Simple Interest Graphs
Read simple interest graphs and compare them for NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-02). Simple interest adds the same amount every year, so graphing interest or total value against time gives a straight line — and the steeper the line, the more interest is earned each year.
You will read interest and total value graphs, get the yearly interest from the gradient, work out the rate from a graph, and set side by side the lines at different interest rates and lengths of time, by hand or with technology — a quick way to see which investment grows faster.
Theory
Simple interest adds the same amount every year, so simple interest graphs are straight lines: the gradient is the interest earned each year, \(Pr\). This Year 12 Standard 1 (NSW) guide shows how to read interest and total value graphs, find the interest per year and the rate from a graph, and compare investments by the steepness of their lines.
Simple interest adds the same amount, \(Pr\), every year, so plotting it against time gives a straight line — simple interest is linear growth.
An interest graph (\(I\) against \(n\)) is a straight line through the origin, because no interest has been earned at the start. A total value graph (\(A\) against \(n\)) is a straight line that starts at the principal \(P\) on the vertical axis.
The gradient of either line is \(\dfrac{\text{rise}}{\text{run}}\) \(=\) the interest earned each year \(=Pr\). A steeper line earns more interest each year; with the same principal, the steeper line has the higher rate.
The interest line passes through the origin with gradient \(Pr\):
The total value line starts at \(P\) and has the same gradient:
The gradient is the interest earned each year:
How to use a simple interest graph
- Identify the graph: an interest line starts at \(0\); a total value line starts at the principal \(P\).
- Read values by going up from the years to the line and across to the vertical axis, checking the scale.
- Find the gradient \(\dfrac{\text{rise}}{\text{run}}\) for the interest per year \(Pr\), then divide by \(P\) for the rate (or by \(r\) for the principal).
- Use technology if allowed: graph each plan in a graphing app or spreadsheet (enter \(I=Prn\) for each plan) and compare the steepness and starting values.
Go up from \(n=3\) to the line, then across. Check with \(I=Prn\).
| \(I\) | \(=\) | \(2400\times0.05\times3\) |
| \(\) | \(=\) | \(120\times3 = 360\) |
The interest after \(3\) years is \(\$360\).
The gradient is the interest per year; the principal is the starting value.
| \(\text{(i) gradient}\) | \(=\) | \(\dfrac{7350-6000}{5-0}\) |
| \(\) | \(=\) | \(\dfrac{1350}{5} = 270\) |
| \(\text{(ii) } r\) | \(=\) | \(\dfrac{270}{6000}\) |
| \(\) | \(=\) | \(0.045 = 4.5\%\) |
(i) \(\$270\) per year. (ii) \(4.5\%\) p.a.
The gradient is the interest per year, \(Pr\).
| \(\text{A per year}\) | \(=\) | \(3000\times0.03 = 90\) |
| \(\text{B per year}\) | \(=\) | \(3000\times0.05 = 150\) |
| \(\text{gap at } n=6\) | \(=\) | \(150\times6-90\times6\) |
| \(\) | \(=\) | \(900-540 = 360\) |
(i) Plan B's line is steeper. (ii) Plan B has earned \(\$360\) more.
The gradient is the yearly interest \(Pr\); solve for \(P\).
| \(\text{gradient}\) | \(=\) | \(\dfrac{336}{4} = 84\) |
| \(P\times0.06\) | \(=\) | \(84\) |
| \(P\) | \(=\) | \(\dfrac{84}{0.06} = 1400\) |
Check: \(1400\times0.06\times4=336\). The principal was \(\$1400\).
Common pitfalls
Frequently asked questions
Why is a simple interest graph a straight line?
Simple interest adds the same amount, P times r, every year. A constant increase each year is linear growth, so the graph of interest or total value against time is a straight line.
What does the gradient of a simple interest graph mean?
The gradient, rise over run, is the interest earned each year, which equals P times r. A steeper line earns more interest each year.
How do you find the interest rate from a simple interest graph?
Find the gradient to get the interest per year, then divide by the principal. For example, a value line from $6000 to $7350 over 5 years has gradient 270, and 270 divided by 6000 is 0.045, or 4.5% per annum.
What is the difference between an interest graph and a total value graph?
An interest graph starts at zero because no interest has been earned at the start. A total value graph starts at the principal. Both have the same gradient, P times r, so for the same investment the two lines are parallel.
How can you tell which investment has the higher rate from a graph?
If the investments have the same principal, the steeper line has the higher rate, because it earns more interest each year. Parallel lines earn the same interest each year.