logo

OCR GCSE Mathematics revision notes

Ratio, proportion and rates of change Revision Notes

Study Ratio, proportion and rates of change with curriculum-aligned Revision Notes resources, practice links, and exam-focused support.

At a glance

revision notes

Resource type

Topic

Ratio, proportion and rates of change

OcrGcseMathematicsRatio, proportion and rates of change

Revision notes

  • Understanding Inverse Proportion Revision Notes

    What is Inverse Proportion?

    Inverse proportion means that as one quantity increases, the other quantity decreases. Crucially, their product remains constant. For example, if you have a fixed amount of work to do, the more workers you have (increasing quantity), the less time it will take to complete the work (decreasing quantity). The total 'work' done remains constant, which is the product of workers and time. This is a fundamental concept for solving problems involving rates and efficiencies.

    The Inverse Proportion Formula

    If 'y' is inversely proportional to 'x', we write y ∝ 1/x. To form an equation, we introduce a constant of proportionality, 'k'. This gives us y = k/x. Alternatively, this can be written as xy = k. The value of 'k' is unique to each problem and represents the constant product of the two inversely proportional quantities. Identifying the correct relationship and setting up the equation is the first key step in solving problems.

    Finding the Constant of Proportionality (k)

    To find 'k', you need a pair of given values for 'x' and 'y' that are related by the inverse proportion. Substitute these values into the equation y = k/x (or xy = k) and solve for 'k'. For instance, if y = 10 when x = 2, then 10 = k/2, so k = 20. Once 'k' is determined, the full relationship (e.g., y = 20/x) is established, allowing you to find other unknown values. This step is critical for making predictions or calculations.

    Using the Formula to Solve Problems

    Once you have found the constant 'k' and established the full inverse proportion equation (e.g., y = 20/x), you can use it to find an unknown value. If you are given a new value for 'x', substitute it into the equation to find 'y'. If you are given a new value for 'y', substitute it into the equation and solve for 'x'. Always ensure your units are consistent throughout the problem and re-read the question to make sure you are calculating what is asked.

  • Understanding Inverse Proportion Revision Notes

    What is Inverse Proportion?

    Inverse proportion means that as one quantity increases, the other quantity decreases, and their product remains constant. If 'y' is inversely proportional to 'x', we write this as y ∝ 1/x. This proportionality can be turned into an equation: y = k/x, where 'k' is the constant of proportionality. Alternatively, this can be written as xy = k. The key characteristic is that the product of the two quantities is always the same fixed value.

    Finding the Constant of Proportionality (k)

    To work with inverse proportion, the first step is usually to find the constant 'k'. You will typically be given a pair of values for 'x' and 'y' that satisfy the relationship. Substitute these values into the equation y = k/x or xy = k. For example, if y = 5 when x = 4, then k = xy = 4 × 5 = 20. Once 'k' is found, the specific relationship between y and x becomes y = 20/x. This constant allows you to predict other values.

    Using the Constant to Solve Problems

    Once you have found the constant of proportionality 'k', you can use it to find any unknown value. If you know 'k' and a new value for 'x', you can find 'y' using y = k/x. If you know 'k' and a new value for 'y', you can find 'x' using x = k/y. Make sure to clearly state your inverse proportion relationship with 'k' before performing calculations. For example, if y = 20/x, and you need to find y when x = 10, then y = 20/10 = 2.

  • Understanding and Applying Inverse Proportion Revision Notes

    Basic Definition of Inverse Proportion

    When two quantities, say x and y, are inversely proportional, it means that as one quantity increases, the other decreases in a way that their product remains constant. This relationship is written as x $\propto$ 1/y. To form an equation, we introduce a constant of proportionality, k, resulting in x = k/y. Rearranging this gives xy = k, which highlights that the product of the two quantities is constant. This constant, k, is crucial for defining the specific relationship between the variables.

    Inverse Proportion to Powers and Roots

    The concept of inverse proportion extends to powers and roots of quantities. For example, if y is inversely proportional to the square of x, we write y $\propto$ 1/x$^2$, which leads to the equation y = k/x$^2$. Similarly, if y is inversely proportional to the square root of x, the relationship is y $\propto$ 1/$\sqrt{x}$, and the equation becomes y = k/$\sqrt{x}$. It's essential to correctly identify the power or root involved from the problem description, as this dictates the form of the proportionality and the resulting equation.

    Steps to Solve Inverse Proportion Problems

    Solving inverse proportion problems generally involves a structured approach. First, establish the proportionality, e.g., y $\propto$ 1/x$^n$. Second, convert this into an equation using a constant of proportionality, k: y = k/x$^n$. Third, use the given initial pair of values for x and y to substitute into the equation and solve for the value of k. Fourth, write out the complete specific formula relating y and x using the determined k. Finally, use this formula to find any other unknown value of x or y. This systematic method ensures accuracy in solving these types of problems.

    Example: Inverse Proportion to a Square

    Consider a case where the time, T, taken to complete a task is inversely proportional to the square of the number of workers, W. If it takes 2 hours for 4 workers, we can find out how long it takes for 8 workers.

    1. T $\propto$ 1/W$^2$
    2. T = k/W$^2$
    3. Use T=2, W=4: 2 = k/4$^2$ $\implies$ 2 = k/16 $\implies$ k = 32.
    4. The full relationship is T = 32/W$^2$.
    5. To find T when W=8: T = 32/8$^2$ = 32/64 = 0.5 hours. This systematic application is key.
  • Understanding Growth and Decay Algebra Revision Notes

    The Core Formula for Growth and Decay

    The fundamental formula for growth and decay is \( A = P(1+r)^n \) which represents the final amount \( A \) after \( n \) time periods. \( P \) is the initial (original) amount, and \( r \) is the rate of growth or decay per period expressed as a decimal. If it's growth, \( r \) is added to 1; if it's decay, \( r \) is subtracted from 1. For example, a 5% increase means \( r = 0.05 \), so \( (1+r) = 1.05 \). A 5% decrease means \( r = 0.05 \), so \( (1-r) = 0.95 \).

    Calculating the Multiplying Factor

    The 'multiplying factor' or 'multiplier' is the crucial part of the formula. For growth, if a quantity increases by X% per period, the multiplying factor is \( (1 + \frac{X}{100}) \). For example, a 7% increase gives a multiplier of \( (1 + \frac{7}{100}) = 1.07 \). For decay, if a quantity decreases by X% per period, the multiplying factor is \( (1 - \frac{X}{100}) \). For example, a 12% decrease gives a multiplier of \( (1 - \frac{12}{100}) = 0.88 \). Always convert percentages to decimals before using them in the formula.

    Applying the Formula to Different Scenarios

    Once you have the initial amount \( (P) \), the correct multiplying factor \( (1+r) \) or \( (1-r) \), and the number of time periods \( (n) \), you can calculate the final amount. Be careful to correctly identify what each part of the problem represents. Time periods must be consistent with the rate; for example, if the rate is annual, then \( n \) must be in years. Always consider whether the problem describes growth (increasing value) or decay (decreasing value) to select the correct form of the multiplier.

Related topics

Study nearby topics next

Revision hubs

Keep studying from the right hub

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