Unit 7: Financial Mathematics

Unit 7: Financial Mathematics - Free Study Resources | Boundless Maths

Topics Covered MCQs (18) Short Answers (6) Long Answers (6) Case Studies (3)

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Topics Covered in Unit 7

Master these 6 important topics

1. Perpetuity

Infinite series of equal payments, present value calculations

2. Sinking Funds

Accumulation of funds for future payments, periodic deposits

3. EMI & Amortization

Equated Monthly Installments, loan repayment schedules

4. CAGR

Compounded Annual Growth Rate for investments

5. Depreciation

Straight line and reducing balance methods

6. Valuation of Bonds

Present value, yield to maturity, bond pricing

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Practice MCQs with Answers

Click "Show Answer" to reveal explanations

Question 1 Perpetuity

The present value of a perpetuity of ₹1000 per year at 8% per annum is:

A ₹8,000

B ₹10,000

C ₹12,500

D ₹15,000

✓ Correct Answer: (C) ₹12,500

Solution:
Formula: PV of Perpetuity = A/r
Where A = Annual payment = ₹1000, r = 8% = 0.08
PV = 1000/0.08 = ₹12,500

Question 2 Sinking Fund

A company wants to accumulate ₹1,00,000 in 5 years by making annual deposits. If interest rate is 10% per annum compounded annually, the annual deposit required is approximately:

A ₹16,380

B ₹18,000

C ₹20,000

D ₹22,500

✓ Correct Answer: (A) ₹16,380

Solution:
Sinking Fund Formula: A = S × r / ((1+r)ⁿ - 1)
S = ₹1,00,000, r = 0.10, n = 5
A = 1,00,000 × 0.10 / ((1.10)⁵ - 1)
A = 10,000 / (1.61051 - 1) = 10,000 / 0.61051 ≈ ₹16,380

Question 3 EMI

The EMI formula for a loan of P at rate r per month for n months is:

A EMI = P × r × (1+r)ⁿ / ((1+r)ⁿ - 1)

B EMI = P / n

C EMI = P × r

D EMI = P × (1+r)ⁿ

✓ Correct Answer: (A) EMI = P × r × (1+r)ⁿ / ((1+r)ⁿ - 1)

Solution:
The standard EMI formula is:
EMI = P × r × (1+r)ⁿ / ((1+r)ⁿ - 1)
Where P = Principal, r = monthly rate, n = number of months

Question 4 EMI

A loan of ₹50,000 is to be repaid in 12 monthly installments at 12% per annum. The monthly interest rate is:

A 0.01 or 1%

B 0.12 or 12%

C 0.05 or 5%

D 0.02 or 2%

✓ Correct Answer: (A) 0.01 or 1%

Solution:
Monthly rate = Annual rate / 12
= 12% / 12 = 1% = 0.01

Question 5 CAGR

An investment grows from ₹10,000 to ₹15,000 in 3 years. The CAGR is approximately:

A 12.5%

B 14.5%

C 16.7%

D 18.2%

✓ Correct Answer: (B) 14.5%

Solution:
CAGR = (Final Value/Initial Value)^(1/n) - 1
= (15,000/10,000)^(1/3) - 1
= (1.5)^(1/3) - 1
= 1.1447 - 1 = 0.1447 ≈ 14.5%

Question 6 CAGR

The formula for CAGR is:

A (Ending Value - Beginning Value) / Beginning Value

B (Ending Value / Beginning Value)^(1/n) - 1

C (Ending Value - Beginning Value) / n

D Ending Value / (Beginning Value × n)

✓ Correct Answer: (B) (Ending Value / Beginning Value)^(1/n) - 1

Solution:
CAGR = (Ending Value / Beginning Value)^(1/n) - 1
Where n is the number of years

Question 7 Depreciation

A machine costs ₹1,00,000 with salvage value ₹10,000 after 10 years. Using straight line method, annual depreciation is:

A ₹9,000

B ₹10,000

C ₹11,000

D ₹12,000

✓ Correct Answer: (A) ₹9,000

Solution:
Annual Depreciation = (Cost - Salvage Value) / Useful Life
= (1,00,000 - 10,000) / 10
= 90,000 / 10 = ₹9,000

Question 8 Depreciation

Using reducing balance method at 10% per annum, the value of a ₹50,000 asset after 2 years is:

