Inverse Proportion: y = k/x

y = k/x and its powers: finding k and the formula, using it both ways, and inverse proportion to x², x³ and √x (Higher).

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024 · 37 real past-paper questions below

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What Inverse Proportion Means (Higher)

When one quantity doubles, the other halves. More workers, fewer days.

Example

4 workers take 6 days. 8 workers take 3 days. The product stays the same: 4 × 6 = 8 × 3 = 24.

Rule

$y \propto \frac{1}{x}$ means $y = \frac{k}{x}$ for some number k. You find k from one pair of values.

Finding the Formula

Most questions give you a pair of values and ask for the formula. Read carefully which power of x goes on the bottom.

Worked example

1
$y = \frac{k}{x}$ — “Inversely proportional to x” puts x on the bottom.
2
$6 = \frac{k}{15}$ — Put in the pair we are given: y = 6 when x = 15.
3
$k = 6 \times 15 = 90$ — Multiply both sides by 15 to get k on its own.
4
$y = \frac{90}{x}$ — Write the formula with k filled in.
5
$y = \frac{90}{5} = 18$ — Now use it: x = 5. A smaller x gives a bigger y, as it should for inverse proportion.

Worked example

1
$I \propto \frac{1}{d ^{2}} \Rightarrow I = \frac{k}{d ^{2}}$ — “Inversely proportional to the square of d” puts d² on the bottom.
2
$12.5 = \frac{k}{4 ^{2}}$ — Put in the pair we are given: I = 12.5 when d = 4.
3
$12.5 = \frac{k}{16}$ — Work out 4² = 16.
4
$k = 12.5 \times 16 = 200$ — Multiply both sides by 16 to get k on its own.
5
$I = \frac{200}{d ^{2}}$ — Write the formula with k filled in.

Common mistake

Stopping at k = 200. The question asks for a formula: I = 200/d².

Key idea

Show every line. Most of the marks are for the working, and finding k earns its own mark.

Using the Formula

The next part gives one value and asks for the other. Use the formula $I = \frac{200}{d ^{2}}$ .

Worked example

1
$I = \frac{200}{2 ^{2}}$ — Find I when d = 2. Put d = 2 in.
2
$I = \frac{200}{4} = 50$ — Work out 2² = 4, then divide.
3
$3.125 = \frac{200}{d ^{2}}$ — Now find d when I = 3.125. Put I in.
4
$d^{2} = 200 \div 3.125$ — d² and I swap places, so d² = 200 ÷ I.
5
$d^{2} = 64$ — Work out 200 ÷ 3.125 = 64.
6
$d = 8$ — Take the square root. A length is positive, so ignore −8.

Negative numbers need care. Keep the bracket round a negative value.

Worked example

1
$f = \frac{k}{h ^{2}}$ — f is inversely proportional to the square of h.
2
$- 2 = \frac{k}{3 ^{2}}$ — Put in f = −2 and h = 3.
3
$k = - 2 \times 9 = - 18$ — Multiply both sides by 9. A negative times a positive is negative.
4
$f = \frac{- 18}{h ^{2}}$ — Write the formula with k filled in.
5
$f = \frac{- 18}{( - 6 ) ^{2}}$ — Now put in h = −6. Keep it in brackets.
6
$(- 6)^{2} = 36$ — A negative number squared is positive.
7
$f = \frac{- 18}{36} = - \frac{1}{2}$ — Divide. The top is negative and the bottom is positive, so f is negative.

Cubes and Square Roots

The same method works for any power. Only the bottom of the fraction changes.

Worked example

1
$y = \frac{k}{x ^{3}}$ — y is inversely proportional to x³.
2
$5 = \frac{k}{2 ^{3}}$ — Put in y = 5 and x = 2.
3
$5 = \frac{k}{8}$ — Work out 2³ = 2 × 2 × 2 = 8.
4
$k = 5 \times 8 = 40$ — Multiply both sides by 8 to get k.
5
$320 = \frac{40}{x ^{3}}$ — Now find x when y = 320.
6
$x^{3} = 40 \div 320 = 0.125$ — x³ and 320 swap places, so x³ = 40 ÷ 320.
7
$x = 3 0.125 = 0.5$ — Take the cube root. Check: 0.5 × 0.5 × 0.5 = 0.125.

Worked example

1
$y = \frac{k}{x}$ — “Inversely proportional to the square root of x” puts √x on the bottom.
2
$5 = \frac{k}{4}$ — Put in y = 5 and x = 4.
3
$5 = \frac{k}{2}$ — Work out √4 = 2.
4
$k = 5 \times 2 = 10$ — Multiply both sides by 2 to get k.
5
$y = \frac{10}{25}$ — Use it: put in x = 25.
6
$y = \frac{10}{5} = 2$ — Work out √25 = 5, then divide.

Inverse with a Bracket

The bracket trick appears here too: “y is inversely proportional to √(x + 4)”. The whole bracket goes on the bottom.

Formula

y = \frac{k}{( x + a ) ^{n}}

Here $k$ is the fixed number you find, $a$ is the number added to $x$ in the question (it can be negative), and $n$ is the power.

Worked example

1
$y = \frac{k}{x + 4}$ — The whole bracket is under the root, on the bottom.
2
$2 = \frac{k}{12 + 4}$ — Put in x = 12 and y = 2. Add under the root first.
3
$2 = \frac{k}{4}$ — Work out √16 = 4.
4
$k = 2 \times 4 = 8$ — Multiply both sides by 4 to get k.
5
$y = \frac{8}{x + 4}$ — Write the formula with k filled in.
6
$y = \frac{8}{60 + 4}$ — Now put in x = 60.
7
$y = \frac{8}{8} = 1$ — Work out √64 = 8, then divide.

Now go the other way, finding x. Watch for a hint such as “x > 2”.

Worked example

1
$y = \frac{k}{( x - 2 ) ^{2}}$ — y is inversely proportional to the square of (x − 2).
2
$3 = \frac{k}{( 7 - 2 ) ^{2}}$ — Put in y = 3 and x = 7.
3
$3 = \frac{k}{25}$ — Work out (7 − 2)² = 5² = 25.
4
$k = 3 \times 25 = 75$ — Multiply both sides by 25.
5
$12 = \frac{75}{( x - 2 ) ^{2}}$ — Now find x when y = 12.
6
$(x - 2)^{2} = 75 \div 12 = 6.25$ — (x − 2)² and 12 swap places, so (x − 2)² = 75 ÷ 12.
7
$x - 2 = 2.5 or x - 2 = - 2.5$ — A square root gives a positive and a negative answer.
8
$x = 4.5 or x = - 0.5$ — Add 2 to each side.
9
$x = 4.5$ — The question says x > 2, so reject −0.5.

Common mistake

Putting only the x on the bottom. $y = \frac{8}{x + 4}$ is not $\frac{8}{x} + 4$ . The whole bracket goes on the bottom.

Letters Instead of Numbers (Higher)

Questions can use letters for the given values. Treat each letter like a number and keep it in k.

Worked example

1
$A = \frac{k}{r ^{2}}$ — A is inversely proportional to r². We are told A = 2p when r = 3.
2
$2 p = \frac{k}{3 ^{2}}$ — Put in A = 2p and r = 3.
3
$2 p = \frac{k}{9}$ — Work out 3² = 9.
4
$k = 2 p \times 9 = 18 p$ — Multiply both sides by 9. The p stays in k.
5
$A = \frac{18 p}{r ^{2}}$ — Write the formula with k filled in.

Worked example

1
$y = \frac{k}{x}$ — y is inversely proportional to √x. We are told y = c³ when x = c², and c is positive.
2
$c^{3} = \frac{k}{c ^{2}}$ — Put in y = c³ and x = c².
3
$c^{3} = \frac{k}{c}$ — √(c²) = c, because c is positive.
4
$k = c^{3} \times c = c^{4}$ — Multiply both sides by c. Add the powers: c³ × c = c⁴.
5
$y = \frac{c ^{4}}{x}$ — Write the formula with k filled in. Check: x = c² gives c⁴ ÷ c = c³.

What Happens to y?

Some questions ask what happens to y when x is halved or multiplied. You do not need k. Put the new x into the formula and compare.

Formula

x \to m x \Rightarrow y \to \frac{y}{m ^{n}}

This is for $y = \frac{k}{x ^{n}}$ . Here $m$ is the number x is multiplied by and $n$ is the power on the bottom.

Worked example

1
$F = \frac{k}{d ^{2}}$ — F is inversely proportional to d². Now d is halved.
2
$d \to \frac{1}{2} d$ — “Halved” means multiplied by ½.
3
$F = \frac{k}{( \frac{1}{2} d ) ^{2}}$ — Put ½d in place of d.
4
$(\frac{1}{2} d)^{2} = \frac{1}{4} d^{2}$ — Square the ½ as well as the d: ½ × ½ = ¼.
5
$F = \frac{k}{\frac{1}{4} d ^{2}} = 4 \times \frac{k}{d ^{2}}$ — Dividing by ¼ is the same as multiplying by 4.
6
$F \times 4$ — So F is multiplied by 4.

Worked example

1
$m = \frac{k}{p}$ — m is inversely proportional to √p. Now p is multiplied by 9.
2
$m = \frac{k}{9 p}$ — Put 9p in place of p.
3
$9 p = 3 p$ — Work out √9 = 3.
4
$m = \frac{1}{3} \times \frac{k}{p}$ — Dividing by 3 is the same as multiplying by ⅓.
5
$m \div 3$ — So m is divided by 3.

Common mistake

Saying F doubles when d is halved. The d is squared, so halving d multiplies F by $2^{2} = 4$ .

Which Method? Read the Question

The words tell you what goes on the bottom.

You see	You write
“inversely … to x”	$y = \frac{k}{x}$
“… to the square of x”	$y = \frac{k}{x ^{2}}$
“… to x³”	$y = \frac{k}{x ^{3}}$
“… to the square root”	$y = \frac{k}{x}$
“directly proportional”	See Direct Proportion
“… to (x + 4)” or “√(x − 2)”	Keep the whole bracket on the bottom
“halved”, “multiplied by 9”	y is divided by mⁿ (m = the multiplier, n = the power)

Key idea

Always find k first, then write the full formula, then use it.

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