A person takes a loan from the bank and repays it in two equal installments of ₹ 729. If the rate of interest for the loan was \(12{1 \over 2}\)%, then for how much money, he took a loan from the bank?

## Answer

Correct Answer : d ) ₹ 1224

Explanation :**Quick Approach**

We know that,

\(12{1 \over 2}\)% \( = {1 \over 8}\)

The ratio of money corresponding to installment on which the rate of interest is applied to the value of the first installment is equal to [1 : (1 + rate of the loan)]

Also, for second installment the ratio of money remaining on which the rate of interest is applied to the value of the second installment is equal to [1 : (1 + rate of the loan)^{2}], and so on for n^{th} installment, the ratio of money remaining on which the rate of interest is applied to the value of n^{th} installment is equal to [1 : (1 + a rate of the loan)^{n}]

Thus, ratio of 1^{st} installment will be = [1 : (1 + 1/8)] = 8 : 9

According to the question,

Value of installment = ₹ 729

Here, 9 units corresponds to 729

So, 1 unit will correspond to 729/9 = 81

And, money corresponding to 1^{st} installment on which the rate of interest is applied = 8 * 81 = ₹ 648

Now, for second installment the ratio will be = [1 : (1 + rate of loan)^{2}] = [1 : (1 + 1/9)^{2}] = 64 : 81

According to the question,

Value of installment = ₹ 729

Here, 81 units correspond to 729

So, 1 unit will correspond to 729/81 = 9

And, money corresponding to 2^{nd} installment on which the rate of interest is applied = 64 * 9 = ₹ 576

Thus, total money he took a loan from the bank = 648 + 576 = ₹ 1224

Hence, (d) is the correct answer.

**Basic Approach**

Let the amount of equal installment be Rs. X

Let P be the principal amount and R be the rate of interest

In the case of equal installment, let the amount for a first and second year is P1, and P2 respectively.

Now,

\(P1= P (1+{R \over 100})\)

Now, from this amount Rs. X will be paid at the end of the first year so the new Principal will be P1 – X

\(P2= (P1 - X) (1+{R \over 100})\)

Since at the end of the second year the amount to be paid will be only the installment of Rs. X

Thus,

\(P2= (P ( 1+{R \over 100}) -X)(1+{R \over 100})=X\)

Or,

\( P ( 1+{R \over 100})^2=X((1+{R \over 100})+1)........(I)\)

Thus, using equation (I) we can say that

\( P ( 1+{12{1 \over 2} \over 100})^2=X((1+{12{1 \over 2} \over 100})+1)\)

\(P({81 \over 64}) = 729(({9 \over 8}) +1)\)

\(P({81 \over 64}) = 729*{17 \over 8}\)

P = 9 * 8 * 17 = ₹ 1224

Hence, (d) is the correct answer.

Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, UPSC CDS etc.

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