school mathematics 9

Kantipur Education Development CouncilMaharajgunj, Chakrapath, Kathmandu, NepalPhone: 01-4720202/4720517Email: [email protected]: www.kedcnepal.com

KED

C

MathematicsMathematicsSCHOOL

Book 9

Approved by Government of Nepal, Ministry of Education, Curriculum Development Centre, Sanothimi, Bhaktapur

Kantipur Education Development Council

Maharajgunj, Chakrapath, Kathmandu, Nepal

Phone: 01-4720202/4720517

Email: [email protected]

Website: www.kedcnepal.com

Kantipur Education Development Council 2058 ALL

RIGHTS RESERVED. No part of this book may be

reproduced or transmitted in any form by any means,

electronic or mechanical, including photocopying and

recording, or by any information storage and retrieval

system, except as may be expressly permitted in writing

by the publisher.

Edited by:

Mahendra Raj Ghimire

Yubaraj Bhandari

Written by:

Hansha Raj Dhungana

Narayan Prasad Dahal

Nava Raj Baral

Edition:

First, 2019, Revised edition, 2020

Layout/Design

Jaydev Koirala

KED

C

Preface

TThis book hopes to develop and deep-root interest in learners in mathematics. The fun in learning mathematics is possible only when there is right co-ordination between the learners and the facilitators. It can helps us raise the curiosity to emancipate the bundle of mysteries of the world.

Mathematics is the most essential discipline among so many subjects. So, the strong foundation is necessary to be built up. The learner will find the content in this book interesting to practice and certainly find it useful in day to day life.

The exercises at design as per the curriculum of CDC of Nepal in a very simple language and in clear an concise form. Complete theory in each chapter, figures, notes, tables are the key component of this book. Simple but different techniques, sufficient project works and practice sets bind both the learners and facilitators for easy preparation of any board's or any examination.

KEDC will like to extent it heartfelt thanks to; Mr. Ghanashyam Dhakal and Mr. Binod Poudel for their emmance help and contribution in producing such wonderful books.

Constructive suggestion for further improvement are heartily apperciated.

Contents

Unit 1: Set 1-191.1. Review 21.2. Venn-diagram 61.3. Application of sets 13Unit 2: Arithmetic 20-602.1. Profit and Loss 212.2 Commission 312.3. Discount 352.4. Taxation 412.5 Dividend 462.6. Home Arithmetic 48Unit 3: Mensuration 61-863.1. Area of Plane Figures 623.2. Area of 4 walls, ceiling & floor 723.3. Area and Volume of solid 78Unit 4: Algebra 87-1624.1 Factorization 884.2. Indices 1054.3. Exponential Equation 1114.4. Ratio and Propertion 1154.5. Algebraic Fractions 1294.6. Equation in two Variables 133

4.7. Simultaneous Equations 1354.8. Quadratic Equation 151Unit 5: Geometry 163-2415.1. Review 1645.2. Congruent Triangles 1785.3. Quadrilateral 1935.4. Mid-point Theorem 2075.5. Similarity 2155.6. Circle 2235.7. Construction 235Unit 6: Trigonometry 242-2596.1. Introduction 2426.2. Trigonometric Ratios 251Unit 7: Statistics 260-2947.1. Introduction 2607.2. Graphical Representation 2657.3. Ogive 2707.4. Histogram 2737.5. Central Tendency 2787.6. Partition Values 289Unit 8 : Probability 295-3058.1. Probability 295Model Question 306

School Mathematics Book - 9 1

Objectives :At the end of the unit, the students will be able to

• define different terms of sets.• establish the relation between two or more than two sets.• find cardinality of the different sets.• perform the operations on sets using Venn-diagram.• illustrate about Venn-diagrams• solve word problems using Venn-diagram.• relate the sets operations with the daily life.

Teaching materials :Chart of Venn-diagrams ,geometrical instruments ,flash cards ,paper cuttings, beads.

Note to the teacher :1. Review the concept of set and its different terminology from previous classes.2. Give the concept of types of set.3. Give the clear concept on operations of different sets. (union, intersection, complement

and difference).4. Discuss and explain various portions of Venn- diagrams.5. Derive the formulae related to cardinality of set using inductive method up to two sets.6. Ask the students to solve the verbal problems by problems solving method under the

guidance of the teacher.

Content : • Review of sets.• Intersection, union, difference and complement of sets.• Simple word problems of two sets using Venn-diagrams and solution of given

problems.Specification Grid

Knowledge(K) Comprehensive(C) Application(A) Higher ability (HA) Total no. of questions Total marks

Each 1 mark Each 2 marks Each of 4 marks Each of 5 marks

Sets -

-

1

-

1

4

Note : At least 4 marks questions are asked from each topic.

1Chapter

SetEstimated period : 8

School Mathematics Book - 92

1.1. Review:Set: Collection of well defined objects is called set. The objects of a set are called its elements or members.For examples

1. The set of natural numbers less than 10.2. The set of vowels.3. The set of Rivers of Nepal.

The elements of a set are enclosed within curly brackets { } separated by commas. A set is denoted by capital letters A,B,C, N, M,Z etc.For examples

1. N = {1, 2, 3, 4,……………………………………. 10}2. V = {a ,e ,i ,o , u}3. R = { Mahakali , Seti , Gandaki , Bagmati }

Specification of sets

We can describe a set by different methods. Some of them are as follows :1. Description method.2. Listing or Roaster form or tabular form.3. The Rule method or set builder form.

1. Description method:

In this method, the set is described in words with common properties of elements.

