)$ $(r)$

$A$ $A$	$=$ $=$	$\color\green{P}(1+\color{blue}{r})^{\large\color{#a300c1}{n}}$	Compound Interest Formula
$431.35$ $431.35$	$=$ $=$	$\color\green{375}(1+\color{blue}{r})^{\large\color{#a300c1}{2}}$	Substitute known values

$\frac{431.35}{375}$ $(431.35)/(375)$	$=$ $=$	${(1 + r)}^{2}$ $(1+r)^2$	Divide both sides by $375$ $375$

$1.15$ $1.15$	$=$ $=$	${(1 + r)}^{2}$ $(1+r)^2$
$1.0725$ $1.0725$	$=$ $=$	$1 + r$ $1+r$	Get square root of both sides
$0.0725$ $0.0725$	$=$ $=$	$r$ $r$	Subtract $1$ $1$ from both sides
$r$ $r$	$=$ $=$	$0.0725$ $0.0725$
$r$ $r$	$=$ $=$	$0.0725$ $0.0725$ $\times 100 %$ $xx100%$	Multiply by $100 %$ $100%$
$r$ $r$	$=$ $=$	$7.25 %$ $7.25%$

$7.25%$ interest rate

Question 3 of 5

How much money would you need to deposit today at an interest rate of

$7.5%$ compounded annually to have $$65{,}000$ in the account after

$8$ years?

Round your answer to two decimal places

$\text(Principal)=$$ (36445.65)

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Compound Interest Formula

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

Label Known Values

Amount Earned

$= A = $65{,}000$

Rate $= r = 0.075$

Number of Years $= n = 8 \text(years)$

Solve for the Principal Amount $(P)$

$A$	$=$	$\color\green{P}(1+\color{blue}{r})^{\large\color{#a300c1}{n}}$	Compound Interest Formula
$65{,}000$	$=$	$\color\green{P}(1+\color{blue}{0.075})^{\large\color{#a300c1}{8}}$	Substitute known values
$65{,}000$	$=$	$P(1.075)^8$	Evaluate

$\frac{65{,}000}{1.075^8}$	$=$	$P$	Divide both sides by $1.075^8$

$\frac{65{,}000}{1.783477826}$	$=$	$P$

$36{,}445.65$	$=$	$P$
$P$	$=$	$36{,}445.65$

$\text{Principal}=$36{,}445.65$

Question 4 of 5

How much money would you need to deposit today at an interest rate of

$6%$ compounded annually to have $$30{,}000$ in the account after

$5$ years?

Round your answer to two decimal places

$\text(Principal)=$$ (22417.75)

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Compound Interest Formula

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

Label Known Values

Amount Earned

$= A = $30{,}000$

Rate $= r = 0.06$

Number of years $= n = 5$ years

Solve for the Principal Amount

$A$	$=$	$\color\green{P}(1+\color{blue}{r})^{\large\color{#a300c1}{n}}$	Compound Interest Formula
$30{,}000$	$=$	$\color\green{P}(1+\color{blue}{0.06})^{\large\color{#a300c1}{5}}$	Substitute known values
$30{,}000$	$=$	$P(1.06)^5$	Evaluate

$\frac{30{,}000}{1.06^5}$	$=$	$P$	Divide both sides by $1.06^5$

$\frac{30{,}000}{1.338225578}$	$=$	$P$

$22{,}417.75$	$=$	$P$
$P$	$=$	$22{,}417.75$

$\text{Principal}=$22{,}417.75$

Question 5 of 5

Two banks receive an investment of

$$20 000$ each with an interest rate of

$7%$ per annum. If one bank uses simple interest and the other bank uses compound interest, what will be the difference in the amount (

$A$ ) after a period of

$3$ years?

$$$ (300.86)

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Simple Interest Formula

$I =$ $P$ $r$ $t$

Compound Interest Formula

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

Label Known Values

Principal Value $= P = $20{,}000$

Interest Rate $= r = 7%$ per annum $= 0.07$ as a decimal

Time Elapsed $= t = 3$ years

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

To understand the concept Simple and Compound interest, let us first try to solve this problem without any formulas.

Simple Interest:

The principal value where the interest is applied stays the same every period.

First year:

		$20 000+(0.07xx20 000)$
	$=$	$20 000+1400$
	$=$	$21 400$

Second year:

		$21 400+(0.07xx20 000)$
	$=$	$21 400+1400$
	$=$	$22 800$

Third year:

		$22 800+(0.07xx20 000)$
	$=$	$22 800+1400$
	$=$	$24 200$

Compound Interest:

The principal value where the interest is applied compounds every period.

First year:

		$20 000+(0.07xx20 000)$
	$=$	$20 000+1400$
	$=$	$21 400$

Second year:

		$21 400+(0.07xx21 400)$
	$=$	$21 400+1498$
	$=$	$22 898$

Third year:

		$22 898+(0.07xx22 898)$
	$=$	$22 898+1602.86$
	$=$	$24 500.86$

Therefore, the difference between the two amounts is

$24 500.86-24 200=$ $300.86$

This time, let’s see more of how Simple and Compound Interest work by using their formulas to solve the problem.

Simple Interest:

$I$	$=$	$P$ $r$ $t$
	$=$	$20 000$ $xx$ $0.07$ $xx$ $3$
	$=$	$4200$

This is the interest earned over the

$3$ years, not the final amount.

$A$ (Final amount)	$=$	$P+I$
	$=$	$20 000+4200$
	$=$	$24 200$

Compound Interest:

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

	$=$	$\color\green{20\:000}\left(1+\frac{\color{blue}{(0.07)}}{\color{#a300c1}{(1)}}\right)^{\large\color{#a300c1}{(1)}\color{#e85e00}{(3)}}$

	$=$	$20 000(1+0.07)^3$
	$=$	$20 000(1.07^3)$
	$=$	$20 000xx1.225043$
	$=$	$24 500.86$

We can confirm again, through the formulas, that the difference between the two amounts is

$24 500.86-24 200=$ $300.86$

$$300.86$

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Quiz summary

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1. Question

Hint

Great Work!

Compound Interest Formula

Label Known Values

2. Question

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Nice Job!

Compound Interest Formula

Label Known Values

3. Question

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Fantastic!

Compound Interest Formula

Label Known Values

4. Question

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Correct!

Compound Interest Formula

Label Known Values

5. Question

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Well Done!

Simple Interest Formula

Compound Interest Formula

Label Known Values

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