Question

Mr. Ravi borrowed Rs.16,000 for 2 years. The rate of interest for the two successive years are \[10\% \] and \[12\% \] respectively. If the repays \[Rs{\text{ }}5600\] at the end of first year, find the amount outstanding at the end of second year.

Answer 1

Hint: The given question is about the rate of interest and after some specific or certain time period, how much our actual money or principle gets interest and after getting or adding interest, how much there is increment in the principle which we will find out using some identities.

Complete step-by-step answer:
 In the question, we had given that a person Mr. Ravi has borrowed \[Rs{\text{ }}16000\] for \[2\] years Therefore \[Rs{\text{ }}16000\] is principle and the rate for keeping this money for \[2\] successive (one after another) Years were \[10\% \] and \[12\% \] respectively. It means in the first year rate of interest \[10\% \] and for second year, it is \[12\% \] also the pays \[Rs{\text{ }}5600\] at the end of first year and we have to find out the amount which he had to repay at the end of second year. We are already having the identity to find the amount after having some rate of interest and some time period and the identity is
\[A = P{\left( {1 + \dfrac{{{r_1}}}{{100}}} \right)^{{t_1}}}{\left( {1 + \dfrac{{{r_2}}}{{100}}} \right)^{{t_2}}}\]
Where \[A\] is the amount which has to be found \[P\] is principle which means actual money borrowed \[{r_1}\] and \[{r_2}\] are two rate of interest on which the amount \[X\] added.
\[{t_1}\] and \[{t_2}\] are two time periods for which time the principle becomes amount after adding specific rate of interest and we are going to substitute the values which are already given to is that \[P = 16000,{\text{ }}{r_1} = 10\% ,{\text{ }}{r_2} = 12\% \]
\[{t_1} = 1\] and \[{t_2} = 1\]
We get,
\[A = 16000{\left( {1 + \dfrac{{10}}{{100}}} \right)^1}{\left( {1 + \dfrac{{12}}{{100}}} \right)^1}\]
\[ \Rightarrow A = 16000\left( {\dfrac{{110}}{{100}}} \right)\left( {\dfrac{{112}}{{100}}} \right)\]
By taking LCM in the brackets
Now we will cancel out the terms and hence after calculation we get
\[A = {\text{16}}\left( {11} \right)\left( {112} \right)\]
\[ \Rightarrow A = 19712\]
This is the amount after \[2\] years but person had already paid the amount after first year which is \[Rs{\text{ }}5600\].Therefore this amount will be deducted from it because we had to find out the amount at the end of second year only. Therefore we get amount left for second year
\[ = Rs\left( {19712 - 5600} \right)\]
\[ \Rightarrow Rs{\text{ }}14112\]
Therefore Mr. Ravi will have to pay \[Rs{\text{ }}14112\] at the end of second year.

Note: The identity used to find out the amount is the identity which can be extended to three, four years too or it can be reduced to one year too. Since the question was about two years therefore we are using it. Where both the time periods are the same one year but rate of interest is different for \[2\] years.