Direct and Inverse Proportions

Mohan makes tea for himself and his sister using 300 mL water, 2 spoons sugar, 1 spoon tea leaves and 50 mL milk. If five guests arrive instead of two people, how much of each item does he now need? And if two students take 20 minutes to arrange chairs for an assembly, how long will five students take to do the same job? Both questions are about one quantity changing because another quantity changes — but notice the two answers move in opposite directions. Tea ingredients must grow as the number of people grows. Time taken must shrink as the number of workers grows. This single chapter gives you the mathematics to handle both kinds of change precisely, instead of guessing. Two quantities $x$ and $y$ are in direct proportion if they increase or decrease together in such a way that their ratio $x/y$ stays a fixed constant $k$ . Two quantities are in inverse proportion if one increases exactly as much as the other decreases, so that their product $xy$ stays a fixed constant $k$ . These two simple tests — is the ratio constant, or is the product constant — are the entire toolkit of this chapter, and they answer Mohan's tea question (direct: ingredients scale with people) and the chair-arranging question (inverse: time shrinks as workers grow). The chapter also warns that not every pair of quantities that rises or falls together is proportional at all — a friend's age and their mother's age both increase every year, yet their ratio keeps changing, so they are in neither kind of proportion. Testing carefully, using the correct relation, rather than assuming a pattern, is the real skill this chapter builds — one used constantly in unitary-method, map-reading, and rate-and-work problems.

Key takeaways

Direct proportion: $x$ and $y$ vary directly if $\dfrac{x}{y} = k$ (a constant), equivalently $x = ky$ ; as one increases, the other increases in the same ratio.
For direct proportion, if $y_1, y_2$ are values of $y$ corresponding to $x_1, x_2$ , then $\dfrac{x_1}{y_1} = \dfrac{x_2}{y_2}$ — the cross-multiplication rule used to find an unknown value.
The graph of two directly proportional quantities is always a straight line through the origin.
Inverse proportion: $x$ and $y$ vary inversely if $xy = k$ (a constant); as one increases, the other decreases so their product never changes.
For inverse proportion, $x_1y_1 = x_2y_2$ , which can be rewritten as $\dfrac{x_1}{x_2} = \dfrac{y_2}{y_1}$ — note the values swap sides compared to the direct-proportion rule.
The graph of two inversely proportional quantities is a smooth falling curve that gets closer and closer to both axes but never touches them.
Increasing or decreasing together is not enough proof of direct proportion — always check that the ratio itself stays constant (a person's age and their mother's age both rise, but their ratio keeps changing).
Direct proportion is also written $x \propto y$ ; inverse proportion is written $x \propto \dfrac{1}{y}$ .
Typical direct-proportion pairs: cost and quantity bought, distance and petrol used, map length and actual distance. Typical inverse-proportion pairs: number of workers and time taken, speed and time for a fixed distance, price per item and number of items buyable with a fixed sum.
Key idea: two quantities are directly proportional when their ratio x over y stays constant (x = ky, graph a straight line through the origin) and inversely proportional when their product xy stays constant (graph a falling curve), and every word problem in this chapter is solved by first deciding which of these two constants — the ratio or the product — is the one that does not change.

The core insight

Two varying quantities are in direct proportion when their ratio $x/y$ stays fixed at a constant $k$ (so $x=ky$ , giving a straight-line graph through the origin), and in inverse proportion when their product $xy$ stays fixed at a constant $k$ (giving a falling curve), and solving any problem in this chapter begins with deciding which of the two — ratio or product — actually stays constant.

What this chapter covers

The chapter opens with everyday situations — tea for extra guests, chairs arranged by more students — where one quantity's change forces a change in another. Section 11.2 defines direct proportion: the constant ratio $x/y=k$ , the cross-multiplication rule $x_1/y_1=x_2/y_2$ , and applications including cost-of-goods, shadow-and-height, and map-scale problems (NCERT Examples 1–6). Section 11.3 defines inverse proportion: the constant product $xy=k$ , the rule $x_1y_1=x_2y_2$ , and applications including pipes-filling-tanks, food-lasting-for-animals, and workers-building-in-days problems (NCERT Examples 7–9). This article keeps NCERT's own worked-example numbers in view and extends them, and adds the two graphs — straight line for direct, falling curve for inverse — that make the difference visible at a glance.

