Calculate principal, amount, or interest rate in compound-interest problems

Ontario Grade 11 Mathematics

Discrete Functions

MCR3U — Expectation C3.3

Have you ever wondered how a savings account grows over time, or how much you originally deposited to end up with a certain balance? Compound interest is at the heart of both questions. Unlike simple interest, which is calculated only on the original amount, compound interest is calculated on the principal plus any interest already earned. This means your balance grows a little faster each period. In this lesson you will use one central formula to find any one of three unknowns — the final amount, the original principal, or the interest rate — as long as the other values are known. You will work with the formula directly, using integer exponents and roots that are already in your Grade 10 toolkit.

What you will learn

Calculate the final amount (A) in a compound-interest problem using the compound-interest formula.
Calculate the original principal (P) when the final amount, interest rate, and time are known.
Calculate the interest rate (i) per compounding period when the principal, amount, and number of periods are known.
Distinguish between the number of compounding periods (n) and time in years, and convert correctly between them.

Prerequisite Bridge: Exponents and Roots

Before using the compound-interest formula, make sure two Grade 10 skills are solid. First, you must be comfortable evaluating a power such as

1.04^{6}

, which means multiplying

1.04

by itself six times. A scientific or graphing calculator handles this with the exponent key (often labelled ^ or

y^x

).

Second, when you need to undo a power — for example, if you know

x^{3} = 1.124864

and want

x

— you take the corresponding root. The

n

-th root of a number

k

is written

k^{1/n}

and can be entered on most calculators as

k

^

(1 \div n)

. For instance,

1.124864^{1/3} \approx 1.04

, so

x = 1.04

. This root-taking skill is the key tool for finding an unknown interest rate later in the lesson.

A power $b^{n}$ means $b$ multiplied by itself $n$ times.
Taking the $n$ -th root undoes an $n$ -th power: if $x^{n} = k$ , then $x = k^{1/n}$ .
Use the calculator's ^ key for both powers and fractional exponents.

The Compound-Interest Formula

Every compound-interest calculation in this course comes from one formula. Let

P

be the principal (the starting amount),

i

be the interest rate per compounding period expressed as a decimal,

n

be the total number of compounding periods, and

A

be the amount (the balance after all compounding). The relationship is

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

.

The formula says: each period, the balance is multiplied by the growth factor

(1 + i)

. After

n

periods that multiplication has happened

n

times, so the balance is

P

multiplied by

(1+i)^n

. This is why compound interest grows faster than simple interest — each period's interest becomes part of the base for the next period.

Before substituting values, always identify which variable is unknown and which three are known. You also need to convert the annual interest rate and time in years into the rate per compounding period and the total number of periods. If the annual rate is

r

(as a decimal) and interest is compounded

m

times per year for

t

years, then

i = r \div m

and

n = m \times t

. Common compounding frequencies are: annually (

m=1

), semi-annually (

m=2

), quarterly (

m=4

), and monthly (

m=12

).

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

$A = P(1+i)^{n}$ is the compound-interest formula for this course.
$i$ is the rate per compounding period, not the annual rate.
$n$ is the total number of compounding periods, not the number of years.
Convert the annual rate and years to a period rate and period count before substituting.
Interest earned equals $A - P$ .

Finding the Amount (A) and the Principal (P)

When

P

,

i

, and

n

are all known, finding

A

is straightforward: substitute and evaluate. For example, if CAD 2 000 is invested at 6% per year compounded monthly for 3 years, then

i = 0.06 \div 12 = 0.005

and

n = 12 \times 3 = 36

. Evaluating

A = 2000 \times (1.005)^{36}

on a calculator gives

(1.005)^{36} \approx 1.19668

, so

A \approx 2000 \times 1.19668 \approx

CAD 2 393.36.

Finding

P

requires rearranging the formula. Start from

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

and divide both sides by

(1+i)^{n}

. This isolates

P

, giving

P = A \div (1+i)^{n}

, which can also be written as

P = A(1+i)^{-n}

. The expression

(1+i)^{-n}

is simply

1 \div (1+i)^{n}

— it 'brings a future amount back' to what it is worth today. There is no new formula to memorize; it is the same equation, rearranged by ordinary algebra.

