A loan of Rs. 21,000 at 10% per annum compounded annually is repaid by the reducing balance method in 2 equal yearly instalments, each paid at the end of a year. Match List-I with List-II.
| List-I | List-II | ||
| (A) | Each instalment | (I) | Rs. 11,000 |
| (B) | Interest part of the first instalment | (II) | Rs. 3,200 |
| (C) | Principal part of the second instalment | (III) | Rs. 12,100 |
| (D) | Total interest paid | (IV) | Rs. 2,100 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
The instalment is \( R = \displaystyle\frac{P\,i\,(1+i)^n}{(1+i)^n - 1} \) with P = 21,000, i = 0.1 and n = 2.
(A) \( R = \displaystyle\frac{21000 \times 0.1 \times 1.21}{0.21} = 12100 \) → (III)
(B) Interest in the first year is on the whole loan: \( 21000 \times 0.1 = 2100 \) → (IV)
(C) The first instalment repays \( 12100 - 2100 = 10000 \), leaving \( 11000 \); the second instalment clears this outstanding principal of 11,000 → (I)
(D) \( nR - P = 2 \times 12100 - 21000 = 3200 \) → (II)