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

💰 Unit 7: Financial Mathematics

Perpetuity, Sinking Funds, EMI & Amortization, CAGR, Depreciation, Bond Valuation

Weightage: 10 Marks in Board Exams
Home › Class 12 Applied Maths › Unit 7: Financial Mathematics
Topics Covered MCQs (18) Short Answers (6) Long Answers (6) Case Studies (3)
📚

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

❓

Practice MCQs with Answers

Click "Show Answer" to reveal explanations

Question 1Perpetuity
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 2Sinking 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 3EMI
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 4EMI
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 5CAGR
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 6CAGR
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 7Depreciation
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 8Depreciation
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 9Depreciation
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 10Bonds
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 11Bonds
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 12Perpetuity
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 13Amortization
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 14Sinking 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 15Bonds
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 16Depreciation
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 17CAGR
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 18EMI
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 1Perpetuity - 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 2Sinking 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) = 40,000 / 0.36049
Annual Deposit = ₹1,10,978 (approximately)
Question 3EMI - 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) = 2,539.4 / 0.2697
EMI = ₹9,414 (approximately)
Question 4CAGR - 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 = (1.6)^0.2 - 1 = 1.0986 - 1
CAGR = 0.0986 or 9.86%
Question 5Depreciation - 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 = 1,80,000 / 8 = ₹22,500
(ii) Book Value after 3 years:
Total Depreciation = 22,500 × 3 = ₹67,500
Book Value = 2,00,000 - 67,500
(i) Annual Depreciation = ₹22,500
(ii) Book Value after 3 years = ₹1,32,500
Question 6Bonds - 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
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 1Sinking 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) = 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 = ₹12,961 | Balance = ₹2,72,175
Year 3: Deposit = ₹1,29,607 | Interest = ₹27,218 | Balance = ₹4,29,000
(a) Annual Deposit = ₹1,29,607
(b) Balance after 3 years = ₹4,29,000
Question 2EMI & 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 = 1% = 0.01 | n = 36 months
(1.01)³⁶ = 1.4308  →  EMI = 5,00,000 × 0.01 × 1.4308 / (1.4308 - 1) = 7,154 / 0.4308
EMI = ₹16,607
(b) Amortization Schedule:
Month 1: Interest ₹5,000 | Principal ₹11,607 | Balance ₹4,88,393
Month 2: Interest ₹4,884 | Principal ₹11,723 | Balance ₹4,76,670
Month 3: Interest ₹4,767 | Principal ₹11,840 | Balance ₹4,64,830
(a) Monthly EMI = ₹16,607
(b) Outstanding after 3 months = ₹4,64,830
Notice: Principal component increases while interest decreases
Question 3Depreciation - 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 = (3,00,000 - 30,000) / 5 = ₹54,000 per year
Year 1: Dep ₹54,000 | BV ₹2,46,000  |  Year 2: Dep ₹54,000 | BV ₹1,92,000
Year 3: Dep ₹54,000 | BV ₹1,38,000  |  Year 4: Dep ₹54,000 | BV ₹84,000  |  Year 5: Dep ₹54,000 | BV ₹30,000
(b) Reducing Balance Method (25%):
Year 1: Dep ₹75,000 | BV ₹2,25,000  |  Year 2: Dep ₹56,250 | BV ₹1,68,750
Year 3: Dep ₹42,188 | BV ₹1,26,562  |  Year 4: Dep ₹31,641 | BV ₹94,921  |  Year 5: Dep ₹23,730 | BV ₹71,191
• 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 4CAGR - 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: (1,60,000/1,00,000)^(1/4) - 1 = (1.6)^0.25 - 1 = 1.1246 - 1 = 12.46%
(b) Year-on-Year: Year 1: 15% | Year 2: 4.35% | Year 3: 20.83% | Year 4: 10.34%
Simple 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 5Bond 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 = 100 × [1 - (1.08)^(-4)] / 0.08 = 100 × [1 - 0.7350] / 0.08 = 100 × 3.3121 = ₹331.21
(b) PV of Face Value:
PV = 1,000 / (1.08)⁴ = 1,000 / 1.3605 = ₹735.03
(c) Bond Price: 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 6Perpetuity - 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: 10,000 / 0.08 = ₹1,25,000
(b) 20-year annuity: 10,000 × [1 - (1.08)^(-20)] / 0.08 = 10,000 × 9.8181 = ₹98,181
(c) 50-year annuity: 10,000 × [1 - (1.08)^(-50)] / 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.

Q1
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
Q2
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
Q3
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
Q4
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.

Q1
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 ≈ 8.8%
Q2
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
Q3
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)
Q4
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%.

Q1
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
Q2
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).
Q3
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
Q4
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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