Question

A sum of money is lent out at compound interest for two years at 20% per annum, compound interest being reckoned yearly. If the same sum of money was lent out at compound interest at the same rate per annum, compound interest being reckoned half-yearly, it would have fetched Rs. 482 more by way of interest in two years. Calculate the sum of money lent out.

Answer 1

Hint:Compound interest is the sum of the principle and the interest. Compound interest is generally found by \[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\]where P is the principle on which interest is to be calculated for a given rate\[r\].
In this question, two conditions have been given relating to the rate of interest for which we need to establish the equations and solve them accordingly to get the value of the principal amount. We first need to calculate the amount from the given principle for the first year, which becomes the principal amount for the second year and then, calculate the amount on the principal amount, obtained from the second year. The compound can also be calculated half-yearly, quarterly and monthly.

Complete step-by-step solution:
Let the principal amount be P.
Substitute \[T = 2years\] and \[R = 20\% = \dfrac{{20}}{{100}}\] in the formula \[A = P{\left( {1 + \dfrac{R}{{100}}} \right)^T}\] to get the relation between A and P as:
\[
  A = P{\left( {1 + \dfrac{R}{{100}}} \right)^T} \\
  A = P{\left( {1 + \dfrac{{20}}{{100}}} \right)^2} - - - - (i) \\
 \]
As, interest per year is the difference in the amount and the principal value so, \[{I_y} = A - P\] . Substituting the value of A from equation (i) in \[{I_y} = A - P\], we get:
\[
  {I_y} = P{\left( {1 + \dfrac{{20}}{{100}}} \right)^2} - P \\
   = P{\left( {1 + \dfrac{1}{5}} \right)^2} - P \\
   = P{\left( {\dfrac{6}{5}} \right)^2} - P \\
   = \dfrac{{36P}}{{25}} - P - - - - (ii) \\
 \]
Now, the rate of interest is imposed half-yearly; hence time would be doubled, and the rate must be halved:
So, substitute \[T = 4years\] and \[R' = \dfrac{{20}}{2}\% = 10\% \] in the equation \[A' = P{\left( {1 + \dfrac{{R'}}{{100}}} \right)^T}\] to determine the relation between A and P as:
 \[
  A' = P{\left( {1 + \dfrac{{R'}}{{100}}} \right)^T} \\
   = P{\left( {1 + \dfrac{{10}}{{100}}} \right)^4} \\
   = P{\left( {1 + \dfrac{1}{{10}}} \right)^4} - - - - (iii) \\
 \]
Again, interest per year is the difference in the amount and the principal value so, \[{I_y}' = A' - P\] . Substituting the value of $A'$ from equation (iii) in \[{I_y}' = A' - P\], we get:
\[
  {I_y}' = P{\left( {\dfrac{{11}}{{10}}} \right)^4} - P \\
   = \dfrac{{121 \times 121P}}{{10000}} - P - - - - (iv) \\
 \]
According to the question, the interest accumulated for the half-yearly rate of interest is more than that accumulated for the yearly rate of interest by 482, so \[{I_y}' = {I_y} + 482\] . Substitute the value of $A$ and $A'$ from the equation (ii) and (iv) in the equation \[{I_y}' = {I_y} + 482\] to determine the value of principal amount.
\[
  {I_y}' = {I_y} + 482 \\
= P\left( {\dfrac{{121 \times 121}}{{10000}}} \right) - P = \left( {\dfrac{{36P}}{{25}} - P} \right) + 482 \\
= P\left( {\dfrac{{121 \times 121}}{{10000}}} \right) = \dfrac{{36P}}{{25}} + 482 - - - - (v) \\
 \]
Now, taking all the P’s terms on either side of the equation as:
\[
= \dfrac{{141641P}}{{10000}} - \dfrac{{36P}}{{25}} = 482 \\
= \dfrac{{\left( {14641 \times 25} \right)P - \left( {36 \times 10000} \right)P}}{{10000 \times 25}} = 482 \\
 \]
Cross-multiplying the terms to get the value of P as:
\[
= P\left( {366025 - 360000} \right) = 482 \times 25 \times 10000 \\
= 6025P = 120500 \\
= P = \dfrac{{120500000}}{{6025}} \\
   = 20000 \\
 \]
Hence, the sum of money lent out\[P = 20000\].

Note: Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other words, interest on interest. It is the result of reinvesting interest, rather than paying it out, so that interest in the next period is then earned on the principal sum plus previously accumulated interest.