Simple and Compound Interest: Formulas, Examples and Shortcuts

Simple interest and compound interest questions appear in almost every campus placement aptitude test, and the distinction between the two formulas explains most of the tricky answer choices.

Both use the same three inputs: principal (P), annual rate (R), and time in years (T). The difference is where the interest accumulates. SI always applies the rate to the original P. CI applies it to an expanding base. One grows linearly; the other grows exponentially. That structural gap is what placement test-setters exploit in every variant they write.

If you’ve already worked through Time and Work aptitude questions, the structure here is similar: one core formula, applied across five question types. Both SI and CI formulas express the rate as a fraction (R divided by 100); if percentage notation feels unsteady, the percentage problem shortcuts guide covers that prerequisite in full.

Simple Interest: Formula and Worked Examples

The Derivation

Interest is a charge per rupee per year. Three steps get you to the formula:

• Charge per rupee per year at rate R%: R/100
• Charge on P rupees per year: P × R/100
• Charge on P rupees over T years: P × R/100 × T

That gives:

• SI = (P × R × T) / 100

Nothing more is hidden in the formula. The three inputs multiply; the 100 normalises the percentage.

Worked Example 1: Find SI and Final Amount

• Given: P = ₹6000, R = 8% per annum, T = 3 years
• Find: SI and the final amount
• Step 1: SI = (P × R × T) / 100 = (6000 × 8 × 3) / 100 = 144000 / 100
• SI = ₹1440
• Amount = P + SI = 6000 + 1440 = ₹7440

Worked Example 2: Find Rate Given SI

In placement papers, the question often reverses the formula: given SI, P, and T, find R.

• Given: SI = ₹1200, P = ₹4000, T = 3 years
• Find: R
• Rearrange SI formula: R = (SI × 100) / (P × T)
• Step 1: R = (1200 × 100) / (4000 × 3) = 120000 / 12000
• R = 10% per annum

The same rearrangement works for finding T: T = (SI × 100) / (P × R).

Compound Interest: Formula and Worked Examples

The Derivation

At the end of year 1, the amount is P × (1 + R/100). In year 2, interest applies to that new base, not to the original P. Repeating this for T years:

• A = P × (1 + R/100)^T
• CI = A − P = P × (1 + R/100)^T − P

The derivation follows directly from the compounding logic described in NCERT Class 8 Mathematics Chapter 8 (Comparing Quantities), which is the standard source most placement test-setters draw from.

Worked Example: Find CI

• Given: P = ₹5000, R = 8% per annum, T = 2 years
• Find: CI and final amount
• Step 1: Compute (1 + 8/100)^2 = (1.08)^2
  • (1.08)^2 = 1 + 2 × 0.08 + (0.08)^2 = 1 + 0.16 + 0.0064 = 1.1664
• Step 2: A = 5000 × 1.1664 = ₹5832
• CI = A − P = 5832 − 5000 = ₹832

SI vs CI Comparison

Factor	Simple Interest	Compound Interest
Interest base	Always original P	P + accumulated interest
Growth shape	Linear	Exponential
Formula	SI = (P × R × T) / 100	CI = P × (1 + R/100)^T − P
Relative return	Lower	Higher (for T > 1 year)

For T = 1 year with the same P and R, SI and CI produce identical results. The gap opens from year 2 onward.

Half-Yearly and Quarterly Compounding

Placement tests add difficulty by changing the compounding frequency. The principle is constant: use the per-period rate and the total number of periods.

Half-Yearly Compounding

If interest compounds every 6 months, the per-period rate is R/2 percent and the number of periods for T years is 2T:

• A = P × (1 + R/200)^(2T)

The most common error here is substituting R unchanged. Always halve the rate and double the periods.

