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WBB • कक्षा X • Mathematics • अध्याय 6
अनुमानित समय: 120 minutes
प्रगति: अध्ययनरत

Compound Interest and Uniform Rate of Increase or Decrease

Compound Interest and Uniform Rate of Increase or Decrease forms Chapter 6 of the WBBSE Class 10 Mathematics curriculum Ganit Prakash, providing an in-depth treatment of non-linear financial arithmetic and exponential growth and decay models. In simple interest, the principal remains unaltered across the entire loan tenure. In contrast, under compound interest, the interest accrued at the end of each specified conversion period is added to the principal to form the new principal for the subsequent period, earning interest upon interest. The total compound amount A after n years compounded annually at r percent per annum is given by A = P(1 + r/100)^n, and the compound interest is CI = A - P. When interest is compounded semi-annually (twice a year), the rate is halved and the number of compounding periods doubles, yielding A = P(1 + r/200)^(2n). When compounded quarterly, the rate is divided by four and the period quadruples, A = P(1 + r/400)^(4n). If the interest rate varies each year, the amount equals the product of successive growth factors. The difference between compound interest and simple interest for two years provides the classic relationship CI - SI = P(r/100)^2. The second half of the chapter extends this geometric progression model to uniform growth and decay. Uniform rate of increase governs human population expansions and microbial cultures using P_n = P_0(1 + r/100)^n, while uniform rate of decrease governs equipment depreciation and asset devaluation using V_n = V_0(1 - r/100)^n, allowing students to project future values and determine historical base values.

क्या आपने कभी सोचा है?

Albert Einstein famously referred to compound interest as the eighth wonder of the world: 'He who understands it, earns it; he who doesn't, pays it.' How does money multiply exponentially when interest earns interest, and how does this exact same mathematical exponential law govern bacterial colony growth, human population censuses, and the inevitable depreciation of factory machinery?

यह अध्याय क्यों महत्वपूर्ण है

Compound interest and uniform growth problems are of paramount importance in the WBBSE Madhyamik examination, consistently accounting for a compulsory 5-mark long question in the commercial arithmetic group. In the real economy, compound interest governs fixed deposits, mutual funds, personal loans, home mortgages, national debt, and pension accumulation. Similarly, uniform rates of growth and depreciation are the mathematical backbone of actuarial science, demography, industrial engineering asset management, and environmental science. Mastering this chapter equips students with computational fluency and essential real-world financial literacy.

अध्ययन से पूर्व (आवश्यक ज्ञान)

  • Foundational mastery of simple interest, principal, rate, and tenure.
  • Binomial expansions and manipulating exponential powers (1 + r/100)^n.
  • Handling fractional and decimal percentages (e.g. 5%, 7.5%, 10%).
  • Linear and quadratic algebraic equations for determining unknown principal or rates.

इस अध्याय के लक्ष्य

  • Articulate the conceptual difference between Simple Interest (constant principal) and Compound Interest (reinvested interest generating 'interest on interest').
  • Derive and apply the compound amount formula A = P(1 + r/100)^n for annual compounding, and compute CI = A - P.
  • Apply conversion period adjustments for half-yearly compounding A = P(1 + r/200)^(2n) and quarterly compounding A = P(1 + r/400)^(4n).
  • Calculate maturity amounts when interest rates vary across successive years: A = P(1 + r_1/100)(1 + r_2/100)...
  • Derive and utilize the landmark difference formulas between CI and SI for 2 years (CI - SI = P(r/100)^2) and 3 years.
  • Model uniform demographic and biological growth: P_n = P_0(1 + r/100)^n, and determine populations n years in the past.
  • Model industrial machine depreciation and vehicle valuation loss: V_n = V_0(1 - r/100)^n, and calculate original purchase costs.

अध्याय रूपरेखा एवं प्रगति

1 Module 1: Concept of Compound Inter...
2 Module 2: Periodic Compounding Inte...
3 Module 3: Difference Between Compou...
4 Module 4: Uniform Rate of Increase...