A ₹40,000

B ₹40,500

C ₹45,000

D ₹48,000

✓ Correct Answer: (B) ₹40,500

Solution:
Book Value = Cost × (1 - r)ⁿ
= 50,000 × (1 - 0.10)²
= 50,000 × (0.90)²
= 50,000 × 0.81 = ₹40,500

Question 9 Depreciation

An asset depreciates from ₹80,000 to ₹64,000 in one year using reducing balance method. The depreciation rate is:

A 15%

B 20%

C 25%

D 30%

✓ Correct Answer: (B) 20%

Solution:
64,000 = 80,000 × (1 - r)
1 - r = 64,000/80,000 = 0.80
r = 1 - 0.80 = 0.20 = 20%

Question 10 Bonds

A bond with face value ₹1,000 pays 8% annual coupon. If required return is 10%, the bond will trade at:

A Premium (above ₹1,000)

B Discount (below ₹1,000)

C Par value (₹1,000)

D Cannot determine

✓ Correct Answer: (B) Discount (below ₹1,000)

Solution:
When coupon rate (8%) < required return (10%), bond trades at discount.
When coupon rate > required return, bond trades at premium.
When coupon rate = required return, bond trades at par.

Question 11 Bonds

A bond with face value ₹10,000 and coupon rate 6% pays annual interest of:

A ₹60

B ₹600

C ₹6,000

D ₹660

✓ Correct Answer: (B) ₹600

Solution:
Annual Coupon Payment = Face Value × Coupon Rate
= 10,000 × 0.06 = ₹600

Question 12 Perpetuity

The present value of perpetuity immediate of ₹500 at 5% is:

A ₹5,000

B ₹10,000

C ₹15,000

D ₹20,000

✓ Correct Answer: (B) ₹10,000

Solution:
PV = A/r = 500/0.05 = ₹10,000

Question 13 Amortization

In an amortization schedule, as time progresses:

A Interest component increases, principal component decreases

B Interest component decreases, principal component increases

C Both components remain constant

D Both components increase

✓ Correct Answer: (B) Interest component decreases, principal component increases

Solution:
In amortization, as the outstanding principal reduces over time, the interest component decreases and the principal repayment component increases, while EMI remains constant.

Question 14 Sinking Fund

If annual deposits of ₹10,000 are made for 4 years at 8% compounded annually, the amount accumulated is approximately:

A ₹40,000

B ₹43,000

C ₹45,061

D ₹48,000

✓ Correct Answer: (C) ₹45,061

Solution:
FV = A × [((1+r)ⁿ - 1) / r]
= 10,000 × [((1.08)⁴ - 1) / 0.08]
= 10,000 × [(1.36049 - 1) / 0.08]
= 10,000 × 4.5061 = ₹45,061

Question 15 Bonds

Current yield of a bond is calculated as:

A Annual Coupon Payment / Current Market Price

B Face Value / Current Market Price

C Annual Coupon Payment / Face Value

D Market Price / Face Value

✓ Correct Answer: (A) Annual Coupon Payment / Current Market Price

Solution:
Current Yield = Annual Coupon Payment / Current Market Price
It shows the annual return based on current market price.

Question 16 Depreciation

Compared to straight line method, reducing balance method results in:

A Higher depreciation in early years

B Lower depreciation in early years

C Same depreciation every year

D No depreciation in early years

✓ Correct Answer: (A) Higher depreciation in early years

Solution:
Reducing balance method (WDV method) results in higher depreciation in early years as it applies a fixed percentage to the reducing book value, while straight line method gives constant depreciation.

Question 17 CAGR

CAGR is most useful for:

A Comparing growth rates of different investments

B Calculating simple interest

C Computing EMI

D Finding present value

✓ Correct Answer: (A) Comparing growth rates of different investments

Solution:
CAGR (Compound Annual Growth Rate) smoothens out volatility and provides a single growth rate, making it ideal for comparing different investments over time.

Question 18 EMI

Each EMI payment consists of:

A Only principal amount

B Only interest amount

C Both principal and interest

D Processing fee

✓ Correct Answer: (C) Both principal and interest

Solution:
Each EMI payment consists of two components: principal repayment and interest. The proportion changes over time with interest decreasing and principal increasing.

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Short Answer Questions with Step-by-Step Solutions

Practice 2-mark and 3-mark questions

Question 1 Perpetuity - 2M

Find the present value of a perpetuity that pays ₹2,000 at the end of each year if the interest rate is 5% per annum.