Historical facts

Set theory is a branch of Mathematical logic that studies sets, which informally are collections of objects. Although any type of object can be collected into a set, set theory is applied most often to objects that are relevant to Mathematics. The language of set theory can be used in the definitions of nearly all Mathematical objects.The modern study of set theory was initiated by Georg Cantor and Richard Dedekind in the 1870s. After the discovery of paradoxes in native set theory, numerous axiom systems were proposed in the early twentieth century of which the Zermelo–Fraenkel axioms, with the axiom of choice, are the best-known.Set theory is commonly employed as a foundational system for Mathematics, particularly in the form of Zermelo–Fraenkel set theory with the axiom of choice. Beyond its foundational role, set theory is a branch of Mathematics in its own right, with an active research community. Contemporary research into set theory includes a diverse collection of topics, ranging from the structure of the real number line to the study of the consistency of large cardinals. John Venn developed the descriptive and easier concept of Venn diagram.


Examples: 1. N is a set of natural numbers. 2. V is a set of vowels of English alphabets.

2. Listing or Roaster form or tabular form:

In this method, we list or enumerate all the elements within two curly braces and are separated with commas.

Examples : 1. N = { 1, 2 ,3 , 4 , 5 ……. 10 } 2. V = { a, e, i ,o , u , }

3. The Rule method or set builder form :

In this method , we specify the set with “defining property” giving a variable. Examples : 1. N = { x : x is a natural number < 10 } 2. P = { x : x ≤ 20 , x is an even number}

Types of sets

1. Empty set or Null set or void set2. Finite set3. Infinite set4. Singleton set 5. Universal set

1. Disjoint sets: Two or more than two sets are said to be disjoint sets, if they do not have common element.Example

If A = {1, 3, 5, 7} and B = {2, 4, 6, 8}Then A and B are disjoint set.

2. Overlapping or joint sets: Two or more than two sets are said to be overlapping sets ,if they have at least one common element.Example

If A = {1, 3, 5, 7, 9} and B = {3, 6, 9, 12} then A and B are overlapping sets because both sets have two common elements 3 and 9.

3. Equal sets: Two finite sets A and B are said to be equal, if both have the same (identical) elements. In symbol, we write A = BExample: A = {2, 3, 5, 7}

B = {2, 3, 5, 7)Here, A and B are equal sets because both sets have same elements.

4. Equivalent sets: Two finite sets A and B are said to be equivalent if they have same number of elements.

1. N = Natural numbers = {1, 2, 3 , 4 , 5 ………. }

2. W= Whole numbers = { 0 , 1, 2 ,3 , 4 , 5………. }

3. Z = Integers = {……-3 , -2 , -1 , 0 , 1, 2 ,3….. }

4. R= Real numbers

Remember

.1.7

.3.5

A

.2.4

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B

A B

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Example:Let, A = {1, 2, 3}

B = {a, b, c} Here, A and B are two equivalent sets because both sets have 3 elements.5. Subsets and super subsets: A set A is said to be a subsets of B if every elements of A is also

element of set B. It is denoted by A ⊂ B. It is read as “A” is contained in B or “B” contains “A”. In this case, B is also called super set of A.This relation is denoted by A ⊂BExample: A = {a, e, i, o ,u } B = {a, e, i, o , u}Then, B ⊂ANote: 1. A null set is a subset of every set. 2. Every set is a subset as well as super set of itself.

6. Power set: The set of all the sub sets of a set A is called the power set of the set A. Example : If A = {1, 2}, then power set of A = { { }, {1} , {2} , {1, 2} }7. Proper sub-set: Set A is proper subset of set B, if every element of set A is an element of set

B and at least an element of B is not an element of A. It is denoted by A ⊂B If N is a set of natural numbers, W is set of whole numbers and Z is set of integers then N ⊂W ⊂Z Note: If number of elements of a set is n then number of subsets = 2n

1. Which of the following collections are sets?a. All rivers of Nepal.b. All triangles of 4 sides.c. All negative integers.d. {a, e, i, o, u}e. All handsome boys of kailali district.

2. Write the following in tabular (listing) form.a. { x:x = 3n, n ∈N}b. { x:x = 3n + 2, n ∈ N}c. { x:x is a vowel alphabet in the word science}d. { x:x = odd prime number, x < 15, x ∈W }

3. Write the following in set builder form.a. A = {1, 4, 9, 16, 25}b. B= {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

Exercise 1.1


c. C = {1, 3, 5, 7, 9}d. D = {Sunday, Monday, … Saturday}

4. a. A set of even numbers between 0 and 15. Express it in listing method and set builder method.

b. If Z = {x:x is an integer, -1 ≤ x ≤ 1}. Express it in description method and listing method.

5. a. If A = {2, 3, 4, 5, 6}, write with symbol whether 2, 6 and 8 are member of set A or not. b. If A = {2, 4, 6} and B = {2, 4, 6, 8, 10}, write with symbol whether A is subset or proper

subset of B.6. a. State whether the sets P = {4, 5, 6, 7} and Q = {2, 4, 6, 8} are equal or equivalent set. Give

reason. b. State whether the set A = {2, 4, 6, 8} and B= {6, 4, 2, 8} are equal or equivalent set. Give

reason.7. a. State whether the sets A= {1, 3, 5, 7} and B = {2, 4, 6, 8} are overlapping or disjoint set.

Give reason. b. State whether the sets M= {a, b, c, d, e} and N = {a, e, i, o, u} are overlapping or disjoint

set. Give reason. c. Prove with example that all equal sets are equivalent sets but all equivalent sets are

not equal sets.8. State whether the following are infinite or finite set.

a. A set of all points of a line.b. A set of all plane figures.c. { x : x ≥ 3, x ∈ N }d. A set of Vowels of English alphabet.

9. Write the possible subsets of the following sets. Also write their power set.a. A = {1,2 }b. B = {a , b , c }c. C = { x , y , z }

10. Show that the number of subsets of the set { k , l, m } is 23.

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1. Show to your teacher 2. a. {3,6,9,…..} b. {5,8,11,…….} c. {e,i} d. {3,5,7,11,13} 3.a. A = {x : x is a square number less than 36 } b. B = { x : x is a even number less than 22 }c. C = { x : x is a odd number less than 11} d. D = { x : x is a day of a week} 4. a. {2,4,6,8,10,12,14} , { x : x is a even number less than 15} b. Z is a set of integers between -1 and 1. Z = {-1,0,1} 5. Show to your teacher. 6.a. P and Q are equivalent sets. b. A and B are equal sets. 7.a. Disjoint b. Overlapping 8.a. Infinite b. Infinite c. Finite d. Finite9.a. { }, {1 },{2 },{1,2 }; 4 b. { }, {a },{b },{c }{a,b }, {a,c },{b,c },{a,b,c }; 8 c. { }, {x },{y },{z}{x,y }, {x,z },{y,z},{x,y,z }; 8


1. Union of sets : The set that is formed by taking elements of the set A or the set B or both is called union of set A and B. In symbol, we denote it by A ∪B and read as ‘A union B’ or ‘A cup B’ Symbolically, A ∪ B = {x:x∈A or x∈B}Example :Let A = {1, 2, 3, 4 ,5} and B = { 4, 5, 6, 7 ,8}, thenA∪B = {1,2,3,4,5} ∪ { 4,5,6,7,8} = {1, 2, 3, 4, 5, 6, 7, 8}The Venn-diagram of A∪B is shown alongside.

2. Intersection of sets : The set that contains common elements of given sets is called intersection set of the given sets. It is denoted by A∩B and read as ‘A intersection B’ or ‘A cap B’ . Symbolically, A∩B = {x:x∈A and x∈B}Example : Let A = {1, 2, 3, 4, 5 } and B= {4, 5 ,6, 7, 8}, then A∩B = {1, 2, 3, 4, 5} ∩ {4, 5 ,6, 7, 8} = {4 , 5}The Venn-diagram of A∩B is shaded.

3. Difference of sets : The set of elements that belong to set A but not in set B is called difference of set A and B. It is denoted by A – B.Similarly, the set of elements that belong to B but not in A is called difference of set B and A.It is denoted by B – A.

A B U

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A∪B

A

A∩B

B U

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1.2. Venn- Diagram : Venn diagrams are efficient way of representing and analyzing sets and performing set operations. The usage of Venn diagrams is just the elaboration of a solving technique. Problems that are solved using Venn diagrams are essentially problems based on sets and set operations. Thus, before we move on to understand Venn- diagrams, we first need to understand the concept of a set operation.Operation of sets :There are following four set operations

a. Union of two or more sets.b. Intersection of sets.c. Difference between two sets.d. Complement of a sets.


Example : Let U = {0, 1, 2, 3, 4, 5, ,6} A = {0, 1, 2, 3} B = {2, 3, 4, 5 ,6}Let us show these information in a Venn diagram.From Venn-diagram,

a. The set of elements belong to both set A and B i.e A ∩ B = {2,3}b. The set of that elements belong to A i.e A = {0, 1, 2 ,3}c. The set of that elements belong to B i.e B = {2, 3, 4, 5, 6 }d. The set of elements that belong to A – only i.e A0 = {0, 1}e. The set of elements that belong to B – only i.e B0= {4, 5 , 6}f. The set of elements that belong to A but not in B = {0, 1}

A

A-B

B U

.0.4.5.6

.1.2.3

g. The set of elements that belong to B but not in A = {4, 5, 6}Mathematically,

A – B = {x:x∈A and x∉B}So, A – B = {0,1}

Similarly,

B – A = {x:x∈B and x∉A}So, B – A = {4, 5, 6}

Note: 1. A – B ≠ B – A 2. If A – B = B – A , It is called symmetrical difference 4. Complement of set: If A is a subset of universal set U, then the complement of the set A is the set of the elements

which belong to U but does not belong to A.Mathematically,

A' = { x:x∈U, x ∉A}

It is denoted by or Ac or A' or A Example : Look at the Venn- diagram along side.

a. What are the elements of set A?b. What are the elements of set U?c. What are the elements of set U – A?d. What is set U – A called?

Here, The set of elements A are A = {1, 2, 3, 4,5} The set of elements U are U = {1, 2, 3, 4, 5, 6, 7, 8}

A

B-A

B U

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Ac

A B U

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A ∩ B


Example 1 : If U = {0,1,2,3,4,5,6,7,8,9,10}, A = {1,2,3,4,5} , B = {4,5,6,7,8}, find A ∪ B and illustrate it in a Venn - diagram.

Solution:Here, U = {0,1,2,3,4,5,6,7,8,9,10}A = {1, 2, 3, 4, 5}B = {4, 5, 6, 7, 8}A ∪ B = ?∴A ∪ B = {1, 2, 3, 4, 5} ∪ {4, 5, 6, 7, 8} = {1,2,3,4,5,6,7,8}In Venn- diagram , we can show it as in figure.The shaded region denotes A ∪ B.

Example 2 : If U = {0, 1, 2, 3…17}, A = {x: x ≤ 15, x ∈ W} and B = {x: x is an odd number, x ≤ 10} Find A ∩ B and illustrate it in a Venn diagram.

Solution:Here,U = {0, 1, 2, 3…17}A = {x: x ≤ 15, x ∈ W} = {0, 1, 2... 15}B = {x: x is an odd number, x < 10} = {1, 3, 5, 7, 9}

A B U

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A∪B

UA

B.0 .2 .4 .6

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16

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17

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A∩B

Then, the complement of set A is = U – A = { 6 ,7 ,8}The elements {6, 7, 8} are also a set that we get by U – A.This set is called complement of set A. It is the subset of universal set U.Note:

1. The complement of set U is φ and complement of φ is U.

2. The complement of complement of a set is the set itself. i. e. A = A5. Symmetric difference :

If A and B are two sets then, their symmetric difference is denoted by A ∆ B is read as ( A delta B) which is the set of all elements of A and B which are not common elements to A and B.It is also written as A ∆ B = (A–B) U (B–A)

Worked Out Examples

A ∆ B A B

U


Now, A ∩B = {0, 1, 2 … 15} ∩{1, 3, 5, 7, 9}∴A ∩ B = {1, 3, 5, 7, 9} = BIn Venn diagram ,the shaded region denotes A ∩ B.

Example 3 : If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} , A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}. Find i. A – B ii. B – A and illustrate it in Venn diagram.

Solution:Here,U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}A = {1, 2, 3, 4, 5}B = {4, 5, 6, 7}A – B =?B – A =?Now, A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7} = {1, 2, 3} B – A = {4, 5, 6, 7} – {1, 2, 3, 4, 5} = {6, 7}

Remember: A - B ≠ B - AExample 4 : If U = {0, 1, 2… 10} , A = { 1, 2, 3, 5, 7, 9} and B = {2, 4, 5, 6, 8}, find

i. A ii. B iii. (A ∪ B) iv. (A ∩ B) v. (A - B) . Also illustrate each in Venn - diagram.

Solution: Here, U = {0, 1, 2… 10}A = {1, 2, 3, 5, 7, 9} B = {2, 4, 5, 6, 8}Now,

i. A = U – A = {0, 1, 2… 10} - {1, 2, 3, 5, 7, 9} = {0, 4, 6, 8, 10}

∴A = {0, 4, 6, 8, 10}

ii. B = U – B = {0, 1, 2… 10} - {2, 4, 5, 6, 8} = {0, 1, 3, 5, 7, 9, 10}

U

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A - B

A B

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B-A

A B

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A B

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A B

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A

B


∴B = {0, 1, 3, 5, 7, 9, 10}iii. (A ∪ B) = U – (A ∪ B) = {0, 1, 2… 10} – {1, 2, 3, 4, 5, 6, 7, 8, 9} = {0, 10}∴(A ∪ B) = {0, 10}

iv. (A ∩ B) = U – (A ∩ B) = {0, 1, 2… 10} - {2, 5} = {0, 1, 3, 4, 6, 7, 8, 9, 10} ∴(A ∩ B) = {0, 1, 3, 4, 6, 7, 8, 9, 10}

v. (A - B) = U - (A - B) (A - B) = {1 , 2 , 3 , 5 , 7, 9 } – { 2 , 4 , 5 , 6 , 8 } = { 1 , 3 , 7 , 9 }∴ A - B = {1, 3, 7, 9}

Example 5: Let A, B and C are subsets of a Universal set U. If U = {1, 2, 3…15}, A = {2, 4, 6, 8, 10, 12}, B = {3, 6, 9, 12, 15} and C = {1, 2, 3, 4, 5, 6}, finda. (A ∩ B) ∪C and b. (A ∪ B) ∩ C

Illustrate the results in Venn diagrams.Solution:

Here,U = {1, 2, 3…15}A = {2, 4, 6, 8, 10, 12}B = {3, 6, 9, 12, 15}C = {1, 2, 3, 4, 5, 6}

U

.0 .10

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A B

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U

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A B

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U

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A B

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A ∩ B

A – B

A ∪ B


a. A ∩ B = {6, 1, 2}(A ∩ B) ∪ C = {1, 2, 3, 4, 5, 6, 1, 2}The shaded portion shows the set (A ∩ B) ∪C

b. A ∪ B = { 2, 3,4,6,8,9,10,12 ,15}(A ∪ B) ∩ C = { 2, 3,4,6,8,9,10,12 ,15} ∩ {1, 2, 3, 4, 5, 6} = { 2 , 3 , 4 , 6}(A ∪ B) ∩ C = ∪ – [(A ∪ B) ∩ C ] = { 1,5,7,8,9,10,11,12,13,14,15}

The shaded portion represent the set (A ∪ B) ∩ C

U

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A

C

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A

C

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Exercise 1.2

1. Write down the sets for the shaded part in the following diagrams:

A B U A B U A B U

A B CU

a.

d. e. f.

b. c.

UA

C

B

(A ∩ B) ∪ C

(A ∪ B) ∩ C

A

BC

U


2. If A = {odd natural numbers less than 9} and B = {prime factors of 15}, find

a. A – B b. B – A c. A ∪ B d. A ∩ B

3. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 3, 5} and B = {2, 4, 6}, find :

a. A ∩ B b. A – B c. A ∪ B d. (A ∩ B)

e. (A ∪ B) f. Ac ∩ Bc g. Ac ∪Bc h. A ∪ (A – B)

4. If U = {1, 3, 5, 7, 9, 11, 13, 15, 17} , A = { 1, 7, 9, 13} , B = { 3, 5, 11, 15} and C = {3, 5, 7, 9}, find

a. Ac ∪B b. (Ac)c c. A ∪ (B – C) d. (A – B) ∩ (A – C)

e. (A ∩ B)∩C f. (A ∪ B) – (A ∩ B)

5. If U = {x: x = 12, x ∈ N} , A = {x:x is a prime number} , B = {x:x is an odd number} and C = {x:x is number divisible by 3}, find :

a.(A ∪ B)c b. (A ∩ B)c c. (A ∩ B)c - (A ∪ B)c

6. Let P, Q and R are the subsets of the Universal set U.

If U = { 1, 2,3…………., 15} , P = { 1 , 3 , 5 , 7 , 9 , 11} Q = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } and

R = { 3 , 6 , 9 , 12 , 15 }. Work out the following set operations and illustrate them in the Venn- diagrams.

a. P ∪Q ∪R b. (P ∪Q) ∩R c. P ∩(Q ∩R) d. P ∪ (Q ∪ R)

e. P ∩ (Q ∪ R) f. P - (Q ∩R) g. P ∩ (Q - R) h. P ∩ Q ∩ R

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wer

s 1.2

1. Show to your teacher 2.a. {7} b. { } c. { 1,3,5,7 } d. {1,3,5} 3.a.{ } b. {1,3,5} c. { 1,2,3,4,5,6,} d. {1,2,3,……….9} e. {7,8,9} f. {7,8,9} g. {1,2,3,4,5,6,7,8,9} h. {1,3,5} 4.a. { 3,5,11,15,17} b. {1,7,9,13} c. {1,7,9,11,13,15} d. {1,13} e. {} f. {1,3,5,7,9,11,13,15} 5.a. {4,6,8,10,12} b. {2,4,6,8,9,10,12} c. {1,2,9} 6.a. { 1,2,3,4,5,6,7,9,11,12,15} b. { 3,6,9} c. {3} d. {8,10,13,14} e. { 2,4,6,8,10,11,12,13,14,15 } f. { 1,5,7,9,11} g. { 6,8,10,12,13,14,15} h. {8,10,13,14}


1.3. Application of setsCardinality relation of sets:

The number of elements present in a set is called cardinality. The cardinal number of a set A is denoted by n (A). Similarly the cardinal number of the set A ∪ B , A ∩ B , A – B , (A - B) are denoted by n (A ∪ B) , n(A ∩ B) , n(A – B) and n (A - B) respectively.

