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 $\frac{0.0075 × (1.0075)^{120}}{(1.0075)^{120}-1}=0.01266$)
|
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 Luxury Car |
List-II 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)
|
List-I Luxury Car |
List-II EMI (in Rs.) |
|
(A) P |
(IV) 25,320 |
|
(B) Q |
(III) 29,118 |
|
(C) R |
(II) 37,980 |
|
(D) S |
(I) 34,182 |
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).