Find the simple interest on $2500$ dollars at $4\%$ a year for $3$ years, and the final amount.

$300$ dollars of interest; $2800$ dollars in total.

The same $2500$ dollars earns $4\%$ a year compounded annually for $3$ years. Find the final amount, and compare it with simple interest.

$2812.16$ dollars, which is $12.16$ dollars more than simple interest.

$8000$ dollars is invested at $5.4\%$ a year compounded monthly for $4$ years. Find the amount and the interest earned.

About $9924.01$ dollars, with $1924.01$ dollars of interest.

An investment of $4000$ dollars, compounded semi-annually, is worth $5013.61$ dollars after $6$ years. Find the annual interest rate, and the interest earned in the 6th year alone.

$3.8\%$ a year compounded semi-annually; about $185.23$ dollars in the 6th year.

Grade 11 · MCR3U · Functions

MCR3U: Simple and Compound Interest

Simple and compound interest are sequences in disguise. Simple interest adds the same amount every year, so the balances form an arithmetic sequence. Compound interest multiplies by the same factor every period, so the balances form a geometric sequence, and that is why it pulls ahead.

How It Works

Simple interest. Interest is paid only on the original principal: $I = Prt$ , where $P$ is the principal, $r$ is the annual rate as a decimal and $t$ is the time in years. The amount is $A = P + Prt$ , and it grows by the same $Pr$ every year.

Compound interest. Interest is added to the balance at the end of each period, and the next period's interest is paid on the new balance. Each period multiplies the balance by $1 + i$ , so after $n$ periods $A = P(1 + i)^n$ .

The rate and the periods. $i$ is the annual rate divided by the number of compounding periods per year, and $n$ is the total number of periods. Annually means $1$ period a year, semi-annually $2$ , quarterly $4$ , monthly $12$ . At $6\%$ a year compounded quarterly for $5$ years, $i = \frac{0.06}{4} = 0.015$ and $n = 4 \times 5 = 20$ .

Present value. To find how much to invest now to have $A$ later, solve the formula for $P$ : $P = \frac{A}{(1 + i)^n} = A(1 + i)^{-n}$ . The same idea, run backward, finds a rate: $i = \left(\frac{A}{P}\right)^{\frac{1}{n}} - 1$ .

Comparing. At the same annual rate, compound interest beats simple interest after the first period, and more frequent compounding earns a little more, because interest starts earning interest sooner. Keep every digit until the end, then round money to the cent.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Find the simple interest on $2500$ dollars at $4\%$ a year for $3$ years, and the final amount.
Show solution
1. $I = Prt = 2500(0.04)(3) = 300$ .
2. $A = 2500 + 300 = 2800$ .
Answer: $300$ dollars of interest; $2800$ dollars in total.
Easy
The same $2500$ dollars earns $4\%$ a year compounded annually for $3$ years. Find the final amount, and compare it with simple interest.
Show solution
1. $i = 0.04$ and $n = 3$ .
2. $A = 2500(1.04)^3 = 2500(1.124864) = 2812.16$ .
3. That is $2812.16 - 2800 = 12.16$ more than simple interest.
Answer: $2812.16$ dollars, which is $12.16$ dollars more than simple interest.
Medium
$8000$ dollars is invested at $5.4\%$ a year compounded monthly for $4$ years. Find the amount and the interest earned.
Show solution
1. $i = \frac{0.054}{12} = 0.0045$ and $n = 12 \times 4 = 48$ .
2. $A = 8000(1.0045)^{48} \approx 9924.01$ .
3. Interest: $9924.01 - 8000 = 1924.01$ .
Answer: About $9924.01$ dollars, with $1924.01$ dollars of interest.
Medium
How much must be invested now at $6\%$ a year compounded quarterly to have $10000$ dollars in $5$ years?
Show solution
1. $i = \frac{0.06}{4} = 0.015$ and $n = 20$ .
2. $P = \frac{10000}{(1.015)^{20}} = 10000(1.015)^{-20} \approx 7424.70$ .
Answer: About $7424.70$ dollars.
Test level
An investment of $4000$ dollars, compounded semi-annually, is worth $5013.61$ dollars after $6$ years. Find the annual interest rate, and the interest earned in the 6th year alone.
Show solution
1. $n = 2 \times 6 = 12$ , so $4000(1 + i)^{12} = 5013.61$ .
2. $(1 + i)^{12} = \frac{5013.61}{4000}$ , so $1 + i = \left(\frac{5013.61}{4000}\right)^{\frac{1}{12}} \approx 1.019$ and $i \approx 0.019$ .
3. The annual rate is $2(0.019) = 0.038$ , or $3.8\%$ .
4. After $5$ years, $n = 10$ : $4000(1.019)^{10} \approx 4828.38$ . The 6th year earned $5013.61 - 4828.38 = 185.23$ .
Answer: $3.8\%$ a year compounded semi-annually; about $185.23$ dollars in the 6th year.

Learn It on Martian Lab

Martian Lab teaches this in the Simple & Compound Interest lesson of its MCR3U Discrete Functions unit. Watch it, then take the Discrete Functions practice tests to see if it stuck.

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More Discrete Functions topics: Arithmetic Sequences and Series · Geometric Sequences and Series · Pascal's Triangle and the Binomial Theorem · Annuities: Future and Present Value

More MCR3U topics: Inverse Functions · Transformations of Functions · Sine Law and the Ambiguous Case · Domain, Range and Function Notation · Simplifying Rational Expressions · Operations with Rational Expressions · Max/Min and Vertex Form · Linear-Quadratic Systems · Quadratic Word Problems · Simplifying Radicals · Exponent Laws and Rational Exponents · Graphing Exponential Functions · Exponential Growth and Decay · Special Angles and the CAST Rule · Cosine Law and 3D Problems · Transformations of Sine and Cosine · Sinusoidal Word Problems

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