Question

A man borrowed Rs. 10000 at 5% per annum compound interest. He repays 35% of the sum borrowed at the end of the first year and 42% of the sum borrowed at the end of the second year. How much must be paid at the end of the third year in order to clear the debt?
(a) Rs. 3398.50
(b) Rs. 3371.50
(c) Rs. 3333.50
(d) Rs. 3307.50

Answer 1

Hint:Find the total amount to be paid at the end of first year by using the formula: $A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}$, where ‘A’ is the total amount accumulated after t years, P is the principal amount of borrowed amount, ‘n’ is the number of times debt is to paid in a year, ‘r’ is the rate of interest and ‘t’ is the duration of time in years. After finding the amount to be paid after 1 year, subtract 35% of 10000 from it. The amount obtained will be the principal amount for second year. Again use the formula $A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}$ to find the amount to be paid at the end of second year. Subtract 42% of 10000 from this amount to get the principal amount for third year. Now, again applying the formula: $A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}$, calculate the amount to be paid after the third year.
Complete step-by-step answer:
For first year,
P = 10000, r = 5%, t = 1 year and n = 1.
Therefore, amount to be paid after 1st year

& {{A}_{1}}=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}} \\\ & \Rightarrow {{A}_{1}}=10000{{\left( 1+\dfrac{5}{100\times 1} \right)}^{1\times 1}} \\\ & =10000\left( 1+\dfrac{5}{100} \right) \\\ & =10000\left( \dfrac{105}{100} \right) \\\ & =100\times 105 \\\ & =10500 \\\ \end{aligned}$$ Now, it is given that he has repaid 35% of the borrowed amount at the end of one year. Therefore, Amount left to be paid after 1 year $\begin{aligned} & =10500-\dfrac{35}{100}\times 10000 \\\ & =10500-3500 \\\ & =7000 \\\ \end{aligned}$ So, Rs. 7000 is the principal amount for 2nd year. For the time gap of first and second year, P = 7000, r = 5%, t = 1 year and n = 1. Therefore, amount to be paid after 2nd year $$\begin{aligned} & {{A}_{2}}=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}} \\\ & \Rightarrow {{A}_{2}}=7000{{\left( 1+\dfrac{5}{100\times 1} \right)}^{1\times 1}} \\\ & =7000\left( 1+\dfrac{5}{100} \right) \\\ & =7000\left( \dfrac{105}{100} \right) \\\ & =70\times 105 \\\ & =7350 \\\ \end{aligned}$$ Now, it is given that he has repaid 42% of the borrowed amount at the end of second year. Therefore, Amount left to be paid after 2nd year $\begin{aligned} & =7350-\dfrac{42}{100}\times 10000 \\\ & =7350-4200 \\\ & =3150 \\\ \end{aligned}$ So, Rs. 3150 is the principal amount for 3rd year. For the time gap of second and third year, P = 3150, r = 5%, t = 1 year and n = 1. Therefore, amount to be paid after 3rd year $$\begin{aligned} & {{A}_{3}}=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}} \\\ & \Rightarrow {{A}_{3}}=3150{{\left( 1+\dfrac{5}{100\times 1} \right)}^{1\times 1}} \\\ & =3150\left( 1+\dfrac{5}{100} \right) \\\ & =3150\left( \dfrac{105}{100} \right) \\\ & =\dfrac{315\times 105}{10} \\\ & =3307.50 \\\ \end{aligned}$$ Therefore, the amount to be paid after 3rd year is Rs. 3307.50. Hence, option (d) is the correct answer. Note: Here, the value of $n$ must be substituted carefully. We have to read the question carefully, as it is given that the rate is compounded annually, therefore, $n=1$ is substituted. Also, we must divide the given rate by 100 and then substitute in the equation.