Interest problems are a guaranteed topic in every aptitude exam. Simple interest is straightforward, but compound interest has a few powerful shortcuts — especially the CI-SI difference formulas — that can turn a 3-minute calculation into a 30-second answer.
Simple Interest (SI)
In simple interest, we earn interest only on the original amount (principal). The interest stays the same every year.
In simple language, if we put Rs 10,000 in a bank at 5% SI for 3 years, we get 5% of 10,000 = Rs 500 every year. After 3 years, total interest = Rs 1,500.
SI Shortcuts
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Doubling time: At R%, the principal doubles in 100/R years (under SI)
- At 10% → doubles in 10 years
- At 12.5% → doubles in 8 years
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If SI = Principal, then R × T = 100
Compound Interest (CI)
In compound interest, we earn interest on the interest too. Each year’s interest gets added to the principal, and next year’s interest is calculated on this new, larger amount.
Quarterly compounding: Rate = R/4, Time = 4n
In simple language, instead of “rate R for n years,” half-yearly means “rate R/2 for 2n half-years.” We’re just changing the unit of time.
Quick Values for (1 + R/100)ⁿ
Memorize these common ones:
| Rate | 2 years | 3 years |
|---|---|---|
| 5% | (1.05)² = 1.1025 | (1.05)³ = 1.157625 |
| 10% | (1.1)² = 1.21 | (1.1)³ = 1.331 |
| 20% | (1.2)² = 1.44 | (1.2)³ = 1.728 |
The CI - SI Difference — The Most Important Shortcut
This is the single most tested concept in CI problems.
Why this works (for 2 years)
SI for 2 years = 2PR/100 CI for 2 years = P(1 + R/100)² - P = P(R/100)² + 2P(R/100) → which is SI + P(R/100)²
The difference P(R/100)² is literally “interest on one year’s interest” — the extra bit CI gives us.
Effective Rate of Interest
When compounding happens more than once a year, the effective annual rate is higher than the stated rate.
Formula: Effective rate = (1 + R/n)ⁿ - 1, where n = number of compounding periods per year.
Example: 10% compounded half-yearly. What’s the effective annual rate?
- = (1 + 0.10/2)² - 1 = (1.05)² - 1 = 1.1025 - 1 = 0.1025 = 10.25%
So 10% compounded half-yearly is equivalent to 10.25% compounded annually.
Installment Problems
When we borrow money and repay in equal installments, each installment covers part of the principal plus interest.
For SI — Equal Installments:
Each installment = Total amount / number of installments (adjusted for interest). The formula is a bit involved, so let’s see it with an example.
For CI — Equal Annual Installments:
If each installment is X, at rate R%, for n years:
P = X/(1 + R/100) + X/(1 + R/100)² + … + X/(1 + R/100)ⁿ
Example: A sum of Rs 10,000 is borrowed at 10% CI and paid back in 2 equal annual installments. Find the installment.
Let installment = X. 10000 = X/1.1 + X/1.21 10000 = X(1/1.1 + 1/1.21) 10000 = X(1.1/1.21 + 1/1.21) 10000 = X(2.1/1.21) X = 10000 × 1.21/2.1 = 12100/2.1 = Rs 5761.90 (approximately)
Worked Examples
Example 1: Find the difference between CI and SI on Rs 8000 for 2 years at 5% per annum.
CI - SI = P × (R/100)² = 8000 × (5/100)² = 8000 × 1/400 = Rs 20
Let’s verify:
- SI = 8000 × 5 × 2 / 100 = 800
- CI: Amount = 8000 × (1.05)² = 8000 × 1.1025 = 8820. CI = 820.
- Difference: 820 - 800 = 20 ✓
Example 2: A sum of money doubles in 5 years at SI. In how many years will it become 5 times?
If it doubles in 5 years, SI in 5 years = P. So PRT/100 = P → RT = 100. With T = 5: R = 20%. For 5 times: Amount = 5P, so SI = 4P. 4P = P × 20 × T / 100 → T = 400/20 = 20 years
Shortcut: If it doubles (2×) in 5 years, for n× it takes (n-1) × 5 years. So 5× takes 4 × 5 = 20 years.
Example 3: The CI on a sum for 2 years is Rs 832 and the SI is Rs 800. Find the rate and the sum.
CI - SI = 832 - 800 = 32. Also, CI - SI for 2 years = P(R/100)². And SI for 2 years = 2PR/100 = 800 → PR = 40000.
From the SI formula: SI for 1 year = 400. CI - SI = R% of (SI for 1 year) = R/100 × 400 = 32. So R = 32 × 100/400 = 8%
From PR = 40000: P = 40000/8 = Rs 5000
Example 4: Find the CI on Rs 15,000 for 1.5 years at 10% compounded half-yearly.
Half-yearly rate = 5%, number of half-years = 3. Amount = 15000 × (1.05)³ = 15000 × 1.157625 = 17364.375. CI = 17364.375 - 15000 = Rs 2364.38 (approximately)
Example 5: A certain sum amounts to Rs 7260 in 2 years and Rs 7986 in 3 years at CI. Find the rate and the sum.
The difference between 3rd year amount and 2nd year amount = interest on the 2nd year amount. 7986 - 7260 = 726 = interest on 7260 for 1 year. Rate = (726/7260) × 100 = 10%
Now: 7260 = P × (1.1)² = 1.21P → P = 7260/1.21 = Rs 6000
Common Exam Patterns
- “Difference between CI and SI for 2 years” → P(R/100)². Instant answer.
- “Sum doubles in n years at SI, when does it triple?” → Triple time = 2n (since SI is linear)
- “Amount after n years at CI, amount after n+1 years” → Difference gives one year’s CI, which reveals the rate
- “Half-yearly / quarterly compounding” → Adjust rate and time accordingly
- “CI for 2 years is X and SI is Y” → CI - SI gives a relation, SI gives another. Solve.
- “Equal installments” → Set up present value equation
Practice Problems
Q1: Find the CI on Rs 12,000 for 3 years at 10% per annum.
Q2: The difference between CI and SI on a certain sum for 2 years at 12% per annum is Rs 90. Find the sum.
Q3: A sum amounts to Rs 2880 in 2 years and Rs 3456 in 3 years at compound interest. Find the principal and the rate.
Answers:
A1: Amount = 12000 × (1.1)³ = 12000 × 1.331 = 15972. CI = 15972 - 12000 = Rs 3972
A2: CI - SI for 2 years = P(R/100)² → 90 = P × (12/100)² = P × 0.0144 → P = 90/0.0144 = Rs 6250
A3: Interest in 3rd year = 3456 - 2880 = 576 = R% of 2880. So R = (576/2880) × 100 = 20%. Amount after 2 years: 2880 = P × (1.2)² = 1.44P → P = 2880/1.44 = Rs 2000