Solution:

Formula: PV of Perpetuity = A/r

Where A = Annual payment = ₹2,000

r = Interest rate = 5% = 0.05

PV = 2,000 / 0.05

Present Value = ₹40,000

Question 2 Sinking Fund - 3M

A company needs to accumulate ₹5,00,000 in 4 years to replace machinery. If the interest rate is 8% per annum compounded annually, find the annual deposit required.

Solution:

Sinking Fund Formula: A = S × r / ((1+r)ⁿ - 1)

Given: S = ₹5,00,000, r = 0.08, n = 4 years

A = 5,00,000 × 0.08 / ((1.08)⁴ - 1)

(1.08)⁴ = 1.36049

A = 40,000 / (1.36049 - 1)

A = 40,000 / 0.36049

Annual Deposit = ₹1,10,978 (approximately)

Question 3 EMI - 3M

Calculate the EMI for a loan of ₹2,00,000 at 12% per annum for 2 years (24 months).

Solution:

EMI = P × r × (1+r)ⁿ / ((1+r)ⁿ - 1)

P = ₹2,00,000

Monthly rate r = 12%/12 = 1% = 0.01

n = 24 months

(1.01)²⁴ = 1.2697

EMI = 2,00,000 × 0.01 × 1.2697 / (1.2697 - 1)

EMI = 2,539.4 / 0.2697

EMI = ₹9,414 (approximately)

Question 4 CAGR - 2M

An investment of ₹50,000 grows to ₹80,000 in 5 years. Calculate the CAGR.

Solution:

CAGR = (Final Value/Initial Value)^(1/n) - 1

Final Value = ₹80,000

Initial Value = ₹50,000

n = 5 years

CAGR = (80,000/50,000)^(1/5) - 1

CAGR = (1.6)^0.2 - 1

CAGR = 1.0986 - 1

CAGR = 0.0986 or 9.86%

Question 5 Depreciation - 3M

A machine costs ₹2,00,000 with salvage value ₹20,000 after 8 years. Calculate: (i) Annual depreciation using straight line method, (ii) Book value after 3 years.

Solution:

(i) Annual Depreciation:

Formula: D = (Cost - Salvage Value) / Useful Life

D = (2,00,000 - 20,000) / 8

D = 1,80,000 / 8 = ₹22,500

(ii) Book Value after 3 years:

Total Depreciation = 22,500 × 3 = ₹67,500

Book Value = Cost - Total Depreciation

Book Value = 2,00,000 - 67,500

(i) Annual Depreciation = ₹22,500
(ii) Book Value after 3 years = ₹1,32,500

Question 6 Bonds - 3M

A bond with face value ₹1,000 and coupon rate 10% has 3 years to maturity. If the required return is 12%, calculate the current yield and comment on whether the bond trades at premium or discount.

Solution:

Annual Coupon = Face Value × Coupon Rate

Annual Coupon = 1,000 × 0.10 = ₹100

Analysis:

Coupon Rate (10%) < Required Return (12%)

Therefore, bond will trade at DISCOUNT

Market Price will be < ₹1,000

To find exact current yield, we need market price

The bond trades at discount because coupon rate (10%) is less than required return (12%). Current yield would be Annual Coupon / Market Price, where Market Price < ₹1,000.

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Long Answer Questions with Complete Solutions

Practice 5-mark questions

Question 1 Sinking Fund - 5M

A company wants to accumulate ₹10,00,000 in 6 years for machinery replacement. The interest rate is 10% per annum compounded annually. (a) Calculate the annual deposit required. (b) Prepare a sinking fund schedule for the first 3 years.

Complete Solution:

(a) Annual Deposit Calculation:

Sinking Fund Formula: A = S × r / ((1+r)ⁿ - 1)

S = ₹10,00,000, r = 0.10, n = 6

(1.10)⁶ = 1.77156

A = 10,00,000 × 0.10 / (1.77156 - 1)

A = 1,00,000 / 0.77156

A = ₹1,29,607 (annually)

(b) Sinking Fund Schedule:

Year 1: Deposit = ₹1,29,607, Interest = 0

Balance = ₹1,29,607

Year 2: Deposit = ₹1,29,607, Interest = 1,29,607 × 0.10 = ₹12,961

Balance = 1,29,607 + 1,29,607 + 12,961 = ₹2,72,175

Year 3: Deposit = ₹1,29,607, Interest = 2,72,175 × 0.10 = ₹27,218

Balance = 2,72,175 + 1,29,607 + 27,218 = ₹4,29,000

(a) Annual Deposit = ₹1,29,607
(b) Balance after 3 years = ₹4,29,000

Question 2 EMI & Amortization - 5M

A person takes a loan of ₹5,00,000 at 12% per annum for 3 years (36 months). (a) Calculate the monthly EMI. (b) Prepare an amortization schedule for the first 3 months showing principal, interest, and outstanding balance.

