Inverse Proportion — When More Means Less

Introduction

Inverse proportion describes situations where one quantity increases while the other decreases in such a way that their product stays constant. In symbols, if \(y\) is inversely proportional to \(x\), we write \[ y \propto \frac{1}{x} \quad \text{or} \quad y = \frac{k}{x}, \] where \(k\) is the constant of proportionality. This means that if you double \(x\), then \(y\) is halved; if you triple \(x\), then \(y\) is divided by three.

You’ll encounter inverse proportion in GCSE Maths and Science contexts: speed–time relationships for a fixed distance, number of workers vs. time to complete a job, intensity of light vs. distance squared, and questions that state directly “\(y\) is inversely proportional to \(x\)”. The hallmark is that increasing one value causes the other to shrink in a predictable, reciprocal way.

Inverse proportion problems are common in exam papers, often disguised in everyday language. For example: - “A job takes 12 hours for 5 workers. How long for 10 workers at the same rate?” - “At constant distance, speed and time vary inversely.” - “The graph of \(y\) against \(x\) shows a curve of the form \(y=\tfrac{k}{x}\).”

In this tutorial you will learn to:

• Recognise when a problem involves inverse proportion.
• Write the relationship as \(y = \tfrac{k}{x}\) and calculate the constant \(k\).
• Use the product rule: \(x \times y = k\).
• Interpret and draw inverse proportion graphs (curves that approach axes but never cross).
• Avoid common mistakes such as treating inverse problems like direct ones.

Key Vocabulary

Understanding the terms and symbols below is essential for working with inverse proportion. They appear frequently in GCSE questions and help you translate word problems into equations.

• Inverse proportion: A relationship where one value increases while the other decreases so that their product remains constant. Written as \(y \propto \tfrac{1}{x}\) or \(y = \tfrac{k}{x}\).
• Constant of proportionality (\(k\)): The fixed product of the two variables. If \(y = \tfrac{k}{x}\), then \(x \times y = k\) for all valid pairs. Example: If 4 workers take 6 days to finish a job, then \(x \times y = 4 \times 6 = 24\). For 8 workers, \(y = \tfrac{24}{8} = 3\) days.
• Reciprocal relationship: Inverse proportion means “reciprocal scaling” — multiplying \(x\) by a factor divides \(y\) by the same factor. Example: If speed doubles, time halves (for a fixed distance).
• Equation of inverse proportion: \[ y = \frac{k}{x}. \] Use any known pair of values to find \(k\), then use the formula to find unknowns.
• Product rule: The defining feature of inverse proportion: \[ x \times y = k. \] This is often quicker to apply than rearranging the formula.
• Graph of inverse proportion: A curve (hyperbola) that gets closer to the axes but never touches them. It slopes downward: as \(x\) grows, \(y\) shrinks.
• Work-rate problems: Situations where more workers mean fewer days, or more machines mean less time, provided the total job stays the same.
• Speed–time link: For a fixed distance, speed and time are inversely proportional: \[ \text{Speed} \times \text{Time} = \text{Distance}. \]
• Not direct proportion: Unlike direct proportion, doubling one variable does not double the other — it halves it. This distinction is a common exam trap.

Core Ideas

Inverse proportion problems are built around the principle that one value increases while the other decreases, keeping the product constant. Understanding these key ideas helps you tackle a wide range of exam questions.

• Definition: If \(y\) is inversely proportional to \(x\), then \[ y = \frac{k}{x}, \] where \(k\) is the constant of proportionality.
• Product Rule: For all pairs of values in inverse proportion, \[ x \times y = k. \] This is often the fastest way to solve questions — once \(k\) is known, the other value is found by dividing.
• Reciprocal Scaling: Multiplying \(x\) by a factor divides \(y\) by the same factor. Example: If 5 workers take 12 days, then 10 workers (double the workforce) take 6 days (half the time).
• Graphs: The graph of \(y = \tfrac{k}{x}\) is a smooth curve (a hyperbola) in the first quadrant for positive values. - It gets closer to the axes but never touches them. - It slopes downward: as \(x\) increases, \(y\) decreases.
• Common Contexts: - Speed & Time for a fixed distance: \( \text{Speed} \times \text{Time} = \text{Distance}\). - Number of workers & Time for a fixed job. - Intensity of light vs. distance (in physics, inverse-square law). - Pressure vs. Volume at constant temperature (Boyle’s Law in science).
• Distinguishing from Direct Proportion: - Direct: Double one, the other doubles (\(y=kx\)). - Inverse: Double one, the other halves (\(y=\tfrac{k}{x}\)). Spotting this difference correctly is essential to avoid using the wrong formula.
• Unit Consistency: As with direct proportion, units must match. Convert carefully (hours ↔ minutes, workers per day ↔ total workers, km/h ↔ m/s) before applying the product rule.

Step-by-Step Method

Solving inverse proportion problems follows a clear routine. These steps will guide you through recognising the situation, setting up the equation, and calculating the unknown values.

1. Check the relationship. Look for phrases such as “inversely proportional”, “more of one means less of the other”, or “product remains constant”. Quick test: multiply given pairs — if the product is constant, it’s inverse proportion.
2. Write the model. Use the equation \[ y = \frac{k}{x} \] or the product rule \[ x \times y = k. \]
3. Find the constant of proportionality \(k\). Substitute a known pair \((x,y)\). Example: If 4 workers take 12 days, \(k = 4 \times 12 = 48\).
4. Use the constant to solve for the unknown. Rearrange \(y = \tfrac{k}{x}\) or \(x = \tfrac{k}{y}\). Example: With \(k=48\), 8 workers take \(48 ÷ 8 = 6\) days.
5. Check the direction of change. Increasing one should decrease the other. If both increased, you may have wrongly treated it as direct proportion.
6. Watch the units. Make sure all values are in the same units (hours not mixed with minutes, kg not mixed with g, etc.) before substituting.
7. Sence-check your answer. Does doubling the input halve the output? If not, go back and re-check your working.

Worked Examples

Let’s apply the step-by-step method to a range of inverse proportion problems, starting with simple cases and moving to higher-level contexts.

Example 1 (Foundation — Workers & Time)

4 workers take 12 days to complete a job. How long will 8 workers take at the same rate?

Step 1: \(k = x \times y = 4 \times 12 = 48\). Step 2: For 8 workers, \(y = \tfrac{48}{8} = 6\). Answer: 6 days.

Example 2 (Foundation — Speed & Time)

A car covers a fixed distance in 4 hours at 60 km/h. How long would it take at 90 km/h?

Step 1: \(k = \text{speed} \times \text{time} = 60 \times 4 = 240\). Step 2: Time = \(240 ÷ 90 = 2.\overline{6}\) hours ≈ 2 hours 40 minutes. Answer: 2 h 40 min.

Example 3 (Foundation — Graph Context)

The table shows values of \(x\) and \(y\): \((2, 30), (3, 20), (5, 12)\). Show that \(y\) is inversely proportional to \(x\).

Check products: - \(2 \times 30 = 60\) - \(3 \times 20 = 60\) - \(5 \times 12 = 60\) Product constant ⇒ inverse proportion. Equation: \(y = 60 ÷ x\).

Example 4 (Higher — Algebraic Form)

\(y\) is inversely proportional to \(x\). When \(x=4\), \(y=15\). Find \(y\) when \(x=10\).

Step 1: \(k = x \times y = 4 \times 15 = 60\). Step 2: For \(x=10\), \(y = 60 ÷ 10 = 6\). Answer: \(y=6\).

Example 5 (Higher — Physics Context: Boyle’s Law)

For a fixed mass of gas, pressure \(P\) is inversely proportional to volume \(V\). If \(P=150\) kPa when \(V=2.0\) L, find \(P\) when \(V=3.5\) L.

Step 1: \(k = P \times V = 150 \times 2.0 = 300\). Step 2: \(P = 300 ÷ 3.5 = 85.7\) kPa (1 d.p.). Answer: 85.7 kPa.

Example 6 (Higher — Graph Shape)

Sketch the graph of \(y = \tfrac{24}{x}\) for \(x=1,2,3,4,6,12\).

Values: (1,24), (2,12), (3,8), (4,6), (6,4), (12,2). Plotting gives a smooth curve approaching the axes but never touching. Key point: The curve decreases steadily and never crosses axes.

Common Mistakes & Fixes

Inverse proportion questions are straightforward once you see the pattern, but many students fall into traps — often by confusing them with direct proportion. Here are the most frequent mistakes and how to avoid them.

