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পশ্চিমবঙ্গ মধ্যশিক্ষা পর্ষদ (WBBSE) • শ্রেণি X • Mathematics • অধ্যায় 6
আনুমানিক সময়: 120 minutes
অগ্রগতি: অধ্যয়নে সক্রিয়

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.

কখনও কি ভেবে দেখেছেন?

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?

অধ্যায়টির গুরুত্ব

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.

পাঠের পূর্বে প্রয়োজনীয় ধারণা

  • 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.

শিখন লক্ষ্যমাত্রা ও ফলাফল

  • 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.

অধ্যায়ের বিষয়সূচি ও রূপরেখা

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...

সম্পূর্ণ তত্ত্ব ও ধারণাগত আলোচনা

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

গুরুত্বপূর্ণ গাণিতিক সূত্র, অভেদ ও উপপাদ্য

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
$$V_n = V_0(1 - r/100)^n$$

সমাধানকৃত উদাহরণ ও প্রয়োগ (Solved Examples)

উদাহরণ 1
Calculate the compound interest on ₹25,000 for 2 years at 8% per annum compounded annually.
ধাপে ধাপে সমাধান / উত্তর:
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.
উদাহরণ 2
Find the compound interest on ₹16,000 for 1.5 years at 10% per annum compounded half-yearly.
ধাপে ধাপে সমাধান / উত্তর:
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.
উদাহরণ 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.
ধাপে ধাপে সমাধান / উত্তর:
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.
উদাহরণ 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.
ধাপে ধাপে সমাধান / উত্তর:
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.
উদাহরণ 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.
ধাপে ধাপে সমাধান / উত্তর:
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.

সাধারণ ভুলত্রুটি ও সতর্কতা (Common Traps)

সাধারণ ভুল ধারণা

Using simple interest formula when compound interest is specified.

সঠিক পদ্ধতি ও সমাধান

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

সাধারণ ভুল ধারণা

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

সঠিক পদ্ধতি ও সমাধান

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

সাধারণ ভুল ধারণা

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

সঠিক পদ্ধতি ও সমাধান

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

সাধারণ ভুল ধারণা

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

সঠিক পদ্ধতি ও সমাধান

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

সাধারণ ভুল ধারণা

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

সঠিক পদ্ধতি ও সমাধান

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

অধ্যায় সারসংক্ষেপ ও গুরুত্বপূর্ণ বিষয়

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

স্ব-মূল্যায়ন অনুশীলন (Check Your Understanding)

মূল ধারণাগত স্পষ্টতা যাচাই করার জন্য অনুশীলন প্রশ্ন। উত্তর দেখার আগে নিজে সমাধান করার চেষ্টা করো।

1
What is the difference between CI and SI on ₹10,000 for 1 year at 10% per annum compounded annually?
উত্তর ও ব্যাখ্যা দেখুন
উত্তর: 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?
উত্তর ও ব্যাখ্যা দেখুন
উত্তর: 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?
উত্তর ও ব্যাখ্যা দেখুন
উত্তর: 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?
উত্তর ও ব্যাখ্যা দেখুন
উত্তর: 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.
উত্তর ও ব্যাখ্যা দেখুন
উত্তর: 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.
অধ্যায় পড়া শেষ হয়েছে?
অনুশীলন শুরু করো

অনলাইন মক টেস্ট দিয়ে প্রস্তুতি যাচাই করো

পশ্চিমবঙ্গ মধ্যশিক্ষা পর্ষদ (WBBSE) পাঠ্যক্রম অনুযায়ী বহু বিকল্পীয় প্রশ্ন (MCQ) সমাধান করো। তাৎক্ষণিক ফলাফল, সঠিক ব্যাখ্যা এবং নিজের স্কোর জেনে নাও।

শ্রেণি 10 Mathematics — সকল অধ্যায়

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

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