Cardinality relation of union of two sets

Let A and B are two overlapping sets in a universal set U, thenn (A ∪ B) = n(A) + n(B) – n(A ∩ B)Proof:

If n (A) = a n (B) = b

n (A ∩ B) = c Then,The number of elements belonging to

A only, n0 (A) = a – cThe number of elements belonging to

B only, n0 (B) = b – cFrom Venn - diagram we get

n (A ∪ B) = n0 (A) + n(A ∩ B) + n0 (B) = a – c + c + b – c= a + b – c

∴n (A ∪B) = n (A) + n (B) - n (A ∩B)Note :

1. If the two sets are disjoint then n (A ∩B) = 0 ∴n (A ∪ B) = n (A) + n (B)

2. If the set B is subset of A. i.e. B ⊂ A. If all elements of B lies on set A.Then, n (A ∪ B) = n (A) + n (B) – n (B) {∵ Set B is common part } = n (A)

∴n (A ∪ B) = n (A) In the same way, if the set A is subset of set B. i.e. A ⊂ B.

Then, n (A ∪ B) = n (A) + n (B) – n (A) {∵ Set A is common part } = n (B )

∴n (A ∪ B) = n (B)

A B

ba

a-c b-c

U

c


The relation given below are equally important to solve problems.1. Number of elements lying in A or B = number of elements within the circles. n (A ∪ B) = n (A) + n (B) - n (A ∩ B) n (A ∪ B) = no(A) + n(A ∩ B) + n0 (B)2. Number of elements lying in A only n0 (A ) = n (A) - n ( A∩ B ) n0 (A ) = n (A ∪ B ) - n (B)3. Number of elements lying in B only n0 (B ) = n (B) - n ( A ∩ B ) n0 (B ) = n (A ∪ B ) - n (A)4. Number of elements lying in exactly only one set n0 ( A ) + n0 (B ) = n (A ∪ B ) - n (A ∩B)5. Number of elements not in A and B n(A ∪ B) = n(U) - n (A ∪ B ) 6. Total numbers of elements under consideration n(U) = n (A ∪ B ) + n(A ∪ B) n(U) = n (A) + n (B) - n (A ∩ B) + n(A ∪ B)7. If the universal set contains only the elements of sets A and B, then n(A ∪ B) = 0. So, n(U) = n (A ∪B )

Example 1 : If n(A) = 60, n(B) = 50 and n (A ∪ B) = 85. Find n (A ∩ B)Solution: Here,

n (A) = 60 , n (B) = 50 , n (A ∪ B) = 85n (A ∩B) = ?

We Know,n (A ∩ B) = n (A) + n (B) – n (A ∪ B) = 60 + 50 – 85 = 110 – 85 = 25

∴n (A ∩B) = 25Example 2 : If n (∪) = 110, n(A) = 75, n(B) = 60 , n (A ∩ B) = 45

Find a. n (A ∪ B) b. n (A ∪ B)

A B

(A ∩B)

n (B)n (A)n (Only B) or n (B–A)

n (Only A) or n (A–B)

U

Worked Out Examples


c. n0 (A) d. n0 (B)

Solution: Here,n (U) = 110n (A) = 75n (B) = 60 n (A ∩ B) = 45

We Know,a. n (A ∪ B) = n (A) + n (B) - n (A ∩B)

= 75 + 60 – 45= 135 – 45= 90

∴ n (A ∪ B) = 90Example 3 : In a survey of 400 people, 100 liked folk songs but not modern songs, 120 liked

modern songs but not folk songs and 50 of them did not like both songs.i. Find the number of people who liked both songs.ii. How many of them liked folk songs?iii. How many of them liked modern songs?iv. Represent the result in Venn-diagram.

Solution : Here,Let U = set of total no. of people F = set of folk song likers M = set of modern song likersTotal number of people n (U) = 400Number of people who liked only folk songs, no (F) = 100Number of people who liked only modern songs, no (M) = 120Number of people who did not like both songs, n (F∪ M) = 50Now, We know,n (F ∪ M) = n (U) – n (F∪ M)

= 400 – 50= 350

Let, the number of people who liked both songs n (F ∩M) be xThen n (F ∪ M) = no(F) + (F ∩M) + no(M)

Or, 350 = 100 + x + 120 Or, x = 350 – 220 ∴x = 130

F M

100 120

U

130

50

b. n (A ∪ B) = n (U) - n (A ∪ B)

= 110 – 90= 20

∴ n (A ∪ B) = 20c. n0 (A) = n (A) - n (A ∩ B)

= 75 – 45= 30

d. no (B) = n (B) - n (A∩B) = 60 – 45 = 25


i. 130 people liked both songs.ii. The number of people who liked folk songs = no(F) + n(F ∩M) = 100 + 130 = 230iii. The number of people who liked modern songs = no(M) + n (F ∩M) = 120 + 130 = 250

Example 4. In a survey conducted among some people of a group it was found that 40% of them liked literature, 65% liked music and 10% of them liked none.

i. Represent the above information in Venn diagram.ii. If there were 30 people who liked both items, find the number of people participated

in the surveySolution:

Let U = set of total no. of people L = set of literature likers M = set of music likersHere, Total number of people, n (U) = 100%Number of people who liked literature, n (L) = 40%Number of people who liked music, n (M) = 65%Number of people who did not like both, n (L∪ M) = 10%

