CAT 2018 Set-1 | Question: 96

Question

John borrowed Rs.$2,10,000$ from a bank at an interest rate of $10\%$ per annum, compounded annually. The loan was repaid in two equal installments, the first after one year and the second after another year. The first installment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each installment, in Rs., is ________

Answer 1

Given that,

• Principle $=210000$
• Rate $ = 10 \%$

The amount after $1$ year on the compound interest.

Apply the formula find the amount on the compound interest $\boxed {\text{A} =\text{P} \left( 1+ \frac{R}{100} \right)^{T} } $

$ \text{A} = 210000 \left( 1+ \frac {10}{100} \right)^{1}$

$  \Rightarrow  \text{A} = 210000 \times \frac {11}{10}$

$ \Rightarrow \text{A} = 21000 \times 11$

$ \Rightarrow  \text{A} = 231000$

Let $x$ be the first installment.

So, due ammount $= \text{Rs.}\;231000-x$

And the due amount will be the principle of second-year $= (231000-x)$, and the rate will be the same as the first year.

Now, amount $\text{A} = (231000-x)  \left( 1+ \frac {10}{100} \right)^{1} $

$ \Rightarrow \text{A} =  (231000-x) \times \frac {11}{10}$

$ \Rightarrow \frac {2541000-11x} {10}$  (This amount is for the second installment)

We know that, installment for year $1 = $ installment for year $2= \text{Rs.}\;x$

$ \frac {2541000-11x}{10} = x$

$ \Rightarrow 2541000 – 11x = 10x $

$ \Rightarrow 2541000 = 21x$

$ \Rightarrow x = 121000 $

$ \therefore $ Each installment $  = \text {Rs.}\; 121000$

Correct Answer $: \text {Rs.}121000 $