30 Interest Questions for Competitive Exams

Simple Interest (SI)

Basic Formula:

SI=P×R×T100\text{SI} = \frac{P \times R \times T}{100}SI=100P×R×T​

जहाँ:

• P = Principal (मूल धन)
• R = Rate of Interest (% प्रति वर्ष)
• T = Time (वर्षों में)

💰 Amount (A):

A=P+SIA = P + SIA=P+SI

⚙️ Formulas Derived:

1. Principal (P) = (SI × 100) / (R × T)
2. Rate (R) = (SI × 100) / (P × T)
3. Time (T) = (SI × 100) / (P × R)

🧮 Example:

Find SI on ₹2000 at 5% for 3 years.
SI = (2000 × 5 × 3)/100 = ₹300

💡 Trick – For fractional time:

If time is 1 year 6 months → T = 1.5 years

💥 When Amount becomes Double / Triple (SI):

If A = 2P → then
P + SI = 2P → SI = P
So, T=100RT = \frac{100}{R}T=R100​

(For Double amount)

For Triple amount →
P + SI = 3P → SI = 2P T=200RT = \frac{200}{R}T=R200​

Compound Interest (CI)

📘 Basic Formula (Annual Compounding):

A=P(1+R100)TA = P \left(1 + \frac{R}{100}\right)^TA=P(1+100R​)T CI=A−PCI = A – PCI=A−P

💰 For Half-Yearly Compounding:

• Rate → R/2
• Time → 2T

A=P(1+R2×100)2TA = P \left(1 + \frac{R}{2 \times 100}\right)^{2T}A=P(1+2×100R​)2T

💰 For Quarterly Compounding:

• Rate → R/4
• Time → 4T

A=P(1+R4×100)4TA = P \left(1 + \frac{R}{4 \times 100}\right)^{4T}A=P(1+4×100R​)4T

Formulas Derived:

1. Principal (P) = A / (1 + R/100)ᵀ
2. Rate (R) = (A/P)(1/T)−1(A/P)^(1/T) – 1(A/P)(1/T)−1 × 100
3. Time (T) = log⁡(A/P)/log⁡(1+R/100)\log(A/P) / \log(1 + R/100)log(A/P)/log(1+R/100)

Example:

Find CI on ₹10,000 for 2 years at 10% p.a.
A = 10000(1 + 10/100)² = 10000 × 1.21 = 12100
CI = 12100 – 10000 = ₹2100

Difference between CI and SI for 2 years:

CI – SI=P(R100)2\text{CI – SI} = P \left(\frac{R}{100}\right)^2CI – SI=P(100R​)2

Difference for 3 years:

CI – SI=P(R100)2(300+R100)\text{CI – SI} = P \left(\frac{R}{100}\right)^2 \left(\frac{300 + R}{100}\right)CI – SI=P(100R​)2(100300+R​)

When Amount Doubles in CI:

(1+R100)T=2(1 + \frac{R}{100})^T = 2(1+100R​)T=2 R=(21/T−1)×100R = (2^{1/T} – 1) \times 100R=(21/T−1)×100

💡 When Amount Triples in CI:

(1+R100)T=3(1 + \frac{R}{100})^T = 3(1+100R​)T=3 R=(31/T−1)×100R = (3^{1/T} – 1) \times 100R=(31/T−1)×100

💡 Shortcut for 2 Years CI (Direct):

CI=P[2R100+R210000]CI = P \left[\frac{2R}{100} + \frac{R^2}{10000}\right]CI=P[1002R​+10000R2​]

Example:
P = 10000, R = 10%
CI = 10000 × [20/100 + 100/10000] = 10000 × 0.21 = ₹2100

💡 Shortcut for 3 Years CI (Direct):

CI=P[3R100+3R210000+R31000000]CI = P \left[\frac{3R}{100} + \frac{3R^2}{10000} + \frac{R^3}{1000000}\right]CI=P[1003R​+100003R2​+1000000R3​]

