Question

Sangeeta invested RS. 20000 in the bank for four years and got back RS. 29282 after 4 years then, what was the rate of interest per annum payable annually?

Answer 1

Hint: We solve this problem by using the compound interest formula because we are given that the amount is payable annually.
The formula for amount in the compound interest is given as
\[\Rightarrow A=P{{\left( 1+\dfrac{R}{100} \right)}^{T}}\]
Where, \['A'\] is the final amount we get after certain period, \['P'\] is the principal amount that we invest initially, \['R'\] is the rate of interest per annum and \['T'\] is the time period of the invest.
By using this formula we can calculate the rate of interest directly.

Complete step by step answer:
We are given that Sangeeta invested RS. 20000/- in the bank.
Let us assume that the principal amount that Sangeeta invested as
\[\Rightarrow P=20000\]
We are given that she gets total of RS. 29282/-
Let us assume that the amount she gets finally as
\[\Rightarrow A=29282\]
We are given that the time period as 4 years.
Let us assume that the time period as
\[\Rightarrow T=4\]
Now, let us assume that the rate of interest per annum as \['R'\]
We know that the formula for amount in the compound interest is given as
\[\Rightarrow A=P{{\left( 1+\dfrac{R}{100} \right)}^{T}}\]
Where, \['A'\] is the final amount we get after certain period, \['P'\] is the principal amount that we invest initially, \['R'\] is the rate of interest per annum and \['T'\] is the time period of the invest.
Now, by using the above formula to given data we get
\[\begin{align}
  & \Rightarrow 29282=20000{{\left( 1+\dfrac{R}{100} \right)}^{4}} \\
 & \Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{4}}=\dfrac{29282}{20000} \\
\end{align}\]
Now let us divide the numerator with 2 in the RHS then we get
\[\Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{4}}=\dfrac{14641}{10000}\]
Here, we can see that there is a power of 4 on LHS.
So, let us convert the RHS also as a power of 4 because we have \[{{11}^{4}}=14641\] and \[{{10}^{4}}=10000\] then we get
\[\Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{4}}={{\left( \dfrac{11}{10} \right)}^{4}}\]
We know that if \[{{a}^{n}}={{b}^{n}}\] then \[a=b\]
Now, by using this theorem to above equation we get
\[\begin{align}
  & \Rightarrow \left( 1+\dfrac{R}{100} \right)=\dfrac{11}{10} \\
 & \Rightarrow \dfrac{R}{100}=\dfrac{11}{10}-1 \\
 & \Rightarrow R=10\% \\
\end{align}\]

Therefore, the rate of interest per annum is 10%.

Note: Students may make mistakes in taking the problem as compound interest.
We are given that the amount is payable annually which means that the interest is added for every year not for 4 years.
If the interest is added every year then it is the compound interest and we have the formula as
\[\Rightarrow A=P{{\left( 1+\dfrac{R}{100} \right)}^{T}}\]
Where, \['A'\] is the final amount we get after certain period, \['P'\] is the principal amount that we invest initially, \['R'\] is the rate of interest per annum and \['T'\] is the time period of the invest.
If the interest is added for 4 years then it is the simple interest and we have the formula as
\[\Rightarrow A=P+\dfrac{P\times T\times R}{100}\]