A person invested Rs $6000$ at beginning of every year at $10\% $ Compound Interest Calculate amount at the
(i) end of Second Year
(ii) end of third year
Answer
615k+ views
Hint:
The formula for the compound interest is ${\text{P = A}}{\left( {1 + \dfrac{r}{{100}}} \right)^t}$ Where P = the final amount A = amount to be paid in initially that is $6000$ r = Rate of interest that is $10\% $ and t = time in the part (i) it is $2$ and in the part (ii) it is $3$ put the values and get the answer.
Complete step by step solution:
As in the question it is given that the person invested Rs $6000$ at beginning of every year at $10\% $ So we have to calculate the Compound Interest at the end of second year or at the end of the third year for this ,
We know the formula of compound interest that is
${\text{P = A}}{\left( {1 + \dfrac{r}{{100}}} \right)^t}$
Where P = the final amount
A = amount to be paid in initially that is $6000$
r = Rate of interest that is $10\% $
and t = time in the part (i) it is $2$ and in the part (ii) it is $3$
Now solving for the Part (i) that is compound interest at the end of Second year
$ = 6000{\left( {1 + \dfrac{{10}}{{100}}} \right)^2}$
As on solving $\dfrac{{10}}{{100}} = 0.1$ hence
$6000{\left( {1.1} \right)^2}$
As the square of $1.1$ is $1.21$
$ = 6000 \times 1.21$
$ = 7260$
Hence at the end of Second year the Amount will be $7260$
Now solving for the Part (ii) that is compound interest at the end of third year
$ = 6000{\left( {1 + \dfrac{{10}}{{100}}} \right)^3}$
As on solving $\dfrac{{10}}{{100}} = 0.1$
Hence, $6000{\left( {1.1} \right)^3}$
As the cube of $1.1$ is $1.331$
$ = 6000 \times 1.331$
$ = 7986$
Hence at the end of third year the Amount will be $7986$.
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.
Simple interest is a quick and easy method of calculating the interest charge on a loan.
The formula for the compound interest is ${\text{P = A}}{\left( {1 + \dfrac{r}{{100}}} \right)^t}$ Where P = the final amount A = amount to be paid in initially that is $6000$ r = Rate of interest that is $10\% $ and t = time in the part (i) it is $2$ and in the part (ii) it is $3$ put the values and get the answer.
Complete step by step solution:
As in the question it is given that the person invested Rs $6000$ at beginning of every year at $10\% $ So we have to calculate the Compound Interest at the end of second year or at the end of the third year for this ,
We know the formula of compound interest that is
${\text{P = A}}{\left( {1 + \dfrac{r}{{100}}} \right)^t}$
Where P = the final amount
A = amount to be paid in initially that is $6000$
r = Rate of interest that is $10\% $
and t = time in the part (i) it is $2$ and in the part (ii) it is $3$
Now solving for the Part (i) that is compound interest at the end of Second year
$ = 6000{\left( {1 + \dfrac{{10}}{{100}}} \right)^2}$
As on solving $\dfrac{{10}}{{100}} = 0.1$ hence
$6000{\left( {1.1} \right)^2}$
As the square of $1.1$ is $1.21$
$ = 6000 \times 1.21$
$ = 7260$
Hence at the end of Second year the Amount will be $7260$
Now solving for the Part (ii) that is compound interest at the end of third year
$ = 6000{\left( {1 + \dfrac{{10}}{{100}}} \right)^3}$
As on solving $\dfrac{{10}}{{100}} = 0.1$
Hence, $6000{\left( {1.1} \right)^3}$
As the cube of $1.1$ is $1.331$
$ = 6000 \times 1.331$
$ = 7986$
Hence at the end of third year the Amount will be $7986$.
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.
Simple interest is a quick and easy method of calculating the interest charge on a loan.
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