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Comparing Quantities - Applications of Compound Interest Formula

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Understanding the Compound Interest Model: The compound interest formula is a mathematical tool used to calculate growth or decay over time. Unlike simple interest where the change is constant, compound interest applies the rate to the new total from the previous period. Visualise this as a staircase where each step is slightly taller than the last, representing the 'interest on interest' effect.

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Population Growth Application: When a population increases at a constant percentage rate, it follows the compound interest formula. If the rate is positive, the population grows exponentially. On a line graph, this appears as a curve that starts slowly and bends upwards more steeply over time.

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Depreciation of Assets: Depreciation is the reduction in the value of an item (like machinery, vehicles, or gadgets) due to wear and tear or age. In this case, the rate RR is negative, so we subtract it in the formula. Visualise this as a downward-sloping curve on a graph that approaches the horizontal axis but never quite reaches zero.

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Bacterial and Microbial Growth: In biological contexts, the count of bacteria often increases at a fixed rate per hour. The formula A=P(1+fracR100)nA = P(1 + \\frac{R}{100})^n is used where nn represents the number of hours. This is visualised as a rapid 'explosion' of data points on a scatter plot, showing how populations can double or triple quickly.

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Appreciation of Property: Appreciation refers to the increase in the value of an asset, such as land or gold. Similar to population growth, the value is calculated using the standard addition formula. A visual representation would be a bar chart where each consecutive year's bar is higher than the previous one.

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Varying Rates of Growth: Sometimes the growth rate changes every year (e.g., 5%5\% in the first year and 8%8\% in the second). Instead of using a power nn, we multiply the principal by different growth factors for each period. Visualise this as a path with segments of different steepness connected together.

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Compounding Periods: While most applications use annual rates, some processes might compound half-yearly. In such cases, the rate is halved (R/2R/2) and the time periods are doubled (2n2n). Visualise a one-year timeline divided into two equal segments, with growth calculated at each midpoint.

📐Formulae

A=P(1+fracR100)nA = P(1 + \\frac{R}{100})^n

A=P(1−fracR100)nA = P(1 - \\frac{R}{100})^n

CI=A−PCI = A - P

A=P(1+fracR1100)(1+fracR2100)A = P(1 + \\frac{R_1}{100})(1 + \\frac{R_2}{100})

A=P(1+fracR2times100)2nA = P(1 + \\frac{R}{2 \\times 100})^{2n}

💡Examples

Problem 1:

The population of a town was 20,00020,000 in the year 20182018. It increased at the rate of 5%5\% per annum. What would be the population in the year 20202020?

Solution:

  1. Identify the given values: Initial population (PP) = 20,00020,000, Rate of growth (RR) = 5%5\%, Time (nn) = 2020−2018=22020 - 2018 = 2 years.\n2. Apply the growth formula: A=P(1+fracR100)nA = P(1 + \\frac{R}{100})^n\n3. Substitute values: A=20,000(1+frac5100)2A = 20,000(1 + \\frac{5}{100})^2\n4. Simplify: A=20,000(1+frac120)2=20,000(frac2120)2A = 20,000(1 + \\frac{1}{20})^2 = 20,000(\\frac{21}{20})^2\n5. Calculate the square: A=20,000timesfrac441400A = 20,000 \\times \\frac{441}{400}\n6. Final calculation: A=50times441=22,050A = 50 \\times 441 = 22,050.

Explanation:

Since the population is increasing, we use the compound interest addition formula. We treat the initial population as the Principal (PP) and the final population as the Amount (AA).

Problem 2:

A refrigerator was bought for Rs 12,00012,000. Its value depreciated at the rate of 10%10\% per annum. Find its value after 11 year.

Solution:

  1. Identify the given values: Price (PP) = Rs 12,00012,000, Rate of depreciation (RR) = 10%10\%, Time (nn) = 11 year.\n2. Apply the depreciation formula: A=P(1−fracR100)nA = P(1 - \\frac{R}{100})^n\n3. Substitute values: A=12,000(1−frac10100)1A = 12,000(1 - \\frac{10}{100})^1\n4. Simplify: A=12,000(frac90100)A = 12,000(\\frac{90}{100})\n5. Final calculation: A=12,000times0.9=10,800A = 12,000 \\times 0.9 = 10,800.

Explanation:

Depreciation means the value is decreasing over time, so we use the subtraction sign in the formula. After one year, the value is 90%90\% of its original price.