Unit 7: Financial Mathematics – Class 12 Applied Maths Free MCQs & Solutions | Boundless Maths

💰 Unit 7: Financial Mathematics

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

⭐ Weightage: 10 Marks in Board Exam
Home › Class 12 Applied Maths › Unit 7: Financial Mathematics

Unit 7 — Financial Mathematics carries 10 marks in the CBSE Class 12 Applied Mathematics board examination, making it one of the most directly rewarding units to prepare thoroughly.

This unit covers six core topics: Perpetuity, where you calculate the present value of an infinite series of equal payments; Sinking Funds, which involve accumulating money through periodic deposits to meet a future target; EMI and Amortization, covering equated monthly instalments and the structure of loan repayment schedules; CAGR (Compounded Annual Growth Rate), used to measure and compare investment growth; Depreciation using both the straight-line method and the reducing-balance method; and Valuation of Bonds, including present value of coupon streams, bond pricing, and the relationship between coupon rates and market yields.

This page includes MCQs with detailed solutions, short-answer questions with full step-by-step workings, long-answer questions covering multi-part problems, and case studies presenting real-world financial scenarios.

All questions and solutions are aligned to the latest CBSE Applied Maths syllabus 2026-27.

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MCQs with Solutions
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Short Answer Qs
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Long Answer Qs
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Case Studies
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 — with key formulas

1. Perpetuity

Infinite series of equal payments; present value calculations

PV = A ÷ r
A = annual payment, r = rate

2. Sinking Funds

Periodic deposits to accumulate a target future amount

A = S·r ÷ [(1+r)ⁿ − 1]
S = target, r = rate, n = periods

3. EMI & Amortization

Equated Monthly Instalments and loan repayment schedules

EMI = P·r·(1+r)ⁿ ÷ [(1+r)ⁿ − 1]
P = principal, r = monthly rate, n = months

4. CAGR

Compounded Annual Growth Rate for comparing investments

CAGR = (FV ÷ PV)^(1/n) − 1
FV = final, PV = initial, n = years

5. Depreciation

Straight-line and reducing-balance (WDV) methods

SLM: D = (Cost − Salvage) ÷ n
RBM: BV = Cost × (1−r)ⁿ

6. Valuation of Bonds

Present value of coupon payments plus face value redemption

Price = C·[1−(1+y)^−n]÷y + F÷(1+y)ⁿ
C = coupon, y = yield, F = face value
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Practice MCQs with Solutions

