Exercise 13.4: Practical Financial Applications

Definition [1 Mark]: A sinking fund is a designated fund created by setting aside periodic payments at compound interest to accumulate a target capital sum by a future date. Objective: Redemption of bonds/debentures or replacement of worn-out capital machinery.

$$\text{EMI} = \frac{P \cdot i}{1 - (1 + i)^{-n}} \quad [1 \text{ Mark}]$$

• (i) Total Flat Interest: $$I = \frac{P \times R \times T}{100} = \frac{2,00,000 \times 10 \times 2}{100} = \mathbf{₹40,000} \quad [1 \text{ Mark}]$$ • (ii) Total Amount $= 2,00,000 + 40,000 = ₹2,40,000$. $$\text{EMI}_{\text{flat}} = \frac{2,40,000}{24} = \mathbf{₹10,000} \quad [1 \text{ Mark}]$$

Explanation [2 Marks]:
• Interest in period $k$ is $I_k = B_{k-1} \times i$. Since outstanding balance $B_{k-1}$ decreases with each installment, the interest component strictly decreases.
• Since $\text{EMI} = I_k + P_k$ is constant, as $I_k$ decreases, the principal component $P_k$ strictly increases over successive installments.

Part (i) Installment Calculation [2 Marks]:
• $P = ₹15,00,000, i = 0.05, n = 6$. $$R = \frac{P \cdot i}{1 - (1+i)^{-6}} = \frac{15,00,000 \times 0.05}{1 - 0.7462} = \frac{75,000}{0.2538} = \mathbf{₹2,95,508}$$

Part (ii) Amortization Schedule (2 Periods) [2 Marks]:
• Period 1:
- Opening Balance $B_0 = ₹15,00,000$
- Installment $R = ₹2,95,508$
- Interest $I_1 = 15,00,000 \times 0.05 = ₹75,000$
- Principal Repayment $P_1 = 2,95,508 - 75,000 = ₹2,20,508$
- Closing Balance $B_1 = 15,00,000 - 2,20,508 = ₹12,79,492$
• Period 2:
- Opening Balance $B_1 = ₹12,79,492$
- Installment $R = ₹2,95,508$
- Interest $I_2 = 12,79,492 \times 0.05 = ₹63,975$
- Principal Repayment $P_2 = 2,95,508 - 63,975 = ₹2,31,533$
- Closing Balance $B_2 = 12,79,492 - 2,31,533 = ₹10,47,959$

• Required Sum $A = 6,00,000 - 60,000 = ₹5,40,000$ [0.5 Mark].
• $i = 0.08, n = 6$. $$R = \frac{5,40,000 \times 0.08}{(1.08)^6 - 1} = \frac{43,200}{1.5869 - 1} = \frac{43,200}{0.5869} = \mathbf{₹73,607} \quad [1.5 \text{ Marks}]$$

• Interest in Period 1: $I_1 = 60,000 \times 0.01 = \mathbf{₹600}$ [1 Mark].
• Principal Repayment in Period 1: $P_1 = \text{EMI} - I_1 = 10,353 - 600 = \mathbf{₹9,753}$ [1 Mark].

• Accumulated Sinking Fund Value: $$A = R \left[ \frac{(1+i)^n - 1}{i} \right] = 4,00,000 \left[ \frac{1.9926 - 1}{0.09} \right] \quad [1 \text{ Mark}]$$ $$A = 4,00,000 \times \frac{0.9926}{0.09} = 4,00,000 \times 11.0289 = \mathbf{₹44,11,560} \quad [1 \text{ Mark}]$$ • Target Liability = ₹50,00,000.
• Conclusion: The fund is insufficient by a deficit of $50,00,000 - 44,11,560 = \mathbf{₹5,88,440}$. [1 Mark]

• In period $k$: $\text{EMI} = I_k + P_k = B_{k-1} \cdot i + P_k$.
• In period $k+1$: $\text{EMI} = I_{k+1} + P_{k+1} = B_k \cdot i + P_{k+1}$.
• Since $\text{EMI}$ is constant: $$B_{k-1} \cdot i + P_k = B_k \cdot i + P_{k+1} \quad [1 \text{ Mark}]$$ • Since $B_k = B_{k-1} - P_k$: $$B_{k-1} \cdot i + P_k = (B_{k-1} - P_k)i + P_{k+1}$$ $$B_{k-1} \cdot i + P_k = B_{k-1} \cdot i - P_k \cdot i + P_{k+1}$$ $$P_k + P_k \cdot i = P_{k+1} \implies P_{k+1} = P_k(1 + i) \quad [1.5 \text{ Marks}]$$ • Hence, the ratio $\frac{P_{k+1}}{P_k} = (1 + i)$ is constant, proving that the principal repayments form a Geometric Progression. [0.5 Mark]

Solution to Q9 [2 Marks]:
• $P = ₹24,00,000, i = 0.0075, n = 120$. $$\text{EMI} = \frac{24,00,000 \times 0.0075}{1 - (1.0075)^{-120}} = \frac{18,000}{1 - 0.4079} = \frac{18,000}{0.5921} = \mathbf{₹30,400.27}$$
• The required monthly EMI is ₹30,400.

Solution to Q10 [2 Marks]:
• Total Payment over 120 months $= 30,400.27 \times 120 = \mathbf{₹36,48,032.40}$.
• Total Interest Cost $= 36,48,032.40 - 24,00,000 = \mathbf{₹12,48,032.40}$.
• Total interest paid is approximately ₹12.48 Lakhs over the 10-year term.