Answer
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Hint: To solve this question, i.e., to find sum of money invested. We will make the equation for the difference given between the interests of third year and first year using compound interest formula. In this we will put all the given values, and then after solving, we will get our required answer.
Complete step-by-step answer:
We have been given that a certain sum of money is invested at the rate of \[10\% \] per annum at compound interest, and the interest is compounded annually.
We have also been given the difference between the interests of third year and first year is ₹\[1,105,\] and we need to find the sum invested.
So, the given rate of interest \[ = {\text{ }}10\% \]
The sum of money is invested annually at compound interest, and the difference between interest of \[3rd\] and \[1st\] year\[ = \] ₹\[1,105\]
So, we will apply the compound interest formula mentioned below.
$\Rightarrow A = P{(1 + \dfrac{R}{{100}})^t}$
where, A \[ = \] final amount
P \[ = \] sum of money invested
R \[ = \] rate of interest
t \[ = \] time period for which the sum is invested
Now, we have been given the difference between interest of 3rd and 1st year, i.e.,
$\Rightarrow P{(1 + \dfrac{R}{{100}})^t} - P(1 + \dfrac{R}{{100}}) = I$
So, using the above formula and putting all the given values, we get
$
\Rightarrow P{(1 + \dfrac{{10}}{{100}})^3} - P(1 + \dfrac{{10}}{{100}}) = 1105 \\
\Rightarrow P[{(\dfrac{{11}}{{10}})^3} - \dfrac{{11}}{{10}}] = 1105 \\
\Rightarrow P[\dfrac{{1331}}{{1000}} - \dfrac{{11}}{{10}}] = 1105 \\
\Rightarrow P[\dfrac{{1331 - 1100}}{{1000}}] = 1105 \\
\Rightarrow P(\dfrac{{231}}{{1000}}) = 1105 \\
\Rightarrow P = \dfrac{{1105 \times 1000}}{{231}} \\
$
$\Rightarrow P = $ ₹$4783.5$
Thus, the sum invested was ₹$4783.5$.
Note: In the solution, we were asked about the compound interest. Compound Interest is the addition of interest to the sum or deposited money. In the question, we were given about the difference between interest of third year and interest of first year, that is why we have used the formula mentioned above in the solution.
Complete step-by-step answer:
We have been given that a certain sum of money is invested at the rate of \[10\% \] per annum at compound interest, and the interest is compounded annually.
We have also been given the difference between the interests of third year and first year is ₹\[1,105,\] and we need to find the sum invested.
So, the given rate of interest \[ = {\text{ }}10\% \]
The sum of money is invested annually at compound interest, and the difference between interest of \[3rd\] and \[1st\] year\[ = \] ₹\[1,105\]
So, we will apply the compound interest formula mentioned below.
$\Rightarrow A = P{(1 + \dfrac{R}{{100}})^t}$
where, A \[ = \] final amount
P \[ = \] sum of money invested
R \[ = \] rate of interest
t \[ = \] time period for which the sum is invested
Now, we have been given the difference between interest of 3rd and 1st year, i.e.,
$\Rightarrow P{(1 + \dfrac{R}{{100}})^t} - P(1 + \dfrac{R}{{100}}) = I$
So, using the above formula and putting all the given values, we get
$
\Rightarrow P{(1 + \dfrac{{10}}{{100}})^3} - P(1 + \dfrac{{10}}{{100}}) = 1105 \\
\Rightarrow P[{(\dfrac{{11}}{{10}})^3} - \dfrac{{11}}{{10}}] = 1105 \\
\Rightarrow P[\dfrac{{1331}}{{1000}} - \dfrac{{11}}{{10}}] = 1105 \\
\Rightarrow P[\dfrac{{1331 - 1100}}{{1000}}] = 1105 \\
\Rightarrow P(\dfrac{{231}}{{1000}}) = 1105 \\
\Rightarrow P = \dfrac{{1105 \times 1000}}{{231}} \\
$
$\Rightarrow P = $ ₹$4783.5$
Thus, the sum invested was ₹$4783.5$.
Note: In the solution, we were asked about the compound interest. Compound Interest is the addition of interest to the sum or deposited money. In the question, we were given about the difference between interest of third year and interest of first year, that is why we have used the formula mentioned above in the solution.
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