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

Module 1: Concept of Compound Interest vs. Simple Interest

1.1 Fundamental Mechanism of Compounding

In Simple Interest, the principal remains strictly fixed at $P$ across all periods. In Compound Interest (চক্রবৃদ্ধি সুদ), at the end of each compounding interval (conversion period), the interest accrued is added to the principal. This accumulated sum becomes the new principal for the subsequent period:

Comparison Parameter Simple Interest (SI) Compound Interest (CI)
Principal Constant throughout the loan term. Increases at the end of each conversion period ($P_{k+1} = A_k$).
Annual Interest Remains identical in every single year. Increases progressively year after year due to interest on interest.
First Year Comparison $ ext{SI}_1 = rac{Pr}{100}$ $ ext{CI}_1 = rac{Pr}{100}$ (Exactly equal to SI for 1st year).
Subsequent Years Growth is linear: $A = P + rac{Prt}{100}$. Growth is exponential: $A = P(1 + rac{r}{100})^n$. $ ext{CI} > ext{SI}$.
1.2 Step-by-Step Derivation of Annual Compound Amount

Let Principal $= P$, annual rate of interest $= r\%$, and time $= n$ years:
• End of 1st year: Interest $I_1 = rac{P \cdot r \cdot 1}{100}$. Amount $A_1 = P + rac{Pr}{100} = P\left(1 + rac{r}{100} ight)$.
• For 2nd year: Principal $P_2 = A_1 = P\left(1 + rac{r}{100} ight)$.
Interest $I_2 = rac{P_2 \cdot r \cdot 1}{100}$. Amount $A_2 = P_2\left(1 + rac{r}{100} ight) = P\left(1 + rac{r}{100} ight)^2$.
• Continuing for $n$ years by mathematical induction:

Compound Amount (সবৃদ্ধিমূল): $A = P\left(1 + rac{r}{100} ight)^n$

Compound Interest (চক্রবৃদ্ধি সুদ): $ ext{CI} = A - P = P\left[\left(1 + rac{r}{100} ight)^n - 1 ight]$

Module 2: Periodic Compounding Intervals & Varying Rates

2.1 Compounding Semi-Annually (Half-Yearly) & Quarterly

When interest is compounded more frequently than once a year, the annual rate $r$ is divided by the number of intervals per year, and the number of years $n$ is multiplied by that number:

Compounding Cycle Effective Rate per Period Total Periods Maturity Amount Formula
Annually (বার্ষিক পর্ব) $r\%$ $n$ $A = P\left(1 + rac{r}{100} ight)^n$
Half-Yearly (৬ মাস অন্তর / অর্ধবার্ষিক) $ rac{r}{2}\%$ $2n$ $A = P\left(1 + rac{r/2}{100} ight)^{2n} = P\left(1 + rac{r}{200} ight)^{2n}$
Quarterly (৩ মাস অন্তর / ত্রৈমাসিক) $ rac{r}{4}\%$ $4n$ $A = P\left(1 + rac{r/4}{100} ight)^{4n} = P\left(1 + rac{r}{400} ight)^{4n}$
2.2 Successive Varying Interest Rates

If the interest rate is $r_1\%$ in the 1st year, $r_2\%$ in the 2nd year, and $r_3\%$ in the 3rd year, the amount becomes:

$$A = P\left(1 + rac{r_1}{100} ight)\left(1 + rac{r_2}{100} ight)\left(1 + rac{r_3}{100} ight)$$

Module 3: Difference Between Compound Interest and Simple Interest

3.1 Derivation of the 2-Year Difference Formula

For $n = 2$ years at rate $r\%$ per annum:
Simple Interest: $ ext{SI} = rac{P \cdot r \cdot 2}{100} = rac{2Pr}{100}$.
Compound Interest: $ ext{CI} = P\left(1 + rac{r}{100} ight)^2 - P = P\left(1 + rac{2r}{100} + rac{r^2}{100^2} ight) - P = rac{2Pr}{100} + P\left( rac{r}{100} ight)^2$.
Subtracting the two:

2-Year Difference: $ ext{CI} - ext{SI} = P\left( rac{r}{100} ight)^2$
3.2 The 3-Year Difference Formula

By expanding $P(1 + r/100)^3 - P$ and subtracting $ rac{3Pr}{100}$:

3-Year Difference: $ ext{CI} - ext{SI} = P\left( rac{r}{100} ight)^2 \left(3 + rac{r}{100} ight)$

Module 4: Uniform Rate of Increase or Decrease (সমহার বৃদ্ধি বা হ্রাস)

