Inverse relationships

Some pairs of quantities move in the same direction — more hours worked, more pay earned. Others move in opposite directions, but not randomly: they trade off in a precise, reciprocal way, so that their product stays fixed. Double one and the other exactly halves; triple one and the other drops to a third. This is an inverse relationship, and it shows up constantly in everyday arithmetic — more workers finishing a wall in fewer days, a faster car covering a fixed distance in less time, more people sharing a bill each paying less. This article isolates that one idea, contrasts it sharply with the direct proportion covered in the main Proportional Reasoning chapter, and builds the habit of recognising which type a word problem is before setting up any equation.

The core insight

Two quantities are inversely related when their product stays constant ( $x \times y = k$ , so $y = k/x$ ), meaning one increases in exactly the ratio the other decreases — the opposite behaviour of direct proportion, where their ratio stays constant.

The core idea

Direct proportion, recalled. Two quantities $x$ and $y$ are directly proportional when their ratio is constant:

\frac{y}{x} = k \quad \Longrightarrow \quad y = kx

where: $k$ is the constant of proportionality, $x$ and $y$ are the two related quantities. As $x$ grows, $y$ grows in the same ratio; as $x$ shrinks, $y$ shrinks with it.

Inverse proportion, defined. Two quantities $x$ and $y$ are inversely proportional when their product is constant instead:

x \times y = k \quad \Longrightarrow \quad y = \frac{k}{x}

where: $k$ is the fixed product (a constant for the given situation), $x$ and $y$ are the two related quantities. Now, as $x$ grows, $y$ must shrink — not by an arbitrary amount, but by exactly the factor needed to keep $x \times y$ unchanged. If $x$ is scaled by a factor $m$ , then $y$ is automatically scaled by $\frac{1}{m}$ : double $x$ and $y$ halves; shrink $x$ to a third and $y$ triples.

Telling the two apart in a word problem. Ask what happens to one quantity as the other increases.

Both increase or decrease together, in the same ratio, → direct proportion, $y = kx$ .
One increases while the other decreases, in a reciprocal way, → inverse relationship, $y = k/x$ .

A classic inverse pair is workers and days: for a fixed amount of work, more workers finish it in fewer days, because (workers) $\times$ (days) stays roughly constant, assuming each worker keeps the same steady rate. Speed and time for a fixed distance behave the same way: (speed) $\times$ (time) = distance, a constant, so a faster speed always means less time. The graph of $y = k/x$ (for positive $x, y$ ) is a smooth curve that falls steeply at first and then flattens out, always getting closer to the axes but never touching them — it never turns negative and never crosses zero, because $x$ and $y$ must both stay positive for these real-world quantities.

Figure

A curve of y equals k over x falling from the top left to the bottom right, showing a point at x equals 6 y equals 15 and another point at x equals 9 y equals 10 both lying on the same curve since their product stays 90

Derivation AWhy the constant-product method always works for a fixed job

The goal is to show, from first principles about "the same fixed amount of work," why workers times days must stay at one constant value, so that the shortcut $x_1 y_1 = x_2 y_2$ used in every inverse-proportion problem is not just a memorised trick but a direct consequence of what "fixed work" means, and why it fails the moment the work itself changes.

Step 1. Define the total work in worker-days. Suppose one worker, working alone at the steady rate every worker shares, would need $W$ days to finish the wall by themselves. Then the whole job amounts to $W$ "worker-days" of effort — a fixed quantity that does not depend on how many people you assign to it, only on how big the wall is.

Step 2. Relate workers and days to that fixed quantity. If $x$ workers share the job equally and each needs $y$ days to finish their share, then the total effort supplied is (number of workers) $\times$ (days each works), because every worker contributes at the same steady rate for the same number of days. So $x \times y = W$ , and since $W$ is fixed once the job is fixed, the product $x \times y$ must equal the same constant $k = W$ no matter how $x$ and $y$ individually change.

Step 3. Confirm with the given numbers. Here $6 \times 15 = 90$ , so the wall is a $90$ worker-day job: $W = 90$ . Assigning 9 workers instead does not change the size of the wall, so the same equation $x \times y = 90$ must still hold, giving $y = 90/9 = 10$ days — matching the worked example below.

