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OCR GCSE Mathematics revision notes

Ratio, proportion and rates of change Revision Notes

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Ratio, proportion and rates of change

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  • Understanding Growth and Decay Algebra Revision Notes

    The Growth and Decay Formula

    Growth and decay problems can be solved using the general formula: Final Value = Initial Value × (Multiplier)^Number of Periods. Here, 'Initial Value' is the starting quantity, 'Final Value' is the quantity after a certain time, 'Multiplier' is the factor by which the quantity changes each period, and 'Number of Periods' is how many times the change occurs. It's important to identify each of these components from the problem description before attempting to solve.

    Calculating the Multiplier for Growth

    When a quantity increases by a percentage 'p%', the multiplier is found by adding the percentage increase to 100% and then dividing by 100. This results in the multiplier (1 + p/100). For instance, if an item increases by 8%, the multiplier is 1 + (8/100) = 1 + 0.08 = 1.08. You would then multiply the initial value by 1.08 for each period of growth.

    Calculating the Multiplier for Decay

    When a quantity decreases by a percentage 'p%', the multiplier is found by subtracting the percentage decrease from 100% and then dividing by 100. This results in the multiplier (1 - p/100). For example, if a car depreciates by 15% each year, the multiplier is 1 - (15/100) = 1 - 0.15 = 0.85. You would then multiply the initial value by 0.85 for each period of decay.

    Applying the Formula Iteratively

    The 'Number of Periods' in the formula is crucial. If a quantity grows or decays for multiple periods (e.g., several years), you raise the multiplier to the power of the number of periods. For example, if something grows at 5% per year for 3 years, the calculation would involve (1.05)^3. This correctly accounts for the compound effect of the growth or decay over time.

  • Higher Methods for Growth and Decay Algebra Revision Notes

    General Formulae for Growth and Decay

    Exponential growth and decay are modelled by the formulae:

    • Growth: $A = P(1 + r)^n$
    • Decay: $A = P(1 - r)^n$

    Where:

    • $A$ = Final amount
    • $P$ = Initial amount
    • $r$ = Rate of growth/decay per period (as a decimal)
    • $n$ = Number of periods

    It is crucial to convert percentage rates to decimals (e.g., 5% becomes 0.05). Ensure the units for 'rate' and 'number of periods' are consistent (e.g., if the rate is annual, $n$ must be in years).

    Solving for Unknowns using Logarithms

    When the number of periods, $n$, is unknown, logarithms are often required to solve the equation. For $A = P(1 + r)^n$:

    1. Divide by $P$: $A/P = (1 + r)^n$
    2. Take the logarithm of both sides: $\log(A/P) = \log((1 + r)^n)$
    3. Use the logarithm power rule: $\log(A/P) = n \log(1 + r)$
    4. Solve for $n$: $n = \frac{\log(A/P)}{\log(1 + r)}$

    Similarly for decay. Remember that logarithms can be base 10 or natural logarithms (base e); the choice does not affect the final value of $n$ as long as the same base is used consistently.

    Finding Unknown Rates or Initial Amounts

    If the rate ($r$) is unknown, you might need to take the $n$-th root: For $A = P(1 + r)^n$:

    1. Divide by $P$: $A/P = (1 + r)^n$
    2. Take the $n$-th root of both sides: $\sqrt[n]{A/P} = 1 + r$
    3. Solve for $r$: $r = \sqrt[n]{A/P} - 1$

    If the initial amount ($P$) is unknown, it's a straightforward division: $P = \frac{A}{(1 + r)^n}$ or $P = A \times (1 + r)^{-n}$. These algebraic manipulations allow full flexibility in solving growth and decay problems.

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Ratio, proportion and rates of change Revision Notes | OCR GCSE Mathematics | ExamCompanion