Direct proportion

Buy 8 notebooks and pay ₹360. Buy 13, and the shopkeeper does not need a new price list — the cost simply scales with the count, in exactly the same ratio every time. That single fact, "the ratio stays fixed," is the whole of direct proportion: whenever doubling one quantity doubles the other, tripling triples it, and so on, a constant $k$ links them by $y = kx$ . It is the mathematical backbone of recipe scaling, map enlargements, constant-speed travel, and any "more in, more out at a fixed rate" situation. This section builds the defining ratio test, the unitary method and the equal-ratios method for solving such problems, and sharpens the boundary against its near-twin, inverse proportion, which uses a constant product instead.

The core insight

Two quantities are in direct proportion exactly when their ratio $y/x$ stays equal to a fixed constant $k$ , so $y = kx$ lets either one be found from the other.

The core idea

Two quantities $x$ and $y$ vary together. They are said to be in direct proportion — or to vary directly — if their ratio never changes as both quantities change:

\dfrac{y}{x} = k

where: $x$ and $y$ are the two related quantities and $k$ is the constant of proportionality, a single fixed number for a given situation (a unit price, a speed, a scale factor). Rearranged, this is the more familiar form

y = kx

which says $y$ is always $k$ times $x$ : double $x$ and $y$ doubles, halve $x$ and $y$ halves.

Testing a table of values. Given several pairs $(x, y)$ , compute $y/x$ for each pair. If every pair gives the same number, the quantities are in direct proportion and that number is $k$ . If even one ratio differs, they are not — "both increasing together" is not enough on its own; the ratio specifically must hold constant.

Two methods for solving problems. The unitary method finds the value of $y$ for a single unit of $x$ first, then scales to whatever is needed. The equal-ratios method instead sets up two corresponding pairs $(x_1, y_1)$ and $(x_2, y_2)$ from the same proportion and cross-multiplies:

\dfrac{y_1}{x_1} = \dfrac{y_2}{x_2} \quad\Longrightarrow\quad x_1 y_2 = x_2 y_1

where: $(x_1, y_1)$ is a known corresponding pair and $(x_2, y_2)$ is the pair with one unknown value to solve for. Both methods always agree, because both rest on the same fixed ratio $k$ .

Similar figures give a geometric flavour of the same idea: if a photograph is enlarged so every length grows by the same factor $k$ , then new-length/old-length $= k$ for every pair of corresponding sides, exactly as $y/x = k$ for any directly proportional pair.

Why this matters beyond one topic. Direct proportion is the simplest possible relationship between two changing quantities, and it is the model you reach for first whenever a real situation has a fixed "rate": a fixed price per item, a fixed speed, a fixed exchange rate, a fixed scale on a map. Recognising that a situation fits $y = kx$ instantly gives you a way to predict any new value once one pair is known — which is exactly why it appears so often in word problems about shopping, travel, and drawing to scale.

Figure

A straight line through the origin on a graph of cost against number of notebooks, with two marked points at eight notebooks three hundred sixty rupees and thirteen notebooks five hundred eighty five rupees, showing the constant ratio forty five rupees per notebook

Derivation ASetting up and solving y = kx from two known pairs

The goal is to show, starting only from the defining condition that the ratio $y/x$ is constant, how a single known pair of values $(x_1, y_1)$ lets you predict $y_2$ for any new $x_2$ — the algebraic engine behind both the unitary method and the cross-multiplication method used throughout this topic.

Step 1. Write the defining condition. Since $x$ and $y$ are in direct proportion, $\dfrac{y}{x} = k$ for every corresponding pair, so in particular $\dfrac{y_1}{x_1} = k$ and $\dfrac{y_2}{x_2} = k$ .

Step 2. Equate the two expressions for $k$ , since both equal the same constant: $\dfrac{y_1}{x_1} = \dfrac{y_2}{x_2}$ .

Step 3. Cross-multiply to clear the fractions: $x_1 y_2 = x_2 y_1$ , so $y_2 = \dfrac{x_2 y_1}{x_1}$ . This single formula solves for any unknown corresponding value once one pair $(x_1, y_1)$ and one new $x_2$ are known.

