Inverse Proportion Study Notes

Graphing Inverse Proportion

Inverse proportion describes a mathematical relationship between two variables where an increase in one variable results in a proportional decrease in the other variable.

The general algebraic equation representing inverse proportion is given by:

$y = \frac{k}{x}$

In this equation, $x$ and $y$ are the two inversely proportional variables, and $k$ represents the constant of proportionality.

When plotted on a Cartesian coordinate plane, an inverse proportion produces a characteristic hyperbola curve located in the first quadrant. As the value on the horizontal axis ($x$) increases, the corresponding value on the vertical axis ($y$) decreases, with the curve approaching the axes asymptotically without ever touching them. For example, as the number of workers installing gas meters increases, the total time required to complete the installation decreases.

Solving Inverse Proportion Questions

Direct proportion problems are typically solved using a divide-then-multiply method (often referred to as the unitary method). Inverse proportion requires the exact opposite sequence of arithmetic operations: TIMES for ONE, then DIVIDE for ALL.

To solve an inverse proportion problem:

1. Multiply the given initial quantity of units by the duration or rate to determine the total time or output required for a single unit ($1$ worker, $1$ teacher, $1$ baker, etc.).
2. Divide this single-unit baseline quantity by the target number of units to find the final time or output.

Example 1: Ploughing a Field

Suppose it takes $2$ farmers $10\,\text{hours}$ to plough a field. To calculate how long it would take $4$ farmers to plough the same field:

1. Multiply by $2$ to calculate how long $1$ farmer would take:

$10 \times 2 = 20\,\text{hours for 1 farmer}$

2. Divide by $4$ to calculate how long $4$ farmers would take:

$20 \div 4 = 5\,\text{hours for 4 farmers}$

Alternatively, observe that $4$ farmers is twice as many as $2$ farmers. Because the number of farmers has doubled, the required time is halved:

$10 \div 2 = 5\,\text{hours}$

Example 2: Decorating Cakes and Algebraic Formulation

Suppose $4$ bakers can decorate $100$ cakes in $5\,\text{hours}$.

To calculate how long it would take $10$ bakers to decorate the same $100$ cakes:

1. Multiply by $4$ to find the time required for $1$ baker:

$4 \times 5 = 20\,\text{hours for 1 baker}$

2. Divide by $10$ to find the time required for $10$ bakers:

$20 \div 10 = 2\,\text{hours for 10 bakers}$

To express this relationship algebraically where $x$ represents the number of bakers and $t$ represents the time in hours taken to decorate $100$ cakes:

1. Write the general inverse proportion equation format:

$t = \frac{k}{x}$

2. Substitute the known values $x = 4$ and $t = 5$ into the equation:

$5 = \frac{k}{4}$

3. Solve for the constant $k$:

$k = 20$

4. Substitute the constant $k = 20$ back into the formula to form the final equation:

$t = \frac{20}{x}$

Verification and Practice Problems

Inverse proportions can be counterintuitive. Answers should always be verified to ensure they make logical sense in context. A useful sanity check is confirming that an increase in workforce or effort results in a decrease in total duration (e.g., more workers must mean less time).

Practice Problem

If $6$ teachers take $20\,\text{hours}$ to mark Year $11$'s Maths exams, how long would it take $15$ teachers?

1. Calculate the total teacher-hours required for $1$ teacher:

$6 \times 20 = 120\,\text{hours for 1 teacher}$

2. Divide the total single-teacher hours by $15$ teachers:

$120 \div 15 = 8\,\text{hours}$