4.1 Uniform Rate of Increase (সমহার বৃদ্ধি)

When physical quantities such as population, vehicle counts, student enrollment, or bacteria proliferate at a uniform percentage rate $r\%$ per period:

Value after $n$ years: $P_n = P_0\left(1 + rac{r}{100} ight)^n$

Value $n$ years ago: $P_0 = rac{P_n}{\left(1 + rac{r}{100} ight)^n}$
4.2 Uniform Rate of Decrease / Depreciation (সমহার হ্রাস বা অবচয়)

Industrial machinery, automobiles, buildings, and electronic devices lose value over time due to wear, tear, and technological obsolescence at an annual depreciation rate $r\%$:

Depreciated value after $n$ years: $V_n = V_0\left(1 - rac{r}{100} ight)^n$

Original purchase cost $n$ years ago: $V_0 = rac{V_n}{\left(1 - rac{r}{100} ight)^n}$

महत्वपूर्ण सूत्र, सर्वसमिकाएँ एवं प्रमेय

Annual Compound Amount
$$A = P(1 + r/100)^n$$
Compound Interest
$$CI = A - P = P[(1 + r/100)^n - 1]$$
Half-Yearly Compound Amount
A = P(1 + r/200)^(2n)
Quarterly Compound Amount
A = P(1 + r/400)^(4n)
Successive Varying Rates
$$A = P(1 + r_1/100)(1 + r_2/100)(1 + r_3/100)$$
2-Year Difference (CI - SI)
$$CI - SI = P * (r / 100)^2$$
3-Year Difference (CI - SI)
$$CI - SI = P(r/100)^2 * (3 + r/100)$$
Uniform Rate of Increase (Future)
$$P_n = P_0(1 + r/100)^n$$
Uniform Rate of Increase (Past)
$$P_0 = P_n / (1 + r/100)^n$$
Uniform Rate of Depreciation
$$V_n = V_0(1 - r/100)^n$$

अवधारणात्मक हल उदाहरण एवं अनुप्रयोग (Solved Examples)

उदाहरण 1
Calculate the compound interest on ₹25,000 for 2 years at 8% per annum compounded annually.
विस्तृत समाधान / उत्तर:
Step 1: Identify given values: Principal P = ₹25,000, Rate r = 8% p.a., Time n = 2 years. Step 2: Apply the compound amount formula: A = P(1 + r/100)^n A = 25000 * (1 + 8/100)^2 A = 25000 * (108/100)^2 = 25000 * (27/25)^2 Step 3: Evaluate arithmetic: A = 25000 * (729 / 625) 25000 / 625 = 40 A = 40 * 729 = ₹29,160. Step 4: Compute Compound Interest: CI = A - P = ₹29,160 - ₹25,000 = ₹4,160. Hence, Compound Amount = ₹29,160 and Compound Interest = ₹4,160.
उदाहरण 2
Find the compound interest on ₹16,000 for 1.5 years at 10% per annum compounded half-yearly.
विस्तृत समाधान / उत्तर:
Step 1: Identify given values and adjust for half-yearly compounding: Principal P = ₹16,000. Annual rate r = 10% p.a. => Rate per half-year = 10/2 = 5%. Time n = 1.5 years = 3/2 years => Number of compounding periods = 2 * 1.5 = 3 periods. Step 2: Apply the half-yearly compound amount formula: A = P(1 + r/200)^(2n) A = 16000 * (1 + 5/100)^3 = 16000 * (21/20)^3 Step 3: Evaluate arithmetic: (21/20)^3 = 9261 / 8000 A = 16000 * (9261 / 8000) = 2 * 9261 = ₹18,522. Step 4: Compute Compound Interest: CI = A - P = ₹18,522 - ₹16,000 = ₹2,522. Hence, Compound Interest = ₹2,522.
उदाहरण 3
If the difference between the compound interest and simple interest on a certain sum of money for 2 years at 5% per annum is ₹150, find the sum.
विस्तृत समाधान / उत्तर:
Step 1: Identify given values: Time n = 2 years, Rate r = 5% p.a., Difference (CI - SI) = ₹150. Step 2: Use the standard 2-year difference formula: CI - SI = P * (r / 100)^2 Step 3: Substitute the known values: 150 = P * (5 / 100)^2 150 = P * (1 / 20)^2 150 = P * (1 / 400) Step 4: Solve for Principal P: P = 150 * 400 = ₹60,000. Hence, the principal sum is ₹60,000.
उदाहरण 4
The present population of a town is 92,610. If the population has been increasing at the rate of 5% per annum, find: (i) the population 3 years ago; (ii) the population after 2 years.
विस्तृत समाधान / उत्तर:
Step 1: Identify parameters: Current population P = 92,610, growth rate r = 5% p.a. Part (i): Let the population 3 years ago be P_0. P = P_0 * (1 + 5/100)^3 92610 = P_0 * (21/20)^3 92610 = P_0 * (9261 / 8000) P_0 = 92610 * (8000 / 9261) = 10 * 8000 = 80,000. Hence, the population 3 years ago was 80,000. Part (ii): Population after 2 years (P_2): P_2 = P * (1 + 5/100)^2 P_2 = 92610 * (21/20)^2 = 92610 * (441 / 400) P_2 = (92610 * 441) / 400 = 40841010 / 400 = 102,102.525 approx 102,103. Hence, population 3 years ago was 80,000 and after 2 years will be approximately 102,103.
उदाहरण 5
A machine was purchased for ₹1,00,000. Its value depreciates at the rate of 10% per annum. Find its value after 3 years and the total depreciation.
विस्तृत समाधान / उत्तर:
Step 1: Identify initial value and depreciation rate: Initial value V_0 = ₹1,00,000, Depreciation rate r = 10% p.a., Time n = 3 years. Step 2: Apply uniform rate of decrease formula: V_3 = V_0 * (1 - r/100)^n V_3 = 100000 * (1 - 10/100)^3 V_3 = 100000 * (9/10)^3 Step 3: Evaluate arithmetic: (9/10)^3 = 729 / 1000 V_3 = 100000 * (729 / 1000) = 100 * 729 = ₹72,900. Step 4: Calculate total depreciation: Total Depreciation = V_0 - V_3 = ₹1,00,000 - ₹72,900 = ₹27,100. Hence, the value after 3 years is ₹72,900 and total depreciation is ₹27,100.

