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Compound Interest - Applications in Growth and Depreciation

Grade 8ICSE

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

🔑Concepts

Growth: This refers to the increase in a quantity over a period of time, such as the increase in the population of a city or the appreciation of land value. Visually, growth is represented by an upward-sloping curve where the slope becomes steeper over time because the increase is calculated on the already increased value of the previous year.

Depreciation: This is the reduction in the value of an asset, such as a car, computer, or machinery, due to wear and tear or age. On a graph, depreciation appears as a downward-sloping curve that starts high and gradually levels out, reflecting that the value decreases by a percentage of its current (reduced) value each year.

The Initial Value (PP): In these applications, the starting population or the original cost of an item is treated as the Principal (PP). This is the baseline from which growth or decay is measured.

Rate of Change (RR): This is the constant percentage at which the value increases (Growth Rate) or decreases (Depreciation Rate). For growth, the rate is positive, and for depreciation, it is negative.

Time Period (nn): This represents the number of years or intervals over which the growth or depreciation occurs. It acts as the exponent in our calculations, representing the number of times the rate is applied.

Variable Growth/Depreciation: Sometimes the rate of growth or depreciation is not constant. If a value grows at R1%R_1\% in the first year and R2%R_2\% in the second year, the final value is calculated by applying these rates successively like a chain reaction.

Scrap Value: In depreciation problems, the value of an item after its useful life is often called the scrap value. This is the final amount (VV) after nn years of depreciation.

📐Formulae

Value after nn years (Growth): V=P(1+R100)nV = P(1 + \frac{R}{100})^n

Value after nn years (Depreciation): V=P(1R100)nV = P(1 - \frac{R}{100})^n

Value after 22 years with variable rates (R1R_1 and R2R_2): V=P(1+R1100)(1+R2100)V = P(1 + \frac{R_1}{100})(1 + \frac{R_2}{100})

Population nn years ago: P=V(1+R100)nP = \frac{V}{(1 + \frac{R}{100})^n}

💡Examples

Problem 1:

The population of a town increases at the rate of 5%5\% per annum. If the present population is 1,60,0001,60,000, find the population after 22 years.

Solution:

Given: Initial population P=1,60,000P = 1,60,000 Growth rate R=5%R = 5\% Time n=2n = 2 years

Using the growth formula: V=P(1+R100)nV = P(1 + \frac{R}{100})^n V=1,60,000(1+5100)2V = 1,60,000(1 + \frac{5}{100})^2 V=1,60,000(1+120)2V = 1,60,000(1 + \frac{1}{20})^2 V=1,60,000(2120)2V = 1,60,000(\frac{21}{20})^2 V=1,60,000×441400V = 1,60,000 \times \frac{441}{400} V=400×441V = 400 \times 441 V=1,76,400V = 1,76,400

So, the population after 22 years will be 1,76,4001,76,400.

Explanation:

This is a growth problem because the population is increasing. We substitute the values into the compound interest formula for growth and solve for the final amount.

Problem 2:

A car was purchased for Rs. 6,00,000Rs.\ 6,00,000. Its value depreciates at the rate of 10%10\% per annum. What will be its value after 33 years?

Solution:

Given: Original price P=6,00,000P = 6,00,000 Depreciation rate R=10%R = 10\% Time n=3n = 3 years

Using the depreciation formula: V=P(1R100)nV = P(1 - \frac{R}{100})^n V=6,00,000(110100)3V = 6,00,000(1 - \frac{10}{100})^3 V=6,00,000(1110)3V = 6,00,000(1 - \frac{1}{10})^3 V=6,00,000(910)3V = 6,00,000(\frac{9}{10})^3 V=6,00,000×7291000V = 6,00,000 \times \frac{729}{1000} V=600×729V = 600 \times 729 V=4,37,400V = 4,37,400

So, the value of the car after 33 years will be Rs. 4,37,400Rs.\ 4,37,400.

Explanation:

This is a depreciation problem. We use the subtraction sign in the formula to account for the reduction in value over time.