📑 Topics on this page (Jump to Topic):
📌 1. Simple Interest
📘 Formulas:
- SI = P × R × T / 100; Amount A = P + SI
- If a sum becomes n times in T years, R = (n − 1) × 100 / T
- In SI the interest is the same every year
✏️ Example: SI on ₹5000 at 8% p.a. for 3 years = 5000 × 8 × 3/100 = ₹1200.
📌 2. Compound Interest
📘 Formulas:
- A = P(1 + R/100)T; CI = A − P
- Half-yearly: rate R/2, time 2T; quarterly: rate R/4, time 4T
- CI − SI for 2 years = P(R/100)²
- CI − SI for 3 years = P(R/100)²(300 + R)/100
- Effective rate for 2 years = 2R + R²/100 %
- Under CI, if a sum doubles in T years it becomes 4 times in 2T and 8 times in 3T
| Rate | 2-year effective % | 3-year effective % |
|---|---|---|
| 5% | 10.25% | 15.7625% |
| 10% | 21% | 33.1% |
| 20% | 44% | 72.8% |
| 4% | 8.16% | 12.4864% |
⚠️ Exam Traps:
- For quarterly compounding divide the rate by 4 and multiply the time by 4 — changing only one is wrong.
- In SI, doubling to quadrupling takes three times as long (n − 1 rule), not twice.
📌 3. Instalments
📘 Formulas:
- Equal annual instalment x under CI: P = x/(1 + r) + x/(1 + r)² + … (r = R/100)
💡 Shortcut Trick: 2 instalments at 10% CI: principal : instalment = 210 : 121.
📌 4. Practice MCQs
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. Simple interest on ₹2500 at 5% per annum for 2 years is: Practice
Correct answer: (B)
Solution: SI = P × R × T/100 = 2500 × 5 × 2/100 = ₹250.
Solution: SI = P × R × T/100 = 2500 × 5 × 2/100 = ₹250.
Q2. Compound interest on ₹1000 at 20% per annum compounded annually for 3 years is: Practice
Correct answer: (C)
Solution: A = 1000 × (120/100)3 = ₹1728; CI = A − P = ₹728.
Solution: A = 1000 × (120/100)3 = ₹1728; CI = A − P = ₹728.
Q3. The difference between CI and SI on a sum at 12% per annum for 2 years is ₹144. The sum is: Practice
Correct answer: (D)
Solution: P = difference × (100/R)² = 144 × (100/12)² = ₹10000.
Solution: P = difference × (100/R)² = 144 × (100/12)² = ₹10000.
Q4. At simple interest a sum becomes 4 times in 10 years. The rate per annum is: Practice
Correct answer: (A)
Solution: R = (n − 1) × 100/T = 3 × 100/10 = 30%.
Solution: R = (n − 1) × 100/T = 3 × 100/10 = 30%.
Q5. Compound interest on ₹10000 at 10% per annum for 1 year, compounded half-yearly, is: Practice
Correct answer: (B)
Solution: Rate = 10/2 = 5%, periods = 2. A = 10000 × (105/100)² = ₹11025; CI = ₹1025.
Solution: Rate = 10/2 = 5%, periods = 2. A = 10000 × (105/100)² = ₹11025; CI = ₹1025.
Q6. At compound interest a sum doubles in 6 years. In how many years will it become 8 times? Practice
Correct answer: (C)
Solution: 2 = (1 + r)6 → 8 = 23 = (1 + r)18 → 18 years.
Solution: 2 = (1 + r)6 → 8 = 23 = (1 + r)18 → 18 years.
Q7. A sum amounts to ₹4800 in 2 years at 10% per annum simple interest. The principal is: Practice
Correct answer: (D)
Solution: A = P(1 + RT/100) → P = 4800 × 100/(100 + 20) = ₹4000.
Solution: A = P(1 + RT/100) → P = 4800 × 100/(100 + 20) = ₹4000.
Q8. The difference between CI and SI on ₹8000 at 10% p.a. for 3 years is: Practice
Correct answer: (A)
Solution: Difference = P(R/100)²(300 + R)/100 = 8000 × (10/100)² × 310/100 = ₹248.
