Compound Interest Formula

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Compound interest formula

Here you will learn about the compound interest formula, including what it is and how to use it.

Students will first learn about the compound interest formula as part of algebra in high school.

What is the compound interest formula?

The compound interest formula is calculated on the principal (original) amount and the interest already accumulated on previous periods.

Take, for example, the amount of money in a savings account.

If you put $\$ 100$ in an account with an annual interest rate of $10 \%,$ the value of the money in the account will increase by $10 \%$ in year one; the new amount will be $\$ 110$ or $110 \%$ of the original.

If a further $10 \%$ was taken for year two, the new amount would be $\$ 121$ or $121 \%$ of the original value, since you are finding $10 \%$ of $\$ 110.$

Compound interest is interest calculated on top of the original amount including any interest accumulated so far. The compound interest formula is:

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

where:

$A$ represents the final amount
$P$ represents the initial principal amount
$r$ is the interest amount (written as a decimal)
$n$ represents the number of times the interest rate is applied over time $t$ (compounding frequency)
$t$ represents the time period (typically number of years)

For example,

Let’s calculate a $3 \%$ increase on an amount $P$ using compound interest over $4$ years. The interest amount written as a decimal is $0.03.$

So, $P \times 0.03$ is the value of the interest only. Add this value to $P$ to find the new amount after the first interval.

This is written as:

\begin{aligned}A&=P+P\times{0.03} \\\\ &=P(1+0.03) \\\\ &=P\times{1.03} \end{aligned}

This is the value after the first interval.

The second interval is the new amount, multiplied by a further $1.03$ and so on for the amount of intervals. This can be expressed using a table.

After $t$ years, the principal amount $P$ is multiplied by the percentage change for $t$ times. This is why the power is expressed as $t.$

So for this example, multiplying $P$ by $1.03^{4}$ would calculate an increase of $3 \%$ compound interest for $4$ years.

What is the compound interest formula?

Common Core State Standards

How does this relate to high school math?

Algebra – Seeing Structure in Expressions (HS.A.SSE.B.3c)
Use the properties of exponents to transform expressions for exponential functions. For example, the expression $1.151t$ can be rewritten as $(1.15/12)12t\approx{1.01212t}$ to reveal the approximate equivalent monthly interest rate if the annual rate is $15 \%.$

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How to use the compound interest formula

In order to use the compound interest formula:

State the formula and the value of each variable.
Substitute the values into the formula.
Solve the equation.

Compound interest examples

Example 1: compound interest – percentage increase

$\$ 8,500$ is invested for $5$ years at $0.3 \%$ per year compound interest. What is the value of the investment after this time?

State the formula and the value of each variable.

Here we use the formula $A=P\left(1+\cfrac{r}{n}\right)^{nt}$ with:

$P=8,500$
$r=0.003$
$n=1$
$t=5$

2Substitute the values into the formula.

Substituting these values into the compound interest formula $A=P\left(1+\cfrac{r}{n}\right)^{nt},$ you get:

A=8500(1+\cfrac{0.003}{1})^{1\times{5}}

3Solve the equation.

\begin{aligned}A&=8500(1+0.003)^5 \\\\ &=8500\times1.003^5 \\\\ &=8500\times1.01509027 \\\\ &=\$8628.27\text{ (2dp)} \end{aligned}

Example 2: compound interest – percentage decrease

A vacuum cleaner is bought for $\$ 399$ and loses $22 \%$ of its value per year compound interest. What is the value of the vacuum cleaner after $2$ years?

State the formula and the value of each variable.

Here we use the formula $A=P\left(1+\cfrac{r}{n}\right)^{nt}$ with:

$P=399$
$r=- \, 0.22$
$n=1$
$t=2$

Substitute the values into the formula.

Substituting these values into the compound interest formula $A=P\left(1+\cfrac{r}{n}\right)^{nt},$ you get:

$A=399\left(1-\cfrac{0.22}{1}\right)^{1\times{2}}$

Solve the equation.

\begin{aligned}A&=399(1-0.22)^{2} \\\\ &=399(0.78)^{2} \\\\ &=399\times0.6084 \\\\ &=\$242.75\text{ (2dp)} \end{aligned}

Example 3: compound interest – compounding periods

A house is valued at $\$ 150,000.$ On average, the house price increases by $2.4 \%$ monthly compounding over $2.5$ years. What is the future value of the house after this time?

State the formula and the value of each variable.

Here we use the formula $A=P\left(1+\cfrac{r}{n}\right)^{nt}$ with:

$P=150,000$
$r=0.024$
$n=12$
$t=2.5$

Substitute the values into the formula.

Substituting these values into the compound interest formula $A=P\left(1+\cfrac{r}{n}\right)^{nt},$ you get:

$A=150000\left(1+\cfrac{0.024}{12}\right)^{12\times{2.5}}$

Notice that the exponent is now $12 \times 2.5$ as the number of months over $2.5$ years $=12\times{2.5}.$

Solve the equation.

\begin{aligned}A&=150000(1+0.002)^{30} \\\\ &=150000(1.002)^{30} \\\\ &=150000\times1.06177 \\\\ &=\$159265.94 \; (2dp) \end{aligned}

Example 4: compound interest – different percentages

A painting is valued at $\$ 2950.$ On average, the value increases by $0.8 \%$ per year for $7$ years compound interest. After this time, the rate of interest is $0.3 \%$ for a further $3$ years. What is the value of the antique after this time?

State the formula and the value of each variable.

Here we use the formula $A=P\left(1+r_{1}\right)^{t_{1}}\left(1+r_{2}\right)^{t_{2}}$ as the time scale $n=1$ with:

$P=2950$
$r_{1}=0.008$
$t_{1}=7$
$r_{2}=0.003$
$t_{2}=3$

Substitute the values into the formula.

Substituting these values into the compound interest formula $A=P(1+r_1)^{t_1}(1+r_2)^{t_2},$ you get:

$A=2,950(1+0.008)^{7}(1+0.003)^{3}$

Solve the equation.

\begin{aligned}A&=2950\times(1.008^7)\times(1.003^3) \\\\ &=2950\times{1.057}\times{1.009} \\\\ &=\$3146.30 \end{aligned}

Example 5: compound interest – writing an answer in standard form

Bacteria reproduce at a rate of $3 \%$ per hour compound interest. If there are $1.2\times{10^7}$ bacteria in a petri dish at hour $0,$ how many bacteria are in the petri dish after $4$ hours?

State the formula and the value of each variable.

Use the formula $A=P(1+r)^{t}$ as the time scale $n=1$ with:

$P=1.2\times{10^7}$
$r=0.03$
$t=4$

Substitute the values into the formula.

Substituting these values into the compound interest formula $A=P(1+r)^{t},$ you get:

$A=(1.2\times{10}^{7})(1+0.03)^{4}$

Solve the equation.

\begin{aligned}A&=\left(1.2\times{10^7}\right)\times{1.03^4} \\\\ &=1.35\times{10^7}\text{ (3sf)} \end{aligned}

Example 6: compound interest – best buy

Two offers exist for the same car.

A car has a value of $\$ 7999.$ Which offers a better value over the period of time stated?

State the formula and the value of each variable.

Here you are calculating compound interest with the formula $A=P(1+r)^{t}$ twice to compare each offer:

Substitute the values into the formula.

Substituting these values into the simple interest formula $A=P(1+r)^{t},$ you get:

Solve the equation.

The best offer is Offer $B,$ because the total amount paid is less.

Teaching tips for compound interest formula

Introduce the topic through real-world examples, including examples that do not involve money. Start off with examples that have compounding frequency of $1$ and then work up to more complex problems.

Be mindful of specific vocabulary words that could keep students from understanding the context, such as ‘semi-annually’, ‘withdrawls’, ‘principal investment’, ‘time interval’ or others that often appear on worksheets.

Easy mistakes to make

Applying the incorrect formula to the question
This is a very common mistake where the simple interest on an amount is calculated instead of using the compound interest formula.

Incorrect percent change due to different time scales
Here is example 3 with this misconception:
A house is valued at \$ 150,000. On average, the house price increases by 2.4 \% per month over a period of 2.5 years compound interest. What is the value of the house after this time?

Here we use the formula A=P(1+\cfrac{r}{n})^{nt} with:
- $P=150,000$
- $r=0.024$
- $n=1$
- $t=2.5$
  
  $\begin{aligned} A&=150000(1.024)^{2.5} \\\\ &=\$159,162.65 \end{aligned}$
  
  Although the answer here looks reasonable, the $2.4 \%$ interest is per month not per year, which is what has been calculated.

Using the percent as the value for $\textbf{r}$
The value for $r$ is the percent divided by $100,$ so it is in decimal form.
Forgetting to convert to a decimal, leads to the incorrect answer.

For example,
When using compound interest to increase $\$ 100$ by $2 \%$ for $5$ years, this calculation is made:

$\begin{aligned}A&=100(1+2)^{5} \\\\ &=100\times243 \\\\ &=\$24,300 \end{aligned}$

Not considering if the value is increasing or decreasing
If a value is depreciating (going down), the value of $r$ is negative. If a value is increasing (going up), the value of $r$ is positive. Confusing this leads to an incorrect answer.

Compound interest practice questions

5 \, g

12.7 \, g

47 \, g

8 \, g

You can use the compound interest formula, since the soap always loses $3 \%$ of its weight, which is the compound interest rate.

$50\times(1-0.03)^{45}=12.7$ (rounded to the nearest tenth)

5,401,788\mathrm{~km}^2

1,100,000\mathrm{~km}^2

4,400,000\mathrm{~km}^2

5,390,000\mathrm{~km}^2

You can use the compound interest formula, since the rate of deforestation is always $0.2 \%.$ Deforestation is the loss of trees, so $r$ is negative and from $2021$ to $2030$ is $9$ years.

$5,500,000\times(1-0.002)^{9}=5,401,788$ (rounded to the nearest whole)

\$ 230.67

\$ 18.93

\$ 249.60

\$ 480.27

$40$ months $=3$ years and $4$ months

$3,000 \times(1+0.025)^{3}-3000=230.67$ (rounded to the nearest hundredth)

Account A has accumulated $\$ 230.67$ in interest after $3$ years ( $4$ months not included as paid annually).

$3,000\times(1+0.002)^{40}-3000=249.60$ (rounded to the nearest hundredth)

After $40$ months Account $B$ has accumulated $\$ 249.60$ in interest.

\$ 249.60-\$ 230.67=\$ 18.93

$2$ hours

$3$ hours

$4$ hours

$5$ hours

Let’s use $b$ for the amount of water in the bucket.

After $1$ hour the bucket is $83 \%$ full.

b\times(1-0.17)^{1}=0.83b

After $2$ hours the bucket is $68.89 \%$ full.

b\times(1-0.17)^{2}=0.6889b

After $3$ hours the bucket is $57.18 \%$ full.

b\times(1-0.17)^{3}=0.5718b\text{ (4dp)}

After $4$ hours the bucket is $47.46 \%$ full. This is less than half.

b\times(1-0.17)^{4}=0.4746b

80.00 \%

60.00 \%

85.74 \%

81.45 \%

Let’s use $c$ for the amount of charge in the battery.

After $2$ hours the battery is $95 \%$ full. In this case $t=1,$ factoring in that each time period is $2$ hours.

c\times(1-0.05)^{1}=0.95c

After $4$ hours, the battery is $90.25 \%$ full. In this case $t=2,$ because $4$ is $2$ time periods of $2$ hours.

c\times(1-0.05)^{2}=0.9025c

After $6$ hours, the battery is $85.74 \%$ full. In this case $t=3,$ because $6$ is $3$ time periods of $2$ hours.

c\times(1-0.05)^{3}=0.8574c\text{ (4dp)}

After $8$ hours the battery is $81.45 \%$ full. In this case $t=4,$ because $8$ is $4$ time periods of $2$ hours.

c\times(1-0.05)^{4}=0.8145c\text{ (4dp)}

401.8 \, L

401.172 \, L

407.08 \, L

407.8 \, L

The pond starts with $400 \, L$ of water. We are dealing with two $6$ month time periods with a different multiplier for each one.

The amount of water is given by $400\times(1+0.01)^{6}\times(1-0.007)^{6}=407.08 \, L$

Compound interest formula FAQs

What is the monthly compound interest formula?

A similar formula that calculates the total monthly interest paid, given the principal amount, interest rate and time period. The value for $n$ is $12,$ since there are $12$ months in a year.

$P \cdot\left(1+\cfrac{r}{12}\right)^{12 t}-P$

What is the daily compound interest formula?

A similar formula that calculates the total daily interest paid, given the principal amount, interest rate and time period. The value for $n$ is $365,$ since there are $365$ days in a year.

$P\cdot\left(1+\cfrac{r}{365}\right)^{365t}-P$

The next lessons are

Compound measures
Types of data
Averages and ranges

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