Question

Thomas took out a loan of 20000rupees from a bank which charges 10 % interest, compounded annually. after two years he paid back 10000 rupees. To settle the loan how much should he pay at the end of three years​

Answer 1

Question:-

Thomas took out a loan of 20000rupees from a bank which charges 10 % interest, compounded annually. After two years he paid back 10000 rupees. To settle the loan how much should he pay at the end of three years?

Required Answer:-

Given:-

• Principle Amount = Rs.20000
• Rate of interest = 10%
• After 2 years, he paid Rs. 10000

To Find:-

• The amount to be paid after 3 years.

Solution:-

First, we have to calculate the amount with compound interest at the end of two years.

We know that:-

$\\ \boxed{ \sf{A = P(1 + \dfrac{R}{100} ){}^{n}}}$

Where,

• A is amount
• P is principle amount
• R is rate of interest
• n is number of years.

Substituting the values:-

$\\ \sf{A = 20000(1 + \dfrac{10}{100}) {}^{2} }$

$\\ \sf{ \implies{A = 20000 \times ( \frac{100 + 10}{100} ){}^{2}}}$

$\\ \sf{ \implies{A = 20000 \times ( 1 + 0.1 ){}^{2}}}$

$\\ \sf{ \implies{A = 20000 \times 1.21}}$

$\\ \sf{ \therefore{A = \bold{24200}}}$

Therefore, amount to be paid after 2 years is Rs.24200

But, he paid Rs.10000

Hence, remaining loan = Rs.(24200-10000) =Rs.14200

Now, we have:-

• Principle amount P' = Rs.14200
• Rate of interest R' = 10%
• Remaining time n' = (3-2) = 1 year

We have to calculate the interest amount after 1 year:-

$\\ \sf{A \prime = P\prime(1 + \dfrac{R \prime}{100}) {}^{n \prime}}$

Substituting the values:-

$\\ \sf{A \prime = 14200(1 + \dfrac{10}{100}) {}^{1}}$

$\\ \sf{ \implies{A\prime = 14200 \times (1 + 0.1)}}$

$\\ \sf{ \implies{A \prime = 14200 \times 1.1}}$

$\\ \sf{ \therefore{A \prime = \red{15620}}}$

Hence, he has to pay Rs.15620 at the end of 3 years.

Answer 2

Answer:

loan amount for the third theatre