Question

Calculate the simple interest on an amount of RS. 8000 at a rate of interest of 5% per annum for 1, 2, 3, …, 6 years. Tabulate the information and plot the interest against the number of years.

Answer 1

Hint: We solve this problem by using the formula of simple interest.
The formula for the simple interest is given as
\[SI=\dfrac{P\times T\times R}{100}\]
Where \[P\] is the principal amount, \[T\] is the time period and \[R\] is the rate of interest per annum.
After finding the simple interest for all years from 1 to 6 we create a table of ordered pairs containing the number of years as x coordinate and simple interest as y co – ordinate so that we can plot them on a graph of a certain scale.

Complete step by step answer:
We are given that the principal amount as
\[\Rightarrow P=8000\]
We are also given that the rate of interest per annum as
\[\Rightarrow R=5%\]
Let us assume that the time period as \[T\]
We know that the formula for the simple interest is given as
\[SI=\dfrac{P\times T\times R}{100}\]
Where, \[P\] is the principal amount, \[T\] is the time period and \[R\] is the rate of interest per annum.
By using the above formula we get the simple interest in terms of time period as
\[\begin{align}
  & \Rightarrow SI=\dfrac{8000\times 5\times T}{100} \\
 & \Rightarrow SI=400T \\
\end{align}\]
Now, let us take the time period as 1 year then we get the simple interest of 1 year as
\[\begin{align}
  & \Rightarrow S{{I}_{1}}=400\times 1 \\
 & \Rightarrow S{{I}_{1}}=400 \\
\end{align}\]
Now, let us take the time period as 2 years then we get the simple interest for 2 years as
\[\begin{align}
  & \Rightarrow S{{I}_{2}}=400\times 2 \\
 & \Rightarrow S{{I}_{2}}=800 \\
\end{align}\]
Now, let us take the time period as 3 years then we get the simple interest for 3 years as
\[\begin{align}
  & \Rightarrow S{{I}_{3}}=400\times 3 \\
 & \Rightarrow S{{I}_{3}}=1200 \\
\end{align}\]
Now, let us take the time period as 4 years then we get the simple interest for 4 years as
\[\begin{align}
  & \Rightarrow S{{I}_{4}}=400\times 4 \\
 & \Rightarrow S{{I}_{4}}=1600 \\
\end{align}\]
Now, let us take the time period as 5 years then we get the simple interest for 5 years as
\[\begin{align}
  & \Rightarrow S{{I}_{5}}=400\times 5 \\
 & \Rightarrow S{{I}_{5}}=2000 \\
\end{align}\]
Now, let us take the time period as 2 years then we get the simple interest for 2 years as
\[\begin{align}
  & \Rightarrow S{{I}_{6}}=400\times 6 \\
 & \Rightarrow S{{I}_{6}}=2400 \\
\end{align}\]
Now, let us create a table containing the order pairs containing number of years as x co – ordinate and simple interest as y co – ordinate as follows

Time period \[\left( T \right)\]	Simple interest \[\left( S{{I}_{i}} \right)\]	Ordered pair\[\left( T,S{{I}_{i}} \right)\]
1	400	\[\left( 1,400 \right)\]
2	800	\[\left( 2,800 \right)\]
3	1200	\[\left( 3,1200 \right)\]
4	1600	\[\left( 4,1600 \right)\]
5	2000	\[\left( 5,2000 \right)\]
6	2400	\[\left( 6,2400 \right)\]

Now, let us consider a scale of the required graph as follows
On X – axis 1 unit equal to 1 year
On Y – axis 1 unit equal to RS. 400
By using the above scale and the coordinate of points in the table we get the graph as

Note:
 Students may make mistakes in finding the simple interest of 2, 3, 4, …, 6 years.
By using the simple interest formula for 2 years we get
\[\begin{align}
  & \Rightarrow S{{I}_{2}}=\dfrac{8000\times 5\times 2}{100} \\
 & \Rightarrow S{{I}_{2}}=800 \\
\end{align}\]
But students may make mistakes in taking the principal value for the time period of 2 years.
The amount after the 1 year is given as
\[\Rightarrow A=8000+400=8400\]
Students may consider this amount of 8400 as the principal value for next year. This is wrong because the simple interest is calculated based on the deposited amount that is 8000.