Question

If a sum of Rs. 40,000 at compound interest of 5 p.m amounted to Rs. 44,100. Find the time for which the money was invented.

Answer 1

Hint: To solve this question, first we will take all the given data from the question and then convert the rate of interest into numeric value by dividing the rate by 100, then we will substitute all the values in the formula of Compound interest $A=P{{\left( 1+\dfrac{r}{n} \right)}^{n}}$, then on solving an equation and on simplification, we will gate the value of n by comparing L.H.S and R.H.S.

Complete step by step answer:
Before we solve the question let us see what is meaning of compound interest and what the components of it are.
Compound interest just simply the addition of interest to the principal sum of a loan or deposits
The components of compound interest are as follows :
(a) Initial Principal Balance – It is the amount of loan or bond or the sum put into an investment.
(b) Interest Rate – The amount charged by a lender for the use of assets expressed as a percentage of the principal.
(c) Number of times interest is applied per time period.
(d) Number of time periods elapsed.

Now, in question, it is given that sum of Rs. 40,000 at compound interest of 5% p.m amounted to Rs.44,100.
So, we can say that the Initial Principal rate equals to Rs. 40,000. Rate of interest equals to 5% per annum and total compound interest calculated for time period ‘n’ equals to Rs. 44,100.
So, rate = $\dfrac{5}{100}$ .
We know that compound interest equals to,
$A=P{{\left( 1+\dfrac{r}{n} \right)}^{n}}$ , where A = compound Interest, P = Initial Principal balance, r = rate of interest, n = Number of times interest is applied per time period.
Putting value sin formula we get
$44,100=40,000{{\left( 1+\dfrac{5}{100} \right)}^{n}}$
On simplifying we get
$\dfrac{44,100}{40,000}={{\left( 1+\dfrac{5}{100} \right)}^{n}}$
On simplification, we get
$\dfrac{441}{400}={{\left( 1+\dfrac{5}{100} \right)}^{n}}$
Again, simplifying, we get
\[\dfrac{441}{400}={{\left( 1+\dfrac{1}{20} \right)}^{n}}\]
Solving bracket, we get
\[\dfrac{441}{400}={{\left( \dfrac{20+1}{20} \right)}^{n}}\]
\[\dfrac{441}{400}={{\left( \dfrac{21}{20} \right)}^{n}}\]
We know that ${{21}^{2}}=441$ and ${{20}^{2}}=400$
So, we can re – write above equation as
\[{{\left( \dfrac{21}{20} \right)}^{2}}={{\left( \dfrac{21}{20} \right)}^{n}}\]
On comparing, both sides L.H.S and R.H.S, we get
n = 2
So, the required time Rs 40,000 at compound interest of 5 p.m amounted to Rs. 44,100 is 2 years.


Note:
To solve such types of questions one must know the formula of compound interest which is $A=P{{\left( 1+\dfrac{r}{n} \right)}^{n}}$ and the meaning of its components. While solving, convert the rate of interest into numeric value by dividing the rate of interest by 100 and solve the equation such a way that we get the value of n by doing less calculation. Try not to make any calculation mistake.