Simple interest

NMAT Practice — Simple Interest (20 Questions)

Formula used: SI = (P × R × T) ÷ 100
Click Show Solution to reveal the step-by-step working in green.

1. Find the simple interest on ₹5,000 at 8% p.a. for 3 years.

SI = (5000 × 8 × 3) ÷ 100 = (120,000) ÷ 100 = ₹1,200.
Answer: ₹1,200

2. Calculate SI on ₹12,000 at 10% p.a. for 2 years.

SI = (12,000 × 10 × 2) ÷ 100 = (240,000) ÷ 100 = ₹2,400.
Answer: ₹2,400

3. Find the SI on ₹7,500 at 6% p.a. for 4 years.

SI = (7,500 × 6 × 4) ÷ 100 = (180,000) ÷ 100 = ₹1,800.
Answer: ₹1,800

4. A sum of ₹15,000 is lent at 9% p.a. for 5 years. Find the SI.

SI = (15,000 × 9 × 5) ÷ 100 = (675,000) ÷ 100 = ₹6,750.
Answer: ₹6,750

5. Find SI if P = ₹2,500, R = 5% p.a., T = 2 years.

SI = (2,500 × 5 × 2) ÷ 100 = (25,000) ÷ 100 = ₹250.
Answer: ₹250

6. A sum yields SI of ₹1,200 in 4 years at 6% p.a. Find the principal.

SI = (P × 6 × 4) ÷ 100 = 1,200 → (24P ÷ 100) = 1,200 → P = (1,200 × 100) ÷ 24 = ₹5,000.
Answer: ₹5,000

7. A loan of ₹8,000 at 7% p.a. gives SI of ₹1,120. Find the time.

SI = (8,000 × 7 × T) ÷ 100 = 1,120 → (560T ÷ 100) = 1,120 → 56T = 1,120 → T = 20 years.
Answer: 20 years

8. Find SI on ₹20,000 at 4% p.a. for 9 months.

Time in years = 9 ÷ 12 = 0.75.
SI = (20,000 × 4 × 0.75) ÷ 100 = (60,000) ÷ 100 = ₹600.
Answer: ₹600

9. The SI on ₹4,000 at 12% p.a. for 2 years is?

SI = (4,000 × 12 × 2) ÷ 100 = (96,000) ÷ 100 = ₹960.
Answer: ₹960

10. Find SI if P = ₹9,500, R = 8% p.a., T = 1 year 6 months.

Time = 1.5 years.
SI = (9,500 × 8 × 1.5) ÷ 100 = (114,000) ÷ 100 = ₹1,140.
Answer: ₹1,140

11. If SI is ₹3,600 on ₹12,000 in 3 years, find the rate %.

SI = (12,000 × R × 3) ÷ 100 = 3,600 → (36,000R ÷ 100) = 3,600 → 360R = 3,600 → R = 10%.
Answer: 10%

12. Find SI on ₹50,000 at 3% p.a. for 5 years.

SI = (50,000 × 3 × 5) ÷ 100 = (750,000) ÷ 100 = ₹7,500.
Answer: ₹7,500

13. The SI on ₹15,000 at 5% p.a. for 4 years is?

SI = (15,000 × 5 × 4) ÷ 100 = (300,000) ÷ 100 = ₹3,000.
Answer: ₹3,000

14. A sum of ₹6,400 at 8% p.a. earns what SI in 18 months?

Time = 1.5 years.
SI = (6,400 × 8 × 1.5) ÷ 100 = (76,800) ÷ 100 = ₹768.
Answer: ₹768

15. If SI = ₹2,000, P = ₹10,000, T = 4 years, find R.

(10,000 × R × 4) ÷ 100 = 2,000 → (40,000R ÷ 100) = 2,000 → 400R = 2,000 → R = 5%.
Answer: 5%

16. Find SI on ₹25,000 at 9% p.a. for 2 years 3 months.

Time = 2.25 years.
SI = (25,000 × 9 × 2.25) ÷ 100 = (506,250) ÷ 100 = ₹5,062.50.
Answer: ₹5,062.50

17. If P = ₹18,000, R = 12% p.a., T = 8 months, find SI.

Time = 8 ÷ 12 = 0.666... years.
SI = (18,000 × 12 × 0.666...) ÷ 100 = (144,000 × 0.666...) ÷ 100 = ₹1,200.
Answer: ₹1,200

18. SI = ₹4,800, P = ₹16,000, R = 8% p.a. Find T.

(16,000 × 8 × T) ÷ 100 = 4,800 → (128,000T ÷ 100) = 4,800 → 1,280T = 4,800 → T = 3.75 years.
Answer: 3.75 years (3 years 9 months)

19. Find SI on ₹30,000 at 7% p.a. for 100 days (use year = 365 days).

Time = 100 ÷ 365 ≈ 0.27397 years.
SI ≈ (30,000 × 7 × 0.27397) ÷ 100 ≈ (57,553.9) ÷ 100 ≈ ₹575.54.
Answer: ≈ ₹575.54

20. If ₹5,000 becomes ₹5,750 in 3 years at SI, find the rate %.

SI = 5,750 − 5,000 = ₹750.
(5,000 × R × 3) ÷ 100 = 750 → (15,000R ÷ 100) = 750 → 150R = 750 → R = 5%.
Answer: 5%

These follow the SI formula. I can also create an MCQ quiz version with automatic scoring if needed.