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Question 1Perpetuity
The present value of a perpetuity of ₹1,000 per year at 8% per annum is:
  • A ₹8,000
  • B ₹10,000
  • C ₹12,500
  • D ₹15,000
✓ Correct Answer: (C) ₹12,500
  • Formula: \(\text{PV} = \dfrac{A}{r}\)
  • Annual payment \(A = ₹1{,}000\), rate \(r = 8\% = 0.08\)
  • \(\text{PV} = \dfrac{1000}{0.08} = \mathbf{₹12{,}500}\)
Question 2Sinking Fund
A company wants to accumulate ₹1,00,000 in 5 years by making annual deposits. If the 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
  • Sinking fund formula: \(A = \dfrac{S \cdot r}{(1+r)^n - 1}\)
  • Given: \(S = ₹1{,}00{,}000,\; r = 0.10,\; n = 5\)
  • \((1.10)^5 = 1.61051\)
  • \(A = \dfrac{1{,}00{,}000 \times 0.10}{1.61051 - 1} = \dfrac{10{,}000}{0.61051} \approx \mathbf{₹16{,}380}\)
Question 3EMI
The EMI formula for a loan of \(P\) at rate \(r\) per month for \(n\) months is:
  • A \(\text{EMI} = \dfrac{P \cdot r \cdot (1+r)^n}{(1+r)^n - 1}\)
  • B \(\text{EMI} = P/n\)
  • C \(\text{EMI} = P \times r\)
  • D \(\text{EMI} = P \times (1+r)^n\)
✓ Correct Answer: (A)
  • The standard EMI formula is derived from the present value of an annuity.
  • \(\text{EMI} = \dfrac{P \cdot r \cdot (1+r)^n}{(1+r)^n - 1}\)
  • Here \(P\) = principal loan amount, \(r\) = monthly interest rate (annual rate ÷ 12), and \(n\) = total number of monthly instalments.
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%
  • Monthly interest rate = Annual rate ÷ 12
  • \(r = \dfrac{12\%}{12} = 1\% = 0.01\)
  • Always convert the annual rate to a monthly rate before applying the EMI formula.
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%
  • \(\text{CAGR} = \left(\dfrac{FV}{PV}\right)^{1/n} - 1\)
  • \(= \left(\dfrac{15{,}000}{10{,}000}\right)^{1/3} - 1 = (1.5)^{0.333} - 1\)
  • \(= 1.1447 - 1 = 0.1447 \approx \mathbf{14.5\%}\)
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)
  • \(\text{CAGR} = \left(\dfrac{\text{Ending Value}}{\text{Beginning Value}}\right)^{1/n} - 1\)
  • Option A gives total return (not annualised). Option C is simple average growth. CAGR accounts for compounding, which is why it differs from a simple average.
Question 7Depreciation
A machine costs ₹1,00,000 with salvage value ₹10,000 after 10 years. Using the straight-line method, annual depreciation is:
  • A ₹9,000
  • B ₹10,000
  • C ₹11,000
  • D ₹12,000
✓ Correct Answer: (A) ₹9,000
  • SLM formula: \(D = \dfrac{\text{Cost} - \text{Salvage Value}}{\text{Useful Life}}\)
  • \(D = \dfrac{1{,}00{,}000 - 10{,}000}{10} = \dfrac{90{,}000}{10} = \mathbf{₹9{,}000}\)
Question 8Depreciation
Using the reducing-balance method at 10% per annum, the book 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
  • RBM formula: \(\text{BV} = \text{Cost} \times (1-r)^n\)
  • \(= 50{,}000 \times (0.90)^2 = 50{,}000 \times 0.81 = \mathbf{₹40{,}500}\)
Question 9Depreciation
An asset depreciates from ₹80,000 to ₹64,000 in one year using the reducing-balance method. The depreciation rate is:
  • A 15%
  • B 20%
  • C 25%
  • D 30%
✓ Correct Answer: (B) 20%
  • \(64{,}000 = 80{,}000 \times (1 - r)\)
  • \(1 - r = \dfrac{64{,}000}{80{,}000} = 0.80\)
  • \(r = 1 - 0.80 = 0.20 = \mathbf{20\%}\)
Question 10Bonds
A bond with face value ₹1,000 pays an 8% annual coupon. If the 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
  • 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
  • Annual Coupon Payment = Face Value × Coupon Rate
  • \(= 10{,}000 \times 0.06 = \mathbf{₹600}\)
Question 12Perpetuity
The present value of a perpetuity immediate of ₹500 at 5% is:
  • A ₹5,000
  • B ₹10,000
  • C ₹15,000
  • D ₹20,000
✓ Correct Answer: (B) ₹10,000
  • \(\text{PV} = \dfrac{A}{r} = \dfrac{500}{0.05} = \mathbf{₹10{,}000}\)
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 is charged on the outstanding principal. As principal reduces each month, the interest portion of the EMI falls.
  • Since total EMI is fixed, the principal repayment component rises correspondingly each month.
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
  • Future value of annuity: \(FV = A \times \dfrac{(1+r)^n - 1}{r}\)
  • \((1.08)^4 = 1.36049\)
  • \(FV = 10{,}000 \times \dfrac{1.36049 - 1}{0.08} = 10{,}000 \times 4.5061 = \mathbf{₹45{,}061}\)
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)
  • \(\text{Current Yield} = \dfrac{\text{Annual Coupon Payment}}{\text{Current Market Price}}\)
  • It shows the annual income return based on what you pay for the bond today — not its face value.
Question 16Depreciation
Compared to the straight-line method, the 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
  • RBM applies a fixed rate to the declining book value. Early on, the book value is highest, so depreciation is largest.
  • SLM charges a constant amount every year regardless of book value.
  • RBM is preferred when assets lose value faster at the start (e.g., vehicles, electronics).
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)
  • CAGR smooths out year-to-year volatility into a single steady-growth equivalent rate.
  • This makes it ideal for comparing two investments over different time periods — even if their yearly returns fluctuate differently.
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
  • Every EMI = Interest portion + Principal portion. The total EMI stays constant throughout the loan tenure.
  • Over time, the interest portion decreases and the principal portion increases as the outstanding balance falls.