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

To find $A$ : substitute $P$ , $i$ , $n$ and evaluate.
To find $P$ : rearrange to $P = A \div (1+i)^{n}$ .
$(1+i)^{-n}$ is just $1 \div (1+i)^{n}$ — no new rule needed.
Always round money to the nearest cent (2 decimal places) at the final step only.

Finding the Interest Rate (i)

Sometimes you know the starting and ending amounts and the number of periods, and you need to find the rate. Start again from

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

. Divide both sides by

P

to get

A \div P = (1+i)^{n}

. You now need to undo the exponent

n

. Raising both sides to the power

1/n

(that is, taking the

n

-th root of both sides) gives

(A \div P)^{1/n} = 1 + i

. Subtracting 1 from both sides then isolates the rate per compounding period:

i = (A/P)^{1/n} - 1

.

Once you have

i

, convert it to an annual rate by multiplying by the number of compounding periods per year,

m

. For example, if

i = 0.015

per month, the annual rate is

0.015 \times 12 = 0.18

, or 18% per year. Always state whether your final rate is per period or per year, and express it as a percentage.

i = \left(\frac{A}{P}\right)^{1/n} - 1

Divide both sides by $P$ , then take the $n$ -th root to isolate $(1+i)$ .
$i = (A/P)^{1/n} - 1$ gives the rate per compounding period.
Multiply $i$ by $m$ (periods per year) to get the annual rate.
Convert the decimal rate to a percentage by multiplying by 100.

Connecting the Formula to Exponential Functions

You may have noticed that

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

looks like the exponential functions studied earlier in MCR3U. If you treat

n

as the input variable and

A

as the output, the formula has exactly the form

f(n) = P \cdot b^{n}

, where the base

b = (1+i)

is a constant greater than 1. This means compound-interest growth is exponential growth — the amount increases by the same multiplicative factor each period.

This connection is useful for checking answers. If you list a few values of

A

as

n

increases, the values should grow by the same ratio each time. If they do not, check your substitution. Recognizing the exponential structure also explains why even a small increase in

i

, or a longer time

n

, can lead to a noticeably larger final amount.

f(n) = P \cdot (1+i)^{n}

$A = P(1+i)^{n}$ is an exponential function with base $(1+i)$ and initial value $P$ .
The amount grows by a factor of $(1+i)$ each compounding period.
A larger base or more periods both increase $A$ exponentially, not just proportionally.

Common Compounding Frequencies

Compounding Term	Periods per Year (m)	Rate per Period (i = r ÷ m)	Periods for t Years (n = m × t)
Annually	1	r ÷ 1 = r	1 × t
Semi-annually	2	r ÷ 2	2 × t
Quarterly	4	r ÷ 4	4 × t
Monthly	12	r ÷ 12	12 × t

Worked example

Example 1 — Finding the Principal

Marcelline wants to have CAD 10 000 in her savings account exactly 5 years from now. The account earns 4.8% per year, compounded quarterly. How much must she deposit today? Round to the nearest cent.

Identify the known and unknown values
The final amount is $A = 10\,000$ . The annual interest rate is 4.8%, so $r = 0.048$ as a decimal. Compounding is quarterly, meaning $m = 4$ times per year. The investment runs for $t = 5$ years. The unknown is the principal $P$ .
Calculate the rate per compounding period
Divide the annual rate by the number of compounding periods per year to get the rate per period $i$ .
$i = \frac{0.048}{4} = 0.012$
Calculate the total number of compounding periods
Multiply the number of compounding periods per year by the number of years to get $n$ .
$n = 4 \times 5 = 20$
Write the rearranged formula for P
Starting from $A = P(1+i)^{n}$ , divide both sides by $(1+i)^{n}$ to isolate $P$ .
$P = \frac{A}{(1+i)^{n}}$
Substitute the known values
Replace $A$ , $i$ , and $n$ with the values found above to set up the calculation.
$P = \frac{10\,000}{(1.012)^{20}}$
Evaluate the growth factor
Using a calculator, raise $1.012$ to the power $20$ . Keep several decimal places to avoid rounding error in the next step.
$(1.012)^{20} \approx 1.26973$
Divide to find P
Divide CAD 10 000 by the growth factor. Round only at this final step.
$P = \frac{10\,000}{1.26973} \approx 7876.60$

Answer: Marcelline must deposit approximately CAD 7 876.60 today.

