What is the difference between amount for 2 years and 3 years for Rs 500 at 10 compound interest

The difference between the compound interest and simple interest on a certain sum of money at 10% per annum for 2 years is Rs.500. Find the sum when the interest is compounded annually.

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The difference between the compound interest and simple interest on a certain sum of money at 10% per annum for 2 years is Rs.500. Find the sum when the interest is compounded annually.
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Answer

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Hint: Let the sum be \[x\] rupees. We know the compound interest \[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\] , we need to find P. we know simple interest formula \[S.I = P \times \dfrac{r}{{100}} \times T\] . We know compound interest is the difference between amount and principal amount. Since the difference between compound and simple interest is given we can find the value of \[x\] .

Complete step-by-step answer:
We know,
\[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\] , where A is amount, R is rate of interest, n is number of times the interest is compounded per year.
\[P = x\] , \[n = 2\] \[r = 10\] , substituting we get,
\[ \Rightarrow A = x{\left( {1 + \dfrac{{10}}{{100}}} \right)^2}\]
\[ \Rightarrow A = x{\left( {1 + \dfrac{1}{{10}}} \right)^2}\]
Taking L.C.M and simplifying we get,
\[ \Rightarrow A = x{\left( {\dfrac{{10 + 1}}{{10}}} \right)^2}\]
\[ \Rightarrow A = x{\left( {\dfrac{{11}}{{10}}} \right)^2}\]
We know that compound interest is the difference between the amount of money accumulated after n years and the principal amount.
\[C.I = A - P\]
\[ \Rightarrow C.I = x{\left( {\dfrac{{11}}{{10}}} \right)^2} - x\] .
Now to find the simple interest we have, \[S.I = P \times \dfrac{r}{{100}} \times T\]
Substituting the known values,
\[ \Rightarrow S.I = x \times \dfrac{{10}}{{100}} \times 2\]
\[ \Rightarrow S.I = x \times \dfrac{1}{{10}} \times 2\]
\[ \Rightarrow S.I = \dfrac{x}{5}\]
Given the difference between compound and simple interest is 500
\[ \Rightarrow C.I - S,I = 500\]
Substituting C.I and S.I we get
\[ \Rightarrow x{\left( {\dfrac{{11}}{{10}}} \right)^2} - x - \dfrac{x}{5} = 500\]
Simple division \[\dfrac{{11}}{{10}} = 1.1\] and \[\dfrac{1}{5} = 0.2\] we get,
\[ \Rightarrow x{(1.1)^2} - x - 0.2x = 500\]
\[ \Rightarrow 1.21x - 1x - 0.20x = 500\]
\[ \Rightarrow 0.21x - 0.20x = 500\]
Taking x as common,
\[ \Rightarrow (0.21 - 0.20)x = 500\]
\[ \Rightarrow 0.01x = 500\]
\[ \Rightarrow x = \dfrac{{500}}{{0.01}}\]
Multiply numerator and denominator by 100.
\[ \Rightarrow x = 50,000\]
That is \[P = 50,000\] .
\[50,000\] Rupees is the sum when the interest is compounded annually.
So, the correct answer is “\[P = 50,000\]”.

Note: Here we used three formulas. Remember the formula for simple interest, compound interest and amount formula. We can also take P as P as it is, and solve for P. Above all we did is substituting the given data in the formula and simplifying. Principal amount is the initial amount you borrow or deposit.

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Updated On: 27-06-2022

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