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Short Answer Questions with Step-by-Step Solutions

2-mark and 3-mark practice 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.
Formula: \(\text{PV} = \dfrac{A}{r}\)
Annual payment \(A = ₹2{,}000\), rate \(r = 5\% = 0.05\)
\(\text{PV} = \dfrac{2{,}000}{0.05}\)
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.
Formula: \(A = \dfrac{S \cdot r}{(1+r)^n - 1}\)
Given: \(S = ₹5{,}00{,}000,\; r = 0.08,\; n = 4\)
\((1.08)^4 = 1.36049\)
\(A = \dfrac{5{,}00{,}000 \times 0.08}{1.36049 - 1} = \dfrac{40{,}000}{0.36049}\)
Annual Deposit ≈ ₹1,10,978
Question 3EMI — 3M
Calculate the EMI for a loan of ₹2,00,000 at 12% per annum for 2 years (24 months).
Formula: \(\text{EMI} = \dfrac{P \cdot r \cdot (1+r)^n}{(1+r)^n - 1}\)
\(P = ₹2{,}00{,}000\), monthly rate \(r = \dfrac{12\%}{12} = 1\% = 0.01\), \(n = 24\)
\((1.01)^{24} = 1.2697\)
\(\text{EMI} = \dfrac{2{,}00{,}000 \times 0.01 \times 1.2697}{1.2697 - 1} = \dfrac{2{,}539.4}{0.2697}\)
EMI ≈ ₹9,414 per month
Question 4CAGR — 2M
An investment of ₹50,000 grows to ₹80,000 in 5 years. Calculate the CAGR.
\(\text{CAGR} = \left(\dfrac{FV}{PV}\right)^{1/n} - 1\)
\(= \left(\dfrac{80{,}000}{50{,}000}\right)^{1/5} - 1 = (1.6)^{0.2} - 1\)
\(= 1.0986 - 1 = 0.0986\)
CAGR ≈ 9.86%
Question 5Depreciation — 3M
A machine costs ₹2,00,000 with salvage value ₹20,000 after 8 years. Calculate: (i) Annual depreciation using the straight-line method, (ii) Book value after 3 years.
(i) Annual Depreciation
\(D = \dfrac{2{,}00{,}000 - 20{,}000}{8} = \dfrac{1{,}80{,}000}{8} = ₹22{,}500\)
(ii) Book Value after 3 years
Total depreciation = \(22{,}500 \times 3 = ₹67{,}500\)
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%, comment on whether the bond trades at premium or discount.
Annual Coupon \(= 1{,}000 \times 10\% = ₹100\)
Coupon Rate (10%) < Required Return (12%)
When the coupon rate is less than the market yield, investors demand a discount to compensate — so the bond's market price will fall below ₹1,000.
The bond trades at DISCOUNT. Market Price < ₹1,000 because coupon rate (10%) < required return (12%).