Complete Solution:

(a) EMI Calculation:

P = ₹5,00,000

Monthly rate r = 12%/12 = 1% = 0.01

n = 36 months

EMI = P × r × (1+r)ⁿ / ((1+r)ⁿ - 1)

(1.01)³⁶ = 1.4308

EMI = 5,00,000 × 0.01 × 1.4308 / (1.4308 - 1)

EMI = 7,154 / 0.4308

EMI = ₹16,607

(b) Amortization Schedule:

Month 1:

Opening Balance = ₹5,00,000

Interest = 5,00,000 × 0.01 = ₹5,000

Principal = 16,607 - 5,000 = ₹11,607

Closing Balance = 5,00,000 - 11,607 = ₹4,88,393

Month 2:

Opening Balance = ₹4,88,393

Interest = 4,88,393 × 0.01 = ₹4,884

Principal = 16,607 - 4,884 = ₹11,723

Closing Balance = 4,88,393 - 11,723 = ₹4,76,670

Month 3:

Opening Balance = ₹4,76,670

Interest = 4,76,670 × 0.01 = ₹4,767

Principal = 16,607 - 4,767 = ₹11,840

Closing Balance = 4,76,670 - 11,840 = ₹4,64,830

(a) Monthly EMI = ₹16,607
(b) Outstanding after 3 months = ₹4,64,830
Notice: Principal component increases while interest decreases

Question 3 Depreciation - 5M

A machine costs ₹3,00,000 with salvage value ₹30,000 after 5 years. Compare the depreciation and book values using: (a) Straight Line Method, (b) Reducing Balance Method at 25% per annum. Show calculations for all 5 years.

Complete Solution:

(a) Straight Line Method:

Annual Depreciation = (Cost - Salvage) / Life

= (3,00,000 - 30,000) / 5 = ₹54,000 per year

Year 1: Dep = ₹54,000, Book Value = ₹2,46,000

Year 2: Dep = ₹54,000, Book Value = ₹1,92,000

Year 3: Dep = ₹54,000, Book Value = ₹1,38,000

Year 4: Dep = ₹54,000, Book Value = ₹84,000

Year 5: Dep = ₹54,000, Book Value = ₹30,000

(b) Reducing Balance Method (25%):

Year 1: Dep = 3,00,000 × 0.25 = ₹75,000, BV = ₹2,25,000

Year 2: Dep = 2,25,000 × 0.25 = ₹56,250, BV = ₹1,68,750

Year 3: Dep = 1,68,750 × 0.25 = ₹42,188, BV = ₹1,26,562

Year 4: Dep = 1,26,562 × 0.25 = ₹31,641, BV = ₹94,921

Year 5: Dep = 94,921 × 0.25 = ₹23,730, BV = ₹71,191

Comparison:
• SLM: Constant depreciation ₹54,000, final BV = ₹30,000
• RBM: Decreasing depreciation, final BV = ₹71,191
• RBM gives higher depreciation in early years
• SLM reaches exact salvage value

Question 4 CAGR - 5M

An investor's portfolio shows the following values: Year 0: ₹1,00,000, Year 1: ₹1,15,000, Year 2: ₹1,20,000, Year 3: ₹1,45,000, Year 4: ₹1,60,000. Calculate: (a) CAGR over 4 years, (b) Year-on-year growth rates, (c) Comment on the difference between CAGR and average growth rate.

Complete Solution:

(a) CAGR Calculation:

CAGR = (Final Value/Initial Value)^(1/n) - 1

= (1,60,000/1,00,000)^(1/4) - 1

= (1.6)^0.25 - 1

= 1.1246 - 1

CAGR = 0.1246 or 12.46%

(b) Year-on-Year Growth:

Year 1: (1,15,000-1,00,000)/1,00,000 = 15%

Year 2: (1,20,000-1,15,000)/1,15,000 = 4.35%

Year 3: (1,45,000-1,20,000)/1,20,000 = 20.83%

Year 4: (1,60,000-1,45,000)/1,45,000 = 10.34%

Average = (15 + 4.35 + 20.83 + 10.34)/4 = 12.63%

(a) CAGR = 12.46%
(b) Annual growth rates: 15%, 4.35%, 20.83%, 10.34%
(c) CAGR (12.46%) smoothens volatility and shows compound effect, while simple average (12.63%) treats each year equally without compounding. CAGR is better for comparing investments.