The core ideas

Direct proportion: the constant ratio

If the cost of 1 kg of sugar is Rs 36, the cost of 3 kg is Rs 108 — three times the sugar costs three times as much. This is the signature of direct proportion: two quantities $x$ and $y$ vary directly if they increase (or decrease) together in such a way that the ratio of their corresponding values stays fixed.

\frac{x}{y} = k \qquad \text{or equivalently} \qquad x = ky. \qquad (1)

where: $k$ is the constant of proportionality (a fixed positive number for a given situation), $x$ and $y$ are the two varying quantities, for example petrol consumed and distance travelled by a car. NCERT's own petrol example fixes the idea: a car uses 4 litres to travel 60 km, and 12 litres (three times as much petrol) takes it 180 km (three times as far); the ratio $x/y = 4/60 = 12/180 = 1/15$ never changes, so $k = 1/15$ , or equivalently $x = \tfrac{1}{15}y$ . Because the ratio stays the same for every pair of corresponding values, any two pairs $(x_1,y_1)$ and $(x_2,y_2)$ satisfy $x_1/y_1 = x_2/y_2$ — this cross-multiplication rule (derived carefully below) is the working tool used to find an unknown fourth value once three are known, and it is exactly how NCERT solves the cost-of-cloth and shadow-and-height problems.

The graph of a direct proportion

Plotting petrol used (x-axis) against distance travelled (y-axis) for the car example gives points that all sit on one straight line, and that line passes through the origin, because when $x=0$ (no petrol used) the distance travelled is also $y=0$ .

Figure

A graph of petrol in litres on the horizontal axis against distance travelled in kilometres on the vertical axis showing marked points that all lie on a single straight line passing through the origin because distance is directly proportional to petrol used with the ratio always one to fifteen

Inverse proportion: the constant product

Zaheeda can reach her school by walking, running, cycling or going by car; the faster she travels, the less time the journey takes. If she doubles her speed, time halves; if she triples her speed, time drops to a third. Two quantities $x$ and $y$ vary in inverse proportion if an increase in $x$ causes a proportional decrease in $y$ , in such a way that their product stays fixed.

x\,y = k. \qquad (2)

where: $k$ is again a constant for the given situation, but now it is the product, not the ratio, that never changes. NCERT's book-buying example makes this concrete: a school has Rs 6000 to spend on textbooks; at Rs 40 each it buys 150 books, and at Rs 50 each it buys only 120 books — but $40 \times 150 = 6000$ and $50 \times 120 = 6000$ , the same product every time, so price and number bought vary inversely with $k = 6000$ . For two pairs of corresponding values $(x_1,y_1)$ and $(x_2,y_2)$ of an inversely proportional pair, $x_1y_1 = x_2y_2$ , which rearranges to $x_1/x_2 = y_2/y_1$ — notice carefully that the y-values swap sides compared with the direct-proportion rule; mixing the two rules up is the single most common error in this chapter, and both rules are proved from scratch below.

The graph of an inverse proportion

Plotting the price of a book (x-axis) against the number of books bought (y-axis) for a fixed budget of Rs 6000 gives a smooth curve that falls steeply at first and then more gently, always hugging closer to the axes without ever reaching them, because neither price nor the number of books bought can ever become exactly zero for a fixed positive budget.

Figure

A graph of the price of one book on the horizontal axis against the number of books that can be bought on the vertical axis showing a smooth falling curve through six marked points that never touches either axis because price and number bought are inversely proportional with their product always six thousand

Telling direct proportion, inverse proportion and neither apart

Every word problem here starts with the same question: as one quantity grows, does the other grow too (in exact step), shrink (in exact reciprocal step), or neither? Read the situation first, not the numbers: more workers finishing a job faster is inverse; more articles purchased costing more money is direct; a person's height and the number of leaves on a nearby tree follow no fixed relation — they are simply unrelated. Then confirm by testing corresponding values: if $x/y$ stays the same for every pair, it is direct; if $xy$ stays the same, it is inverse; if neither survives more than one pair, the quantities are not in any simple proportion, and this chapter's formulas do not apply.

Derivation ADeriving the cross-multiplication rule for direct proportion

The goal of this derivation is to show, starting only from the definition of direct proportion, exactly why any two pairs of corresponding values $(x_1,y_1)$ and $(x_2,y_2)$ of two directly proportional quantities must satisfy $x_1/y_1 = x_2/y_2$ , and why this single equation is enough to find an unknown fourth value whenever the other three are known — which is the method NCERT uses throughout Examples 1 to 6 of this chapter, including the cloth-cost and map-scale problems.