We know,n (L ∪ M) = n (U) - n (L∪ M)

= 100% - 10%= 90%

Also, n (L ∩ M) = n (L) + n (M) – n(L ∪ M) = 40% + 65% - 90% = 105% - 90% = 15%iii. Let the number of people who participated in the survey be x Now,

15% of x = 3015x100 = 30

∴x = 30 × 100

15 = 200

Hence, 200 people participated in the survey.Example 5 : In a group of 950 students, the ratio of students who like Mathematics and

Science is 4:5. If 100 of them like both subjects and 150 of them like none of the subjects, then by drawing a Venn-diagram, find how many of themi. Like only Mathematics

L M

25% 50%

U

15%

10%


i. 130 people liked both songs.ii. The number of people who liked folk songs = no(F) + n(F ∩M) = 100 + 130 = 230iii. The number of people who liked modern songs = no(M) + n (F ∩M) = 120 + 130 = 250

Example 4. In a survey conducted among some people of a group it was found that 40% of them liked literature, 65% liked music and 10% of them liked none.

i. Represent the above information in Venn diagram.ii. If there were 30 people who liked both items, find the number of people participated

in the surveySolution:

Let U = set of total no. of people L = set of literature likers M = set of music likersHere, Total number of people, n (U) = 100%Number of people who liked literature, n (L) = 40%Number of people who liked music, n (M) = 65%Number of people who did not like both, n (L∪ M) = 10%

We know,n (L ∪ M) = n (U) - n (L∪ M)

= 100% - 10%= 90%

Also, n (L ∩ M) = n (L) + n (M) – n(L ∪ M) = 40% + 65% - 90% = 105% - 90% = 15%iii. Let the number of people who participated in the survey be x Now,

15% of x = 3015x100 = 30

∴x = 30 × 100

15 = 200

Hence, 200 people participated in the survey.Example 5 : In a group of 950 students, the ratio of students who like Mathematics and

Science is 4:5. If 100 of them like both subjects and 150 of them like none of the subjects, then by drawing a Venn-diagram, find how many of themi. Like only Mathematics

ii. Like only ScienceSolution:

Here,Let U = set of total no. of students M = set of mathematics likers S = set of science likersTotal number of students, n (U) = 950Let constant be xNumber of students who like Mathematics, n (M) = 4xNumber of students who like Science, n (S) = 5xNumber of students who like both subjects, n (M ∩ S) = 100Number of students who did not like both subjects, n (M ∪ S) = 150

From Venn-diagram4x – 100 +100 + 5x -100 + 150 = 950Or, 9x = 950 – 50Or, 9x = 900Or, x = 100a. no(M) = 4x – 100 = 4 × 100 – 100 = 300 b. no(S) = 5x – 100

= 5 × 100 – 100 = 400

M S

4x - 100 5x - 100

U

100

150

Exercise 1.3

1. a. If A = {a, b, c} and B = {c, d, e, f} ,verify that n (A ∩ B) = n (A) + n (B) – n (A ∪ B). b. If A = {2, 3, 4, 5} and B = {4, 5, 6, 7}, verify that n (A ∪ B) = n (A) + n (B) – n (A ∩ B).2. a. A and B are two subsets of a universal set ‘U’ where n (U) = 43, n (A) = 25 , n (B) = 18

and n (A ∩ B) = 7i. Draw a Venn-diagram to illustrate the above information.ii. Find the value of n (A ∪ B)

b. If n (A) = 45, n (B) = 65, n (A ∪ B) = 85 then findi. The value of n (A ∩ B) ii. The value of n0(B) iii. Show the above information in Venn-diagram.


3. a. Out of 100 participants in a picnic, 70 drank milk, 40 drank soup and 20 drank both.i. Represent the above information in a Venn-diagram.ii. Find the number of participants who drank neither of the drinks by using Venn-

diagram . b. In a class of 55 students, 15 students liked Maths but not English and 18 students liked

English but not Maths. If 5 students did not like both,how many students liked both subjects? Represent the above information in a Venn-diagram.

4. a. In a survey of mobile and telephone, it was found that 200 people had telephone, 25 had both, 135 did not have telephone and 300 did not have mobile.i. How many people participated in the survey? ii. How many people had mobile only?

b. In a group of 150 people, 65 like tea only and 60 like milk only. If each people like at least one of the two drinks.i. Draw a Venn-diagram to illustrate the above information.ii. How many people like both drinks? iii. How many people like tea? iv. How many people like milk ?

5. a. In a group of students, 50% like tea, 70% like coffee, 10% don’t like both and 120 like both. By using Venn diagram, find the total number of students.

b. In an examination, 40% of the students passed in Mathematics only, 30% passed in Science only and 10% students failed in both subjects. If 200 students passed in Science, find the total number of students by drawing Venn diagram.

6. a. 50 students in a classroom like Nepali, History or both subjects. Out of them 20 like both subjects. If the ratio of number of students who like Nepali and History is 3:2. Using Venn diagram findi. How many students like Nepali ?ii. How many like History only?

b. In a group of 54 people, every people like music or dance or both. If the ratio of people who like music only and dance only is 5:4 and the number who like both is 18, find the number of people who like dance by using Venn-diagram.

7. a. In a community of 700 people, the ratio of people who like to eat orange only, apple and neither orange nor apple is 1:5:1. Using a Venn-diagram find the number of people who don’t like both of them.

b. In a survey 628 children like only Mathematics and 70 don’t like Mathematics at all.

Also 1828 children like science but 20 like none of them.

i. Show the above data in a Venn diagram.ii. How many children like both type of subjects?


Model Question Group C (4 × 1 = 4 )

1. Out of 120 students who appeared in an examination, the number of students who passed in Mathematics only is twice the number of students who passed in Science only. If 50 students passed in both subjects and 40 students failed in both subjects then

a. Find the number of students who passed in Mathematics. (70)b. Find the number of students who passed in Science. (60)c. Show the result in a Venn- diagram.