Step 4. See why the method breaks if the job changes size. If the wall were made twice as large, $W$ would double to $180$ , and the old equation $x \times y = 90$ would no longer apply — a reminder that the constant $k$ is only fixed for a given, unchanged amount of work, not a universal number.

Worked example

Problem. 6 workers can build a wall in 15 days. If 9 workers work at the same steady rate, roughly how many days would they take to build the same wall?

Step 1. Decide the type of relationship. More workers on the same fixed amount of work should finish faster — as one quantity (workers) rises, the other (days) should fall. That reciprocal behaviour is the signature of an inverse relationship, so

\text{workers} \times \text{days} = k \quad \text{(constant, for a fixed amount of work)}

Step 2. Find the constant $k$ from the given data: $k = 6 \times 15 = 90$ .

Step 3. Use the same constant with 9 workers: $9 \times \text{days} = 90 \implies \text{days} = \dfrac{90}{9} = 10$ .

Answer: 9 workers take 10 days — fewer days than 6 workers, exactly as an inverse relationship predicts.

Competitive stretch

A car travelling at 40 km/h covers a fixed distance in 6 hours. At what speed must it travel to cover the same distance in 4 hours? Since speed $\times$ time = distance (constant) for a fixed distance, the constant is $40 \times 6 = 240$ km. With the new time, $\text{speed} \times 4 = 240 \implies \text{speed} = \dfrac{240}{4} = 60$ km/h. Speed rose from 40 to 60 km/h exactly as time fell from 6 to 4 hours, keeping the product fixed at 240 km throughout — the same reciprocal pattern as the workers-and-days example, just with a different pair of quantities.

Common traps

Using the direct-proportion cross-multiplication ( $\frac{a}{b} = \frac{c}{d}$ ) on an inverse situation. For an inverse relationship it is the products that must be equal ( $a \times b = c \times d$ ), not the ratios; applying $\frac{6}{15} = \frac{9}{x}$ to the workers-and-days problem gives the wrong answer, $x = 22.5$ , instead of the correct 10.
Skipping the "which direction should it move" check. Before calculating, ask in words whether the second quantity should logically go up or down. If more workers should mean fewer days but the arithmetic gives more days, the wrong relationship type was used.
Assuming every "one up, one down" situation is inverse proportion. True inverse proportion needs the product to stay exactly constant, not merely a general decreasing trend.
Mixing up the vocabulary. "Directly proportional" ( $y = kx$ ) and "inversely proportional" ( $y = k/x$ ) describe structurally different relationships and cannot be used interchangeably.

Ask the tutor: Not sure whether a word problem is direct or inverse proportion — ask the AI tutor to walk through the wording with you before setting up the equation.

How is inverse proportion different from something just decreasing

Not every decreasing pair is inversely proportional — inverse proportion requires the product of the two quantities to stay exactly constant at every stage. For example, your age going up while the years left in a race countdown go down is a decrease, but there is no fixed product linking them. Workers and days for a fixed job pass the test: however you split the workers, workers times days always lands on the same number, because the same fixed amount of work is being shared out.

Why does cross-multiplying like a direct proportion give the wrong answer here

Cross-multiplication, $\frac{a}{b} = \frac{c}{d}$ , is the correct move only when the ratio between the two quantities is fixed — that is direct proportion. Inverse proportion fixes the product, not the ratio, so the correct setup is $a \times b = c \times d$ . Using cross-multiplication on the workers-and-days problem gives 22.5 days, but the true answer is 10 days: cross-multiplication silently assumes more workers should take more days, the opposite of what actually happens.

Is inverse proportion tested in this chapter or the next one

In the CBSE Class 8 syllabus, this chapter, Proportional Reasoning, is deliberately restricted to direct proportion, so that idea is fully mastered first. Inverse proportion is developed in full one chapter later, in Proportional Reasoning-2, where you will meet more word problems (workers-days, speed-time, sharing a fixed task) and practice setting up and solving them. Here, the goal is only to recognise the core contrast between the two types.

Inverse relationships

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads