Direct and inverse variation

Some pairs of quantities march in step — spend more hours at work, earn more money. Others move in opposite directions — put more workers on a job, and it finishes sooner. Direct and inverse variation are the two simplest, most common patterns connecting two changing quantities, and almost every "if this many, then how many" word problem you meet in school reduces to one of them. The whole skill is deciding, before you touch a calculator, which pattern the situation follows: do the two quantities rise and fall together, or does one grow only as the other shrinks? Once that decision is made, the arithmetic is short — either a constant ratio or a constant product carries you straight to the answer. This section builds that decision-making skill alongside the two working formulas, the unitary and arrow methods used to apply them, and the traps that catch students who reach for a formula before understanding the situation.

The core insight

Two quantities vary directly when their ratio $\dfrac{y}{x}$ stays constant, and vary inversely when their product $xy$ stays constant — spotting which one applies to a word problem, before calculating, is the real skill.

The core idea

Direct variation describes quantities that grow or shrink together, at a fixed rate. If you double one, the other doubles too; halve one, and the other halves.

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

where: $x$ and $y$ are the two varying quantities and $k$ is the constant of variation (the fixed rate connecting them, such as price per metre). Because the ratio never changes, any two pairs of matching values satisfy

\frac{y_1}{x_1} = \frac{y_2}{x_2}

Wages and hours worked, distance travelled and petrol used at a steady rate, and the number of identical items bought and their total cost are everyday examples of direct variation.

Inverse variation describes quantities that move in opposite directions so their product never changes — as one goes up, the other must come down by just enough to keep the product fixed.

xy = k \quad \Longleftrightarrow \quad y = \frac{k}{x}

where: $k$ is again a fixed constant (the total amount of work or resource being shared out), and $x_1y_1 = x_2y_2$ for any two matching pairs of values. The number of workers on a job and the days it takes to finish (same total work), or the number of taps filling a tank and the time taken, are classic inverse-variation situations. The figure below contrasts the two shapes of behaviour side by side: in direct variation the ratio line stays flat, while in inverse variation the product stays flat even though each quantity swings in the opposite direction to the other.

Figure

Two panels comparing direct variation where y over x is constant and inverse variation where x times y is constant

Two solving tools carry these ideas into arithmetic. The unitary method finds the value for a single unit first (cost of 1 metre, work done by 1 worker in 1 day), then scales that single-unit value up or down — this is the natural route for direct variation. The arrow method chains several proportional steps in sequence (5 pens cost so much, so 1 pen costs so much, so 8 pens cost so much), which keeps multi-step problems organised without losing track of the quantities involved.

Derivation ACross-multiplying the direct-variation ratio

Starting from the definition that two quantities vary directly whenever their ratio is a fixed constant, show that the unitary-method shortcut (find the value of one unit, then multiply) and the cross-multiplied ratio equation $y_1 x_2 = y_2 x_1$ are exactly the same rule written two different ways, so either can be used interchangeably to solve a direct-variation word problem.

Step 1. By definition of direct variation, $y = kx$ for some fixed constant $k$ , so every matching pair of values obeys $\dfrac{y}{x} = k$ . Taking two pairs, $(x_1, y_1)$ and $(x_2, y_2)$ , both give the same $k$ :

\frac{y_1}{x_1} = k = \frac{y_2}{x_2}

Step 2. Cross-multiply the equality $\dfrac{y_1}{x_1} = \dfrac{y_2}{x_2}$ to clear the fractions:

y_1 x_2 = y_2 x_1

This single equation lets you solve for any one of the four values once the other three are known — it is the ratio-equation form of direct variation.

Step 3. Now check this matches the unitary method. Find the constant rate per unit: $k = \dfrac{y_1}{x_1}$ (the "value of one unit"). To find $y_2$ for a new $x_2$ , the unitary method multiplies: $y_2 = k \times x_2 = \dfrac{y_1}{x_1} \times x_2$ . Rearranging this gives $y_2 x_1 = y_1 x_2$ , identical to the cross-multiplied equation from Step 2. The unitary method and the ratio equation are the same rule; the ratio equation is just the faster way to write it down when you already have two complete pairs of values.

Worked example

If 12 metres of cloth cost ₹900, find the cost of 18 metres of the same cloth, and then find how many metres could be bought for ₹1,650.

Step 1. Decide the pattern: cost and length vary directly, since more cloth at the same rate costs proportionally more. So $\dfrac{\text{cost}}{\text{length}} = k$ stays fixed.

Step 2. Unitary method for the first part: cost of 1 metre $= \dfrac{900}{12} = ₹75$ . Cost of 18 metres $= 75 \times 18 = ₹1350$ .

Step 3. Check with the ratio equation from the derivation: $\dfrac{900}{12} = \dfrac{x}{18}$ , so $x = \dfrac{900 \times 18}{12} = ₹1350$ . Both methods agree.

Step 4. For the second part, use the same constant rate, ₹75 per metre, in reverse: length $= \dfrac{1650}{75} = 22$ metres. The cost of 18 metres is ₹1,350, and ₹1,650 buys 22 metres of the same cloth.

Competitive stretch

A hostel has food supplies to last 45 days for 40 students. If 10 more students join, how long will the same stock last, assuming everyone eats the same amount daily? Here students and days vary inversely, because the total food consumed (students × days) is fixed: $40 \times 45 = 50 \times d$ , so $1800 = 50d$ and $d = 36$ days. Notice the new number of students is $40 + 10 = 50$ , not $10$ — a common misreading of "10 more join."

Common traps

Assuming every proportion problem is direct. Always ask first: does the second quantity increase or decrease as the first increases? Guessing wrong flips the entire calculation.
Using the direct-variation ratio for an inverse situation, writing $\dfrac{x_1}{x_2} = \dfrac{y_1}{y_2}$ when the correct relation is $x_1y_1 = x_2y_2$ .
Forgetting the constant-rate assumption. Time-and-work and cost-and-quantity problems assume every unit (worker, metre, tap) behaves identically at a steady rate; if the problem says otherwise, simple variation does not apply directly.
Losing track of the running total in multi-step arrow-method problems — always carry the same unit forward consistently rather than restarting the chain partway through.

Ask the tutor: if a word problem has you unsure whether it is direct or inverse variation, describe the real-life situation to the AI tutor and it will help you reason out the pattern before you calculate.

How do I decide direct or inverse without getting confused

Ask what happens physically, not mathematically, first: if the second quantity is being shared out or spread thinner as the first grows (more people sharing one job, more taps sharing one tank), it is inverse. If the second quantity is simply accumulating alongside the first at a steady rate (more items, more total cost), it is direct. Picture the real situation before writing any equation.

Why does the inverse-variation equation use multiplication instead of a ratio

Because what stays fixed in inverse variation is the product of the two quantities (like total work, or total tank volume), not their ratio. Since $xy = k$ is fixed, any two pairs must satisfy $x_1y_1 = x_2y_2$ . Writing a ratio equation like the direct-variation one would silently assume the wrong pattern and give the wrong answer.

Can a problem involve more than two quantities varying together

Not in this chapter — you are expected to handle relationships between exactly two related quantities at a time, using direct or inverse variation on each pair, or chaining several two-quantity steps together with the arrow method. Situations depending on two or more independent quantities at once (joint variation) come later, so if a problem seems to need that, re-read it — it is very likely two ordinary steps in sequence.

Direct and inverse variation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads