Question

Simple interest on a certain sum of money for 4 years at \[4\% \] per annum exceeds the compound interest on the same sum for 3 years at \[5\% \] per annum by Rs. 228. Find the sum.

Answer 1

Hint:
Here, we need to find the sum of money. First, we will find the simple and compound interest using the respective formulae. Then, we will use the given information to form an equation. Finally, we will solve this equation to get the value of \[P\], and hence, the sum of money.
Formula used: We will use the following formulas:
1) The simple interest is given by \[S.I. = \dfrac{{P \times R \times T}}{{100}}\], where \[P\] is the principal amount, \[R\] is the rate of interest, and \[T\] is the time period.
2) The amount is the sum of the principal amount and the interest. It is given by the formula \[A = P + S.I.\], where \[P\] is the principal amount and \[S.I.\] is the simple interest.
3) The amount \[A\] of an investment after \[t\] years is given by \[A = P{\left( {1 + \dfrac{r}{n}} \right)^{nt}}\], where \[P\] is the amount invested, \[n\] is the number of compounding periods in a year and \[r\] is the interest rate compounded annually.
4) The compound interest on a sum of money is given by \[C.I. = A - P\], where \[A = P{\left( {1 + \dfrac{r}{n}} \right)^{nt}}\] is the final amount and \[P\] is the amount invested.

Complete step by step solution:
Let \[P\] be the sum of money.
We will use the formula for simple interest to find the simple interest the sum of money for 4 years at \[4\% \] per annum.
 The simple interest is given by \[S.I. = \dfrac{{P \times R \times T}}{{100}}\], where \[P\] is the principal amount, \[R\] is the rate of interest, and \[T\] is the time period.
Substituting \[T = 4\] and \[R = 4\] in the formula, we get
\[ \Rightarrow S.I. = \dfrac{{P \times 4 \times 4}}{{100}}\]
Multiplying the terms, we get
\[ \Rightarrow S.I. = \dfrac{{16P}}{{100}} = \dfrac{{4P}}{{25}}\]
Now, the amount \[A\] of an investment after \[t\] years is given by \[A = P{\left( {1 + \dfrac{r}{n}} \right)^{nt}}\], where \[P\] is the amount invested, \[n\] is the number of compounding periods in a year and \[r\] is the interest rate compounded annually.
We will use this formula to calculate the amount if the sum of money is compounded annually for 3 years at \[5\% \] per annum.
Since the sum is compounded annually, the number of compounding periods in a year is 1.
Substituting \[n = 1\], \[t = 3\] and \[R = 5\% \] in the formula, we get
\[ \Rightarrow A = P{\left( {1 + \dfrac{{5\% }}{1}} \right)^{1 \times 3}}\]
Simplifying the expression, we get
\[ \Rightarrow A = P{\left( {1 + \dfrac{5}{{100}}} \right)^3}\]
Adding the terms in the expression, we get
\[\begin{array}{l} \Rightarrow A = P{\left( {\dfrac{{105}}{{100}}} \right)^3}\\ \Rightarrow A = {\left( {\dfrac{{21}}{{20}}} \right)^3}P\end{array}\]
Applying the exponent on the base, we get
\[ \Rightarrow A = \dfrac{{9261}}{{8000}}P\]
The compound interest on a sum of money is given by \[C.I. = A - P\], where \[A = P{\left( {1 + \dfrac{r}{n}} \right)^{nt}}\] is the final amount and \[P\] is the amount invested.
Substituting \[A = \dfrac{{9261}}{{8000}}P\] in the formula, we get
Therefore, we get
\[ \Rightarrow C.I. = \dfrac{{9261}}{{8000}}P - P\]
Taking the L.C.M. and subtracting the terms, we get
\[\begin{array}{l} \Rightarrow C.I. = \dfrac{{9261P - 8000P}}{{8000}}\\ \Rightarrow C.I. = \dfrac{{1261P}}{{8000}}\end{array}\]
Now, it is given in the question the simple interest on a certain sum of money for 4 years at \[4\% \] per annum exceeds the compound interest on the same sum of money for 3 years at \[5\% \] per annum by Rs. 228.
We can write this in mathematical form as
\[S.I. - C.I. = {\rm{Rs}}{\rm{. }}228\]
Substituting \[S.I. = \dfrac{{4P}}{{25}}\] and \[C.I. = \dfrac{{1261P}}{{8000}}\] in the equation, we get
\[ \Rightarrow \dfrac{{4P}}{{25}} - \dfrac{{1261P}}{{8000}} = 228\]
We can observe that this is a linear equation in one variable in terms of \[P\].
We will simplify and solve this equation to find the sum of money.
The L.C.M. of the denominators 25 and 8000 is 8000.
Taking the L.C.M., we get
\[ \Rightarrow \dfrac{{1280P - 1261P}}{{8000}} = 228\]
Subtracting the like terms, we get
\[ \Rightarrow \dfrac{{19P}}{{8000}} = 228\]
Multiplying both sides of the equation by 8000, we get
\[\begin{array}{l} \Rightarrow \dfrac{{19P}}{{8000}} \times 8000 = 228 \times 8000\\ \Rightarrow 19P = 1824000\end{array}\]
Dividing both sides of the equation by 19, we get
\[ \Rightarrow \dfrac{{19P}}{{19}} = \dfrac{{1824000}}{{19}}\]
Thus, we get
\[ \Rightarrow P = {\rm{Rs}}{\rm{. }}96000\]

\[\therefore\] The sum of money is Rs. 96,000.

Note:
We have formed a linear equation in one variable in terms of \[P\] in the solution. A linear equation in one variable is an equation that can be written in the form \[ax + b = 0\], where \[a\] is not equal to 0, and \[a\] and \[b\] are real numbers. For example, \[x - 100 = 0\] and \[100P - 566 = 0\] are linear equations in one variable \[x\] and \[P\] respectively. A linear equation in one variable has only 1 solution.