Simple Interest (S.I.) and Compound Interest (C.I.) – Complete Short Notes

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1. Basic Concepts

Interest is the extra money paid for using borrowed money.

Term	Meaning
Principal (P)	The sum of money lent or invested
Rate (R)	Interest rate per annum (%)
Time (T)	Period for which the money is borrowed (in years)
Amount (A)	Total money after adding interest (A = P + Interest)

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2. Simple Interest (S.I.)

Formula:

$$S\mathrm{.}I\mathrm{.}=\frac{P\times R\times T}{100}$$$$A=P+S.I.$$

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Example:

P = ₹5000, R = 10%, T = 2 years

$$S.I. = (5000×10×2)/100 = ₹1000$$
$$A=5000+1000=\text{₹}6000$$

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3. Key Points about S.I.

• Interest is same every year.

• Rate (R) always in % per annum.

• Time (T) in years.

  • For months: T=months/12

  • For days: 

    $T = \text{days}/365$

• When rate changes every year, calculate interest separately for each year.

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4. Conversion of Units

Time Given	Convert To Years
6 months	0.5 year
9 months	0.75 year
18 months	1.5 year

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5. Finding Each Variable

Principal

$P = \frac{100×S.I.}{R×T}$

Rate	$R = \frac{100×S.I.}{P×T}$

Time	$T = \frac{100×S.I.}{P×R}$

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6. Important Relationships

Given	Find	Formula
Difference between Amount and Principal	S.I.	A − P
If Amount and S.I. known	P = A − S.I.
S.I. same for equal P, R, T	Always constant

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7. Compound Interest (C.I.)

In C.I., interest is added to principal after each year — interest earns interest.

Formula:

$$A = P \left(1 + \frac{R}{100}\right)^T$$
$$C\mathrm{.}I\mathrm{.}=A-P$$

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Example:

P = ₹5000, R = 10%, T = 2 years

$$A = 5000(1 + 10/100)^2 = 5000(1.1)^2 = 6050$$
$$C\mathrm{.}I\mathrm{.}=6050-5000=\text{₹}1050$$

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8. Compound Interest Table Method

Year	Opening	Interest	Closing
1	5000	500	5500
2	5500	550	6050

Interest keeps increasing each year.

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9. When Interest Compounded More Than Once a Year

Compounding Type	Formula

Half-yearly

$A = P(1 + \frac{R}{200})^{2T}$

Quarterly

$A = P(1 + \frac{R}{400})^{4T}$

Monthly

$A = P(1 + \frac{R}{1200})^{12T}$

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Example:

P = ₹8000, R = 10% p.a., T = 1 year, quarterly compounding

$$A = 8000(1 + 10/400)^4 = 8000(1.025)^4 ≈ ₹8205$$
$$C\mathrm{.}I\mathrm{.}=\text{₹}205$$

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10. Difference Between S.I. and C.I.

Feature            Simple Interest            Compound Interest Interest             Same every year            Increases every year Formula    $P×R×T/100$ $P(1+R/100)^T − P$ Amount $P + S.I.$ $P(1+R/100)^T$ Growth Type             Linear            Exponential

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11. Difference Between C.I. and S.I.

When Time = 2 years,

$$C\mathrm{.}I\mathrm{.}-S\mathrm{.}I\mathrm{.}=P{\left(\frac{R}{100}\right)}^2$$

Example:
P = 10000, R = 10%, T = 2 years

$$Diff=10000\times (10\mathrm{/}100{)}^2=100$$ → C.I. is ₹100 more than S.I.

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12. Successive Year Method (Without Formula)

If you don’t want to use power formulas:

Year 1 Interest = (P×R)/100
Year 2 Interest = (P + Year1Interest)×R/100
... and so on.

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13. Finding Rate (R) in C.I.

If P, A, and T given:

$$R = \left[\left(\frac{A}{P}\right)^{1/T} - 1\right] × 100$$

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14. Finding Time (T)

$$T=\frac{\log (A\mathrm{/}P)}{\log (1+R\mathrm{/}100)}$$

(Logarithm method used in advanced problems)

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15. Depreciation (Negative Interest)

If value decreases by R% per year:

$$A=P{(1-\frac{R}{100})}^T$$

Example:
Value = ₹100000, depreciation 10% yearly, 2 years later:

$$A = 100000(0.9)^2 = ₹81000$$

16. Population Growth and Decay (Same Concept)

Growth: $P_n = P(1 + r/100)^n$
Decay: $P_n = P(1 - r/100)^n$

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17. Compound Interest for Fractional Time

If time = 2½ years,
C.I. = C.I. for 2 years + S.I. on 2-year interest for remaining ½ year.

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Example:

P = ₹8000, R = 10%, T = 2½ years
C.I. (2 years) = 8000[(1.1)^2 − 1] = ₹1680
Interest on ₹1680 for ½ year = (1680×10×½)/100 = ₹84
→ Total C.I. = ₹1764

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18. Relation Between S.I. and C.I. for 2 Years

$$C\mathrm{.}I\mathrm{.}=S\mathrm{.}I\mathrm{.}+\frac{S\mathrm{.}I\mathrm{.}\times R}{\mathrm{100}}$$

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19. Shortcut Tricks

✅ For small time and low rates, difference between S.I. and C.I. ≈ negligible.
✅ For equal time and rate, ratio of C.I. to S.I. always > 1.
✅ Successive percentage formula applies for compound growth.
✅ Remember: ( (1 + R/100)^n = 1 + nR/100 + n(n−1)R²/20000 + ... )

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20. Typical Questions

Q1. Find S.I. on ₹5000 for 3 years at 8%.
→ S.I. = (5000×8×3)/100 = ₹1200

Q2. Find C.I. on ₹5000 for 3 years at 8%.
→ A = 5000(1.08)^3 = 5000×1.2597 = ₹6298.5
→ C.I. = ₹1298.5

Q3. Find difference between C.I. and S.I. for 2 years at 10% on ₹2000.
→ Diff = 2000×(10/100)² = ₹20

Q4. Find rate if ₹1000 becomes ₹1210 in 2 years (C.I.).
→ (A/P) = 1.21 = (1 + R/100)² → R = 10%

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21. Practice Formula Summary

Type	Formula
S.I.	(P×R×T)/100
C.I.	P[(1+R/100)^T − 1]
Difference (2 years)	P(R/100)²
Quarterly	P(1+R/400)^(4T) − P
Half-yearly	P(1+R/200)^(2T) − P
Depreciation	P(1−R/100)^T
Population Growth	P(1+R/100)^T

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22. Quick Revision Checklist

☑ Definition of P, R, T
☑ S.I. and C.I. formulas
☑ Half-yearly, quarterly compounding
☑ Difference between S.I. and C.I.
☑ Depreciation and population formulas
☑ Fractional year concept
☑ Successive growth rule

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✅ In One Line:

“Simple Interest grows linearly; Compound Interest grows exponentially — always remember (1 + R/100)^T formula.”