Step 1. Start from the definition. Two quantities $x$ and $y$ are in direct proportion precisely when their ratio stays fixed at some constant $k$ for every pair of corresponding values that occurs in the situation, so $\dfrac{x}{y} = k$ for all such pairs, which can equally be written $x = ky$ .

Step 2. Apply the definition to the first pair of values, $x_1$ and $y_1$ . Since the ratio equals $k$ for every pair, in particular $\dfrac{x_1}{y_1} = k$ .

Step 3. Apply the very same definition to a second, different pair of values, $x_2$ and $y_2$ , drawn from the same directly proportional relationship. Because $k$ does not change from one pair to another, $\dfrac{x_2}{y_2} = k$ as well.

Step 4. Both Step 2 and Step 3 give an expression equal to the same constant $k$ , so those two expressions must equal each other: $\dfrac{x_1}{y_1} = \dfrac{x_2}{y_2}$ . Cross-multiplying gives $x_1 y_2 = x_2 y_1$ , an equivalent form that is sometimes more convenient for solving.

Step 5. Use the rule to find an unknown. Suppose $x_1, y_1, x_2$ are known and $y_2$ is unknown. Rearranging Step 4 gives $y_2 = \dfrac{x_2 y_1}{x_1}$ . Checking against NCERT's cloth example, where 5 metres of cloth cost Rs 210 ( $x_1=5, y_1=210$ ) and we want the cost of 2 metres ( $x_2=2$ ): $y_2 = \dfrac{2 \times 210}{5} = 84$ , exactly the value NCERT obtains.

Step 6. Note the boundary of the method. The rule holds only because $k$ was assumed fixed throughout — if the ratio ever changes between pairs (as with a friend's age against their mother's age), the two quantities are not in direct proportion, and the rule cannot be applied.

Go deeper · Derivation B · 1 minDeriving the product rule for inverse proportion

Derivation BDeriving the product rule for inverse proportion

The goal of this derivation is to show, from the definition of inverse proportion alone, why any two pairs of corresponding values $(x_1,y_1)$ and $(x_2,y_2)$ of two inversely proportional quantities must satisfy $x_1y_1 = x_2y_2$ , and to obtain the equivalent ratio form $x_1/x_2 = y_2/y_1$ that NCERT uses in the pipes, hostel-food and wall-building examples — while making clear exactly how this rule differs from the direct-proportion rule derived above, since confusing the two is the chapter's most common mistake.

Step 1. Start from the definition. Two quantities $x$ and $y$ are in inverse proportion precisely when their product stays fixed at some constant $k$ for every pair of corresponding values, so $xy = k$ for all such pairs; equivalently $x = k/y$ or $y = k/x$ .

Step 2. Apply the definition to the first pair, $x_1$ and $y_1$ : since the product equals $k$ for every pair, $x_1 y_1 = k$ .

Step 3. Apply the same definition to a second pair, $x_2$ and $y_2$ , from the same inversely proportional relationship: because $k$ is the same constant throughout, $x_2 y_2 = k$ as well.

Step 4. Both Step 2 and Step 3 equal the same constant $k$ , so they equal each other: $x_1 y_1 = x_2 y_2$ . This is the product rule, and it looks very different from the direct-proportion cross-multiplication rule — here it is a product on each side, not a ratio.

Step 5. Convert to a ratio form for convenience. Dividing both sides of $x_1y_1 = x_2y_2$ first by $x_2$ and then by $y_1$ gives $\dfrac{x_1}{x_2} = \dfrac{y_2}{y_1}$ — notice that $y_2$ and $y_1$ have swapped sides compared with the direct-proportion rule $x_1/y_1=x_2/y_2$ ; this swap is exactly what makes inverse proportion "inverse."

Step 6. Check against NCERT's wall-building example: 15 workers build a wall in 48 hours ( $x_1 = 48$ hours, $y_1 = 15$ workers), and we want the workers $y_2$ needed to finish in $x_2 = 30$ hours. From Step 4, $48 \times 15 = 30 \times y_2$ , so $y_2 = \dfrac{48\times15}{30} = 24$ workers — exactly NCERT's answer, confirming the rule and its direction of use.

Worked examples

Board level - Cost of cloth by direct proportion, extending NCERT Example 1

The cost of 5 metres of a particular cloth is Rs 210. Find the cost of 8 metres and of 20 metres of the same cloth.