Conceptual Tricks for Exams

Concept	Trick
Double in CI	(1 + R/100)ᵗ = 2
Triple in CI	(1 + R/100)ᵗ = 3
Find Principal	Divide amount by growth factor
Find Rate (using A & P)	(A/P)^(1/T) – 1
Find CI faster (2 yrs)	Add SI + P(R/100)²
Find CI faster (3 yrs)	Add SI + P(R/100)²(300+R)/10000
Convert months to years	6 months = 0.5 years, 9 months = 0.75 years

30 Interest Questions with Solutions for Competitive Exams

🔹 LEVEL 1: BASIC (10 Questions)

(Focus: Formula understanding & basic calculation)

1. Find the Simple Interest on ₹2,000 at 5% per annum for 3 years.
Formula: SI = (P × R × T) / 100
Solution: (2000 × 5 × 3)/100 = ₹300

2. Find the amount after 2 years on ₹5,000 at 6% simple interest.
Amount = P + SI
SI = (5000 × 6 × 2)/100 = ₹600
Amount = 5000 + 600 = ₹5600

3. What will be the Simple Interest on ₹1,200 at 10% for 1 year 6 months?
T = 1.5 years
SI = (1200 × 10 × 1.5)/100 = ₹180

4. The rate of interest is 12% per annum. In how many years will ₹1,000 double?
Let SI = P
P = ₹1000
SI = (P × R × T)/100 ⇒ 1000 = (1000 × 12 × T)/100
T = 8.33 years (approx. 8 years 4 months)

5. Find the rate if ₹5000 amounts to ₹6000 in 4 years under SI.
SI = 6000 – 5000 = 1000
R = (SI × 100)/(P × T) = (1000 × 100)/(5000 × 4) = 5%

6. A sum of ₹2,500 earns ₹250 in 2 years at SI. Find the rate.
R = (250 × 100)/(2500 × 2) = 5%

7. Find the principal if SI = ₹400, R = 10%, T = 4 years.
P = (SI × 100)/(R × T) = (400 × 100)/(10 × 4) = ₹1000

8. Find SI for ₹10,000 at 7% for 9 months.
T = 9/12 = 0.75 years
SI = (10000 × 7 × 0.75)/100 = ₹525

9. A person borrowed ₹4,000 at 5% p.a. for 2 years. Find total amount.
SI = (4000 × 5 × 2)/100 = ₹400
Amount = 4000 + 400 = ₹4400

10. The SI on ₹900 for 3 years is ₹243. Find the rate.
R = (SI × 100)/(P × T) = (243 × 100)/(900 × 3) = 9%

🔸 LEVEL 2: MEDIUM (10 Questions)

(Focus: Compound Interest and comparison)

11. Find the Compound Interest on ₹10,000 for 2 years at 10% p.a.
Formula: A = P(1 + R/100)ⁿ
A = 10000(1 + 10/100)² = 10000 × 1.21 = ₹12,100
CI = 12100 – 10000 = ₹2,100

12. Find CI for ₹8,000 at 5% p.a. for 3 years.
A = 8000(1 + 5/100)³ = 8000 × 1.157625 = ₹9,261
CI = ₹1,261

13. Find CI for ₹5,000 at 4% p.a. for 2 years, compounded half-yearly.
R = 4/2 = 2% per half-year, n = 2 × 2 = 4
A = 5000(1 + 2/100)⁴ = 5000 × 1.08243 = ₹5,412.15
CI = ₹412.15

14. Find CI on ₹10,000 for 2 years at 10% p.a. compounded annually and compare with SI.
CI = ₹2,100 (from Q11)
SI = (10000 × 10 × 2)/100 = ₹2,000
Difference = ₹100

15. Find CI on ₹20,000 at 10% per annum for 1 year compounded quarterly.
R = 10/4 = 2.5%, n = 4
A = 20000(1 + 2.5/100)⁴ = 20000 × 1.10381289 = ₹22,076.26
CI = ₹2,076.26