• Treating inverse problems as direct proportion. Mistake: Doubling \(x\) and then doubling \(y\). Fix: Remember: in inverse proportion, doubling one halves the other. Check by multiplying: \(x \times y\) must stay constant.
• Forgetting the product rule. Mistake: Trying to set up a unitary method like in direct proportion. Fix: Use \(x \times y = k\). It’s usually quicker and avoids confusion.
• Not converting units consistently. Mistake: Mixing hours with minutes, or litres with millilitres, leading to wrong constants. Fix: Convert all quantities into the same unit before multiplying.
• Forgetting to sense-check. Mistake: Getting a longer time when more workers are added. Fix: Stop and think: more workers should mean less time, not more.
• Graph misinterpretation. Mistake: Expecting a straight line through the origin. Fix: Inverse proportion graphs are curves (hyperbolas) approaching the axes, not straight lines.
• Mixing up constants. Mistake: Calculating \(k = \tfrac{y}{x}\) instead of \(k = x \times y\). Fix: Memorise: direct = divide, inverse = multiply.
• Ignoring restrictions. Mistake: Using negative or zero values of \(x\) in real-world contexts (e.g. “0 workers finish a job in … ?”). Fix: Only use realistic positive values that make sense in the situation.

Practice Questions — Foundation

These questions focus on the basics of inverse proportion. Use the product rule \(x \times y = k\) to solve them, and always check that increasing one value decreases the other.

1. 6 workers take 15 days to finish a job. How long will 10 workers take, working at the same rate?
2. A cyclist covers a journey in 5 hours at 12 km/h. How long would the same journey take at 20 km/h?
3. A machine makes 120 parts in 8 hours. How many hours would it take to make the same parts if its speed increased so it made 240 parts in 8 hours?
4. If \(y\) is inversely proportional to \(x\) and \(y=30\) when \(x=2\), find \(y\) when \(x=5\).
5. A car travels a fixed distance. If the average speed is 80 km/h, the journey takes 3 hours. How long would the journey take at 120 km/h?
6. A table shows values of \(x\) and \(y\): \((2, 18), (3, 12), (6, 6)\). Show that \(x\) and \(y\) are inversely proportional and write the equation linking them.

Practice Questions — Higher

These questions explore inverse proportion in more demanding contexts: algebraic forms, graphs, and applied scenarios. Use the model \(y=\tfrac{k}{x}\) or the product rule \(x \times y = k\) to solve them.

1. \(y\) is inversely proportional to \(x\). When \(x=4\), \(y=18\). (a) Find the constant \(k\). (b) Write the equation for \(y\) in terms of \(x\). (c) Find \(y\) when \(x=9\).
2. A job takes 24 hours for 12 workers. (a) Find how many worker-hours the job requires. (b) How long would it take 18 workers to finish the same job?
3. A car covers 300 km. At 75 km/h it takes 4 hours. (a) Show that speed and time are inversely proportional. (b) Find the time if the car’s speed is 120 km/h.
4. The pressure \(P\) of a gas varies inversely with volume \(V\). When \(V=2.5\) L, \(P=96\) kPa. (a) Find the constant \(k\). (b) Work out \(P\) when \(V=4\) L. (c) Rearrange to express \(V\) in terms of \(P\).
5. The graph of \(y=\tfrac{60}{x}\) is drawn. (a) Calculate the values of \(y\) for \(x=5,10,15,20\). (b) Sketch the curve and describe its shape. (c) What happens to \(y\) as \(x\) becomes very large?
6. In an experiment, intensity \(I\) is inversely proportional to distance squared (\(d^2\)). When \(d=2\) m, \(I=25\). (a) Show the relationship in the form \(I=\tfrac{k}{d^2}\). (b) Find \(I\) when \(d=5\) m.
7. A truckload of bricks takes 15 hours to move using 4 trucks. (a) How long would it take using 10 trucks? (b) If 6 trucks are used instead, how long would it take?

Challenge Questions

These problems push inverse proportion into more complex, multi-step contexts. They combine algebra, graphs, and real-life scenarios, just like top-grade GCSE exam questions.