सामान्य गलतियाँ एवं परीक्षक के जाल (Examiner Traps)

सामान्य भ्रम / गलत उत्तर

Using simple interest formula when compound interest is specified.

सही वैज्ञानिक तथ्य

Always use A = P(1 + r/100)^n for compound interest; never use I = PRT/100.

सामान्य भ्रम / गलत उत्तर

Forgetting to halve the rate and double the periods for half-yearly compounding.

सही वैज्ञानिक तथ्य

When compounded half-yearly, replace r with r/2 and n with 2n: (1 + r/200)^(2n).

सामान्य भ्रम / गलत उत्तर

Confusing Amount (A) with Compound Interest (CI).

सही वैज्ञानिक तथ्य

A is the total maturity value. To find interest, always subtract principal: CI = A - P.

सामान्य भ्रम / गलत उत्तर

Using (1 + r/100) instead of (1 - r/100) for machine depreciation.

सही वैज्ञानिक तथ्य

Depreciation means loss of value: use V_n = V_0(1 - r/100)^n.

सामान्य भ्रम / गलत उत्तर

Multiplying by (1 + r/100)^n to find past population.

सही वैज्ञानिक तथ्य

To find population n years AGO, DIVIDE by (1 + r/100)^n: P_0 = P_n / (1 + r/100)^n.

Concept Map: Compound Interest & Uniform Growth (WBBSE Class 10 Ganit Prakash)

Compound Interest & Uniform Growth (চক্রবৃদ্ধি সুদ ও সমহার) WBBSE Class 10 Mathematics • Chapter 6 • Compounding Cycles & Exponential Rates 1. Annual & Periodic Compounding • Annual: A = P(1 + r/100)^n, CI = A - P• Half-yearly: A = P(1 + r/200)^(2n)• Quarterly: A = P(1 + r/400)^(4n)• Successive rates: A = P(1 + r1/100)(1 + r2/100)... 2. Difference Between CI & SI • 1 Year difference (annual compounding) = 0• 2 Years difference: CI - SI = P * (r / 100)^2• 3 Years difference: P*(r/100)^2 * (3 + r/100)• Rapid principal determination from (CI - SI) 3. Uniform Rate of Increase • Population growth: P_n = P_0 * (1 + r/100)^n• Past population (n yrs ago): P_0 = P_n / (1 + r/100)^n• Bacteria count / school enrollment growth• Positive exponential compounding model 4. Uniform Rate of Decrease • Machine depreciation: V_n = V_0 * (1 - r/100)^n• Original cost (n yrs ago): V_0 = V_n / (1 - r/100)^n• Vehicle resale & scrap metal devaluation• Negative compounding depreciation model