16. A sum of ₹12,000 amounts to ₹13,272 in 2 years compounded annually. Find the rate.
A = P(1 + R/100)²
13272 = 12000(1 + R/100)²
(1 + R/100)² = 1.106
1 + R/100 = 1.05
R = 5%

17. At what rate will ₹2,000 amount to ₹2,420 in 3 years at CI?
A = P(1 + R/100)³
1 + R/100 = (2420/2000)^(1/3) = (1.21)^(1/3) ≈ 1.065
R ≈ 6.5%

18. The difference between CI and SI on ₹5,000 for 2 years at 5% is?
CI – SI = P(R/100)² = 5000(5/100)² = ₹12.5

19. Find CI on ₹40,000 at 8% for 2 years compounded half-yearly.
R = 8/2 = 4%, n = 4
A = 40000(1 + 4/100)⁴ = 40000 × 1.169858 = ₹46,794.32
CI = ₹6,794.32

20. If the CI on ₹10,000 in 2 years is ₹2,100, find the rate.
A = 12,100 → (1 + R/100)² = 1.21
R = 10%

🔹 LEVEL 3: ADVANCED (10 Questions)

(Focus: Mixed CI-SI, time, rate, and tricky concepts)

21. The difference between CI and SI on a sum for 2 years at 10% p.a. is ₹100. Find the principal.
CI – SI = P(R/100)²
100 = P(10/100)²
P = ₹10,000

22. A sum of money becomes ₹1,210 in 2 years and ₹1,331 in 3 years under CI. Find the rate.
Ratio = 1331 / 1210 = 1 + R/100
R = (1331/1210 – 1) × 100 = 10%

23. A sum amounts to ₹6,600 in 2 years and ₹6,930 in 3 years at CI. Find the principal and rate.
A₂ / A₁ = 6930 / 6600 = 1.05 → R = 5%
P = 6600 / (1.05)² = 6600 / 1.1025 = ₹5,985

24. Find the rate if a sum doubles itself in 8 years at CI.
(1 + R/100)⁸ = 2 → 1 + R/100 = 2^(1/8) = 1.0905 → R = 9.05%

25. A sum triples in 12 years at CI. Find the rate.
(1 + R/100)¹² = 3
R = (3^(1/12) – 1) × 100 = 9.6%

26. The CI on a certain sum for 2 years at 10% is ₹420. Find the sum.
CI = P[(1 + R/100)² – 1]
420 = P(1.21 – 1) = P(0.21)
P = 420 / 0.21 = ₹2,000

27. A sum becomes ₹4,840 in 2 years and ₹5,324 in 3 years at CI. Find the rate and principal.
Ratio = 5324 / 4840 = 1.1 → R = 10%
P = 4840 / (1.1)² = 4840 / 1.21 = ₹4,000

28. At what rate will ₹12,000 amount to ₹13,398 in 2 years compounded annually?
13398 = 12000(1 + R/100)²
(1 + R/100)² = 1.1165
R ≈ 5.7%

29. A sum of money doubles in 5 years at CI. Find the rate and the time to become 8 times.
(1 + R/100)⁵ = 2 → R = 14.87%
For 8 times: (1 + R/100)ⁿ = 8 → n = 15 years

30. A sum of ₹25,000 is lent at 8% for the first year and 10% for the second year, compounded annually. Find the amount after 2 years.
A = 25000(1 + 8/100)(1 + 10/100) = 25000 × 1.08 × 1.10 = ₹29,700

Key Formulas Summary

Concept	Formula
Simple Interest	SI = (P × R × T) / 100
Amount (SI)	A = P + SI
Compound Interest	A = P(1 + R/100)ⁿ
CI	CI = A – P
Difference (2 years)	CI – SI = P(R/100)²
Difference (3 years)	CI – SI = P(R/100)²(300 + R)/10000