Table of Contents
Here we will learn about the graph of simple interest vs. the number of years. On a graph paper, take the time or the number of years on the x-axis, simple interest amount on the y-axis. Get the simple steps to draw a graph of simple interest vs time and solved example questions in the below sections.
How to draw a Graph of Simple Interest vs. Number of Years?
Go through the following sections to find the steps to draw a graph of simple interest vs the number of years. You have to find the relation between the simple interest amount, time period for a constant interest rate.
- Let us take the total amount and number of years to calculate the simple interest.
- Treat the constant interest rate of 1%.
- Calculate the interest amount for every year and write it as a table.
- Get the points and plot them on the graph.
- Join those points to get the required graph line.
- To read the numbers from the graph, substitute the time and find the total amount and draw on the graph.
Example Questions & Answers
Example 1.
Draw the graph between simple interest vs Number of years on amount Rs. 2000 at the rate of interest of 4% per annum?
Solution:
Given that,
Amount taken for interest p = Rs. 2000/-
Rate of interest r = 4%
Percentage of simple interest = (prt) / 100
= (2000 * 4 * t) / 100
= 80t
Consider 80t as a simple function.
By putting t = 1, 2, 3 successively, we will get the corresponding values of 80t. We get the table given below.
| t | 1 | 2 | 3 |
|---|---|---|---|
| 80t | 80 | 160 | 240 |
Along the x-axis: Take 1 small square = 1 unit.
Along the y-axis: Take 1 small square = 25 units.
Plot the points A (1, 80), B (2, 160), C (3, 240) on a graph paper.
Join these points to get the graph.
Example 2.
Simple interest on a certain sum is $ 50 per year. Then, S = 50 t, where t is the number of years.
1. Draw a graph of the above function.
2. From the graph find the value of S, when (a) t = 5 (b) t = 6?
Solution:
Given sample function is S = 50 t
Put t = 1, 2, 3, 4 successively and getting the corresponding value of S. we get the table given below.
| t | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| S | 50 | 100 | 150 | 200 |
Along the x-axis: Take 1 small square = 1 unit.
Along the y-axis: Take 1 small square = 25 units.
Plot the points A (1, 50), B (2, 100), C (3, 150), D (4, 200) on a graph.
Join the points to get a graph line.
Reading off from the graph of simple interest vs. a number of years:
(a) On the x-axis, take the point L at t = 5.
Draw LP ⊥ x-axis, meeting the graph at P.
Clearly, PL = 250 units.
Therefore, t = 5 ⇒ S = 250.
(b) On the x-axis, take the point M at t = 6
Draw MQ ⊥ x-axis, meeting the graph at Q.
So, MQ = 300 units.
Therefore, t = 6 ⇒ S = 300.
Example 3.
Draw the graph between simple interest vs Number of years on amount Rs. 1500 at the rate of interest of 2% per annum. From the graph find the value of Simple interest, when (a) number of years = 5 years 6 months (b) t = 8 years 9 months?
Solution:
Given that,
Amount taken for interest p = Rs. 1500
Rate of interest r = 2%
Percentage of simple interest = (ptr) / 100
= (1500 * 2 * t) / 100
= 30 t
Let us take S = 30t as the function.
Find the value of simple interest, when the number of years is 1, 2, 3. Write them on a table.
| t | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| S | 30 | 60 | 90 | 120 |
Along the x-axis: Take 1 small square = 1 unit.
Along the y-axis: Take 1 small square = 25 units.
Plot the points A (1, 30), B (2, 60), C (3, 90), D (4, 120) on the coordinate graph.
Join the points ABCD to get the required graph line.
Reading off from the graph of simple interest vs. a number of years:
(a) On the x-axis, take the point L at t = 5 years 6 months.
Draw LP ⊥ x-axis, meeting the graph at P.
Clearly, PL = 165 units.
Therefore, t = 5 ⇒ S = 165.
(b) On the x-axis, take the point M at t = 8 years 9 months
Draw MQ ⊥ x-axis, meeting the graph at Q.
So, MQ = 262.5 units.
Therefore, t = 6 ⇒ S = 262.5.