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1
In simple interest, principal is constant; in compound interest, interest earned is reinvested into the principal.
मुख्य बिंदु 2
Annual Compound Amount: A = P(1 + r/100)^n; Compound Interest: CI = A - P.
मुख्य बिंदु 3
Half-Yearly Compounding: A = P(1 + r/200)^(2n); Quarterly Compounding: A = P(1 + r/400)^(4n).
मुख्य बिंदु 4
Successive varying interest rates: A = P(1 + r_1/100)(1 + r_2/100)(1 + r_3/100).
मुख्य बिंदु 5
Difference between CI and SI for 2 years: CI - SI = P(r/100)^2.
मुख्य बिंदु 6
Difference between CI and SI for 3 years: CI - SI = P(r/100)^2 * (3 + r/100).
मुख्य बिंदु 7
Uniform rate of increase: P_n = P_0(1 + r/100)^n; past population P_0 = P_n / (1 + r/100)^n.
मुख्य बिंदु 8
Uniform rate of decrease (depreciation): V_n = V_0(1 - r/100)^n; original cost V_0 = V_n / (1 - r/100)^n.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

1
What is the difference between CI and SI on ₹10,000 for 1 year at 10% per annum compounded annually?
उत्तर एवं व्याख्या देखें
उत्तर: For 1 year with annual compounding, CI equals SI. Hence, the difference is ₹0.
2
If the rate of interest is 8% p.a. compounded half-yearly, what is the rate per conversion period?
उत्तर एवं व्याख्या देखें
उत्तर: Since there are 2 half-years in a year, the rate per conversion period is 8/2 = 4%.
3
What is the formula for the difference between CI and SI for 2 years?
उत्तर एवं व्याख्या देखें
उत्तर: CI - SI = P * (r / 100)^2.
4
A motorcycle bought for ₹60,000 depreciates at 10% per annum. What is its value after 1 year?
उत्तर एवं व्याख्या देखें
उत्तर: V_1 = 60000 * (1 - 10/100) = 60000 * 0.9 = ₹54,000.
5
If the population of a village increases from 1,000 to 1,210 in 2 years, find the annual rate of growth.
उत्तर एवं व्याख्या देखें
उत्तर: 1210 = 1000(1 + r/100)^2 => (1 + r/100)^2 = 121/100 => 1 + r/100 = 11/10 => r/100 = 1/10 => r = 10% p.a.
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कक्षा 10 Mathematics के सभी अध्याय

अध्याय 1: Quadratic Equations in One Variable अध्याय 2: Simple Interest अध्याय 3: Theorems related to Circle अध्याय 4: Rectangular Parallelopiped or Cuboid अध्याय 5: Ratio and Proportion अध्याय 6: Compound Interest and Uniform Rate of Increase or Decrease अध्याय 7: Theorems Related to Angles in a Circle अध्याय 8: Right Circular Cylinder अध्याय 9: Quadratic Surd अध्याय 10: Theorems Related to Cyclic Quadrilateral अध्याय 11: Construction of Circumcircle and Incircle of a Triangle अध्याय 12: Sphere अध्याय 13: Variation अध्याय 14: Partnership Business अध्याय 15: Theorems Related to Tangent to a Circle अध्याय 16: Right Circular Cone अध्याय 17: Construction of Tangent to a Circle अध्याय 18: Similarity अध्याय 19: Problems Related to Different Solid Objects अध्याय 20: Trigonometry: Concept of Measurement of Angle अध्याय 21: Construction: Determination of Mean Proportional अध्याय 22: Pythagoras Theorem अध्याय 23: Trigonometric Ratios and Trigonometric Identities अध्याय 24: Trigonometric Ratios of Complementary Angle अध्याय 25: Application of Trigonometric Ratios: Heights and Distances अध्याय 26: Statistics: Mean, Median, Ogive, Mode

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