Ratio and Proportion Class 10 ICSE — Bansal Chapter 7

Written by Greya LakshmiPublished on 18 September 2026Updated on 28 September 2026

Chapter 7 of the Bansal Class 10 ICSE Mathematics book is titled Ratio and Proportion Including Properties and Uses. It covers ratio and proportion class 10 icse topics in detail. You will study the definition of ratio, increase and decrease in a ratio, and the six composed ratio types. The chapter then defines proportion and continued proportion. It finishes with five properties of proportion and their direct applications. This page walks through each section in the book’s own order.

ICSE Class 10 Mathematics Bansal Chapter 7 Ratio and Proportion Including Properties and Uses PDF

What is inside Chapter 7

25 pages · 9 sections · 23 worked examples · 90 exercise questions

Section	Title
7.1	Introduction
7.2	Ratio
7.3	Increase (or decrease) in a ratio
7.4	Commensurable and incommensurable quantities
7.5	Composition of Ratios
7.6	Proportion
7.7	Continued proportion
7.8	Some Important Properties of Proportion
7.9	Direct Applications

EXERCISE 7(A): 30 questions
EXERCISE 7(B): 20 questions
EXERCISE 7(C): 15 questions
EXERCISE 7(D): 25 questions

7.1 Introduction

You first met ratio and proportion in earlier classes. The book assumes you know the basics from Class 8 and Class 9. Chapter 7 returns to these ideas in more detail.

The chapter does more than compare two quantities. It lays down the rules that govern ratios and their terms. It then introduces proportion, continued proportion, and the properties derived from a proportion.

These properties become working tools. Section 7.9 shows how to use them to solve equations and prove identities. The exercises test each tool in turn. This chapter is one chapter in the Class 10 ICSE Maths book collection.

7.2 Ratio

A ratio compares two quantities of the same kind and the same units. For quantities $a$ and $b$ , with $b \neq 0$ , the ratio is the quotient $\frac{a}{b}$ , written $a:b$ . It is read as “a is to b”. A ratio has no unit.

The quantities $a$ and $b$ are the terms of the ratio. The first term $a$ is the antecedent. The second term $b$ is the consequent. The second term can never be zero. In the ratio $b:a$ , the term $a$ must also be non-zero.

A ratio stays unchanged when both terms are multiplied or divided by the same non-zero number. A ratio must be in lowest terms, so the terms are co-prime with HCF 1. Thus $3:7$ is in lowest terms, but $4:20$ is not. Order matters: $a:b = b:a$ only when $a = b$ .

One standard problem gives a ratio of two expressions and asks for $x:y$ . Suppose $3x+4y : 5x+7y = 11:19$ . Write the ratio as a fraction and cross-multiply.

$\frac{3x+4y}{5x+7y} = \frac{11}{19}$

$19(3x+4y) = 11(5x+7y)$

$57x + 76y = 55x + 77y$

$2x = y \Rightarrow x:y = 1:2$

A second type gives $x:y$ and asks for the value of another expression. Take $x:y = 4:9$ . You need $5x+2y : 3x+7y$ . Three routes lead to the same answer.

Route 1: divide each term by $y$ .

$\frac{5x+2y}{3x+7y} = \frac{5\left(\frac{x}{y}\right)+2}{3\left(\frac{x}{y}\right)+7} = \frac{5\times\frac{4}{9}+2}{3\times\frac{4}{9}+7} = \frac{\frac{38}{9}}{\frac{75}{9}} = \frac{38}{75}$

Route 2: substitute $x = \frac{4y}{9}$ .

$\frac{5x+2y}{3x+7y} = \frac{5\times\frac{4y}{9}+2y}{3\times\frac{4y}{9}+7y} = \frac{\frac{38y}{9}}{\frac{75y}{9}} = \frac{38}{75}$

Route 3: use the k method, $x = 4k$ , $y = 9k$ .

$\frac{5x+2y}{3x+7y} = \frac{5(4k)+2(9k)}{3(4k)+7(9k)} = \frac{38k}{75k} = \frac{38}{75}$

So $5x+2y : 3x+7y = 38:75$ . Every route gives the same ratio.

One common error spoils an otherwise correct solution. A ratio $x:y = 2:3$ does not mean $x = 2$ and $y = 3$ . Ages make the reason clear. If two people are 15 and 24, their ratio is $5:8$ , yet their ages are not 5 and 8. The ratio fixes a comparison, not actual values. Write $x = 2k$ and $y = 3k$ instead. Every pair like (4, 6) fits the ratio.

The section also covers adding or subtracting a quantity from both terms of a ratio. Work-and-days problems reduce to a similar equation. Some produce a quadratic, solved by the methods in Chapter 6 of this book.

7.3 Increase (or decrease) in a ratio

A ratio can express a change in a quantity. If a price rises from ₹20 to ₹24, the original to new ratio is $20:24 = 5:6$ . The price increased in the ratio $5:6$ . If it falls from ₹24 to ₹20, the original to new ratio is $24:20 = 6:5$ .

The general rule is short. If a quantity changes in the ratio $a:b$ , the new quantity equals $\frac{b}{a}$ times the original. So new $= \frac{b}{a} \times$ original.

Apply the rule to a fare problem. A bus fare rises in the ratio $4:5$ , and the new fare is ₹250. The original fare was $\frac{4}{5} \times 250 = ₹200$ . The increase in the fare is $250 - 200 = ₹50$ .

Board problems apply this idea to counts that change. Suppose officers and soldiers are in the ratio $2:7$ . In a battle, 5 officers and 12 soldiers are killed, and the new ratio is $1:4$ . Let the original counts be $2x$ and $7x$ . After the battle they are $2x - 5$ and $7x - 12$ .

$\frac{2x-5}{7x-12} = \frac{1}{4}$

$4(2x-5) = 7x-12 \Rightarrow 8x - 20 = 7x - 12 \Rightarrow x = 8$

Original officers $= 2 \times 8 = 16$ . Original soldiers $= 7 \times 8 = 56$ .

The section also records a fact about equal ratios. If $\frac{a}{b} = \frac{c}{d}$ , then each ratio equals $\frac{a+c}{b+d}$ . This sum property helps when a chain of equal ratios is given.

7.4 Commensurable and incommensurable quantities

Two quantities are commensurable when their ratio can be written exactly as a ratio of two integers. Otherwise they are incommensurable. The two quantities must be of the same kind and unit.

Test $1\frac{3}{4}$ and $2\frac{1}{3}$ . Their ratio is $\frac{7}{4} : \frac{7}{3}$ .

$1\frac{3}{4} : 2\frac{1}{3} = \frac{7}{4} : \frac{7}{3} = \frac{7}{4} \times \frac{3}{7} = \frac{3}{4} = 3:4$

The ratio is $3:4$ , a ratio of integers. So the quantities are commensurable.

Now compare $\sqrt{5}$ and 2. The ratio $\sqrt{5}:2$ can never be written as a ratio of two integers. These are incommensurable quantities. The idea is conceptual. It tells you which ratios can be simplified exactly.

7.5 Composition of Ratios

This section defines six composed ratio types. Each type is built from the basic ratio $a:b$ . The table summarises them with one numeric example each.

Type	Formula for $a:b$	Example
Compound	$a:b$ and $c:d$ give $(a \times c):(b \times d)$	$2:3$ and $4:5$ give $8:15$
Duplicate	$a^2:b^2$	$2:3$ gives $4:9$
Triplicate	$a^3:b^3$	$2:3$ gives $8:27$
Sub-duplicate	$\sqrt{a}:\sqrt{b}$	$9:16$ gives $3:4$
Sub-triplicate	$\sqrt[3]{a}:\sqrt[3]{b}$	$27:64$ gives $3:4$
Reciprocal	$\frac{1}{a}:\frac{1}{b} = b:a$	$3:5$ gives $5:3$

The compound ratio multiplies antecedents and consequents separately. For $a:b$ and $c:d$ , it is $(a \times c):(b \times d)$ . For three ratios, multiply all three antecedents and all three consequents. The duplicate, triplicate and other ratios are special cases of this idea.

Board questions combine several types in one compound ratio. Find the ratio compounded of the duplicate ratio of $3:2$ , the reciprocal ratio of $25:36$ , and the sub-triplicate ratio of $8:27$ .

$\text{Duplicate of } 3:2 = 9:4$

$\text{Reciprocal of } 25:36 = 36:25$

$\text{Sub-triplicate of } 8:27 = 2:3$

$(9 \times 36 \times 2):(4 \times 25 \times 3) = 648:300 = 54:25$

Compound ratios can also be built from algebraic expressions. After cancelling common factors, such compounds sometimes simplify to $1:1$ . Exercise 7(A) tests this idea.

7.6 Proportion

Four non-zero quantities $a, b, c, d$ are in proportion when $a:b = c:d$ . The statement is written $a:b::c:d$ and read “a is to b as c is to d”.

The four quantities are the terms of the proportion. The extremes are the end terms $a$ and $d$ . The means are the middle terms $b$ and $c$ . The product rule connects them: $a \times d = b \times c$ . Product of extremes equals product of means.

The rule comes from cross-multiplication. Since $\frac{a}{b} = \frac{c}{d}$ , multiplying by $b \times d$ gives $ad = bc$ . This single equation solves most proportion problems.

In $a:b = c:d$ , the fourth term $d$ is the fourth proportional. The quantities $a$ and $b$ must share the same kind and units. The quantities $c$ and $d$ may form their own same-kind pair. For example, $5\text{ kg}:15\text{ kg} = ₹75:₹225$ .

Find the fourth proportional to $4, 10$ and $7$ . Set $4:10 = 7:x$ .

$4x = 10 \times 7 = 70 \Rightarrow x = 17.5$

The fourth proportional is 17.5.

7.7 Continued proportion

Three non-zero quantities of the same kind are in continued proportion when $a:b = b:c$ . The general chain extends this: $\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{d}{e}$ and so on.

In $a:b = b:c$ , the middle quantity $b$ is the mean proportional between $a$ and $c$ . The third quantity $c$ is the third proportional to $a$ and $b$ . Three finding patterns follow.

Fourth proportional: set $a:b = c:x$ , so $x = \frac{bc}{a}$ .
Mean proportional: set $a:x = x:c$ , so $x^2 = ac$ .
Third proportional: set $a:b = b:x$ , so $x = \frac{b^2}{a}$ .

Here are the patterns with numbers. The fourth proportional to $5, 12$ and $8$ is $x = \frac{12 \times 8}{5} = 19.2$ . The mean proportional between 9 and 16 is $x = \sqrt{144} = 12$ . The third proportional to 8 and 12 is $x = \frac{144}{8} = 18$ .

A board pattern combines mean and third proportionals. Suppose 12 is the mean proportional between $x$ and $y$ , and 96 is the third proportional to them.

$xy = 12^2 = 144 \quad \text{and} \quad y^2 = 96x$

Substitute $x = \frac{144}{y}$ into the second equation.

$y^3 = 96 \times 144 = 13824 \Rightarrow y = 24, \ x = 6$

Another classic problem asks for the number to subtract from four given numbers so the remainders stay in proportion. Subtract $x$ from $13, 17, 29, 41$ .

$\frac{13-x}{17-x} = \frac{29-x}{41-x}$

$(13-x)(41-x) = (17-x)(29-x)$

$533 - 54x + x^2 = 493 - 46x + x^2 \Rightarrow x = 5$

Check: $8:12 = 24:36 = 2:3$ .

7.8 Some Important Properties of Proportion

Starting from $a:b = c:d$ , five derived proportions are used throughout this chapter. Each follows from a simple operation on the fractions. The table states them all.

Property	Result from $a:b = c:d$	How to derive
Invertendo	$b:a = d:c$	Take reciprocals of both sides
Alternendo	$a:c = b:d$	Cross-multiply, then divide by $c \times d$
Componendo	$(a+b):b = (c+d):d$	Add 1 to each side
Dividendo	$(a-b):b = (c-d):d$	Subtract 1 from each side
Componendo–dividendo	$(a+b):(a-b) = (c+d):(c-d)$	Divide the componendo result by the dividendo result

Invertendo takes reciprocals. Since $\frac{a}{b} = \frac{c}{d}$ , flipping both sides gives $\frac{b}{a} = \frac{d}{c}$ . Alternendo uses cross-multiplication: from $ad = bc$ , divide by $c \times d$ to get $\frac{a}{c} = \frac{b}{d}$ .

Componendo adds 1 to each side. This turns $\frac{a}{b} = \frac{c}{d}$ into $\frac{a+b}{b} = \frac{c+d}{d}$ . Dividendo subtracts 1 in the same way. Dividing the two results gives componendo–dividendo: $\frac{a+b}{a-b} = \frac{c+d}{c-d}$ .

These properties form the backbone of Exercise 7(C). Every question there uses them to prove identities or solve for $x$ . Board-tagged years confirm their importance. Exercise 7(C) includes questions from 2000, 2007 and 2013. Exercise 7(D) adds board questions from 2010 to 2015. Componendo–dividendo is the property most often examined. You can verify the current Class 10 Mathematics syllabus on the official CISCE website.

