topic badge

7.04 Interest rates

Lesson

Introduction

While interest rates are usually presented and calculated using yearly rates, we can also use different time periods like days, weeks, or months when accumulating interest. When an interest rate is presented using one time period but is calculated using another, these conversions can help with the simple interest calculations.

Convert interest rate periods

Converting interest rate periods works the same way that most conversions do. When converting from one time period to another, we calculate how many of the new time period can fit into one of the old time periods. We can then divide the interest percentage by this number to fully convert the interest rate.

Suppose that an investment has a simple interest rate of 6\% p.a., accumulating monthly. What is the monthly interest rate?

We are given that the interest rate per annum is 6\%. This is a yearly interest rate and we want to find the monthly one.

Since there are 12 months in a year, we can divide the yearly interest rate of 6\% by 12 to find the monthly interest rate. This conversion can presented as the working out:

\text{Monthly interest rate $$}=6\% \text{ per } 12 \text{ months}

Which can be simplified to give us:

\text{Monthly interest rate $$}=0.5\% \text{ per } \text{ months}

As expected, the monthly interest rate was \dfrac{1}{12} of the yearly interest rate, since a month is equal to \dfrac{1}{12} of a year.

We can use this same conversion method to convert between any two interest rate periods.

For our conversions, we can assume that there will be:

  • 12 months in a year

  • 52 weeks in a year

  • 365 days in a year

  • 7 days in a week

  • 2 weeks in a fortnight

Other time periods like half-years and quarters carry their conversion factors in their names, being \dfrac{1}{2}and \dfrac{1}{4} of a year respectively.

Notice that there are no direct conversions between weeks and months or days and months. For these conversions, we need to convert from one period into years, then from years into the other period. For example: since a week is \dfrac{1}{52} of a year and a year is 12 months, we can combine these two conversions to find that a week is \dfrac{12}{52} of a month.

Examples

Example 1

Peter takes out a loan which earns interest at a flat rate of 0.18\% per week. What is the equivalent yearly simple interest rate? Assume there are 52 weeks in a year.

Worked Solution
Create a strategy

Multiply the weekly rate by the number of weeks in a year.

Apply the idea
\displaystyle \text{Equivalent annual rate}\displaystyle =\displaystyle 0.18\% \times 52Multiply the weekly rate by 52
\displaystyle =\displaystyle 9.36\%Evaluate the multiplication

Example 2

Calculate the simple interest earned on an investment of \$6050 at 0.7\% per quarter for 3 years.

Worked Solution
Create a strategy

We can use the simple interest formula: I=Prn. Use the fact that there are 4 quarters in a year.

Apply the idea

To find the number of quarters in 3 years, we can multiply 3 by 4 to get 3\times 4=12.

\displaystyle I\displaystyle =\displaystyle PrnWrite the formula
\displaystyle =\displaystyle 6050 \times 0.7\% \times 12Substitute P,\,r,\, and n
\displaystyle =\displaystyle \$508.20Evaluate the multiplication
Idea summary

When converting from one time period to another, we calculate how many of the new time period can fit into one of the old time periods. We can then divide the interest percentage by this number to fully convert the interest rate.

For conversions, we can assume that there will be:

  • 12 months in a year

  • 52 weeks in a year

  • 365 days in a year

  • 7 days in a week

  • 2 weeks in a fortnight

Appreciation and depreciation

Appreciation occurs when an item increases in value by some percentage. This is equivalent to accruing interest on an investment, where the value of the item is equivalent to the investment total.

Depreciation occurs when an item decreases in value by some percentage. This can be calculated the same way as appreciation is calculated, except that the percentage amount is subtracted from the original value.

Examples

Example 3

A camera valued at \$400 depreciates at a rate of \$32 per year. Calculate the amount the camera will be worth after:

a

One year.

Worked Solution
Create a strategy

Subtract the depreciation amount from the value of the camera.

Apply the idea
\displaystyle \text{Camera value}\displaystyle =\displaystyle 400-32Subtract the values
\displaystyle =\displaystyle \$368Evaluate the subtraction
b

Two years.

Worked Solution
Create a strategy

Multiply the depreciation amount by the number of years, and subtract it from the camera value.

Apply the idea
\displaystyle \text{Camera value}\displaystyle =\displaystyle 400-32 \times 2Write the equation
\displaystyle =\displaystyle 400-64Evaluate the multiplication
\displaystyle =\displaystyle \$336Evaluate the subtraction
c

Ten years.

Worked Solution
Create a strategy

Multiply the depreciation amount by the number of years, and subtract it from the camera value.

Apply the idea
\displaystyle \text{Camera value}\displaystyle =\displaystyle 400-32 \times 10Write the equation
\displaystyle =\displaystyle 400-320Evaluate the multiplication
\displaystyle =\displaystyle \$80Evaluate the subtraction
Idea summary

Appreciation occurs when an item increases in value by some percentage.

Depreciation occurs when an item decreases in value by some percentage.

Outcomes

VCMNA304

Solve problems involving simple interest.

What is Mathspace

About Mathspace