Statistics & Applications

Financial Math

Medium

applied

A man plans to take a housing loan of Rs 99,53,000 from a bank costing 18% per annum compounded monthly. The loan is to be paid back in 30 years in equal monthly installments (EMI). The EMI by reducing balance method is:

[Given $(1.015)^{-360} = 0.0047$ ]

Rs 1,82,500

Rs 1,50,000

Rs 1,75,000

Rs 2,04,500

✅ Correct Option: 2

The loan amount is Rs 99,53,000 with an annual interest rate of 18% compounded monthly, to be repaid over 30 years.

The monthly interest rate is:

$r = \frac{18\%}{12} = 1.5\% = 0.015$

The number of monthly installments is:

$n = 30 \times 12 = 360$ months

The EMI formula for reducing balance method is:

$EMI = \frac{P \times r}{1 - (1+r)^{-n}}$

where $P$ is the principal, $r$ is the monthly interest rate, and $n$ is the number of months.

Substituting the values:

$EMI = \frac{99,53,000 \times 0.015}{1 - (1.015)^{-360}}$

Using the given value $(1.015)^{-360} = 0.0047$ :

$EMI = \frac{99,53,000 \times 0.015}{1 - 0.0047}$

Calculating the numerator:

$99,53,000 \times 0.015 = 1,49,295$

Calculating the denominator:

$1 - 0.0047 = 0.9953$

Therefore:

$EMI = \frac{1,49,295}{0.9953}$

$EMI = 1,50,000$

The monthly EMI is Rs 1,50,000.

2025: 19 May Shift 1

Which of the following is NOT correct about "Sinking Fund"?

2025: 21 May Shift 2

Mr. X wishes to purchase a flat for Rs. 44,65,000 with a down payment of Rs. 10,00,000 and balance in equated monthly installments (EMI) for 25 years. If the bank charges 6 % per annum compounded monthly, the EMI is:

[Given: $(1.005)^{300} =4.4650$ ]

2025: 14 May Shift 1

A person wishes to purchase a house for ₹ 39,65,000 with a down payment of ₹ 5,00,000 and balance in equal monthly installments (EMI) for 25 years. If bank charges 6% per annum (compounded monthly), then EMI on reducing balance payment method is:

[Given $(1.005)^{300} = 4.465$ ]

2025: 15 May Shift 1

Mr. X invested Rs. 4,00,000 in shares for 5 years. The value of this investment was Rs. 4,50,000 at the end of the second year, Rs. 490000 at the end of the third year and on maturity, the final value stood at Rs. 6,00,000. The compound annual growth rate of this investment is:

[Given that: $(1.5)^{1/5} = 1.084]$

2025: 2 June Shift 1

The amount of money needed to ensure for a prize of ₹ 5000 at the begining of each year indefinitely if money is worth 5% compounded annually is:

2026: 12 May Shift 2

Identify the formula used for List I from List II

Match List-I with List-II

List-I	List-II
(A) Present value of perpetuity payable at the end of each payment period	(I) $R+\dfrac{R}{i}$ (₹R payable at the beginning of each payment period, $i$ = interest per rupee per payment period)
(B) Present value of perpetuity payable at the beginning of each payment period	(II) $\left(\dfrac{V_n}{V_0}\right)^{1/n}-1$ , $V_n$ = Final value of investment, $V_0$ = Beginning value of the investment, $n$ = number of years
(C) Computation of CAGR	(III) $\dfrac{C-s}{n}$ , $C$ = Original cost, $s$ = Scrap value, $n$ = number of years
(D) Annual Depreciation	(IV) $\dfrac{R}{i}$ (₹R payable at the end of each period, $i$ = interest per rupee per payment period)

Choose the correct answer from the options given below:

A man plans to take a housing loan of Rs 99,53,000 from a bank costing 18% per annum compounded monthly. The loan is to be paid back in 30 years in equal monthly installments (EMI). The EMI by reducing balance method is: [Given (1.015)−360=0.0047(1.015)^{-360} = 0.0047(1.015)−360=0.0047]

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

A man plans to take a housing loan of Rs 99,53,000 from a bank costing 18% per annum compounded monthly. The loan is to be paid back in 30 years in equal monthly installments (EMI). The EMI by reducing balance method is:

[Given $(1.015)^{-360} = 0.0047$ ]