CUET Mathematics: Statistics & ApplicationsFinancial Math. Free, no login required.

Q1:

2026: 12 May Shift 2

Financial Math

Medium

A car whose cost is ₹9,00,000 will depreciate to a scrap value of ₹1,20,000 in 12 years. Which of the following statements are correct ?

(A) The annual depreciation is ₹65,000

(B) The book value of the car at the end of the fourth year is ₹6,40,000

(C) The book value of the car at the end of the sixth year is ₹5,00,000

(D) The book value of the car at the end of the fifth year is ₹5,75,000

Choose the correct answer from the options given below:

Answer options
Option 2
Correct Answer
Explanation for 2026: 12 May Shift 2 MAT question 1

Q2:

2026: 12 May Shift 2

Financial Math

Easy

Mr. Bean purchased a laptop for ₹75,000. The Laptop is estimated to have a scrap value of ₹10,000 after a span of 8 years. The book value of the laptop at the end of 5 years is

Answer options
Option 4
Correct Answer
Explanation for 2026: 12 May Shift 2 MAT question 2

Q3:

2026: 12 May Shift 2

Financial Math

Easy

Rama takes a personal loan for a car worth ₹5,00,000 at an interest rate of 7% per annum to be repaid by equal monthly instalments (EMI) in 5 years. Using flat rate method, EMI will be:-

Answer options
Option 2
Correct Answer
Explanation for 2026: 12 May Shift 2 MAT question 3

Q4:

2026: 12 May Shift 2

Financial Math

Easy

Identify the formula used for List I from List II

Match List-I with List-II

List-IList-II
(A) Present value of perpetuity payable at the end of each payment period(I) R+RiR+\dfrac{R}{i} (₹R payable at the beginning of each payment period, ii = interest per rupee per payment period)
(B) Present value of perpetuity payable at the beginning of each payment period(II) (VnV0)1/n1\left(\dfrac{V_n}{V_0}\right)^{1/n}-1, VnV_n = Final value of investment, V0V_0 = Beginning value of the investment, nn = number of years
(C) Computation of CAGR(III) Csn\dfrac{C-s}{n}, CC = Original cost, ss = Scrap value, nn = number of years
(D) Annual Depreciation(IV) Ri\dfrac{R}{i} (₹R payable at the end of each period, ii = interest per rupee per payment period)

Choose the correct answer from the options given below:

Answer options
Option 3
Correct Answer
Explanation for 2026: 12 May Shift 2 MAT question 4

Q5:

2025: 3 June Shift 2

Financial Math

Medium

Applied

Ram invested Rs.20,000 in a mutual fund in the year 2012. The value of the mutual fund increased to Rs.32,000 in the year 2017, then the compound annual growth rate of his investment is (Given that (1.6)15=1.098(1.6)^{\frac{1}{5}} = 1.098)

Answer options
Option 3
Correct Answer
Explanation for 2025: 3 June Shift 2 MAT question 5

Q6:

2025: 3 June Shift 2

Financial Math

Easy

Applied

Mr. Vishnu has an initial investment of Rs.80,000 in an investment plan. After 3 years, it has grown to Rs.1,00,000, then his rate of return is

Answer options
Option 2
Correct Answer
Explanation for 2025: 3 June Shift 2 MAT question 6

Q7:

2025: 3 June Shift 2

Financial Math

Medium

Applied

Choose the correct statement about Sinking Fund?

Answer options
Option 3
Correct Answer
Explanation for 2025: 3 June Shift 2 MAT question 7

Q8:

2025: 3 June Shift 2

Financial Math

Medium

Applied

Choose the correct statement about CAGR(compound annual growth rate)?

Answer options
Option 4
Correct Answer
Explanation for 2025: 3 June Shift 2 MAT question 8

Q9:

2025: 3 June Shift 2

Financial Math

Easy

Applied

A machine costing Rs.2,00,000 has a useful life of 5 years.The estimated scrap value is Rs.20,000. By using straight line method, the annual depreciation is

Answer options
Option 3
Correct Answer
Explanation for 2025: 3 June Shift 2 MAT question 9

Q10:

2025: 3 June Shift 1

Financial Math

Medium

Applied

The average cost function for a commodity is given by AC=0.05x25x+1000+3000xAC = 0.05x^2 - 5x + 1000 + \frac{3000}{x} in terms of output x. The fixed cost is

Answer options

Q11:

2025: 3 June Shift 1

Financial Math

Medium

Applied

Match List-I with List-II

List-IList-II
(A) Perpetuity(I) A person deposits a fixed amount every year in his bank account to renovate his house after 10 yrs.
(B) EMI(II) A person depositis an amount regularly in his bank account and withdraws in case of need.
(C) Sinking Fund(III) A fixed amount is debited from the bank account of a person, every month, against a personal loan.
(D) Saving Account(IV) A person purchased a house and rents it out.

