
A man borrows Rs 12500 from a bank at 20% compound interest at the end of every year he pays Rs 2000 as part repayment. How much does he still owe to the bank after three such instalments?
(a). Rs. 15600
(b). Rs. 12864
(c). Rs. 12000
(d). Rs. 14320
Answer
595.2k+ views
- Hint: We will calculate compound interest for each year and then subtract the instalment he pays at the end of each year for three years by using the formula\[A=P{{\left( 1+\dfrac{r}{n} \right)}^{(n)(t)}}\], where A is the amount, P is the principle, R is the rate of interest, T is the time of investment and CI is the compound interest.
Complete step-by-step solution -
Given that a man borrows Rs 12500 from a bank at 20% compound interest at the end of every year he pays Rs 2000 as part repayment.
Amount which he has to pay after one year,
\[\begin{align}
& A=12500{{\left( 1+\dfrac{20}{100} \right)}^{1}} \\
& \Rightarrow A=12500\left( \dfrac{6}{5} \right) \\
& \Rightarrow A=15000 \\
\end{align}\]
Amount which he has to pay after one year is Rs 15000
After the payment 1st payment the amount he owes is = 15000 – 2000
i.e. Rs 13000
So, the principle for second year will be Rs 13000
Now the Amount for second year will be,
\[A=P{{\left( 1+\dfrac{r}{n} \right)}^{(n)(t)}}\]
\[\begin{align}
& A=13000{{\left( 1+\dfrac{20}{100} \right)}^{1}} \\
& \Rightarrow A=13000\left( \dfrac{6}{5} \right) \\
& \Rightarrow A=15600 \\
\end{align}\]
Amount which he has to pay after second year is Rs 15600
After the payment 2nd payment the amount he owes is = 15600 – 2000
i.e. Rs 13600
So, the principle for third year will be Rs 13600
Now the Amount for third year will be,
\[A=P{{\left( 1+\dfrac{r}{n} \right)}^{(n)(t)}}\]
\[\begin{align}
& A=13600{{\left( 1+\dfrac{20}{100} \right)}^{1}} \\
& \Rightarrow A=13600\left( \dfrac{6}{5} \right) \\
& \Rightarrow A=16320 \\
\end{align}\]
Amount which he has to pay after third year is Rs 16320
After the payment 3rd payment the amount he owes is = 16320 – 2000
i.e. Rs 14320 option (d)
Note: The possibility of error in this question can be calculation mistakes including the calculations done for every year separately. Always go for calculating amount and compound interest for every year separately then arrive at the result.
Complete step-by-step solution -
Given that a man borrows Rs 12500 from a bank at 20% compound interest at the end of every year he pays Rs 2000 as part repayment.
Amount which he has to pay after one year,
\[\begin{align}
& A=12500{{\left( 1+\dfrac{20}{100} \right)}^{1}} \\
& \Rightarrow A=12500\left( \dfrac{6}{5} \right) \\
& \Rightarrow A=15000 \\
\end{align}\]
Amount which he has to pay after one year is Rs 15000
After the payment 1st payment the amount he owes is = 15000 – 2000
i.e. Rs 13000
So, the principle for second year will be Rs 13000
Now the Amount for second year will be,
\[A=P{{\left( 1+\dfrac{r}{n} \right)}^{(n)(t)}}\]
\[\begin{align}
& A=13000{{\left( 1+\dfrac{20}{100} \right)}^{1}} \\
& \Rightarrow A=13000\left( \dfrac{6}{5} \right) \\
& \Rightarrow A=15600 \\
\end{align}\]
Amount which he has to pay after second year is Rs 15600
After the payment 2nd payment the amount he owes is = 15600 – 2000
i.e. Rs 13600
So, the principle for third year will be Rs 13600
Now the Amount for third year will be,
\[A=P{{\left( 1+\dfrac{r}{n} \right)}^{(n)(t)}}\]
\[\begin{align}
& A=13600{{\left( 1+\dfrac{20}{100} \right)}^{1}} \\
& \Rightarrow A=13600\left( \dfrac{6}{5} \right) \\
& \Rightarrow A=16320 \\
\end{align}\]
Amount which he has to pay after third year is Rs 16320
After the payment 3rd payment the amount he owes is = 16320 – 2000
i.e. Rs 14320 option (d)
Note: The possibility of error in this question can be calculation mistakes including the calculations done for every year separately. Always go for calculating amount and compound interest for every year separately then arrive at the result.
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