Question 5 Bond Valuation - 5M

A bond with face value ₹1,000 pays 10% annual coupon and has 4 years to maturity. The required rate of return is 8%. Calculate: (a) Present value of coupon payments, (b) Present value of face value, (c) Bond price, (d) Comment if it trades at premium or discount.

Complete Solution:

Annual Coupon = 1,000 × 0.10 = ₹100

Required return (r) = 8% = 0.08

(a) PV of Coupon Payments:

PV = C × [1 - (1+r)^(-n)] / r

= 100 × [1 - (1.08)^(-4)] / 0.08

= 100 × [1 - 0.7350] / 0.08

= 100 × 0.2650 / 0.08

= 100 × 3.3121 = ₹331.21

(b) PV of Face Value:

PV = FV / (1+r)ⁿ

= 1,000 / (1.08)⁴

= 1,000 / 1.3605

= ₹735.03

(c) Bond Price:

Bond Price = PV of Coupons + PV of Face Value

= 331.21 + 735.03

= ₹1,066.24

(a) PV of Coupons = ₹331.21
(b) PV of Face Value = ₹735.03
(c) Bond Price = ₹1,066.24
(d) Bond trades at PREMIUM because coupon rate (10%) > required return (8%), hence price (₹1,066.24) > face value (₹1,000)

Question 6 Perpetuity - 5M

Compare the present values of: (a) A perpetuity paying ₹10,000 annually at 8%, (b) An annuity paying ₹10,000 annually for 20 years at 8%, (c) An annuity paying ₹10,000 annually for 50 years at 8%. Comment on the relationship between annuity and perpetuity.

Complete Solution:

(a) Perpetuity PV:

PV = A / r

= 10,000 / 0.08

= ₹1,25,000

(b) Annuity for 20 years:

PV = A × [1 - (1+r)^(-n)] / r

= 10,000 × [1 - (1.08)^(-20)] / 0.08

= 10,000 × [1 - 0.2145] / 0.08

= 10,000 × 9.8181

= ₹98,181

(c) Annuity for 50 years:

= 10,000 × [1 - (1.08)^(-50)] / 0.08

= 10,000 × [1 - 0.0213] / 0.08

= 10,000 × 12.2335

= ₹1,22,335

(a) Perpetuity PV = ₹1,25,000
(b) 20-year annuity PV = ₹98,181
(c) 50-year annuity PV = ₹1,22,335
Comment: As annuity period increases, its PV approaches perpetuity PV. The 50-year annuity (₹1,22,335) is 97.9% of perpetuity value (₹1,25,000), showing that distant cash flows have negligible present value.

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Case Study Based Questions

Real-world application problems

Case Study 1 EMI & Amortization

Home Loan Planning

Rajesh wants to buy a house worth ₹50,00,000. He has ₹10,00,000 as down payment and needs to take a loan for the remaining amount. The bank offers a home loan at 9% per annum for 20 years (240 months). He wants to understand his EMI obligations and loan repayment structure.

What is the loan amount Rajesh needs to borrow?

A ₹30,00,000

B ₹40,00,000

C ₹45,00,000

D ₹50,00,000

✓ Correct Answer: (B) ₹40,00,000

Solution:
Loan Amount = House Value - Down Payment
= ₹50,00,000 - ₹10,00,000 = ₹40,00,000

What is the monthly interest rate for this loan?

A 0.75%

B 0.9%

C 1.0%

D 1.5%

✓ Correct Answer: (A) 0.75%

Solution:
Monthly rate = Annual rate / 12
= 9% / 12 = 0.75% = 0.0075

The approximate monthly EMI for this loan is:

A ₹30,000

B ₹35,984

C ₹40,000

D ₹45,000

✓ Correct Answer: (B) ₹35,984

Solution:
EMI = P × r × (1+r)ⁿ / ((1+r)ⁿ - 1)
P = ₹40,00,000, r = 0.0075, n = 240
(1.0075)²⁴⁰ = 6.0092
EMI = 40,00,000 × 0.0075 × 6.0092 / 5.0092
≈ ₹35,984

In the first month's EMI, the interest component is approximately:

A ₹20,000

B ₹25,000

C ₹30,000

D ₹35,000

✓ Correct Answer: (C) ₹30,000

Solution:
First month interest = Principal × Monthly rate
= ₹40,00,000 × 0.0075 = ₹30,000
Principal component = EMI - Interest
= ₹35,984 - ₹30,000 = ₹5,984

Case Study 2 CAGR & Depreciation

Investment and Asset Management

A company invested ₹20,00,000 in a business venture. After 4 years, the investment value grew to ₹28,00,000. Simultaneously, the company purchased machinery worth ₹15,00,000 which depreciates at 15% per annum using the reducing balance method.

