An automobile dealer wishes to buy four luxury cars of different brands given in the table below with some down payment and balance in equal monthly installments (EMI) for 10 years. The bank charges 9% interest per annum compounded monthly.
\(\left(\text{Given } \displaystyle\frac{0.0075 \times (1.0075)^{120}}{(1.0075)^{120} - 1} = 0.01266\right)\)
| Luxury Car | Price of the Car (in Rs.) | Down payment (in Rs.) |
| P | 25,00,000 | 5,00,000 |
| Q | 35,00,000 | 12,00,000 |
| R | 45,00,000 | 15,00,000 |
| S | 42,00,000 | 15,00,000 |
Match List-I with List-II
| List-I | List-II | ||
| Luxury Car | EMI (in Rs.) | ||
| (A) | P | (I) | 34,182 |
| (B) | Q | (II) | 37,980 |
| (C) | R | (III) | 29,118 |
| (D) | S | (IV) | 25,320 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
The loan for each car is its price minus the down payment. With i = 0.09/12 = 0.0075 per month and n = 120 months, the EMI formula (reducing balance) gives
\( \displaystyle \text{EMI} = \text{Loan} \times \frac{i(1 + i)^n}{(1 + i)^n - 1} = \text{Loan} \times 0.01266 \)
(A) P
This matches (IV).
Loan = 25,00,000 − 5,00,000 = 20,00,000
EMI = 2000000 × 0.01266 = 25,320
(B) Q
This matches (III).
Loan = 35,00,000 − 12,00,000 = 23,00,000
EMI = 2300000 × 0.01266 = 29,118
(C) R
This matches (II).
Loan = 45,00,000 − 15,00,000 = 30,00,000
EMI = 3000000 × 0.01266 = 37,980
(D) S
This matches (I).
Loan = 42,00,000 − 15,00,000 = 27,00,000
EMI = 2700000 × 0.01266 = 34,182
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).