Choose the correct answer from the options given below:

Answer options

Q12:

2025: 3 June Shift 1

Financial Math

Medium

Applied

The effective rate, which is equivalent to a nominal rate of 12% compounded semi-annually, is

Answer options

Q13:

2025: 3 June Shift 1

Financial Math

Easy

Applied

Ajesh purchased a printer ₹ 15,000. The printer is estimated to have a scrap value of ₹ 3,000 after a span of 6 years. Then the book value of the printer at the end of 3 years will be:-

Answer options

Q14:

2025: 3 June Shift 1

Financial Math

Medium

Applied

A company purchased a machine for ₹ 15,00,000 and its effective life is estimated to be 10 years. A sinking fund is created for replacing the machine at the end of its effective life when its scrap value is ₹ 2,42,000. What amount company should provide, at the end of every year out of profits for the sinking fund if it accumulates an interest of 5% per annum? [Given(1.05)¹⁰=1.629]

Answer options

Q15:

2025: 3 June Shift 1

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(1.015)^{-360} = 0.0047]

Answer options

Q16:

2025: 3 June Shift 1

Financial Math

Medium

Applied

A startup company invested ₹ 5,00,000 in shares for 4 years. The value of the investment was ₹ 5,50,000 at the end of first year, ₹ 5,25,000 at the end of third year, and on maturity, the final value stood ₹ 6,25,000. The CAGR on the investment will be :- [Given : (1.25)14=1.06(1.25)^{\frac{1}{4}} = 1.06]

Answer options

Q17:

2025: 2 June Shift 1

Financial Math

Easy

Applied

The effective rate per annum equivalent to a nominal rate of 8% compounded semi-annually is

Answer options

Q18:

2025: 2 June Shift 1

Financial Math

Easy

Applied

A machine costing Rs 50,000 has a useful life of 4 years.The estimated scrap value is Rs 10,000 . The rate of depreciation per annum is:

Answer options

Q19:

2025: 2 June Shift 1

Financial Math

Medium

Applied

If CAGR stands for Compound Annual Growth Rate, F.V stands for final value of an investment, P.V stands for present value of an investment and n is the number of years then

Answer options

Q20:

2025: 2 June Shift 1

Financial Math

Medium

Applied

A machine costing ₹ 3,00,000 will have its scrap value of ₹ 50,000. The company at present plans to put ₹ 36,650 per annum at the end of each year in a sinking fund at the rate 5% per annum for the replacement of the machine after its useful life. Suppose the new machine will cost ₹ 4,00,000 at that time, then the useful life (approx.) of the machine is : [Given: (1.4775)1/8=1.05(1.4775)^{1/8} = 1.05]

Answer options

Q21:

2025: 2 June Shift 1

Financial Math

Medium

Applied

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:

Answer options

Q22:

2025: 2 June Shift 1

Financial Math

Medium

Applied

Maneesh took a loan of ₹ 9,00,800 from bank at an interest rate of 6% per annum for 10 years. If she has to pay the loan back with the help of equal monthly installments (EMI). Then, the EMI using reduced balance method is (approx):

[Given: (1.005)120=0.5496(1.005)^{-120}=0.5496]

Answer options

Q23:

2025: 30 May Shift 2

Financial Math

Medium

Applied

The effective rate of return equivalent to a nominal rate of 12% per annum compounded quarterly is: [Given that: (1.03)41.1255(1.03)^4 ≈ 1.1255]

Answer options

Q24:

2025: 30 May Shift 2

Financial Math

Medium

Applied

An investment of ₹ 3,00,000 becomes ₹ 4,50,000 in 5 years, then the compound annual growth rate (CAGR) is equal to:

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

Answer options

Q25:

2025: 30 May Shift 2

Financial Math

Medium

Applied

At what rate of interest will the present value of a perpetuity of ₹ 600 payable at the end of every 3 months be ₹ 18,000?

Answer options

Q26:

2025: 30 May Shift 2

Financial Math

Medium

Applied

The annual depreciation of a car is ₹ 40,000. If the scrap value of the car after 15 years is ₹ 50,000, then the original cost of the car using linear method is

Answer options

Q27:

2025: 30 May Shift 2

Financial Math

Medium

Applied

Ram wishes to purchase a house for ₹ 15,00,000 and made a down payment of ₹ 5,00,000. If he can amortize the balance at 9% per annum compounded monthly for 25 years, then his EMI is:

[Given (1.0075)3009.41(1.0075)^{300} ≈ 9.41]

Answer options

Q28:

2025: 30 May Shift 2

Financial Math

Medium

Applied

The cost of a property appreciates by 10% of the previous month every month. If in end march 2024 it was ₹ 13.31 lakh, when was it ₹ 10 lakh?

Answer options

Q29:

2025: 30 May Shift 1

Financial Math

Medium

Applied

A piece of machinery is bought for Rs. 50,000. In the first year, it depreciates by 15%, and in each subsequent year, the depreciation rate increases by 5% from the previous year. The value of machinery after 3 years will be:

Answer options

Q30:

2025: 30 May Shift 1

Financial Math

Medium

Applied

A man takes a personal loan worth Rs.3,00,000 at an interest rate of 6% per annum compounded monthly to be repaid by equal monthly installments in 3 years, then the EMI using flat rate method will be:-

Answer options