Question

At what rate will a sum of Rs.\[4000\] yield a compound interest of Rs.\[410\] in \[2\] years?

Answer 1

Hint: To solve this question we should have the knowledge of compound interest. We should know how to apply the formula of compound interest on real life problem statements. The questions from this topic always demand the conversion of problem statements into mathematical form. We have to just identify the required values from the given question and substitute the values in the formula of compound interest.

Complete step-by-step solution:
This question requires the knowledge of compound interest and how to evaluate compound interest by substituting the values in the formula. We should know how to convert this real life problem statement into mathematical form. We just need to know what is given in the question and just apply the required concepts. We will now move on to the discussion of the above concepts one by one.
Compound Interest: Compound interest is the addition of interest to the principal sum of a loan or deposit. In simple words, compound interest can also be defined as the interest on interest. In the case of simple annual interest it is the amount per period, multiplied by the number of periods per year.
For e.g. When you deposit money in a savings account, you will usually receive interest based on the amount that you deposited.
The above example is a real life example.
Compound interest is calculated by multiplying the initial principal amount by one plus the annual interest rate raised to the number of compound periods minus one. The total initial amount of the loan is then subtracted from the resulting value.
The formula for compound interest is given by,
\[A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}\]
Here,
\[A=\] final amount
\[P=\] initial principal balance
\[r=\] interest rate
\[n=\] number of times interest applied per time period
\[t=\] number of time periods elapsed
After revising all the required concepts, let us move to solve the main question.
In the given problem,
\[P=4000\]
\[A=4000+410=4410\]
\[t=2\]
\[n=1\]
Substituting the above values in above formula we get,
 \[4410=4000{{\left( 1+\dfrac{r}{1} \right)}^{1\times 2}}\]
\[\begin{align}
  & \Rightarrow \dfrac{441}{400}={{(1+r)}^{2}} \\
 & \Rightarrow \sqrt{\dfrac{441}{400}}=(1+r) \\
 & \Rightarrow 1+r=\dfrac{21}{20} \\
 & \Rightarrow r=\dfrac{21}{20}-1 \\
 & \Rightarrow r=\dfrac{1}{20} \\
\end{align}\]
Let, \[R\] be the rate of interest.
$ \Rightarrow R=r\times 100 $
$ \Rightarrow R=\dfrac{1}{20}\times 100 $
$ \Rightarrow R=5% $
From this we can conclude that the rate of interest is \[5%\].

Note: This question completely depends on the concept of compound interest. The concept of compound interest is derived from the basic concept of simple interest. The knowledge of converting real life problem statements into mathematical form is really important while solving these types of questions. It is just an application based question which requires the substitution of values in the formula.