1. What is Compound Interest?
Compound Interest (CI) is the interest calculated on the original principal as well as the interest accumulated during previous periods.
In simple interest, interest is calculated only on the original principal. In compound interest, previously earned interest is added to the principal for calculating future interest.
Important Terms
- Principal (P): Original amount invested or borrowed.
- Rate (R): Rate of interest per period.
- Time (T): Number of years or compounding periods.
- Amount (A): Principal plus compound interest.
- Compound Interest (CI): Amount minus principal.
2. Main Compound Interest Formula
If interest is compounded annually:
CI = A − P
Therefore:
CI = P[(1 + R/100)T − 1]
3. Understanding the Formula
| Symbol | Meaning |
|---|---|
| P | Principal |
| R | Rate per annum |
| T | Time in years |
| A | Final Amount |
| CI | Compound Interest |
4. Solved Examples
``` A = 10000(1 + 10/100)2
= 10000 × (1.1)2
= ₹12,100
CI = 12,100 − 10,000
= ₹2,100 ```
``` A = 5000 × (120/100)2
= 5000 × 1.44
= ₹7,200
CI = 7,200 − 5,000
= ₹2,200 ```
5. Half-Yearly Compounding
If interest is compounded half-yearly, interest is calculated twice a year. Therefore, the rate for each half-year is half the annual rate and the number of periods is doubled.
Rate per period = R/2
Number of periods = 2T
A = P(1 + R/200)2T
``` Half-yearly rate = 10/2 = 5%
Number of periods = 2×2 = 4
A = 10000(1.05)4
= ₹12,155.06 approximately ```
6. Quarterly Compounding
When interest is compounded quarterly, there are four compounding periods in one year.
Rate per quarter = R/4
Number of periods = 4T
A = P(1 + R/400)4T
``` Quarterly rate = 8/4 = 2%
Number of periods = 4
A = 8000(1.02)4
≈ ₹8,659.87 ```
7. General Compounding Formula
A = P(1 + R/(100n))nT
| Compounding | Rate per Period | Periods in T Years |
|---|---|---|
| Annually | R | T |
| Half-yearly | R/2 | 2T |
| Quarterly | R/4 | 4T |
8. Compound Interest for Fractional Time
In competitive-examination problems, a common case is 2 years 6 months or 3 years 6 months. When the compounding period is annual, the completed years are compounded first, followed by simple interest on the amount for the remaining fraction of a year.
``` Amount after 2 years:
20000 × (1.10)2 = ₹24,200
Interest for remaining 6 months:
24200 × 10 × 1/2 /100 = ₹1,210
Final Amount = 24,200 + 1,210
= ₹25,410 ```
9. Difference Between CI and SI for 2 Years
For two years, there is a very useful shortcut for finding the difference between compound interest and simple interest.
Or:
CI − SI = PR²/10,000
``` Difference = 10000 × 10²/10000
= ₹100 ```
10. Difference Between CI and SI for 3 Years
For three years, the difference can be obtained by calculating both interests or by using the expansion below.
CI = P[3R/100 + 3R²/10,000 + R³/1,000,000]
CI − SI = P[3R²/10,000 + R³/1,000,000]
``` A = 10000(1.10)3
= ₹13,310
CI = 13,310 − 10,000
= ₹3,310 ```
11. Depreciation
When the value of an article decreases by a fixed percentage every year, the process is called depreciation.
A = P(1 − R/100)T
``` Value = 50000(1 − 10/100)2
= 50000 × (0.9)2
= ₹40,500 ```
12. Population Growth
Population increasing by a fixed percentage every year can be treated like compound growth.
PT = P(1 + R/100)T
``` Population = 50000(1.04)2
= 50000 × 1.0816
= 54,080 ```
13. Successive Percentage Change
Compound interest is closely related to successive percentage changes because each percentage change is applied to the updated value.
Net increase = a + b + ab/100
Increase by a% followed by decrease by b%:
Net change = a − b − ab/100
``` Net increase = 20 + 10 + (20×10)/100
= 30 + 2
= 32% ```
14. Complete Formula Sheet
| Concept | Formula |
|---|---|
| Annual CI Amount | A = P(1+R/100)T |
| Compound Interest | CI = A − P |
| CI Direct Formula | P[(1+R/100)T−1] |
| Half-Yearly | A=P(1+R/200)2T |
| Quarterly | A=P(1+R/400)4T |
| n times/year | A=P(1+R/100n)nT |
| 2-Year CI − SI | PR²/10,000 |
| Depreciation | P(1−R/100)T |
| Growth | P(1+R/100)T |
15. Solved Revision Examples
A = 20000(1.05)2
= ₹22,050
CI = 22,050 − 20,000
= ₹2,050
Difference = 8000 × 10²/10000
= ₹80
Value = 80000(0.95)2
= ₹72,200
16. Practice Questions
- Find the compound interest on ₹10,000 at 10% per annum for 2 years. ```
- Find the amount on ₹8,000 at 5% per annum compounded annually for 3 years.
- Find the CI on ₹20,000 at 10% per annum for 2 years.
- Find the difference between CI and SI on ₹15,000 at 8% per annum for 2 years.
- Find the amount on ₹10,000 at 12% per annum compounded half-yearly for 1 year.
- Find the amount on ₹20,000 at 8% per annum compounded quarterly for 1 year.
- A machine worth ₹60,000 depreciates at 10% per year. Find its value after 2 years.
- A town has a population of 40,000. It grows at 5% per year. Find its population after 2 years.
- A sum of ₹5,000 is invested at 20% compound interest for 2 years. Find the CI.
- A value increases by 10% every year. Find the total percentage increase after 2 years. ```
17. Answers
18. Quick Exam Shortcuts
A = P(1 + R/100)T.
Shortcut 2: Compound Interest = Amount − Principal.
Shortcut 3: For half-yearly compounding, divide the rate by 2 and multiply the number of years by 2.
Shortcut 4: For quarterly compounding, divide the rate by 4 and multiply the number of years by 4.
Shortcut 5: For 2 years:
CI − SI = PR²/10,000.
Shortcut 6: Growth uses:
(1 + R/100)T
while depreciation uses:
(1 − R/100)T. ```
A = P(1 + R/100)T