Worked Example: Half-Yearly

• Given: P = ₹8000, R = 10% per annum, T = 1.5 years, compounded half-yearly
• Find: CI
• Step 1: Number of periods = 2 × 1.5 = 3; per-period rate = 10/2 = 5%
• Step 2: A = 8000 × (1.05)^3
  • (1.05)^2 = 1.1025
  • (1.05)^3 = 1.05 × 1.1025 = 1.157625
  • A = 8000 × 1.157625 = ₹9261
• CI = 9261 − 8000 = ₹1261

Quarterly Compounding

If interest compounds every 3 months, the per-period rate is R/4 percent and the number of periods for T years is 4T:

• A = P × (1 + R/400)^(4T)

Worked Example: Quarterly

• Given: P = ₹10000, R = 8% per annum, T = 1 year, compounded quarterly
• Find: CI
• Step 1: Number of periods = 4 × 1 = 4; per-period rate = 8/4 = 2%
• Step 2: A = 10000 × (1.02)^4
  • (1.02)^2 = 1.0404
  • (1.02)^4 = (1.0404)^2 = 1.08243216
  • A = 10000 × 1.08243216 = ₹10824.32
• CI ≈ ₹824.32

The CI-SI Difference Shortcuts

Placement papers test this shortcut directly: given P and R, find the difference between CI and SI without computing both from scratch.

2-Year Shortcut: Derivation

Let r = R/100.

• SI for 2 years = 2Pr
• CI for 2 years = P(1+r)^2 − P = P(1 + 2r + r^2) − P = 2Pr + Pr^2
• Difference = CI − SI = Pr^2 = P × (R/100)^2

3-Year Shortcut: Derivation

• SI for 3 years = 3Pr
• CI for 3 years = P[(1+r)^3 − 1] = P[3r + 3r^2 + r^3]
• Difference = CI − SI = P[3r^2 + r^3] = Pr^2(3 + r) = P × (R/100)^2 × (3 + R/100)

Verification with P = ₹5000, R = 10%

• 2-year check:

  • SI = (5000 × 10 × 2)/100 = ₹1000
  • CI = 5000 × (1.10)^2 − 5000 = 5000 × 1.21 − 5000 = ₹1050
  • Difference = ₹50
  • Formula: 5000 × (0.10)^2 = 5000 × 0.01 = ₹50 ✓

• 3-year check:

  • SI = (5000 × 10 × 3)/100 = ₹1500
  • CI = 5000 × (1.10)^3 − 5000 = 5000 × 1.331 − 5000 = ₹1655
  • Difference = ₹155
  • Formula: 5000 × (0.10)^2 × (3 + 0.10) = 5000 × 0.01 × 3.1 = ₹155 ✓

A note on the approximation sometimes printed in textbooks: the shortcut CI ≈ SI + SI × R/100 is only valid for 2-year scenarios. It is not a general formula. The exact expressions above apply for any P and R.

Placement Test Question Bank

These four question types are the most frequently tested patterns across campus placement evaluation tests. Work through each before looking at the solution.

Q1: Find SI

• Question: A sum of ₹6000 is invested at 6% per annum SI for 4 years. Find the SI and final amount.
• Step 1: SI = (P × R × T) / 100 = (6000 × 6 × 4) / 100 = 144000 / 100
• SI = ₹1440
• Amount = 6000 + 1440 = ₹7440

Q2: Find Principal from CI-SI Difference

• Question: The CI and SI on a principal at 12% per annum for 2 years differ by ₹57.60. Find the principal.
• Step 1: Difference = P × (R/100)^2
• Step 2: 57.60 = P × (12/100)^2 = P × 0.0144
• P = 57.60 / 0.0144 = ₹4000

Q3: Half-Yearly Compounding

• Question: Find the CI on ₹4000 at 10% per annum for 1 year, compounded half-yearly.
• Step 1: Number of periods = 2; per-period rate = 5%
• Step 2: A = 4000 × (1.05)^2 = 4000 × 1.1025 = ₹4410
• CI = 4410 − 4000 = ₹410

Q4: Find Rate from SI

• Question: At what annual rate does ₹4000 yield SI of ₹1200 over 3 years?
• Step 1: R = (SI × 100) / (P × T) = (1200 × 100) / (4000 × 3) = 120000 / 12000
• R = 10% per annum

For additional practice sets drawn from real placement papers, IndiaBix Compound Interest maintains a wide bank of CI questions with verified solutions.

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