Test Yourself1. Define: Power set and cardinality. 2

2. If P = {9, 11, 13, 15} and Q = {10, 11, 12, 13, 14} . Find P ∩ Q and Q – P. 2

3. If ∪ = {1, 2, 3, 4, 5, 6, 7, 8} , A = {1, 3, 4, 5} and B = {1, 2, 3, 4} prove that

(A ∪ B) = A ∩B 2

4. In a group of 200 students who like games, 120 of them like cricket and 105 like football. By drawing Venn-diagram find 4

a. How many students like both the games?b. How many students like only cricket ?.

F. M. 4

F.M. 10

Ans

wer

s 1.3

2. a. ii. 7 b. i. 25 ii. 40 3. a.i. 10 ii. 15 b. 17 4.a. i. 335 ii. 10 b. ii. 25 iii. 90 iv. 85 5.a. 400 b. 400 6.a.i. 42 ii. 8 b. 34 7. a. 100 b. ii. 40


Specific Objectives: At the end of this unit, the students will be able to

• tell the general concept of profit and loss.• find profit or loss with given cost price and selling price of any article.• find profit or loss percent with given cost price and selling price of any article.• find selling price or cost price if profit or loss percent are given.

Teaching Materials : Simple market bill , Chart , price list etc.

Note to the teacher :1. Review the fundamental concept of profit and loss.2. Give the clear concept of profit and loss by giving many examples in a practical way.3. Demonstrate some examples to give the concept of percentage of profit or loss.4. Illustrate some examples to find cost price and selling price giving profit or loss

percent .

Content :• Profit and profit percentage.• Loss and loss percentage.

Specification GridCognitive

Domain/TopicKnowledge

(K)Comprehensive

(C)Application

(A)Higher ability

(HA)Total no. of questions

Total marks

Each 1 mark Each 2 marks Each of 4 marks

Each of 5 marks

Profit and lossCommissionDiscountTaxationDividendHousehold Arithmetic

1

2

1

1

5

14


2Chapter

ArithmeticEstimated period : 25


Review2.1 Profit and Loss : Pratap buys a pen for Rs 65 and sells it for Rs 70. We say he makes a profit of Rs (70 – 65) = Rs 5. If he sells the same pen for Rs 50, we say he has a loss of Rs (65 – 50) = Rs 15

A shopkeeper first purchases an article and sells it. The price which the shopkeeper pays for an article is called the cost prices. In the above example Rs 65 is the cost price. The price for which the shopkeeper sells an article is called the selling price. In the above example Rs 70 and Rs 50 are the selling prices. In short we write C.P. for cost price and S.P. for selling price.

If S.P is more than C.P., there will be a profit.

∴Profit = SP – CP

If SP is less than CP, there will be a loss.

∴Loss = CP – SP Profit and loss percentFrom above, we know that

Profit = SP – CP when SP > CPLoss = CP – SP when CP > SP

When profit and loss amounts are expressed in terms of the percentage of cost price, it is called profit percent or loss percent.

∴Profit percent = Actual profit

CP × 100%

∴Loss percent = Actual loss

CP × 100%When profit % is given :

SP = CP + profitor, SP = CP + profit % of CP

or, SP = CP + p

100 × CP

or, SP = CP P100( )1 +

or, SP = CP (100 + P

100 )

∴CP = SP × 100100 + P

When loss% is given: SP = CP – lossor, SP = CP – loss% of CP

or, SP = CP – L

100 × CP


or, SP = CP L100( )1 -

or, SP = CP (100 - L

100 )

∴CP = SP × 100100 - L

Now, simply we can write SP = CP + profit % of CP SP = CP – loss% of CP

Example 1 : A shopkeeper bought a watch for Rs750 and sold it for Rs 825. Find the profit or loss percentage.

Solution : Here,Cost price (CP) = Rs 750Selling Price (SP) = Rs 825As, SP > CP, there will be profit.Profit = SP – CP = Rs 825 – Rs 750 = Rs 75

Now, Profit % = Profit amount

C.P. × 100 %

= 75750 × 100%

= 10%∴Profit = 10%

Example 2 : The cost price of an article is Rs1200. If it is sold at a loss of 8%, find the selling price of an article.

Solution : Here, Cost price (CP) = Rs 1200 Loss = 8%Selling price (SP) = ?

1st method: Loss amount = 8% of CP

= 8

100 × 1200

Worked Out Examples


= Rs 96 We know, SP = CP – Loss = 1200 – 96

∴ SP = Rs 11042nd Method :

SP = CP – loss% of CP = 1200 – 8% of 1200

= 1200 - 8

100 × 1200

= 1200 – 96 = Rs 1104

∴ SP = Rs. 1104Example 3 : A shopkeeper sold a radio for Rs 960 at 20% loss. Find the cost price of the

radio.Solution : Here,

Selling price (SP) = Rs 960 Loss = 20% Cost price(CP) = ?Let C.P. be Rs xWe know, SP = CP – loss% of CPor, 960 = x – 20% of x

or, 960 = x - 20100 × x

or, 960 = 4x5

or, x = 960 × 5

4 ∴CP = Rs 1200

Example 4 : A shopkeeper sells 20 kg of sugar for the cost of 22 kg, find his gain percent.Solution : Here,

Let CP of 1kg sugar be Rs x.∴CP of 22 kg sugar = Rs 22 x According to question,∴SP of 20kg sugar = CP of 22 kg sugar∴SP of 20kg sugar = Rs 22 x

∴SP of 1kg sugar = Rs 22x20 = Rs

11x10


∴Profit in 1kg sugar = SP of 1 kg sugar – CP of 1kg sugar

= Rs (11x10 - x)

= Rs (11x - 10x

10 )

= Rs x10

Now profit% = profit in 1 kg sugar

CP of 1kg sugar × 100%

= x10x × 100% =

x10 ×

1x × 100%

= 10% ∴Profit = 10%

Example 5 : A merchant sold a radio for Rs 342 at 10% loss. At what price should he sell it to gain 20%?

