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Ratio and Proportion — chapter notes

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1. Ratio and its kinds

A ratio a:ba : b (b≠0b \ne 0) compares two quantities of the same kind; it is unchanged when both terms are multiplied or divided by the same non-zero number, so always give it in lowest terms.

From a:ba : b (with a,b>0a, b > 0):

  • duplicate ratio a2:b2a^2 : b^2; triplicate ratio a3:b3a^3 : b^3;
  • sub-duplicate ratio a:b\sqrt{a} : \sqrt{b}; sub-triplicate ratio a3:b3\sqrt[3]{a} : \sqrt[3]{b};
  • reciprocal (inverse) ratio b:ab : a;
  • compound ratio of a:ba : b and c:dc : d: ac:bdac : bd.

Example: the compound ratio of the duplicate ratio of 3:43 : 4 and the sub-duplicate ratio of 16:2516 : 25 is (9×4):(16×5)=9:20(9 \times 4) : (16 \times 5) = 9 : 20.

Six kinds of ratio built from $a : b$.
Kind of ratioOf a:ba : bExample
Duplicatea2:b2a^2 : b^22:5→4:252 : 5 \to 4 : 25
Triplicatea3:b3a^3 : b^32:5→8:1252 : 5 \to 8 : 125
Sub-duplicatea:b\sqrt{a} : \sqrt{b}16:49→4:716 : 49 \to 4 : 7
Sub-triplicatea3:b3\sqrt[3]{a} : \sqrt[3]{b}8:125→2:58 : 125 \to 2 : 5
Reciprocalb:ab : a2:5→5:22 : 5 \to 5 : 2
Compound (with c:dc : d)ac:bdac : bd2:52 : 5 and 3:4→6:20=3:103 : 4 \to 6 : 20 = 3 : 10
Find each ratio first, then multiply term by term.
StepResult
Duplicate ratio of 3:43 : 49:169 : 16
Sub-duplicate ratio of 16:2516 : 254:54 : 5
Compound ratio(9×4):(16×5)=36:80(9 \times 4) : (16 \times 5) = 36 : 80
Lowest terms (divide by 44)9:209 : 20
Sides $2 : 3$, areas $4 : 9$: areas are in the **duplicate ratio** of the sides.

Formulas

  • Compound ratio
    (a:b) compounded with (c:d)=ac:bd(a : b) \text{ compounded with } (c : d) = ac : bd

Key terms

Duplicate ratio

The ratio of the squares of the terms: a2:b2a^2 : b^2.

Sub-duplicate ratio

The ratio of the square roots of the terms: a:b\sqrt{a} : \sqrt{b}.

Compound ratio

The product of two or more ratios, term by term: ac:bdac : bd.

2. Proportion and the fourth proportional

Four numbers a,b,c,da, b, c, d are in proportion, written a:b=c:da : b = c : d or a:b::c:da : b :: c : d, when ab=cd\dfrac{a}{b} = \dfrac{c}{d}.

  • aa and dd are the extremes; bb and cc are the means.
  • Test: product of extremes == product of means, ad=bcad = bc.
  • dd is the fourth proportional to a,b,ca, b, c: d=bcad = \dfrac{bc}{a}.

Example: 1.2,2,4.8,81.2, 2, 4.8, 8 are in proportion since 1.2×8=9.6=2×4.81.2 \times 8 = 9.6 = 2 \times 4.8; the fourth proportional to 5,7,155, 7, 15 is 7×155=21\dfrac{7 \times 15}{5} = 21.

In $4 : 7 = 12 : 21$, extremes $\times$ extremes $=$ means $\times$ means.
TermNameValue
Firstextreme44
Secondmean77
Thirdmean1212
Fourthextreme2121
Product of extremes4×21=844 \times 21 = 84
Product of means7×12=847 \times 12 = 84
Pairs in the same ratio lie on one line through the origin.
Equal cross-products mean the numbers are in proportion.
TestWorkingResult
Product of extremes1.2×8=9.61.2 \times 8 = 9.6
Product of means2×4.8=9.62 \times 4.8 = 9.6equal, so in proportion
Fourth proportional to 5,7,155, 7, 155:7=15:x5 : 7 = 15 : x, x=7×155x = \dfrac{7 \times 15}{5}2121

Formulas

  • Test for proportion
    a:b=c:d  ⟺  ad=bca : b = c : d \iff ad = bc

Key terms

Extremes

The first and fourth terms of a proportion.

Means

The second and third terms of a proportion.

Fourth proportional

The number xx with a:b=c:xa : b = c : x, that is x=bcax = \dfrac{bc}{a}.

3. Continued proportion, mean and third proportionals

a,b,ca, b, c are in continued proportion when a:b=b:ca : b = b : c, so b2=acb^2 = ac.

  • b=acb = \sqrt{ac} is the mean proportional between aa and cc.
  • c=b2ac = \dfrac{b^2}{a} is the third proportional to aa and bb.
  • a,b,c,da, b, c, d are in continued proportion when a:b=b:c=c:da : b = b : c = c : d; then a:d=a3:b3a : d = a^3 : b^3 (the triplicate ratio of a:ba : b).

Example: the mean proportional between 0.320.32 and 0.50.5 is 0.16=0.4\sqrt{0.16} = 0.4; the third proportional to 55 and 7.57.5 is 56.255=11.25\dfrac{56.25}{5} = 11.25.

Continued proportion: one equation, two formulas.
Equal areas: $12 \times 3 = 6 \times 6$, so $6$ is the mean proportional between $12$ and $3$.
Both answers come from $b^2 = ac$.
FindWorkingAnswer
Mean proportional between 0.320.32 and 0.50.50.32×0.5=0.16\sqrt{0.32 \times 0.5} = \sqrt{0.16}0.40.4
Third proportional to 55 and 7.57.57.525=56.255\dfrac{7.5^2}{5} = \dfrac{56.25}{5}11.2511.25
Four numbers in continued proportion: a common ratio of $1.5$.

Formulas

  • Mean proportional
    b=acb = \sqrt{ac}
  • Third proportional
    c=b2ac = \dfrac{b^2}{a}

Key terms

Continued proportion

Three numbers with a:b=b:ca : b = b : c, i.e. b2=acb^2 = ac.

Mean proportional

The middle term ac\sqrt{ac} of a:b=b:ca : b = b : c.

Third proportional

The number c=b2ac = \dfrac{b^2}{a} with a:b=b:ca : b = b : c.

4. Invertendo and alternendo

If a:b=c:da : b = c : d, then

  • invertendo: b:a=d:cb : a = d : c (invert both ratios);
  • alternendo: a:c=b:da : c = b : d (interchange the means).

Each new proportion still has equal cross-products, so it is a true proportion.

Each new proportion still has equal cross-products.
PropertyProportionCross-products
Given7:4=21:127 : 4 = 21 : 127×12=84=4×217 \times 12 = 84 = 4 \times 21
Invertendo4:7=12:214 : 7 = 12 : 214×21=84=7×124 \times 21 = 84 = 7 \times 12
Alternendo7:21=4:127 : 21 = 4 : 127×12=84=21×47 \times 12 = 84 = 21 \times 4

Key terms

Invertendo

a:b=c:d⇒b:a=d:ca : b = c : d \Rightarrow b : a = d : c.

Alternendo

a:b=c:d⇒a:c=b:da : b = c : d \Rightarrow a : c = b : d.

5. Componendo, dividendo, and componendo and dividendo

If a:b=c:da : b = c : d, then

  • componendo: a+bb=c+dd\dfrac{a + b}{b} = \dfrac{c + d}{d} (add 11 to both sides);
  • dividendo: a−bb=c−dd\dfrac{a - b}{b} = \dfrac{c - d}{d} (subtract 11 from both sides);
  • componendo and dividendo: a+ba−b=c+dc−d\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d} (divide the two results).

The steps are reversible: if a+ba−b=c+dc−d\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}, then a:b=c:da : b = c : d.

The new proportions from $7 : 4 = 21 : 12$.
PropertyProportionCross-products
Componendo11:4=33:1211 : 4 = 33 : 1211×12=132=4×3311 \times 12 = 132 = 4 \times 33
Dividendo3:4=9:123 : 4 = 9 : 123×12=36=4×93 \times 12 = 36 = 4 \times 9
Componendo and dividendo11:3=33:911 : 3 = 33 : 911×9=99=3×3311 \times 9 = 99 = 3 \times 33
Why componendo and dividendo works.

Formulas

  • Componendo and dividendo
    ab=cd  ⟺  a+ba−b=c+dc−d\dfrac{a}{b} = \dfrac{c}{d} \iff \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}

Key terms

Componendo

ab=cd⇒a+bb=c+dd\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow \dfrac{a + b}{b} = \dfrac{c + d}{d}.

Dividendo

ab=cd⇒a−bb=c−dd\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow \dfrac{a - b}{b} = \dfrac{c - d}{d}.

Componendo and dividendo

ab=cd⇒a+ba−b=c+dc−d\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}, and conversely.

6. Using the properties: equations, ratios and proofs

  • Equations: when one side is (sum) ÷\div (difference), apply componendo and dividendo to both sides. Example: 2x+92x−9=53\dfrac{2x + 9}{2x - 9} = \dfrac{5}{3} gives 4x18=4\dfrac{4x}{18} = 4, so x=18x = 18.
  • Square roots: x+4+x−4x+4−x−4=2\dfrac{\sqrt{x + 4} + \sqrt{x - 4}}{\sqrt{x + 4} - \sqrt{x - 4}} = 2 gives x+4x−4=3\dfrac{\sqrt{x + 4}}{\sqrt{x - 4}} = 3; square and solve: x=5x = 5.
  • Finding a ratio: 5x+3y5x−3y=73\dfrac{5x + 3y}{5x - 3y} = \dfrac{7}{3} gives 10x6y=104\dfrac{10x}{6y} = \dfrac{10}{4}, so x:y=3:2x : y = 3 : 2.
  • Proofs (k-method): put ab=cd=k\dfrac{a}{b} = \dfrac{c}{d} = k, so a=bka = bk, c=dkc = dk (or b=akb = ak, c=ak2c = ak^2 for continued proportion), and simplify each side separately.
The curve meets the line $y = \dfrac{5}{3}$ at $x = 18$.
The coefficients $5$ and $3$ must be divided out at the end.
StepWorking
Componendo and dividendo10x6y=7+37−3=104\dfrac{10x}{6y} = \dfrac{7 + 3}{7 - 3} = \dfrac{10}{4}
Simplifyxy=104×610=32\dfrac{x}{y} = \dfrac{10}{4} \times \dfrac{6}{10} = \dfrac{3}{2}
Answerx:y=3:2x : y = 3 : 2
Check with x=3x = 3, y=2y = 215+615−6=219=73\dfrac{15 + 6}{15 - 6} = \dfrac{21}{9} = \dfrac{7}{3}
Square only after the roots stand alone.
StepWorking
Componendo and dividendox+4x−4=2+12−1=3\dfrac{\sqrt{x + 4}}{\sqrt{x - 4}} = \dfrac{2 + 1}{2 - 1} = 3
Square both sidesx+4x−4=9\dfrac{x + 4}{x - 4} = 9, so x+4=9x−36x + 4 = 9x - 36
Solve8x=408x = 40, so x=5x = 5
Check9+19−1=42=2\dfrac{\sqrt{9} + \sqrt{1}}{\sqrt{9} - \sqrt{1}} = \dfrac{4}{2} = 2
A numerical check is not a proof, but it catches slips.
ValuesLHS (a+b)2(c+d)2\dfrac{(a + b)^2}{(c + d)^2}RHS abcd\dfrac{ab}{cd}
b=3b = 3, d=5d = 5, k=2k = 2: a=6a = 6, c=10c = 1081225=925\dfrac{81}{225} = \dfrac{9}{25}1850=925\dfrac{18}{50} = \dfrac{9}{25}
b=6b = 6, d=9d = 9, k=13k = \dfrac{1}{3}: a=2a = 2, c=3c = 364144=49\dfrac{64}{144} = \dfrac{4}{9}1227=49\dfrac{12}{27} = \dfrac{4}{9}

Key terms

k-method

Writing every term through a common ratio kk (e.g. a=bka = bk, c=dkc = dk) to prove a relation.

7. Direct applications of proportion

Write the proportion, use product of extremes == product of means, solve, and check in the original ratios.

  • Scales and recipes: 33 cm for 2020 km means 7.57.5 cm stands for 20×7.53=50\dfrac{20 \times 7.5}{3} = 50 km.
  • Changing ratios: numbers 2k2k and 5k5k with 33 subtracted from each in the ratio 1:41 : 4 give k=3k = 3.
  • Making numbers proportional: adding 33 to each of 3,6,7,123, 6, 7, 12 gives 6,9,10,156, 9, 10, 15.
  • Areas and volumes: areas of similar figures are in the duplicate ratio, and volumes of cubes in the triplicate ratio, of their lengths.

Not in the ICSE 2027 syllabus: applications of proportion beyond direct simple applications; this chapter keeps to direct ones.

Five steps for a direct application.
The $x^2$ terms always cancel.
StepWorking
Equation(3+x)(12+x)=(6+x)(7+x)(3 + x)(12 + x) = (6 + x)(7 + x)
Expand36+15x+x2=42+13x+x236 + 15x + x^2 = 42 + 13x + x^2
Solve2x=62x = 6, so x=3x = 3
Check6×15=90=9×106 \times 15 = 90 = 9 \times 10
Only $k = 3$ gives $1 : 4$: the numbers are $6$ and $15$.
kkNumbers 2k,5k2k, 5kAfter subtracting 33Equal to 1:41 : 4?
224,104, 101:71 : 7no
336,156, 153:123 : 12yes
448,208, 205:175 : 17no
5510,2510, 257:227 : 22no

Key terms

Direct application

A problem solved by writing one proportion and using product of extremes = product of means.

Exam tips

How marks are awarded

  • Proportionals and kinds of ratio (1–2 marks): proportion or rule written 1; answer in lowest terms 1.
  • Componendo and dividendo equation (3–4 marks): property applied to both sides 1; simplified 1; answer 1; check 1.
  • Proof (3–4 marks): substitution a=bka = bk, c=dkc = dk 1; each side simplified 1–2; conclusion 1.
  • 10-mark structured question: parts (a), (b), (c) usually 3 + 3 + 4; show all working.

Common mistakes

  1. Doubling the terms instead of squaring them for a duplicate ratio.
  2. Using the average instead of the mean proportional ac\sqrt{ac}.
  3. Writing the third proportional in the wrong order (b2÷ab^2 \div a, not a2÷ba^2 \div b).
  4. Calling alternendo invertendo, or applying componendo and dividendo to one side only.
  5. Forgetting the negative square root when a ratio of squares appears.
  6. Leaving coefficients in when finding x:yx : y, or leaving a ratio unsimplified.