Anand obtained a loan of Rs $125000$ from the Allahabad Bank for buying a computer. The bank charges compound interest at $8\% $ per annum, compounded annually. How much should he pay after $3$ years?
Answer
630.6k+ views
Hint: Use the formula given below, to find the amount paid by Anand after the 3 years.
The amount$\left( A \right) = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$, where P is the principal amount taken by Anand, $r$ is the rate of interest, and $n$ is the time in years.
Substitute the values and simplify to get the desired result.
Complete step by step solution:
It is given in the problem that Anand obtained a loan of Rs. $125000$ from Allahabad Bank for buying a computer and the bank charges the compound interest of $8\% $ per annum, compounded annually.
We have to find the amount that Anand has to pay after the time period of $3$ years.
Assume that the amount taken by Anand be $P$ at the interest of $r$ and he will pay back to the bank after $n$ years.
Given,
The principal amount taken by Anand from the bank $\left( P \right) = Rs.125000$
Interest rate charges $\left( r \right) = 8\% $per annum
Time $\left( n \right) = 3$years
To find the amount, that Anand has to pay after the 3 years, use the formula:
The amount $\left( A \right) = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$, where P is the principal amount taken by Anand, $r$ is the rate of interest and $n$ is the time in years.
Substitute the values $P = 125000$, $r = 8$, and $n = 3$ into the equation:
$A = 125000{\left( {1 + \dfrac{8}{{100}}} \right)^3}$
Simplify the above equation to get the amount.
\[A = 125000{\left( {1 + \dfrac{2}{{25}}} \right)^3}\]
\[A = 125000{\left( {\dfrac{{27}}{{25}}} \right)^3}\]
\[A = 125000\left( {\dfrac{{27}}{{25}} \times \dfrac{{27}}{{25}} \times \dfrac{{27}}{{25}}} \right)\]
\[A = 8 \times 27 \times 27 \times 27\]
\[A = 157464\]
Therefore, the amount paid by Anand after the three years is Rs.$157464$.
Note: In case of simple interest, we only have to pay the interest on the principal amount but in case of compound interest, we have to pay the interest on the principal as well as the interest too. Therefore, the simple interest is the cheapest way to take a loan.
The amount$\left( A \right) = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$, where P is the principal amount taken by Anand, $r$ is the rate of interest, and $n$ is the time in years.
Substitute the values and simplify to get the desired result.
Complete step by step solution:
It is given in the problem that Anand obtained a loan of Rs. $125000$ from Allahabad Bank for buying a computer and the bank charges the compound interest of $8\% $ per annum, compounded annually.
We have to find the amount that Anand has to pay after the time period of $3$ years.
Assume that the amount taken by Anand be $P$ at the interest of $r$ and he will pay back to the bank after $n$ years.
Given,
The principal amount taken by Anand from the bank $\left( P \right) = Rs.125000$
Interest rate charges $\left( r \right) = 8\% $per annum
Time $\left( n \right) = 3$years
To find the amount, that Anand has to pay after the 3 years, use the formula:
The amount $\left( A \right) = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$, where P is the principal amount taken by Anand, $r$ is the rate of interest and $n$ is the time in years.
Substitute the values $P = 125000$, $r = 8$, and $n = 3$ into the equation:
$A = 125000{\left( {1 + \dfrac{8}{{100}}} \right)^3}$
Simplify the above equation to get the amount.
\[A = 125000{\left( {1 + \dfrac{2}{{25}}} \right)^3}\]
\[A = 125000{\left( {\dfrac{{27}}{{25}}} \right)^3}\]
\[A = 125000\left( {\dfrac{{27}}{{25}} \times \dfrac{{27}}{{25}} \times \dfrac{{27}}{{25}}} \right)\]
\[A = 8 \times 27 \times 27 \times 27\]
\[A = 157464\]
Therefore, the amount paid by Anand after the three years is Rs.$157464$.
Note: In case of simple interest, we only have to pay the interest on the principal amount but in case of compound interest, we have to pay the interest on the principal as well as the interest too. Therefore, the simple interest is the cheapest way to take a loan.
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