Compound Interest
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Solved Examples
Study MaterialSolved Examples – Compound Interest
Solved examples help students understand the practical application of Compound Interest concepts in competitive examinations. These examples are designed from basic to advanced level and cover important questions frequently asked in SSC, Banking, Railway, CDS, NDA, UPSC, CAT, Defence, and placement aptitude tests.
Topics Covered in Solved Examples
- Basic Compound Interest Problems
- Amount Calculation
- Half-Yearly Compounding
- Quarterly Compounding
- Difference Between SI and CI
- Variable Rate Problems
- Depreciation Applications
- Advanced Compound Interest Concepts
Example 1: Basic Compound Interest Problem
Question: Find the Compound Interest on ₹10000 at 10% per annum for 2 years.
Solution:
Using formula:
Amount = P(1 + R/100)n
= 10000(1 + 10/100)2
= 10000(1.1)2
= 10000 × 1.21
= ₹12100
Compound Interest:
= 12100 − 10000
= ₹2100
Therefore:
Compound Interest = ₹2100
Example 2: Finding Amount
Question: Find the amount on ₹8000 at 5% compound interest for 3 years.
Solution:
Amount:
= 8000(1 + 5/100)3
= 8000(1.05)3
= 8000 × 1.157625
= ₹9261
Therefore:
Amount = ₹9261
Example 3: Finding Compound Interest Directly
Question: Find the Compound Interest on ₹12000 at 20% per annum for 2 years.
Solution:
Amount:
= 12000(1.2)2
= 12000 × 1.44
= ₹17280
CI:
= 17280 − 12000
= ₹5280
Therefore:
Compound Interest = ₹5280
Example 4: Half-Yearly Compounding
Question: Find the amount on ₹10000 at 10% compounded half-yearly for 1 year.
Solution:
Half-yearly rate:
= 10/2
= 5%
Number of periods:
= 2
Amount:
= 10000(1 + 5/100)2
= 10000(1.05)2
= 10000 × 1.1025
= ₹11025
Therefore:
Amount = ₹11025
Example 5: Quarterly Compounding
Question: Find the amount on ₹16000 at 8% compounded quarterly for 1 year.
Solution:
Quarterly rate:
= 8/4
= 2%
Number of periods:
= 4
Amount:
= 16000(1.02)4
= 16000 × 1.082432
≈ ₹17318.91
Therefore:
Amount ≈ ₹17318.91
Example 6: Difference Between SI and CI
Question: Find the difference between Simple Interest and Compound Interest on ₹5000 at 10% for 2 years.
Solution:
Using shortcut:
Difference:
= P(R/100)2
= 5000(10/100)2
= 5000 × 1/100
= ₹50
Therefore:
Difference = ₹50
Example 7: Finding Principal
Question: A sum amounts to ₹14520 in 2 years at 10% compound interest. Find the principal.
Solution:
Amount:
= P(1.1)2
14520 = P × 1.21
P = 14520/1.21
= ₹12000
Therefore:
Principal = ₹12000
Example 8: Variable Rate Problem
Question: Find the amount on ₹10000 if rates are 10%, 20%, and 25% for three consecutive years.
Solution:
Amount:
= 10000 × 1.10 × 1.20 × 1.25
= 10000 × 1.65
= ₹16500
Therefore:
Amount = ₹16500
Example 9: Depreciation Problem
Question: The value of a machine decreases by 10% every year. Find its value after 2 years if present value is ₹50000.
Solution:
Value after depreciation:
= 50000(1 − 10/100)2
= 50000(0.9)2
= 50000 × 0.81
= ₹40500
Therefore:
Value after 2 years = ₹40500
Example 10: Compound Interest for 3 Years
Question: Find the Compound Interest on ₹20000 at 15% per annum for 3 years.
Solution:
Amount:
= 20000(1.15)3
= 20000 × 1.520875
= ₹30417.50
CI:
= 30417.50 − 20000
= ₹10417.50
Therefore:
Compound Interest = ₹10417.50
Example 11: Difference Between CI and SI for 3 Years
Question: Find the difference between CI and SI on ₹10000 at 10% for 3 years.
Solution:
SI:
= (10000 × 10 × 3)/100
= ₹3000
CI Amount:
= 10000(1.1)3
= ₹13310
CI:
= 13310 − 10000
= ₹3310
Difference:
= 3310 − 3000
= ₹310
Therefore:
Difference = ₹310
Example 12: Compound Interest with Fractional Time
Question: Find the amount on ₹8000 at 10% compound interest for 2½ years.
Solution:
Amount after 2 years:
= 8000(1.1)2
= 8000 × 1.21
= ₹9680
For remaining ½ year:
SI:
= (9680 × 10 × 1)/(100 × 2)
= ₹484
Final amount:
= 9680 + 484
= ₹10164
Therefore:
Amount = ₹10164
Example 13: Finding Rate
Question: A sum becomes ₹12100 in 2 years under compound interest. If principal is ₹10000, find the rate.
Solution:
Using formula:
12100 = 10000(1 + R/100)2
1.21 = (1 + R/100)2
1 + R/100 = 1.1
R = 10%
Therefore:
Rate = 10%
Example 14: Advanced Growth Problem
Question: The population of a town increases by 5% annually. If present population is 40000, find population after 2 years.
Solution:
Population after 2 years:
= 40000(1.05)2
= 40000 × 1.1025
= 44100
Therefore:
Population after 2 years = 44100
Example 15: Advanced Half-Yearly Problem
Question: Find the Compound Interest on ₹20000 at 12% compounded half-yearly for 1 year.
Solution:
Half-yearly rate:
= 12/2
= 6%
Periods:
= 2
Amount:
= 20000(1.06)2
= 20000 × 1.1236
= ₹22472
CI:
= 22472 − 20000
= ₹2472
Therefore:
Compound Interest = ₹2472
Important Exam Tips
- Memorize all CI formulas.
- Learn important square and cube values.
- Carefully identify compounding frequency.
- Use direct amount formula for faster solving.
- Practice SI vs CI comparison regularly.
- Simplify powers and percentages early.
- Verify calculations carefully.
Practicing solved examples regularly improves conceptual clarity, logical thinking, and calculation speed in solving Compound Interest questions in competitive examinations.