The CAGR of the investment is approximately:

A 8.8%

B 10%

C 12%

D 15%

✓ Correct Answer: (A) 8.8%

Solution:
CAGR = (Final/Initial)^(1/n) - 1
= (28,00,000/20,00,000)^(1/4) - 1
= (1.4)^0.25 - 1
= 1.0878 - 1 = 0.0878 ≈ 8.8%

The book value of machinery after 1 year is:

A ₹12,00,000

B ₹12,75,000

C ₹13,50,000

D ₹14,00,000

✓ Correct Answer: (B) ₹12,75,000

Solution:
Book Value = Cost × (1 - r)ⁿ
= 15,00,000 × (1 - 0.15)¹
= 15,00,000 × 0.85 = ₹12,75,000

The book value of machinery after 4 years is approximately:

A ₹7,50,000

B ₹7,82,063

C ₹8,50,000

D ₹9,00,000

✓ Correct Answer: (B) ₹7,82,063

Solution:
Book Value = 15,00,000 × (0.85)⁴
= 15,00,000 × 0.52200625
= ₹7,82,063 (approximately)

After 4 years, the net asset position (investment value - machinery depreciation) is:

A ₹20,17,937

B ₹25,00,000

C ₹30,00,000

D ₹35,82,063

✓ Correct Answer: (D) ₹35,82,063

Solution:
Net Position = Investment Value + Machinery Value
= ₹28,00,000 + ₹7,82,063
= ₹35,82,063

Case Study 3 Bond Valuation & Perpetuity

Fixed Income Investment Portfolio

An investor is considering two investment options: Option A is a corporate bond with face value ₹10,000, coupon rate 12% per annum, 5 years to maturity, and market yield of 10%. Option B is a perpetual bond (console) paying ₹1,200 annually with the same market yield of 10%.

The annual coupon payment from Option A is:

A ₹1,000

B ₹1,200

C ₹1,500

D ₹2,000

✓ Correct Answer: (B) ₹1,200

Solution:
Annual Coupon = Face Value × Coupon Rate
= ₹10,000 × 12% = ₹1,200

Option A bond will trade at:

A Discount

B Premium

C Par value

D Cannot determine

✓ Correct Answer: (B) Premium

Solution:
Coupon Rate (12%) > Market Yield (10%)
When coupon rate exceeds market yield, bond trades at premium (above face value).

The present value of Option B (perpetuity) is:

A ₹10,000

B ₹12,000

C ₹15,000

D ₹20,000

✓ Correct Answer: (B) ₹12,000

Solution:
PV of Perpetuity = Annual Payment / Yield
= ₹1,200 / 0.10 = ₹12,000

If the market yield increases to 12%, the value of Option B will:

A Increase to ₹14,400

B Remain ₹12,000

C Decrease to ₹10,000

D Decrease to ₹8,000

✓ Correct Answer: (C) Decrease to ₹10,000

Solution:
New PV = 1,200 / 0.12 = ₹10,000
When market yield increases, bond/perpetuity value decreases (inverse relationship).

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💰 Unit 7: Financial Mathematics

Topics Covered in Unit 7

1. Perpetuity

2. Sinking Funds

3. EMI & Amortization

4. CAGR

5. Depreciation

6. Valuation of Bonds

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💰 Unit 7: Financial Mathematics

Topics Covered in Unit 7

1. Perpetuity

2. Sinking Funds

3. EMI & Amortization

4. CAGR

5. Depreciation

6. Valuation of Bonds

Practice MCQs with Answers

💰 Master Financial Mathematics!

Short Answer Questions with Step-by-Step Solutions

📊 Master All Question Types!

Long Answer Questions with Complete Solutions

🎯 Ace Your Board Exams!

Case Study Based Questions

🏆 Complete Your Exam Preparation!

Explore Other Units

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