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Long Answer Questions with Complete Solutions

5-mark practice 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.
(a) Annual Deposit
\(A = \dfrac{S \cdot r}{(1+r)^n - 1}\) where \(S = ₹10{,}00{,}000,\; r = 0.10,\; n = 6\)
\((1.10)^6 = 1.77156\)
\(A = \dfrac{10{,}00{,}000 \times 0.10}{1.77156 - 1} = \dfrac{1{,}00{,}000}{0.77156} \approx ₹1{,}29{,}607\)
(b) Sinking Fund Schedule (first 3 years)
Year 1: Deposit ₹1,29,607 | Interest ₹0 | Closing Balance ₹1,29,607
Year 2: Deposit ₹1,29,607 | Interest ₹12,961 | Closing Balance ₹2,72,175
Year 3: Deposit ₹1,29,607 | Interest ₹27,218 | Closing 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.
(a) EMI Calculation
\(P = ₹5{,}00{,}000,\; r = 1\% = 0.01,\; n = 36\)
\((1.01)^{36} = 1.4308\)
\(\text{EMI} = \dfrac{5{,}00{,}000 \times 0.01 \times 1.4308}{1.4308 - 1} = \dfrac{7{,}154}{0.4308} \approx ₹16{,}607\)
(b) Amortization Schedule
Month 1: Interest \(= 5{,}00{,}000 \times 0.01 = ₹5{,}000\) | Principal \(= 16{,}607 - 5{,}000 = ₹11{,}607\) | Balance ₹4,88,393
Month 2: Interest \(= 4{,}88{,}393 \times 0.01 = ₹4{,}884\) | 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: interest falls while principal repayment rises each month.
Question 3Depreciation — 5M
A machine costs ₹3,00,000 with salvage value ₹30,000 after 5 years. Compare depreciation and book values under (a) Straight-Line Method and (b) Reducing-Balance Method at 25% per annum, for all 5 years.
(a) Straight-Line Method
\(D = \dfrac{3{,}00{,}000 - 30{,}000}{5} = ₹54{,}000 \text{ per year}\)
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 ✓ (reaches salvage value exactly)
(b) Reducing-Balance Method (r = 25%)
\(\text{BV}_n = 3{,}00{,}000 \times (0.75)^n\)
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; reaches exact salvage ₹30,000. RBM: depreciation decreases each year; final BV ₹71,191 (higher than SLM). RBM charges more in early years — useful for assets that lose value quickly.
Question 4CAGR — 5M
An investor's portfolio shows: 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 CAGR vs simple average growth rate.
(a) CAGR
\(\text{CAGR} = \left(\dfrac{1{,}60{,}000}{1{,}00{,}000}\right)^{1/4} - 1 = (1.6)^{0.25} - 1 = 1.1246 - 1 = 12.46\%\)
(b) Year-on-Year Growth Rates
Year 1: \(\dfrac{1{,}15{,}000 - 1{,}00{,}000}{1{,}00{,}000} = 15\%\)  |  Year 2: \(\dfrac{5{,}000}{1{,}15{,}000} = 4.35\%\)  |  Year 3: \(20.83\%\)  |  Year 4: \(10.34\%\)
Simple average \(= \dfrac{15 + 4.35 + 20.83 + 10.34}{4} = 12.63\%\)
(a) CAGR = 12.46%  |  (b) Annual rates: 15%, 4.35%, 20.83%, 10.34%. (c) CAGR (12.46%) accounts for compounding — it is slightly lower than the simple average (12.63%) and is the preferred measure for comparing investments because it smooths out volatility.
Question 5Bond Valuation — 5M
A bond with face value ₹1,000 pays a 10% annual coupon and has 4 years to maturity. The required rate of return is 8%. Calculate (a) PV of coupon payments, (b) PV of face value, (c) bond price, (d) state whether it trades at premium or discount.
Annual Coupon \(C = 1{,}000 \times 10\% = ₹100\), yield \(y = 8\% = 0.08\), \(n = 4\)
(a) PV of Coupon Payments (annuity)
\(PV_C = C \times \dfrac{1-(1+y)^{-n}}{y} = 100 \times \dfrac{1-(1.08)^{-4}}{0.08}\)
\((1.08)^{-4} = 0.7350 \Rightarrow PV_C = 100 \times \dfrac{0.2650}{0.08} = 100 \times 3.3121 = ₹331.21\)
(b) PV of Face Value
\(PV_F = \dfrac{1{,}000}{(1.08)^4} = \dfrac{1{,}000}{1.3605} = ₹735.03\)
(c) Bond Price
\(\text{Price} = 331.21 + 735.03 = ₹1{,}066.24\)
(a) PV of Coupons = ₹331.21  |  (b) PV of Face Value = ₹735.03  |  (c) Bond Price = ₹1,066.24  |  (d) PREMIUM — because coupon rate (10%) > required return (8%), so price exceeds face value.
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.
(a) Perpetuity PV
\(PV = \dfrac{10{,}000}{0.08} = ₹1{,}25{,}000\)
(b) 20-year annuity PV
\(PV = 10{,}000 \times \dfrac{1-(1.08)^{-20}}{0.08} = 10{,}000 \times 9.8181 = ₹98{,}181\)
(c) 50-year annuity PV
\(PV = 10{,}000 \times \dfrac{1-(1.08)^{-50}}{0.08} = 10{,}000 \times 12.2335 = ₹1{,}22{,}335\)
(a) ₹1,25,000  |  (b) ₹98,181  |  (c) ₹1,22,335. Comment: As the annuity period grows, its PV converges towards the perpetuity value. The 50-year annuity reaches 97.9% of the perpetuity PV — confirming that cash flows far in the future contribute almost nothing in present-value terms.
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Case Study Based Questions

Real-world application problems

Case Study 1EMI & Amortization

Home Loan Planning

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

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
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%
Monthly rate \(= \dfrac{9\%}{12} = 0.75\% = 0.0075\)
Q3
The approximate monthly EMI is:
  • A ₹30,000
  • B ₹35,984
  • C ₹40,000
  • D ₹45,000
✓ Correct Answer: (B) ₹35,984
  • \(P = ₹40{,}00{,}000,\; r = 0.0075,\; n = 240\)
  • \((1.0075)^{240} = 6.0092\)
  • \(\text{EMI} = \dfrac{40{,}00{,}000 \times 0.0075 \times 6.0092}{6.0092 - 1} \approx \mathbf{₹35{,}984}\)
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
  • First month interest \(= 40{,}00{,}000 \times 0.0075 = ₹30{,}000\)
  • Principal component \(= 35{,}984 - 30{,}000 = ₹5{,}984\)
  • In the early months of a home loan, most of the EMI goes towards interest because the outstanding principal is at its highest.
Case Study 2CAGR & Depreciation

Investment and Asset Management

A company invested ₹20,00,000 in a business venture. After 4 years, the investment grew to ₹28,00,000. Simultaneously, the company purchased machinery worth ₹15,00,000 that 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%
\(\text{CAGR} = \left(\dfrac{28{,}00{,}000}{20{,}00{,}000}\right)^{1/4} - 1 = (1.4)^{0.25} - 1 = 1.0878 - 1 \approx 8.8\%\)
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
\(\text{BV} = 15{,}00{,}000 \times (1 - 0.15)^1 = 15{,}00{,}000 \times 0.85 = \mathbf{₹12{,}75{,}000}\)
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
\(\text{BV} = 15{,}00{,}000 \times (0.85)^4 = 15{,}00{,}000 \times 0.52201 \approx \mathbf{₹7{,}82{,}063}\)
Q4
After 4 years, the combined asset position (investment value + machinery book value) is:
  • A ₹20,17,937
  • B ₹25,00,000
  • C ₹30,00,000
  • D ₹35,82,063
✓ Correct Answer: (D) ₹35,82,063
Net Position = Investment Value + Machinery BV = ₹28,00,000 + ₹7,82,063 = ₹35,82,063
Case Study 3Bond Valuation & Perpetuity

Fixed Income Investment Portfolio

An investor is considering two options. Option A is a corporate bond with face value ₹10,000, coupon rate 12%, 5 years to maturity, and market yield of 10%. Option B is a perpetual bond paying ₹1,200 annually at 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
Annual Coupon \(= 10{,}000 \times 12\% = \mathbf{₹1{,}200}\)
Q2
Option A bond will trade at:
  • A Discount
  • B Premium
  • C Par value
  • D Cannot determine
✓ Correct Answer: (B) Premium
Coupon Rate (12%) > Market Yield (10%) → bond trades at premium (price > 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
\(\text{PV} = \dfrac{1{,}200}{0.10} = \mathbf{₹12{,}000}\)
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
  • New PV \(= \dfrac{1{,}200}{0.12} = ₹10{,}000\)
  • Key principle: bond/perpetuity value and market yield move in opposite directions. When yield rises, price falls.

📝 Exam Tips for Unit 7 — Financial Mathematics

Common mistakes examiners flag every year

✅ Tip 1 — Always convert the annual rate to a monthly rate before applying the EMI formula. The most common error in EMI questions is using the annual rate (e.g., 12%) directly as \(r\) instead of dividing by 12 first to get the monthly rate (1% = 0.01). Every mark for the EMI calculation depends on this first step being correct.

🔒 More Tips — one per sub-topic — are included in the Question Bank Common mistakes for Perpetuity, Sinking Funds, CAGR, Depreciation & Bonds · All 8 units covered
Get the Question Bank → Exam Tips for All Units

❓ Common Questions on Unit 7 — Financial Mathematics

What is the marks weightage of Unit 7 in Class 12 Applied Maths? ▼
Unit 7 Financial Mathematics carries 10 marks in the CBSE Class 12 Applied Mathematics board examination. This makes it one of the most rewarding units — covering 6 distinct topics, each of which can appear as an MCQ, short answer, long answer, or case study question.
What topics are covered in Unit 7 Financial Mathematics? ▼
Unit 7 covers six topics: Perpetuity (present value of infinite payment streams), Sinking Funds (periodic savings to meet a future target), EMI and Amortization (loan repayment and schedule construction), CAGR (compounded annual growth rate for investments), Depreciation using both straight-line and reducing-balance methods, and Valuation of Bonds (pricing based on coupon rates and market yields).
What is the EMI formula and how is it applied? ▼
The standard EMI formula is \(\text{EMI} = \dfrac{P \cdot r \cdot (1+r)^n}{(1+r)^n - 1}\), where \(P\) = principal, \(r\) = monthly interest rate (annual rate ÷ 12), and \(n\) = number of monthly instalments. The most common mistake is forgetting to divide the annual rate by 12 before substituting.
What is the difference between straight-line and reducing-balance depreciation? ▼
Under the straight-line method (SLM), depreciation is constant every year: \(D = (\text{Cost} - \text{Salvage}) \div n\). Under the reducing-balance method (RBM), a fixed percentage is applied to the declining book value each year, giving higher depreciation in early years. SLM always reaches the exact salvage value; RBM typically leaves a higher residual value.
When does a bond trade at a premium vs a discount? ▼
A bond trades at a premium (price > face value) when the coupon rate > market yield. It trades at a discount (price < face value) when the coupon rate < market yield. It trades at par (price = face value) when both rates are equal. This inverse relationship between bond price and yield is a key concept in Unit 7.
What is the difference between CAGR and simple average growth rate? ▼
The simple average adds up annual growth rates and divides by the number of years — it ignores compounding. CAGR accounts for compounding and represents the single steady rate at which the investment would have grown from start to end. CAGR is preferred when comparing investments because it reflects the true compound effect of growth over time.

📚 All Units — Class 12 Applied Mathematics

Unit 1: Numbers, Quantification & Numerical Applications6 marks Unit 2: Algebra10 marks Unit 3: Calculus15 marks Unit 4: Probability Distributions10 marks Unit 5: Inferential Statistics5 marks Unit 6: Index Numbers & Time-Based Data9 marks Unit 7: Financial Mathematics10 marks — You are here Unit 8: Linear Programming15 marks

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