1. A factory has 6 machines that complete an order in 20 hours. (a) How many machine-hours does the order require? (b) How long would it take if 15 machines worked at the same rate? (c) How many machines would be needed to finish the job in 5 hours?
2. Two quantities \(x\) and \(y\) are inversely proportional. When \(x=8\), \(y=15\). (a) Find the equation linking \(x\) and \(y\). (b) Find \(y\) when \(x=20\). (c) Find \(x\) when \(y=60\).
3. A car travels 240 km. At 60 km/h the journey takes 4 hours. (a) Find the constant of proportionality between speed and time. (b) Calculate the journey time at 80 km/h. (c) Sketch the graph of time against speed for this journey.
4. The pressure \(P\) of a gas varies inversely with its volume \(V\). If \(P=150\) kPa when \(V=2\) L: (a) Find the equation relating \(P\) and \(V\). (b) Work out \(P\) when \(V=7.5\) L. (c) What happens to \(P\) as \(V\) increases without bound?
5. A curve is defined by \(y=\tfrac{120}{x}\). (a) Find \(y\) when \(x=3, 6, 12, 24\). (b) Sketch the graph. (c) Describe what happens to \(y\) when \(x\) becomes very large.
6. A group of students is asked to paint 200 chairs. If 10 students take 8 hours: (a) How long would 20 students take? (b) If the time limit is 4 hours, how many students are needed? (c) Verify your answers using the product rule.
7. In physics, the brightness \(B\) of a light source varies inversely with the square of the distance \(d\) from the source. (a) Write \(B\) in terms of \(k\) and \(d\). (b) If \(B=40\) when \(d=2\), find \(k\). (c) Find \(B\) when \(d=5\). (d) Interpret what happens to brightness as \(d\) doubles.
8. The time \(T\) needed to download a file is inversely proportional to the internet speed \(S\). If it takes 12 minutes at 20 Mbps: (a) Write down the relationship between \(T\) and \(S\). (b) Calculate the time at 50 Mbps. (c) Find the speed required to download the file in 4 minutes.

Quick Revision Sheet

This one-page summary captures everything you need to remember about inverse proportion. Use it as a rapid checklist before exams.

Key Formulas

• Model: \(y \propto \tfrac{1}{x}\) ⇔ \(y = \tfrac{k}{x}\).
• Product Rule: \(x \times y = k\) (constant for all pairs).
• Constant of Proportionality: \(k = x \times y\).

How to Spot Inverse Proportion

• Phrases like: “inversely proportional”, “more of one means less of the other”, “fixed total”, “at this rate”.
• Multiplying values gives a constant product (e.g. 2×30=60, 3×20=60).
• Graphs are downward curves (hyperbolas), not straight lines, approaching axes but never touching.

Fast Method (5 Steps)

1. Confirm: does one go up as the other goes down?
2. Write \(x \times y = k\).
3. Find \(k\) using a known pair.
4. Use \(y = k/x\) or \(x = k/y\) to find the unknown.
5. Sence-check: doubling one should halve the other.

Common Contexts

• Work problems: More workers → less time.
• Speed-time (fixed distance): Faster speed → shorter time.
• Gas laws: Pressure and volume inverse at constant temperature.
• Physics (light, gravity, sound): Often involve inverse-square laws.

Graph Facts

• Curve in first quadrant if \(x,y > 0\).
• As \(x\) increases, \(y\) decreases.
• Curve never crosses axes (asymptotes at \(x=0\) and \(y=0\)).

Common Traps

• Confusing with direct proportion (\(y=kx\)).
• Forgetting to multiply instead of divide to find \(k\).
• Not converting all units consistently.
• Accepting answers where both values increase together.

Mini Quick Practice

1. 5 workers take 18 days. How long for 15 workers? Answer: \(5\times18=90\). \(90 ÷ 15 = 6\) days.
2. A car covers a distance in 4 h at 60 km/h. How long at 80 km/h? Answer: \(60\times4=240\). \(240 ÷ 80 = 3\) h.

Answers

Foundation

1. \(k = 6 \times 15 = 90\). For 10 workers: \(90 ÷ 10 = \mathbf{9 \text{ days}}\).
2. \(k = 12 \times 5 = 60\). For 20 km/h: \(60 ÷ 20 = \mathbf{3 \text{ hours}}\).
3. \(120\) parts in 8 h ⇒ rate = 15 parts/h. Doubling to 240 parts in 8 h means speed doubled, so time halved. \(\mathbf{4 \text{ hours}}\).
4. \(k = 2 \times 30 = 60\). For \(x=5\): \(y = 60 ÷ 5 = \mathbf{12}\).
5. Distance = \(80 \times 3 = 240\). Time at 120 km/h = \(240 ÷ 120 = \mathbf{2 \text{ hours}}\).
6. Products: \(2 \times 18 = 36\), \(3 \times 12 = 36\), \(6 \times 6 = 36\). Equation: \(y = \tfrac{36}{x}\).