Step 1. Check the type of proportion. More cloth costs proportionally more money at a fixed rate per metre, so length and cost vary directly; the working relation is $\dfrac{x_1}{y_1} = \dfrac{x_2}{y_2}$ , where $x$ is length in metres and $y$ is cost in rupees.

Step 2. Find the cost of 8 metres. With $x_1=5, y_1=210, x_2=8$ : $\dfrac{5}{210} = \dfrac{8}{y_2}$ , so $y_2 = \dfrac{8\times210}{5} = 336$ .

Step 3. Find the cost of 20 metres, reusing the same constant ratio. With $x_1=5, y_1=210, x_3=20$ : $y_3 = \dfrac{20\times210}{5} = 840$ .

Step 4. Sanity-check using the constant $k$ . Here $k = x/y = 5/210 = 1/42$ , so the rate is Rs 42 per metre; indeed $8\times42=336$ and $20\times42=840$ , matching both answers. The cost of 8 metres is Rs 336, and the cost of 20 metres is Rs 840.

Board level - Workers and time by inverse proportion, NCERT Example 9

If 15 workers can build a wall in 48 hours, how many workers are needed to build the same wall in 30 hours?

Step 1. Check the type of proportion. Finishing the same wall in fewer hours needs more workers, so the number of hours and the number of workers vary inversely; the working relation is $x_1y_1 = x_2y_2$ , where $x$ is hours and $y$ is workers.

Step 2. Set up the equation with $x_1=48, y_1=15, x_2=30$ : $48\times15 = 30\times y_2$ .

Step 3. Solve for $y_2$ : $y_2 = \dfrac{48\times15}{30} = \dfrac{720}{30} = 24$ .

Step 4. Sanity-check: reducing the time from 48 hours to 30 hours (a factor of $48/30 = 1.6$ ) should scale the workers up by the same factor, $15\times1.6=24$ , confirming the answer without redoing the division. 24 workers are required to finish the wall in 30 hours.

Higher-order thinking (HOTS / olympiad-style) - a canal dug by a changing number of workers over a changing length

18 workers, all working at the same steady rate, can dig a 100 m canal in 6 days. If only 12 workers are available, and the canal now needs to be 150 m long, how many days will they take, assuming every worker digs at the same rate as before?

Step 1. Recognise that this problem mixes both kinds of proportion at once — a situation NCERT's single-relation examples do not show directly. The total amount of digging work needed is directly proportional to the length of canal (more length needs more total work), while the number of days needed for a fixed amount of work is inversely proportional to the number of workers (more workers finish faster). The safe way through is to first find the total work in a fixed unit, worker-days, and then split it up.

Step 2. Find the total work needed for the original 100 m canal in worker-days: 18 workers for 6 days is $18\times6=108$ worker-days of work to dig 100 m.

Step 3. Use direct proportion to scale the work to the new 150 m length. Work needed is directly proportional to length, so $\dfrac{108}{100} = \dfrac{w}{150}$ , giving $w = \dfrac{108\times150}{100} = 162$ worker-days for the 150 m canal.

Step 4. Use inverse proportion to convert the 162 worker-days into a number of days for 12 workers. Number of days and number of workers vary inversely for a fixed total work, so $12 \times d = 162$ , giving $d = \dfrac{162}{12} = 13.5$ .

The 12 workers will take 13.5 days, that is, 13 days and half a day, to dig the 150 m canal. Timing note: budget 3 to 4 minutes for a combined problem like this. The one idea that saves time is converting to a single steady unit — worker-days, or man-days — the moment you see both length and workforce changing together; trying to apply one proportion rule directly to two changing quantities at once is the usual way students get stuck or invert a ratio by mistake.

Common traps

The most frequent slip is applying the direct-proportion rule $x_1/y_1=x_2/y_2$ to a situation that is actually inverse, or the reverse — always decide the type from the situation before writing any equation. Students also forget to match units before setting up a ratio or product — mixing kilograms with grams, or hours with minutes, gives a wrong constant $k$ ; convert everything to one consistent unit first, exactly as NCERT converts two and a half kilograms to 2500 grams before comparing with sheets weighing 40 grams. Another trap appears in map-scale problems: the scale ratio (map distance to actual distance) must be kept in the same order throughout, or the answer comes out inverted, giving an impossibly tiny or huge actual distance. Finally, in mixed problems like the HOTS example above, jumping straight from workers and length to days without first fixing the total work in worker-days very often produces the reciprocal of the correct answer.

Common trapThe single biggest misconception: assuming increasing or decreasing together always means proportional

Seeing two quantities rise together, or one rise while the other falls, is not by itself proof of direct or inverse proportion — it only becomes proportion when the ratio (for direct) or the product (for inverse) of every pair of corresponding values stays exactly fixed. A person's age and their mother's age both increase every year, which looks like direct proportion at first glance, but the ratio of their ages keeps changing as time passes (a baby and a 30-year-old mother have a very different ratio than a 20-year-old and a 50-year-old mother), so the two are in neither direct nor inverse proportion at all. Before writing $x_1/y_1=x_2/y_2$ or $x_1y_1=x_2y_2$ for any new problem, always test the constant with at least two pairs of values from the given information, not just one.

You should be able to…

State the definitions of direct proportion ( $x/y=k$ ) and inverse proportion ( $xy=k$ ) precisely, including what stays constant in each case.
Decide, from the wording of a real situation, whether two quantities are in direct proportion, inverse proportion, or neither, before writing any equation.
Apply the cross-multiplication rule $x_1/y_1 = x_2/y_2$ to find an unknown value in a direct-proportion problem.
Apply the product rule $x_1y_1 = x_2y_2$ , and its ratio form $x_1/x_2 = y_2/y_1$ , to find an unknown value in an inverse-proportion problem.
Sketch and interpret the graph of a direct proportion (a straight line through the origin) and of an inverse proportion (a falling curve that never touches the axes).
Solve map-scale problems by treating map distance and actual distance as a direct proportion.
Solve work-and-time and pipes-and-tanks style problems by treating workers or pipes and time as an inverse proportion.
Convert all quantities to the same unit before setting up a ratio or a product, to avoid a wrong constant of proportionality.
Combine direct and inverse proportion in a single multi-step problem, such as work that depends on both the length of a job and the number of workers.

Ask the tutor: Want to try a few more mixed direct-and-inverse word problems together, step by step, until spotting the right rule becomes automatic?

How do I decide whether a word problem is direct proportion or inverse proportion

Read the situation, not the numbers, first. Ask yourself: if one quantity increases, does the other quantity also increase in step, or does it decrease? If more of the first quantity naturally needs more of the second — more articles bought costs more money, more cloth costs more rupees, more petrol covers more distance — it is direct proportion, and the ratio $x/y$ stays constant. If more of the first quantity naturally needs less of the second — more workers finish a job faster, higher speed takes less time, a higher price buys fewer items with the same budget — it is inverse proportion, and the product $xy$ stays constant. Only after deciding the type should you write the equation and solve.

Why does the direct-proportion rule use x1 over y1 equals x2 over y2, but the inverse-proportion rule swaps the y values around

This comes straight from the two definitions. Direct proportion means the ratio $x/y$ is always the same constant $k$ , so equating that ratio for two pairs naturally gives $x_1/y_1 = x_2/y_2$ , with matching subscripts on each side. Inverse proportion means the product $xy$ is always the same constant $k$ , so equating $x_1y_1$ with $x_2y_2$ and then dividing both sides by $x_2$ and by $y_1$ swaps which y goes with which x, giving $x_1/x_2 = y_2/y_1$ . The swap is not a typing mistake — it is a direct algebraic consequence of starting from a product instead of a ratio, and it is exactly why mixing the two rules up gives a wrong answer.

Is saying x is proportional to y the same as saying x equals k times y

For direct proportion, yes — $x \propto y$ and $x = ky$ mean exactly the same thing, with $k$ being the fixed constant of proportionality, and $k$ can be found from any single known pair of corresponding values by computing $k = x/y$ . Be careful, though, with inverse proportion: there $x \propto \dfrac{1}{y}$ means $x = \dfrac{k}{y}$ , which is the same as $xy = k$ , not $x=ky$ . The symbol $\propto$ always needs you to check which kind of proportionality is meant — direct, using $y$ itself, or inverse, using $1/y$ — before you can write down the correct equation.

Can two quantities be neither directly nor inversely proportional to each other

Yes, and this happens often. Height and weight of different people generally increase together in a rough sense, but not in any fixed ratio — a taller person is not always proportionally heavier by the same constant $k$ , so height and weight are in neither direct nor inverse proportion. Similarly, a tree's height and the number of leaves on it have no fixed relationship at all. The test is strict: unless the ratio $x/y$ or the product $xy$ stays genuinely fixed across every pair of values, do not apply either proportion rule, even if the quantities appear to move together or oppositely in general.