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.

(Given $\frac{0.0075 × (1.0075)^{120}}{(1.0075)^{120}-1}=0.01266$)

Match List-I with List-II

Choose the correct answer from the options given below:

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Given data:

Luxury Cars:

P: Price = 25,00,000, Down payment = 5,00,000

Q: Price = 35,00,000, Down payment = 12,00,000

R: Price = 45,00,000, Down payment = 15,00,000

S: Price = 42,00,000, Down payment = 15,00,000

Bank charges 9% per annum compounded monthly → monthly interest rate:

$i = 0.09 / 12 = 0.0075$

Tenure: 10 years → n = 10*12 = 120 months

Loan amount for each car (Principal for EMI calculation):

P: 25,00,000 − 5,00,000 = 20,00,000

Q: 35,00,000 − 12,00,000 = 23,00,000

R: 45,00,000 − 15,00,000 = 30,00,000

S: 42,00,000 − 15,00,000 = 27,00,000

EMI formula:

$EMI = \frac{P \cdot i \cdot (1+i)^n}{(1+i)^n - 1}$

Approximate EMI values (matching with given options):

P (20,00,000) → EMI ≈ 25,320 → (A)-(IV)

Q (23,00,000) → EMI ≈ 29,118 → (B)-(III)

R (30,00,000) → EMI ≈ 37,980 → (C)-(II)

S (27,00,000) → EMI ≈ 34,182 → (D)-(I)

Reasoning: EMI increases with principal for the same interest rate and tenure. Therefore, the smallest loan (P) corresponds to the smallest EMI (25,320), and the largest loan (R) corresponds to the largest EMI (37,980).