7.9 Direct Applications

Componendo–dividendo turns a ratio equation into a linear equation. Solve $\frac{4x+9y}{4x-9y} = \frac{5}{3}$ for $x:y$ . Apply the property to both sides.

$\frac{4x+9y+4x-9y}{4x+9y-4x+9y} = \frac{5+3}{5-3}$

$\frac{8x}{18y} = \frac{8}{2} = 4 \Rightarrow \frac{x}{y} = 4 \times \frac{18}{8} = 9$

So $x:y = 9:1$ . Cross-multiplication gives the same answer. From $3(4x+9y) = 5(4x-9y)$ , you get $12x + 27y = 20x - 45y$ . Then $8x = 72y$ , so $x:y = 9:1$ . The section also solves a 2008 board problem by applying alternendo first, then componendo–dividendo.

The k method proves identities in four steps. Prove that $5a+3b : 5a-3b = 5c+3d : 5c-3d$ , given $a:b = c:d$ .

Step 1: set $\frac{a}{b} = \frac{c}{d} = k$ .

Step 2: write $a = bk$ and $c = dk$ .

Step 3: substitute into the left side.

$\frac{5a+3b}{5a-3b} = \frac{5bk+3b}{5bk-3b} = \frac{b(5k+3)}{b(5k-3)} = \frac{5k+3}{5k-3}$

Step 4: substitute into the right side.

$\frac{5c+3d}{5c-3d} = \frac{5dk+3d}{5dk-3d} = \frac{d(5k+3)}{d(5k-3)} = \frac{5k+3}{5k-3}$

Both sides reduce to the same expression, so the identity holds.

A radical pattern also appears in board papers. When $x = \frac{\sqrt{3a+2b} + \sqrt{3a-2b}}{\sqrt{3a+2b} - \sqrt{3a-2b}}$ , componendo–dividendo gives $\frac{x+1}{x-1} = \frac{\sqrt{3a+2b}}{\sqrt{3a-2b}}$ . Squaring and applying the property again leads to $bx^2 - 3ax + b = 0$ . Exercise 7(C) questions 11 to 13 and Exercise 7(D) questions 22 to 26 practise this method.

What to practise from this chapter

Chapter 7 closes with four exercise blocks. Together they hold 90 questions. Board questions appear throughout, with year tags from 1992 to 2015 visible in the exercises.

Exercise 7(A) — 30 questions on ratio basics, substitution methods and composed ratios.
Exercise 7(B) — 20 questions on proportion, mean and third proportional, and k method proofs.
Exercise 7(C) — 15 questions on the properties of proportion and solving for $x$ .
Exercise 7(D) — 25 questions of mixed revision, including board questions.

For the full subject list, see the ICSE Class 10 books page.

Frequently asked questions

What is the difference between a ratio and a proportion?

A ratio compares two quantities of the same kind, such as $a:b$ . A proportion states that two ratios are equal, $a:b = c:d$ . A ratio has two terms. A proportion has four terms linked by the rule that the product of extremes equals the product of means.

How do I find the mean proportional between two numbers?

Set $a:x = x:c$ . Then $x^2 = ac$ , so $x = \sqrt{ac}$ . For example, the mean proportional between 9 and 16 is $\sqrt{144} = 12$ .

When should I use componendo and dividendo?

Use it when an equation or identity has a fraction of the form $\frac{a+b}{a-b}$ . Applying it gives $\frac{a+b}{a-b} = \frac{c+d}{c-d}$ , which often reduces the expression to a simpler linear form. It is the standard tool for the radical equations in Exercise 7(C) and Exercise 7(D).

What is the k method in ratio and proportion?

Set each equal ratio to a constant $k$ , such as $\frac{a}{b} = \frac{c}{d} = k$ . Then write $a = bk$ and $c = dk$ , substitute into both sides of the identity, and simplify. It is used to prove proportion identities.

Why is x:y = 2:3 not the same as x = 2 and y = 3?

A ratio fixes a comparison, not the actual values. The pair $x = 2, y = 3$ is one of infinitely many that satisfy the ratio. The correct approach is to write $x = 2k$ and $y = 3k$ .

Reference: ICSE Class 10 Mathematics textbook material, chapter 7: Ratio and Proportion Including Properties and Uses — CISCE-aligned.

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