Inverse proportion

A tap that fills a tank faster does not need less water — it needs less time. A team with more workers does not need less work done — it needs fewer days. In both cases one quantity going up forces another to come down, and the two changes balance each other so precisely that their product never changes. This exact, opposite-but-balanced relationship is inverse proportion, one of the two central ideas of the CBSE Class 8 chapter Proportional Reasoning-2 (the other is pie charts). Inverse proportion shows up constantly in daily life — more painters finish a wall faster, a faster car covers the same trip in less time, more people sharing a fixed prize each get a smaller share. The Class 8 syllabus asks you to recognise such situations, tell them apart from direct proportion, and solve word problems using one clean rule: keep the product of the two quantities constant. This section builds that rule from scratch, tests it on the classic workers-and-days and speed-and-time problems, and flags the traps that cost the most marks.

The core insight

Two quantities are in inverse proportion when their product stays constant, so solve such problems by equating products ( $x_1y_1 = x_2y_2$ ), never ratios.

The core idea

Two quantities $x$ and $y$ vary inversely if, as $x$ changes, $y$ changes in the opposite direction in a way that keeps their product fixed:

x \times y = k \qquad (k \text{ a constant})

where: $x$ and $y$ are the two varying quantities (for example, workers and days) and $k$ is the constant of proportionality, the fixed value that their product always equals. As $x$ grows, $y$ must shrink by exactly the factor needed to keep $xy$ unchanged — double $x$ and $y$ halves; triple $x$ and $y$ becomes a third.

Comparing two pairs of values. If $(x_1, y_1)$ and $(x_2, y_2)$ are two corresponding pairs of an inversely proportional relationship, both pairs give the same product $k$ , so

x_1 y_1 = x_2 y_2 \implies \dfrac{x_1}{x_2} = \dfrac{y_2}{y_1}

where: $x_1, y_1$ is one known pair of values and $x_2, y_2$ is the pair with one value known and the other to be found. Notice the ratio flips — this is the single biggest difference from direct proportion. In direct proportion the ratios $\dfrac{y_1}{x_1}$ and $\dfrac{y_2}{x_2}$ are equal (same-sided comparison); in inverse proportion, $x_1$ compares to $x_2$ the way $y_2$ compares to $y_1$ (cross-sided, flipped comparison). Mixing the two up — using the direct-proportion cross-ratio for an inverse situation — is the most common error in this topic, so the safest habit is to skip ratios altogether and simply write $x_1y_1 = x_2y_2$ every time.

Reading a table of values. Given a table, you can test both possibilities directly: compute $\dfrac{y}{x}$ for every pair — if it is the same number throughout, the quantities are in direct proportion. Compute $x \times y$ for every pair instead — if that is constant, they are in inverse proportion. A table need not be either; always check both conditions rather than assuming one must hold just because the numbers "look proportional."

Where inverse proportion appears. Three families of situations recur throughout the syllabus:

Workers and days: more workers finish the same job in fewer days, provided every worker works at the same steady rate and the total amount of work does not change.
Speed and time: for a fixed distance, $\text{distance} = \text{speed} \times \text{time}$ is constant, so a faster speed always means less time.
Sharing a fixed quantity: dividing a fixed total (a prize, a task, a length of rope) among more people means a smaller share for each.

The figure below contrasts the two proportion types on the same axes: direct proportion is a straight line through the origin, while inverse proportion is a curve that drops steeply and never touches either axis.

Figure

Two graphs side by side showing direct proportion as a straight rising line through the origin and inverse proportion as a curve that falls steeply and approaches but never touches the axes

Derivation ASolving an inverse-proportion problem by keeping the product constant

The goal is a reusable three-step method that turns any inverse-proportion word problem — workers and days, speed and time, or sharing a fixed quantity — into a short calculation, by treating the constant product $xy = k$ as the single fact that carries all the information, avoiding the ratio confusion that direct-proportion habits create.

Step 1. Identify the two quantities and confirm they vary inversely — that is, check that their product represents something fixed (total work, or total distance) rather than checking whether their ratio is fixed. This confirms $x \times y = k$ is the correct relationship to use, not the direct-proportion ratio rule.

Step 2. Use the one known pair of corresponding values, $(x_1, y_1)$ , to compute the constant: $k = x_1 \times y_1$ . This number is fixed for the whole problem — it does not change as $x$ and $y$ individually change.

Step 3. Use the new known value, $x_2$ , and the same constant $k$ , to solve for the unknown $y_2$ : $x_2 \times y_2 = k \implies y_2 = \dfrac{k}{x_2}$ . Because $k$ was fixed by the first pair, this single equation determines the unknown quantity directly, without ever needing to write a ratio. Sanity check every answer: if $x_2 > x_1$ , then $y_2$ must come out smaller than $y_1$ , and vice versa — a workers-and-days answer where more workers take more days signals an arithmetic slip, almost always from copying the direct-proportion ratio method by mistake.

Worked example

$6$ workers can build a wall in $15$ days. How many days will $9$ workers take to build the same wall, working at the same steady rate?

Step 1. Recognise the inverse proportion: more workers finish the job in fewer days, and (workers) $\times$ (days) stays constant because it represents the fixed total amount of work.

Step 2. Find the constant using the known pair: $k = 6 \times 15 = 90$ , representing $90$ "worker-days" of total work.

Step 3. Solve for the new number of days, $d$ , using $9$ workers: $9 \times d = 90 \implies d = \dfrac{90}{9} = 10$ . So $9$ workers will take 10 days. Check: $6 \times 15 = 90$ and $9 \times 10 = 90$ — the product is constant. Also sensible: more workers ( $9 > 6$ ) took fewer days ( $10 < 15$ ), confirming the direction of change is correct.

Competitive stretch

A car covers a fixed distance in $5$ hours at $60$ km/h. At $75$ km/h instead, how long does the same trip take?

Step 1. Speed and time vary inversely for a fixed distance, since $\text{speed} \times \text{time} = \text{distance}$ is constant.

Step 2. Find the constant (the fixed distance): $k = 60 \times 5 = 300$ km.

Step 3. Solve for the new time $t$ : $75 \times t = 300 \implies t = \dfrac{300}{75} = 4$ . At $75$ km/h, the car takes 4 hours. Check: $60 \times 5 = 300$ and $75 \times 4 = 300$ ; higher speed ( $75 > 60$ ) gave less time ( $4 < 5$ ), as expected. Notice both examples follow the same shape: find the fixed total from one known pair, then divide by the new value of the other quantity.

Common traps

Using the direct-proportion cross-ratio method. Writing $\dfrac{x_1}{x_2} = \dfrac{y_1}{y_2}$ instead of $x_1y_1 = x_2y_2$ is the single most common error — inverse proportion is solved by keeping the product constant, not the ratio.
Assuming every "workers and time" problem is inverse. It is inverse only when the total amount of work stays fixed and every worker keeps the same steady rate; a problem that changes the work amount too is not simple inverse proportion.
Ignoring the implicit "equal rate" assumption, which every standard inverse-proportion word problem quietly relies on.
Not checking the direction of change. More workers should give fewer days, and higher speed should give less time — an answer going the wrong way signals a mistake, usually the ratio-versus-product mix-up above.

Ask the tutor: Not sure whether your word problem is direct or inverse proportion, or want a second pair of eyes on a workers-and-days calculation — ask the AI tutor to walk through it with you.

How do I quickly tell if a word problem is direct or inverse proportion?

Ask what stays fixed. If a total quantity is fixed — total work, total distance, a total prize being shared — and one rate-like quantity trading off against another, it is usually inverse: more of one means less of the other, and their product equals that fixed total. If instead the quantities simply grow together, like the cost of buying more identical items, it is direct: their ratio stays fixed. When unsure, test both conditions on the numbers given — compute $x \times y$ and $y/x$ for a couple of pairs and see which one stays constant.

Why does the ratio flip in the inverse-proportion formula but not in direct proportion?

In direct proportion, $y$ scales the same way as $x$ , so $\dfrac{y_1}{x_1} = \dfrac{y_2}{x_2}$ compares like with like on both sides. In inverse proportion, $y$ scales the opposite way — when $x$ doubles, $y$ halves — so comparing $x_1$ to $x_2$ directly against $y_1$ to $y_2$ would be comparing two quantities moving in opposite directions, which gives the wrong answer. Flipping one ratio, to get $\dfrac{x_1}{x_2} = \dfrac{y_2}{y_1}$ , restores the match. It is simplest, though, to skip the ratio form altogether and just use $x_1y_1 = x_2y_2$ , since that equation is true regardless of which value is larger.

Can a problem be neither direct nor inverse proportion?

Yes, and Class 8 word problems sometimes include this as a check on understanding. If neither $y/x$ nor $x \times y$ stays constant across the given values, the relationship is simply not a proportion of either type, and no proportion formula applies — the problem may need a different method entirely, such as a straightforward addition or a different formula for the situation described. Always verify the constant condition on the actual numbers before applying either rule; do not assume a word problem must be one type just because both types are being studied together.

Inverse proportion

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads