3.2 Simple Interest & Compound Interest (Compounding Cycles & Differences)

Simple Interest & Compound Interest Mechanics

Interest calculations form a fundamental component of financial mathematics tested extensively in the SBI PO examination. Understanding how money accumulates linearly under Simple Interest (SI) versus exponentially under Compound Interest (CI) is critical for solving comparison problems, non-annual compounding cycles, and equal loan repayment installments.

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1. Core Principles & Mathematical Formulas

Simple Interest (SI)

Simple Interest is calculated solely on the principal sum throughout the loan tenure. The annual interest earned remains strictly constant.

• Formula: $SI = \frac{P \times R \times T}{100}$
• Total Amount ($A$): $A = P + SI = P \left(1 + \frac{R \times T}{100}\right)$ Where $P$ = Principal, $R$ = Annual Rate of Interest (%), $T$ = Time in years.

Compound Interest (CI)

Compound Interest incorporates interest earned in preceding periods into the principal base for subsequent compounding cycles. Interest earns interest.

• Total Accumulated Amount ($A$): $A = P \left(1 + \frac{R}{100}\right)^T$
• Compound Interest ($CI$): $CI = A - P = P \left[ \left(1 + \frac{R}{100}\right)^T - 1 \right]$

Non-Annual Compounding Cycles

When compounding occurs more frequently than once a year, the nominal annual rate ($R%$) is divided by the compounding frequency ($n$), and the total number of periods increases to $n \times T$.

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2. CI - SI Differences & Pascal's Ratio Shortcuts

CI - SI Difference Formulas

Direct difference formulas eliminate long multi-step calculations in SBI PO Prelims & Mains:

• 2-Year Difference ($D_2$): $D_2 = CI_2 - SI_2 = P \left(\frac{R}{100}\right)^2$
• 3-Year Difference ($D_3$): $D_3 = CI_3 - SI_3 = P \left(\frac{R}{100}\right)^2 \left(3 + \frac{R}{100}\right) = P \left(\frac{R}{100}\right)^2 \left(\frac{300 + R}{100}\right)$
• Ratio of 3-Year to 2-Year Difference: $\frac{D_3}{D_2} = 3 + \frac{R}{100} = \frac{300 + R}{100}$

Pascal's Triangle / Ratio Method for CI

Instead of raising decimal multipliers to power $T$, calculate interest on interest step-by-step and apply Pascal's coefficients:

• For 2 Years: Ratio is 2 : 1
  • $CI = 2(x) + 1(y)$, where $x = R% \text{ of } P$, and $y = R% \text{ of } x$.
• For 3 Years: Ratio is 3 : 3 : 1
  • $CI = 3(x) + 3(y) + 1(z)$, where $x = R% \text{ of } P$, $y = R% \text{ of } x$, and $z = R% \text{ of } y$.
• For 4 Years: Ratio is 4 : 6 : 4 : 1

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3. Step-by-Step Worked Math Examples

Worked Example 1: Pascal's Ratio Method for 3-Year CI

Problem: Find the compound interest on ₹10,000 for 3 years at 10% per annum compounded annually using the ratio shortcut.

Step-by-Step Solution:

1. Calculate 1st stage interest ($x$): $10% \text{ of } 10,000 = ₹1,000$.
2. Calculate 2nd stage interest ($y$): $10% \text{ of } 1,000 = ₹100$.
3. Calculate 3rd stage interest ($z$): $10% \text{ of } 100 = ₹10$.
4. Apply 3-Year Pascal Ratio 3 : 3 : 1: $CI = 3(x) + 3(y) + 1(z) = 3(1,000) + 3(100) + 1(10)$
5. $CI = 3,000 + 300 + 10 = ₹3,310$.

Worked Example 2: Finding Principal from CI-SI Difference

Problem: The difference between compound interest and simple interest on a sum of money for 2 years at 8% per annum is ₹128. Find the principal sum.

Step-by-Step Solution:

1. Use the 2-Year Difference formula: $D_2 = P \left(\frac{R}{100}\right)^2$.
2. Substitute known values: $128 = P \left(\frac{8}{100}\right)^2$.
3. Simplify rate fraction: $\frac{8}{100} = \frac{2}{25} \implies \left(\frac{2}{25}\right)^2 = \frac{4}{625}$.
4. Equation becomes: $128 = P \times \frac{4}{625}$.
5. Solve for $P$: $P = \frac{128 \times 625}{4} = 32 \times 625 = ₹20,000$.

Worked Example 3: Equal Annual Installments under Simple Interest

Problem: What annual installment will discharge a debt of ₹4,600 due in 4 years at 10% simple interest per annum?

Step-by-Step Solution:

1. Let each annual installment be ₹$x$.
2. Installment paid at the end of Year 4 earns no interest (settles debt immediately) $= x$.
3. Installment paid at end of Year 3 earns interest for 1 year $= x + \left(\frac{x \times 10 \times 1}{100}\right) = 1.10x$.
4. Installment paid at end of Year 2 earns interest for 2 years $= x + \left(\frac{x \times 10 \times 2}{100}\right) = 1.20x$.
5. Installment paid at end of Year 1 earns interest for 3 years $= x + \left(\frac{x \times 10 \times 3}{100}\right) = 1.30x$.
6. Sum of all installment discharge values $= x + 1.10x + 1.20x + 1.30x = 4.60x$.
7. Equate to total debt due: $4.60x = 4,600 \implies x = \frac{4,600}{4.60} = ₹1,000$.

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4. Advanced Equal Installment Formulas for Loan Repayments

• Compound Interest Equal Installment Formula: If a loan of principal $P$ is repaid in $n$ equal annual installments of ₹$x$ at $R%$ per annum compounded annually: $P = \frac{x}{\left(1 + \frac{R}{100}\right)^1} + \frac{x}{\left(1 + \frac{R}{100}\right)^2} + \dots + \frac{x}{\left(1 + \frac{R}{100}\right)^n}$

• Simple Interest Equal Installment Formula: For debt $A$ due in $n$ years at $R%$ simple interest, annual installment $x$ is: $A = n x + \frac{x \times R}{100} \times \left[ \frac{(n - 1) n}{2} \right]$