Check: Verify by computing the amount forward:

7876.60 \times (1.012)^{20} \approx 7876.60 \times 1.26973 \approx 9999.99 \approx 10\,000

. The tiny difference is due to rounding mid-calculation, confirming the answer is correct.

Worked example

Example 2 — Finding the Annual Interest Rate

Theo invests CAD 3 500 in a GIC (Guaranteed Investment Certificate). After 3 years of monthly compounding, the account holds CAD 4 214.33. What annual interest rate did the GIC earn? Round the annual rate to two decimal places.

Identify the known and unknown values
The principal is $P = 3500$ , the final amount is $A = 4214.33$ , compounding is monthly so $m = 12$ , and the term is $t = 3$ years. The unknown is the annual interest rate $r$ .
Calculate the total number of compounding periods
Multiply the compounding frequency per year by the number of years.
$n = 12 \times 3 = 36$
Rearrange the formula to isolate the growth factor
Start from $A = P(1+i)^{n}$ . Divide both sides by $P$ so that $(1+i)^{n}$ is alone on one side.
$\frac{A}{P} = (1+i)^{n}$
Take the n-th root of both sides
To undo the exponent $n = 36$ , raise both sides to the power $\frac{1}{36}$ . This leaves $(1+i)$ by itself on the right side, because $\left((1+i)^{36}\right)^{1/36} = (1+i)^{1} = 1+i$ .
$\left(\frac{A}{P}\right)^{1/36} = 1 + i$
Compute the ratio A divided by P
Divide the final amount by the principal to find the total growth factor over all 36 periods.
$\frac{4214.33}{3500} \approx 1.20409$
Raise the ratio to the power 1/36
Enter $1.20409$ ^ $(1 \div 36)$ on your calculator to take the 36th root and find $(1+i)$ .
$(1.20409)^{1/36} \approx 1.00520$
Solve for i, then convert to an annual rate
Subtract 1 to get the monthly rate $i$ . Then multiply by $m = 12$ to get the annual rate $r$ as a decimal.
$i = 1.00520 - 1 = 0.00520, \quad r = 0.00520 \times 12 = 0.0624$
Express the annual rate as a percentage
Multiply the decimal annual rate by 100 to convert to a percentage.
$r = 0.0624 \times 100 = 6.24\%$

Answer: The GIC earned an annual interest rate of approximately 6.24%, compounded monthly.

Check: Verify forward:

i = 0.0624 \div 12 = 0.0052

and

n = 36

, giving

A = 3500 \times (1.0052)^{36} \approx 3500 \times 1.20409 \approx 4214.33

. This matches the given amount, confirming the rate is correct.

Common mistakes and how to avoid them

Using the annual interest rate directly as i instead of dividing by the number of compounding periods per year.

Correction: Always divide the annual rate by m (the compounding frequency) before substituting into the formula. For example, 6% compounded monthly gives i = 0.06 ÷ 12 = 0.005, not 0.06.

Using the number of years as n instead of the total number of compounding periods.

Correction: Multiply the number of years by m to get n. For 5 years compounded quarterly, n = 4 × 5 = 20, not 5.

Rounding the growth factor to too few decimal places mid-calculation, causing the final answer to be off by several dollars.

Correction: Keep at least 5 decimal places in intermediate results and round only the final dollar answer to the nearest cent.

Forgetting to subtract 1 after taking the n-th root when solving for the interest rate, and reporting the root itself as the rate.

Correction: The n-th root gives (1 + i), so you must subtract 1 to isolate i. The correct step is i = (A/P)^(1/n) − 1.

Confusing the amount A with the interest earned, and reporting A − P as the final amount or P as the interest.

Correction: A is the total balance (principal plus all interest). The interest earned is A − P. Make sure you solve for the quantity the question actually asks for.

Lesson summary

The compound-interest formula $A = P(1+i)^{n}$ connects the principal $P$ , the amount $A$ , the rate per period $i$ , and the number of periods $n$ .
Before substituting, convert the annual rate to a per-period rate using $i = r \div m$ , and convert years to total periods using $n = m \times t$ .
To find $P$ , rearrange the formula to $P = A \div (1+i)^{n}$ .
To find $i$ , rearrange to $i = (A/P)^{1/n} - 1$ , then multiply by $m$ to get the annual rate.
Round money to the nearest cent only at the final step; keep extra decimal places in intermediate calculations.
The compound-interest formula is an exponential function — the amount grows by the same multiplicative factor $(1+i)$ every compounding period.

Check your understanding

Question 1

CAD 5 000 is invested at 3% per year compounded semi-annually for 4 years. Which values of i and n should be substituted into

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

?

$i = 0.03$ , $n = 4$
$i = 0.015$ , $n = 8$
$i = 0.03$ , $n = 8$
$i = 0.015$ , $n = 4$

Show answer and explanation

i = 0.015

,

n = 8

Semi-annual compounding means m = 2. The rate per period is i = 0.03 ÷ 2 = 0.015, and the total number of periods is n = 2 × 4 = 8. Option B is correct.

Question 2

You want to find the principal

P

needed to grow to CAD 8 000 in 6 years at 5% per year compounded annually. Which expression gives

P

?

$P = 8000 \times (1.05)^{6}$
$P = 8000 - (1.05)^{6}$
$P = 8000 \div (1.05)^{6}$
$P = (8000 \div 1.05)^{6}$

Show answer and explanation

P = 8000 \div (1.05)^{6}

Rearranging

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

by dividing both sides by

(1+i)^{n}

gives

P = A \div (1+i)^{n} = 8000 \div (1.05)^{6}

. Option C is correct.

Question 3

An investment of CAD 2 000 grows to CAD 2 662.00 after 10 years of annual compounding. Which calculation correctly finds the annual interest rate?

$(2662 - 2000)^{1/10} - 1$
$(2662 \div 2000)^{10} - 1$
$(2662 \div 2000)^{1/10} - 1$
$(2000 \div 2662)^{1/10} - 1$

Show answer and explanation

(2662 \div 2000)^{1/10} - 1

With n = 10 annual periods, the formula

i = (A/P)^{1/n} - 1

becomes

(2662 \div 2000)^{1/10} - 1 = (1.331)^{0.1} - 1 \approx 0.029

, or about 2.9% per year. Option C is correct.

Question 4

A GIC pays 4.8% per year compounded monthly. After 2 years, the account balance is CAD 4 400.00. Which of the following is closest to the original principal deposited?

CAD 3 990.00
CAD 4 000.00
CAD 4 200.00
CAD 4 085.00

Show answer and explanation

CAD 4 000.00

Here

i = 0.048 \div 12 = 0.004

and

n = 24

. Then

(1.004)^{24} \approx 1.10000

, so

P = 4400 \div 1.10000 \approx

CAD 4 000.00. Option B is correct.

Key terms

Principal (P): The original amount of money deposited or borrowed before any interest is added.
Amount (A): The total balance after interest has been added over all compounding periods; sometimes called the future value.
Interest: The extra money earned (or owed) as a percentage of the principal. In compound interest, previously earned interest also earns interest.
Compounding period: The regular time interval at which interest is calculated and added to the balance, such as monthly or quarterly.
Interest rate per period (i): The annual interest rate divided by the number of compounding periods per year. This is the rate used directly in the formula.
Number of periods (n): The total count of compounding periods over the entire investment term, equal to the number of periods per year multiplied by the number of years.
Growth factor: The expression $(1+i)$ , which represents how many times larger the balance becomes after one compounding period.
n-th root: The number that, when raised to the power n, gives a specified value. Written as $k^{1/n}$ , it is the inverse operation of raising to the power n.

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation C3.3. It is a study resource, not an official curriculum publication.

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