Ratio Class 10 ICSE — A. Das Gupta Chapter 8

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

Chapter 8 of A. Das Gupta’s ICSE Class 10 Mathematics book explains ratio class 10 icse concepts from the ground up. It starts with continued proportion, then builds to the mean proportional, the third proportional, a key property about sums of antecedents and consequents, and the five standard transformation properties. Each section below follows the book’s own order and numbering. For the full set of chapters in this series, see the Class 10 Maths book pages.

ICSE Class 10 Mathematics A. Das Gupta Chapter 8 Ratio PDF

What is inside Chapter 8

21 pages · 5 sections · 57 exercise questions

Section	Title
6	Continued Proportion
7	Mean Proportional (or Geometric Mean)
8	Third Proportional
9	An Important Property
10	Properties of Ratio and Proportion

EXERCISE 8(A): 47 questions
EXERCISE 8 (B): 10 questions

6 Continued Proportion

The book says terms $a, b, c, d, e, f$ are in continued proportion when each adjacent pair shares one common ratio:

$\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{d}{e} = \frac{e}{f}$

The chain idea matters. Each middle term is the mean proportional of its two neighbours. So $b$ is the mean proportional between $a$ and $c$ , and $d$ is the mean proportional between $c$ and $e$ .

Worked pattern: take 2, 4, 8. Check $\frac{2}{4} = \frac{4}{8} = \frac{1}{2}$ . The single ratio $\frac{1}{2}$ repeats, so these three terms are in continued proportion. The same test extends to longer chains.

7 Mean Proportional (or Geometric Mean)

Let $a, b, c$ be in continued proportion. Then $\frac{a}{b} = \frac{b}{c}$ . Cross-multiply to get $b^2 = ac$ . Taking the positive square root gives:

$b = \sqrt{ac}$

The book calls $b$ the mean proportional or geometric mean between $a$ and $c$ . This is the central formula of the section.

Worked pattern: find the mean proportional between 9 and 16.

$b = \sqrt{9 \times 16} = \sqrt{144} = 12$

Verify: $\frac{9}{12} = \frac{12}{16} = \frac{3}{4}$ . The continued-proportion chain holds.

8 Third Proportional

If $a : b = b : c$ , then $c$ is the third proportional to $a$ and $b$ . The book also notes that $c$ is the fourth proportional to $a, b, b$ .

Worked pattern: find the third proportional to 4 and 12.

$4 : 12 = 12 : c \quad \Rightarrow \quad 4c = 144 \quad \Rightarrow \quad c = 36$

Check: $\frac{4}{12} = \frac{12}{36} = \frac{1}{3}$ . The ratios match.

Derivation: from $a : b = b : c$ , cross-multiply to get $ac = b^2$ , so $c = \frac{b^2}{a}$ . For our numbers, $c = \frac{12^2}{4} = 36$ .

9 An Important Property

If $\frac{a}{b} = \frac{c}{d} = \frac{e}{f}$ , then each of these equal ratios also equals $\frac{a+c+e}{b+d+f}$ . The proof uses substitution.

Let $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$ . Then $a = bk$ , $c = dk$ , and $e = fk$ . Substitute into the sum:

$\frac{a+c+e}{b+d+f} = \frac{bk+dk+fk}{b+d+f} = \frac{k(b+d+f)}{b+d+f} = k$

So each ratio equals the sum of antecedents over the sum of consequents.

Worked pattern: $\frac{2}{3} = \frac{4}{6} = \frac{6}{9}$ . Sum of antecedents: $2 + 4 + 6 = 12$ . Sum of consequents: $3 + 6 + 9 = 18$ . So $\frac{12}{18} = \frac{2}{3}$ , which matches the original ratio.

Edge case: the sum of consequents must not be zero, or the division is undefined.

10 Properties of Ratio and Proportion

This section lists five transformation properties. Each starts from $a : b :: c : d$ and produces a new valid ratio. The table summarises them.

Property	Starting form	Result
Invertendo	$\frac{a}{b} = \frac{c}{d}$	$\frac{b}{a} = \frac{d}{c}$
Alternendo	$\frac{a}{b} = \frac{c}{d}$	$\frac{a}{c} = \frac{b}{d}$
Componendo	$\frac{a}{b} = \frac{c}{d}$	$\frac{a+b}{b} = \frac{c+d}{d}$
Dividendo	$\frac{a}{b} = \frac{c}{d}$	$\frac{a-b}{b} = \frac{c-d}{d}$
Componendo and Dividendo	$\frac{a}{b} = \frac{c}{d}$	$\frac{a+b}{a-b} = \frac{c+d}{c-d}$

Each property has a one-line proof move:

Invertendo: take reciprocals of both sides.
Alternendo: cross-multiply to $ad = bc$ , then divide both sides by $cd$ .
Componendo: add 1 to both sides.
Dividendo: subtract 1 from both sides.
Componendo and Dividendo: divide the componendo result by the dividendo result.

Edge case: Componendo and Dividendo needs $a \neq b$ . The final step divides by $(a - b)$ , so $a - b = 0$ puts zero in the denominator.

The book closes with a compact restatement of all five rules in fraction form, each labelled with its property name. Other chapters in this collection are listed on the ICSE Class 10 books page. You can also verify the current Class 10 Mathematics syllabus on the official CISCE syllabus page.

What to practise from this chapter

Two exercise blocks close the chapter. Each targets a different skill set.

EXERCISE 8(A) — 47 questions. It covers unit conversion before forming a ratio, combined ratios such as $A : B : C$ , share division, age problems, mixture problems, coin and note problems, and comparing and arranging ratios. Board-tagged questions from 1992 and 1994 appear in this block.
EXERCISE 8 (B) — 10 questions. It covers duplicate, triplicate, sub-duplicate, sub-triplicate, reciprocal and compounded ratios, and solving for $x$ when a given ratio equals one of these forms.

Two mistakes trip students up most often:

Mistake	Correct rule	How to check
Forming a ratio before converting units	Convert both quantities to the same unit first	60 paise : Rs 1.35 must become 60 : 135 paise
Applying componendo and dividendo when $a = b$	The result has $(a - b)$ in a denominator	Skip the rule when the two terms are equal

Frequently asked questions

How do you compare two ratios such as 7 : 9 and 11 : 16?

Convert both to the same denominator. $\frac{7}{9} = \frac{112}{144}$ and $\frac{11}{16} = \frac{99}{144}$ . Since 112 is greater than 99, $7 : 9$ is the larger ratio.

What is the mean proportional between two numbers and how do you find it?

The mean proportional between $a$ and $c$ is $\sqrt{ac}$ . For 9 and 16, it is $\sqrt{9 \times 16} = 12$ . Verify by checking $\frac{9}{12} = \frac{12}{16}$ .

When can I use componendo and dividendo, and what condition must hold?

Use it when you start from $a : b :: c : d$ . The rule requires $a \neq b$ , because $(a - b)$ appears in the denominator of the result.

How do you convert a fractional ratio like 1/2 : 1/3 : 1/4 into whole numbers?

Multiply every term by the LCM of the denominators. The LCM of 2, 3, 4 is 12, so the ratio becomes $6 : 4 : 3$ .

What is the difference between a duplicate ratio and a compounded ratio?

A duplicate ratio squares each term of one ratio. A compounded ratio multiplies two or more ratios together, term by term. For example, the duplicate of $3 : 4$ is $9 : 16$ , while the compound of $3 : 4$ and $2 : 5$ is $6 : 20 = 3 : 10$ .

Reference: ICSE Class 10 Mathematics textbook material, chapter 8: Ratio — CISCE-aligned.

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