Last Updated: May 2026
Simple and Compound Interest is a guaranteed IPMAT 2027 quant topic — typically 1–2 questions in IIM Indore IPMAT and JIPMAT. The chapter rewards formula fluency: master five core formulas and you handle nearly every interest question in under a minute.
Core Formulas — Memorise These Five
| Quantity | Formula |
|---|---|
| Simple Interest (SI) | SI = P × R × T / 100 |
| Compound Interest Amount (annual) | A = P (1 + R/100)^T |
| Compound Interest Amount (n compounds/year) | A = P (1 + R/(100·n))^(n·T) |
| CI for 2 years (annual) | CI = P × R(R + 200) / 10000 |
| CI − SI for 2 years | P × (R/100)² |
Important Identity — CI vs SI Difference
For 2 years: CI − SI = P × (R/100)²
For 3 years: CI − SI = P × (R/100)² × (3 + R/100)
These two identities convert what looks like a six-step calculation into a one-step plug-in. They appear in nearly every IPMAT paper.
Compounding Frequency — When R is Annual
- Annually — n = 1, applied once per year
- Half-yearly — divide rate by 2, multiply time by 2
- Quarterly — divide rate by 4, multiply time by 4
- Monthly — divide rate by 12, multiply time by 12
Quick check: ₹1000 at 10% per annum compounded half-yearly for 1 year → A = 1000 × (1.05)² = ₹1102.50.
Worked Example 1 — Direct CI
Problem: Find compound interest on ₹8000 at 5% per annum for 3 years.
Solution: A = 8000 × (1.05)³ = 8000 × 1.157625 = 9261. CI = A − P = 9261 − 8000 = ₹1261.
Worked Example 2 — CI − SI Trick
Problem: The difference between CI and SI on ₹4000 at 8% per annum for 2 years is?
Solution: CI − SI = P × (R/100)² = 4000 × (0.08)² = 4000 × 0.0064 = ₹25.60.
Worked Example 3 — Effective Rate
Problem: What is the effective annual rate equivalent to 12% per annum compounded quarterly?
Solution: Effective rate = (1 + 0.12/4)⁴ − 1 = (1.03)⁴ − 1 = 1.1255 − 1 = 12.55%.
Worked Example 4 — Time Doubling
Problem: In how many years will a sum double itself at 10% per annum compound interest?
Solution: 2P = P × (1.1)^T → (1.1)^T = 2. Take log: T = log 2 / log 1.1 ≈ 0.301 / 0.0414 ≈ 7.27 years.
For SI: 2P = P + P × R × T / 100 → T = 100/R years (so 10 years at 10%). For CI, the rough rule is the “Rule of 72”: doubling time ≈ 72/R years.
Five IPMAT Trick Patterns
- Mixing SI and CI — Same principal earns different rates SI vs CI; find the principal
- Instalment problems — equal instalments, find original principal
- Equivalence statements — “SI on P at R% for T = CI on Q at S% for U”
- Compounding frequency change — convert nominal rate to effective rate
- Population growth — same formula as CI, often disguised
30 Practice Problems — IPMAT SI and CI
[cg_quiz id=”cg-ipm-si-ci-2027″]
Frequently Asked Questions
When are SI and CI equal?
For exactly one year (T = 1) at the same rate, SI and CI are equal because compounding has not yet occurred. For T > 1, CI is always greater than SI at the same rate.
What is the Rule of 72?
An estimate for doubling time at compound interest: T ≈ 72/R years. At 8% per annum a sum doubles in roughly 9 years; at 12% in roughly 6 years. Useful for quick mental approximations.
How is half-yearly compounding handled?
Halve the rate (R/2) and double the time periods (2T). For 8% per annum, half-yearly: 4% per period for 2T periods. The total amount becomes A = P (1 + R/200)^(2T).
Why is CI − SI for 3 years P × (R/100)² × (3 + R/100)?
Expand A = P(1 + R/100)³ and subtract SI = 3PR/100 + P. The cubic terms collapse to P × (R/100)² × (3 + R/100). Memorise it; on the IPMAT clock you will not have time to derive.
Continue Your IPMAT 2027 Prep
- IPMAT Profit Loss Discount 2027
- IPMAT Time Speed Distance and Work 2027
- IPM Gurukul Courses
- IPMAT Mock Test
Bottom line: Memorise the five formulas, the two CI−SI shortcuts, and the four compounding-frequency conversions. Practise 30 problems and SI/CI will become an automatic 30-second answer in your IPMAT 2027 attempt.