Solution: Difference = P(R/100)²(300 + R)/100 = 8000 × (10/100)² × 310/100 = ₹248.
Q9. A loan is repaid in two equal annual instalments of ₹1210 at 10% per annum compound interest. The loan amount is: Practice
Correct answer: (B)
Solution: P = x/(1 + r) + x/(1 + r)² = 1210/1.1 + 1210/1.21 = ₹2100.
Solution: P = x/(1 + r) + x/(1 + r)² = 1210/1.1 + 1210/1.21 = ₹2100.
Q10. The effective interest percent for 2 years at 25% per annum compound interest is: Practice
Correct answer: (C)
Solution: 2R + R²/100 = 50 + 6.25 = 56.25%.
Solution: 2R + R²/100 = 50 + 6.25 = 56.25%.
Q11. Simple interest on ₹2500 at 5% per annum for 4 years is: Practice
Correct answer: (D)
Solution: SI = P × R × T/100 = 2500 × 5 × 4/100 = ₹500.
Solution: SI = P × R × T/100 = 2500 × 5 × 4/100 = ₹500.
Q12. The difference between CI and SI on ₹2000 at 10% per annum for 2 years is: Practice
Correct answer: (A)
Solution: Difference = P(R/100)² = 2000 × (10/100)² = ₹20.
Solution: Difference = P(R/100)² = 2000 × (10/100)² = ₹20.
Q13. At simple interest a sum becomes 4 times in 5 years. The rate per annum is: Practice
Correct answer: (B)
Solution: R = (n − 1) × 100/T = 3 × 100/5 = 60%.
Solution: R = (n − 1) × 100/T = 3 × 100/5 = 60%.
Q14. A sum amounts to ₹2360 in 3 years at 6% per annum simple interest. The principal is: Practice
Correct answer: (C)
Solution: A = P(1 + RT/100) → P = 2360 × 100/(100 + 18) = ₹2000.
Solution: A = P(1 + RT/100) → P = 2360 × 100/(100 + 18) = ₹2000.
Q15. A loan is repaid in two equal annual instalments of ₹441 at 5% per annum compound interest. The loan amount is: Practice
Correct answer: (D)
Solution: P = x/(1 + r) + x/(1 + r)² = 441/1.05 + 441/1.1 = ₹820.
Solution: P = x/(1 + r) + x/(1 + r)² = 441/1.05 + 441/1.1 = ₹820.
Q16. The effective interest percent for 2 years at 4% per annum compound interest is: Practice
Correct answer: (A)
Solution: 2R + R²/100 = 8 + 0.16 = 8.16%.
Solution: 2R + R²/100 = 8 + 0.16 = 8.16%.
Q17. Compound interest on ₹16000 at 5% per annum compounded annually for 3 years is: Practice
Correct answer: (B)
Solution: A = 16000 × (105/100)3 = ₹18522; CI = A − P = ₹2522.
Solution: A = 16000 × (105/100)3 = ₹18522; CI = A − P = ₹2522.
Q18. The difference between CI and SI on ₹2500 at 12% per annum for 2 years is: Practice
Correct answer: (C)
Solution: Difference = P(R/100)² = 2500 × (12/100)² = ₹36.
Solution: Difference = P(R/100)² = 2500 × (12/100)² = ₹36.
Q19. Compound interest on ₹20000 at 20% per annum for 1 year, compounded half-yearly, is: Practice
Correct answer: (D)
Solution: Rate = 20/2 = 10%, periods = 2. A = 20000 × (110/100)² = ₹24200; CI = ₹4200.
Solution: Rate = 20/2 = 10%, periods = 2. A = 20000 × (110/100)² = ₹24200; CI = ₹4200.
Q20. A sum amounts to ₹6250 in 5 years at 5% per annum simple interest. The principal is: Practice
Correct answer: (A)
Solution: A = P(1 + RT/100) → P = 6250 × 100/(100 + 25) = ₹5000.
Solution: A = P(1 + RT/100) → P = 6250 × 100/(100 + 25) = ₹5000.