Step 4. Recover the unitary method as a special case. Put $x_1 = 1$ : then $y_2 = x_2 y_1$ , i.e. "find the value for one unit ( $y_1$ when $x_1 = 1$ ), then multiply by $x_2$ ." The unitary method is simply this general formula applied with the first quantity set to a single unit — both routes always give the same answer because both come from the one constant ratio $k$ .

Worked example

A car travels $210$ km in $3$ hours at constant speed. Find (a) the distance covered in $7$ hours, and (b) the time needed to travel $560$ km, at the same speed.

Step 1 — Recognise the proportion and find $k$ . Distance and time are directly proportional at constant speed, with $k$ equal to the speed:

k = \dfrac{210}{3} = 70\ \text{km/h}

Step 2 — Distance in 7 hours (unitary method). One hour covers $70$ km, so $7$ hours cover

70 \times 7 = 490\ \text{km}

Step 3 — Time for 560 km (equal ratios). Using $\dfrac{210}{3} = \dfrac{560}{t}$ , cross-multiply: $210\,t = 3 \times 560 = 1680$ , so $t = \dfrac{1680}{210} = 8$ hours.

Step 4 — Verify. Check the ratio stays at $70$ throughout: $\dfrac{210}{3} = 70$ , $\dfrac{490}{7} = 70$ , $\dfrac{560}{8} = 70$ . All three pairs give the same constant, confirming the answers.

The car covers 490 km in 7 hours, and takes 8 hours to cover 560 km.

Competitive stretch

The same car is driven a further $2$ hours beyond the $560$ km point, still at $70$ km/h. Total time is $8 + 2 = 10$ hours, so total distance is $70 \times 10 = \mathbf{700\ km}$ — a quick reminder that once $k$ is known, any new distance–time pair can be generated instantly, without repeating the whole method.

Common traps

Assuming "both increase" means direct proportion. Only a constant ratio qualifies — always check $y/x$ across several pairs, not just that both quantities are growing.
Cross-multiplying with mismatched quantities, e.g. writing $\dfrac{8}{13} = \dfrac{x}{360}$ instead of keeping notebooks-with-notebooks and cost-with-cost on matching sides.
Skipping the unit-rate step and guessing the scale factor mentally, which invites arithmetic slips when the ratio is not a whole number.
Confusing direct with inverse proportion. Direct proportion keeps the ratio $y/x$ fixed; inverse proportion (a later topic) keeps the product $xy$ fixed — state the defining condition first to avoid mixing them under exam pressure.

Ask the tutor: Want to practise telling direct from inverse proportion with a few mixed word problems, or see the unitary method worked through a trickier non-whole-number ratio?

How is direct proportion different from a simple ratio?

A ratio compares two specific numbers, like $8 : 360$ . Direct proportion is a statement about how pairs of numbers behave together — it says the ratio $y/x$ stays the same constant $k$ for every corresponding pair you could pick, not just the one you started with. So $8 : 360$ becomes useful for prediction only once you know it stays fixed as the quantities change; that constancy is what "direct proportion" adds on top of a single ratio.

Do I have to use the unitary method, or can I always cross-multiply instead?

Either method always gives the same answer, because both come from the same constant ratio $k$ — the derivation above shows the unitary method is really cross-multiplication with the first pair fixed at one unit. Use whichever feels faster for the numbers in front of you: unitary method when finding "cost per unit" is natural (like ₹45 per notebook), cross-multiplication when the numbers do not divide evenly and you would rather avoid a messy unit value.

How do I quickly tell if a word problem is direct proportion and not inverse?

Ask what happens physically as one quantity grows. If more of one thing genuinely requires or produces proportionally more of the other at a fixed rate — more notebooks costs more money, more hours at constant speed covers more distance — it is direct. If instead more of one thing means less is needed of the other for the same fixed total, like more workers taking less time to finish a job, that is inverse proportion, a different topic with a constant product $xy$ instead of a constant ratio.

Direct proportion

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads