1. Formula
SI = P × R × T / 100; Amount A = P + SI. T in years (months ÷ 12).
2. Derived results
| Case | Result |
|---|---|
| Sum doubles in T years | R = 100/T |
| Sum becomes n times in T years | R = 100(n − 1)/T |
| Doubles in T years; time to become n times | (n − 1) × T |
| SI is 1/x of the principal | R × T = 100/x |
| Amount A₁ after T₁, A₂ after T₂ | SI per year = (A₂ − A₁)/(T₂ − T₁) |
3. Solved examples
₹5,000 at 8% for 3 years → SI = ₹1,200.
Sum doubles in 8 years → R = 12.5%. It triples in 2 × 8 = 16 years.
Amount ₹1,240 in 2 years, ₹1,360 in 5 years → SI/year = 40; P = 1240 − 80 = ₹1,160.
Sum lent partly at 5% and partly at 8%; total ₹10,000, yearly interest ₹620 → alligation: (8 − 6.2) : (6.2 − 5) = 1.8 : 1.2 = 3 : 2 → ₹6,000 at 5%.
If the rate rises by 2%, interest on ₹2,500 for 3 years rises by 2500 × 2 × 3/100 = ₹150.
Solved examples by question type
Each type that comes in the ASO paper, worked step by step. Try it yourself first, then read the solution.
Basic level
Example 1. Find the simple interest on ₹2,000 at 4% per annum for 3 years.
(a) ₹480 (b) ₹80 (c) ₹240 (d) ₹2,240
Answer: (c) ₹240
Solution: SI = P×R×T/100 = 2,000×4×3/100 = ₹240.
Example 2. Find the simple interest on ₹12,000 at 6% per annum for 4 years.
(a) ₹720 (b) ₹14,880 (c) ₹5,760 (d) ₹2,880
Answer: (d) ₹2,880
Solution: SI = P×R×T/100 = 12,000×6×4/100 = ₹2,880.
Example 3. ₹2,500 amounts to ₹3,000 in 4 years at simple interest. The rate per annum is:
(a) 4.17% (b) 20% (c) 5% (d) 7%
Answer: (c) 5%
Solution: SI = 500. R = 100 × 500/(2,500 × 4) = 5%.
ASO exam level
Example 4. A sum of ₹1,200 amounts to ₹1,344 in 3 years at simple interest. If the rate is increased by 2%, it would amount to:
(a) ₹1,416 (b) ₹1,440 (c) ₹1,368 (d) ₹1,364
Answer: (a) ₹1,416
Solution: SI = 144 → R = 4%. New R = 6% → SI = 216 → Amount = ₹1,416.
Example 5. A sum was put at simple interest for 3 years. Had it been put at 2% higher rate, it would have fetched ₹240 more. The sum is:
(a) ₹4,000 (b) ₹5,000 (c) ₹8,000 (d) ₹12,000
Answer: (a) ₹4,000
Solution: P × 2 × 3/100 = 240 → P = ₹4,000.
Example 6. A sum doubles itself in 10 years at simple interest. In how many years will it become 4 times?
(a) 30 years (b) 40 years (c) 34 years (d) 26 years
Answer: (a) 30 years
Solution: Doubling: R = 100/10 = 10%. For 4 times, SI = 3P → T = 3 × 100/10 = 30 years.
Example 7. A sum at simple interest amounts to ₹4,640 in 2 years and ₹5,600 in 5 years. The sum is:
(a) ₹3,700 (b) ₹3,680 (c) ₹4,200 (d) ₹4,000
Answer: (d) ₹4,000
Solution: SI for 3 years = 960 → 1 year = 320. P = 4,640 − 2×320 = ₹4,000.
Topic-wise practice for Simple interest (Quant): Sum becomes n times, rate and time, difference in interest. Useful for OPSC ASO, OSSC, OSSSC and other Odisha exams.
Question types covered
- Basic SI = PRT/100
- Find principal, rate or time
- Sum becomes n times in t years
- Difference in SI for two rates or times
- Amount after given years
- Money lent in parts at different rates