Compound interest inquiry
Mathematical inquiry processes: Explore; set conditions; test different cases; conjecture and reason. Conceptual field of inquiry: Compound interest and depreciation; percentages.
In the orientation phase of the inquiry, students will ask about the meaning of 'compound interest' and how it is calculated. They might suggest values for a and b and even make claims about the truth of the statement.
In a structured inquiry the teacher explains the concept if required, demonstrates the use of the compound interest formula, and then sets a condition such as a - b = 1. In this case, the second option in the prompt will always give the greater compound interest (see the table).
Students might then set their own conditions with greater differences between a and b or with b > a.
In the case of a > b, the prompt is false. The compound interest is always greater when the number of years is more than the percentage.
However, students have argued that even though the total interest is greater, it is lower if calculated as the mean interest per annum. Using the first line of the table as an example, compound interest is worth 0.02 units of currency per year when a is the rate of interest, but only 0.01005 units of currency per year when b is the rate of interest.
A note on the variables
In the formula for compound interest, the rate of interest is normally denoted by r and the number of years by t. Additionally, n can stand for the number of times interest is applied each year. The prompt uses a and b to avoid confusion; a refers both to the rate (in the first part of the prompt) and time (in the second part), and vice versa for b.
Lines of inquiry
1. Posing problems
A line of inquiry that developed in a year 9 classroom involves posing problems. One student, for example, asked: In the case when a - b = 1, how many years it would take for the compound interest in the second column (see the table above) to be at least twice as much as in the first. Using a spreadsheet, she showed that it takes 331 years. Even when a - b = 10, it takes 45 years.
When students try to explain the result, they might start with a principal amount of 100 units of currency with a = 10 and b = 1. This gives 100 x (1.1)1 = 110 and 100 x (1.01)10 = 110.46 (rounded to 2 decimal places). In the first case, the 10% is applied once to the principal amount; in the second case, the 1% is applied ten times to amounts that increase each time.
Is the prompt still false if we substitute the word 'depreciation' for 'compound interest'? Using the same figures as above (a principal amount of 100 units of currency with a = 10 and b = 1), we get 100 x (0.9)1 = 90 and 100 x (0.99)10 = 90.44 (rounded to 2 decimal places). Thus, the depreciation is greater in the first case because the percentage is applied to the full amount, whereas in the second case it is applied to amounts that decrease each time.