Higher

1. (a) \(k = 4 \times 18 = 72\). (b) \(y = \tfrac{72}{x}\). (c) \(y = 72 ÷ 9 = \mathbf{8}\).
2. (a) \(12 \times 24 = 288\) worker-hours. (b) For 18 workers: \(288 ÷ 18 = \mathbf{16 \text{ hours}}\).
3. (a) \(75 \times 4 = 300\). Distance constant. (b) Time at 120 = \(300 ÷ 120 = \mathbf{2.5 \text{ h}}\).
4. (a) \(k = 96 \times 2.5 = 240\). (b) \(P = 240 ÷ 4 = \mathbf{60 \text{ kPa}}\). (c) \(V = \tfrac{240}{P}\).
5. (a) \(y(5)=12,\; y(10)=6,\; y(15)=4,\; y(20)=3\). (b) Curve slopes down, approaches axes. (c) As \(x\to\infty,\; y\to0\).
6. (a) \(I=\tfrac{k}{d^2}\). \(k=25 \times 2^2=100\). (b) \(I=100 ÷ 25 = \mathbf{4}\).
7. Total work = \(4 \times 15 = 60\) truck-hours. (a) \(60 ÷ 10 = \mathbf{6 \text{ hours}}\). (b) \(60 ÷ 6 = \mathbf{10 \text{ hours}}\).

Challenge

1. (a) \(6 \times 20 = 120\) machine-hours. (b) \(120 ÷ 15 = \mathbf{8 \text{ h}}\). (c) \(120 ÷ 5 = \mathbf{24 \text{ machines}}\).
2. (a) \(k=8 \times 15=120\). Equation: \(y=\tfrac{120}{x}\). (b) \(y=120 ÷ 20=6\). (c) \(x=120 ÷ 60=2\).
3. (a) \(60 \times 4 = 240\). Constant product. (b) Time at 80 = \(240 ÷ 80=3\). (c) Curve decreasing, approaches axes.
4. (a) \(P=\tfrac{300}{V}\). (b) \(P=300 ÷ 7.5=40\). (c) As \(V\to\infty\), \(P\to0\).
5. (a) Values: \(40,20,10,5\). (b) Downward curve. (c) \(y\) approaches 0.
6. (a) \(10 \times 8=80\). (b) \(80 ÷ 20=4\). (c) For 4 h: \(80 ÷ 4=20\) students. Verified.
7. (a) \(B=\tfrac{k}{d^2}\). \(k=40 \times 4=160\). (b) \(B=160 ÷ 25=6.4\). (c) Brightness quarters when distance doubles.
8. (a) \(T=\tfrac{k}{S}\). \(k=20 \times 12=240\). (b) \(T=240 ÷ 50=4.8\) min. (c) For 4 min: \(S=240 ÷ 4=60\) Mbps.

Conclusion & Next Steps

Inverse proportion is all about balance: as one value rises, the other falls, keeping their product constant. Whether it’s workers finishing a job, cars travelling at different speeds, or gases under pressure, the rule is the same: \[ x \times y = k. \] Once you know the constant \(k\), every other pair of values is just a divide away. The graph is always a curve that approaches but never touches the axes, showing the reciprocal link.

Here’s what you’ve mastered:

• Identifying inverse proportion: Look for “fixed total” or “more of one, less of the other”.
• Using the product rule: \(k = x \times y\) gives you the key to unlock any pair of values.
• Writing equations: \(y = \tfrac{k}{x}\) or rearranging to \(x = \tfrac{k}{y}\).
• Graph skills: Recognising and sketching the hyperbolic shape of \(y = \tfrac{k}{x}\).
• Sense-checking answers: If increasing one variable increases the other, something has gone wrong.

Exam strategy recap:

1. Decide: direct or inverse? (Check if doubling one halves or doubles the other).
2. Write the correct formula: \(y=kx\) (direct) or \(y=\tfrac{k}{x}\) (inverse).
3. Calculate the constant \(k\) from one data pair.
4. Use the formula to find the unknown.
5. Check: does the result make logical sense?

Next topics to study:

• Proportion in context: Mixed problems involving recipes, scale models, density, and “best buy” questions.
• Direct vs. Inverse Proportion (Comparisons): Practice switching between them — a common exam skill.
• Compound proportion: Multi-step problems (e.g. “If 4 men paint 3 walls in 2 days, how long for 6 men to paint 5 walls?”).
• Graphs and functions: Comparing \(y=kx\) and \(y=\tfrac{k}{x}\) to strengthen algebra-graph links.