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Number - Exponential Growth and Decay

Grade 9IGCSE

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

🔑Concepts

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Exponential growth occurs when a quantity increases by a constant percentage r%r\% over equal time intervals nn. The growth is cumulative, meaning the increase is calculated on the updated value from the previous period.

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Exponential decay occurs when a quantity decreases by a constant percentage r%r\% over time. A common application is depreciation, where the value of an asset like a car or machine reduces over time.

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The multiplier kk represents the scale factor for each time period. For growth, k=1+r100k = 1 + \frac{r}{100}. For decay, k=1−r100k = 1 - \frac{r}{100}.

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Compound interest is a specific type of exponential growth where interest is earned on both the initial principal and the accumulated interest from previous periods.

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If the growth or decay happens over nn years, the initial amount PP is multiplied by the multiplier kk exactly nn times, represented as P×knP \times k^n.

📐Formulae

A=P(1+r100)nA = P \left(1 + \frac{r}{100}\right)^n

A=P(1−r100)nA = P \left(1 - \frac{r}{100}\right)^n

k=1±r100k = 1 \pm \frac{r}{100}

💡Examples

Problem 1:

A sum of 5000 is invested in a savings account that pays 3%3\% compound interest per annum. Calculate the total amount in the account after 55 years.

Solution:

Initial Principal P=5000P = 5000 Interest rate r=3r = 3 Time period n=5n = 5 Using the formula: A=5000(1+3100)5A = 5000 \left(1 + \frac{3}{100}\right)^5 A=5000×(1.03)5A = 5000 \times (1.03)^5 A=5000×1.159274...A = 5000 \times 1.159274... A=5796.37A = 5796.37 (to 2 decimal places)

Explanation:

To find the final amount, identify the initial value, the growth rate, and the number of periods. The multiplier for a 3%3\% increase is 1.031.03.

Problem 2:

A car is purchased for 18000. It depreciates at a rate of 12%12\% per year. Find the value of the car after 33 years.

Solution:

Initial value P=18000P = 18000 Depreciation rate r=12r = 12 Time period n=3n = 3 Using the decay formula: A=18000(1−12100)3A = 18000 \left(1 - \frac{12}{100}\right)^3 A=18000×(0.88)3A = 18000 \times (0.88)^3 A=18000×0.681472A = 18000 \times 0.681472 A=12266.50A = 12266.50 (to 2 decimal places)

Explanation:

Since the value is decreasing, we use the decay formula. A 12%12\% decrease means the car retains 88%88\% of its value each year, so the multiplier is 0.880.88.

Problem 3:

The population of a town is 2500025000. It grows by 2%2\% each year. Calculate the population after 1010 years, giving your answer to the nearest hundred.

Solution:

Initial population P=25000P = 25000 Growth rate r=2r = 2 Time period n=10n = 10 A=25000×(1.02)10A = 25000 \times (1.02)^{10} A=25000×1.21899...A = 25000 \times 1.21899... A=30474.8...A = 30474.8... Population ≈30500\approx 30500 (to the nearest hundred)

Explanation:

Use the exponential growth formula. After calculating the final value, round it as requested by the problem constraints.