Solution : Here, 1st part :

Selling price(S.P) = Rs 342Loss = 10%Let CP = Rs xNow, SP = CP – loss% of CPor, 342 = x – 10% of x

or, 342 = x – 10100 × x =

9x10

or, x = 342 × 10

9 ∴ CP = Rs 380

Again 2nd part : CP = Rs 380Gain = 20%SP = ?

we know, SP = CP + gain% of CP = 380 + 20% of 380

= 380 + 20100 × 380

= Rs 456∴ To gain 20%, the radio should be sold at Rs 456.


Example 6 : A merchant sold two pens for Rs100 each. If he gained 25% on the one and lost 25% on the other. Find the gain or loss percent on the total outlay.

Solution : Here, For the 1st pen, SP = Rs 100 Gain = 25%Let CP = Rs xWe have, SP = CP + gain% of CPor, 100 = CP + 25% of CP

or, 100 = x + 25100 × x

or, 100 = 5x4

or, x = 100×4

5 ∴x = Rs 80

For the 2nd pen, SP = Rs 100 Loss = 25% Let CP = Rs yThen, SP = CP – loss% of CP

or, 100 = y - 25100 × y

or, 100 = 4y -y

4

or, 100 = 3y4

or, y = Rs 4003

Total CP = CP of 1st pen + CP of 2nd pen

= Rs 80 + Rs 4003

= Rs 240 + 400

100 × = Rs 6403

Since total CP is more than total SP, Loss = total CP – total SP

= Rs 6403 - Rs 200

= Rs 403

Now, loss% = Loss amount

C.P. × 100


= 403

6403

× 100

= 403 ×

3640 × 100

= 6.25∴Loss = 6.25 %

Example 7 : Dealer Ram Bahadur bought 50 caps for Rs 6000. He sold 15 caps for Rs140 each and 15 caps for Rs 160 each. At what price per cap must he sell the remaining to gain 15% on his outlay?

Solution: Here,CP of 50 caps = Rs 6000

∴CP of 1 cap = Rs 600050 = Rs 120

SP of 1st 15 caps = Rs 15 × 140 = Rs 2100and SP of 2nd 15 caps = Rs 15 × 160 = Rs 2400Total SP of 30 caps = Rs (2100+ 2400) = Rs 4500Gain in total outlay = 15%∴Total SP = Total CP + gain% of Total CP

= 6000 + 15100 × 6000

= Rs 6000 + 900 = Rs 6900∴SP of remaining 20 caps = Rs 6900 – Rs 4500 = Rs 2400

∴SP of remaining 1 cap = Rs 2400640 = Rs 120

So, he has to sell the remaining caps at Rs 120 per cap.Example 8 : Tenzing bought two watches for Rs 400. If he sold them to gain 5% on one and

loss 5% on the other. Calculate his final gain or loss percent if the selling price of both watches is the same.

Solution : Here,Let the CP of 1st watch be Rs x .Then, CP of 2nd watch = Rs ( 400 – x)For the 1st watch, CP = Rs x Gain = 5% ∴SP = CP + gain % of CP

= x + 5% of x = x + 5

100 × x


= Rs 21x20

For the 2nd watch, CP = Rs (400 – x ) Loss = 5 %

∴SP = CP – Loss % of CP = (400 – x) – 5% of ( 400 – x )

= (400 – x) – 5

100 (400 – x )

= 8000 - 20x - 400 + x

20 =

7600-19x20

According to question,SP of 1st watch = SP of 2nd watch

or, 21x20 =

7600-19x20

or, 21 x + 19 x = 7600or, 40 x = 7600

or, x = 760040

∴x = Rs 190

Now SP of 1st watch = Rs 21 x20

= Rs 21 × 190

20 = Rs 199.5 ∴Total SP of two watches = Rs 199.5 × 2 = Rs 399

As the total SP < total CP,Loss = CP – SP = Rs 400 – Rs 399 = Re. 1

Now, loss% = Loss amount

C.P. × 100

= 1

400 × 100

= 0.25 ∴ Final loss is 0.25 %

Example 9 : A gift item when sold at a profit of 5% yields Rs 50 more than when sold at a loss of 5%. Find cost price of the gift item.

Solution: Here,Let CP of the gift item = Rs x


SP at 5% profit = CP + profit % of CP = x + 5% of CP

= x + 5

100 × x

∴SP1 = Rs 21 x20

And SP at 5% loss = CP – loss% of CP = x - 5% of x

= x – 5

100 × x

∴SP2 = Rs 19 x20

According to question,SP1 – SP2 = Rs 50

or, 21 x20 -

19 x20 = 50

or, 2x20 = 50

or, x = Rs 500∴ Cost price of the gift item = Rs 500

Example 10 : The manufacturing price of a table fan is Rs 1440. It is sold at a profit of 40 % to a whole seller. The whole seller sold it at a profit of 25% to a retailer. If the retailer sold it to a customer at 30 % profit. How much does it cost to the customer?

Solution: Here,The problem can be summarized as

P = 40 %Here, cost of manufacturer (CP) = Rs 1440

Profit = 40 % SP of manufacturer = CP + profit % of CP = 1440 + 40 % of 1440 = 1440 + 576 = Rs 2016

But, SP of manufacturer = CP of whole seller ∴ For whole seller, CP = Rs 2016

CP = Rs1440 P = 25% P = 30% CP =?

Manufacturer whole seller Retailer customer


1. a. If S.P. is greater than C. P. then what is the result ? b. If C.P. is greater than S. P. then what is the result ? c. Find the selling price of an article which is bought for Rs x and sold at y % profit. d. If loss is L% and CP is Rs x, find the selling price. e. If the profit is P% and selling price is SP, find cost price. f. If the loss is x % and selling price is Rs y, find the cost price.2. Find the unknown values to complete the table below.

S.N. Name of goods Cost price Selling price profit/loss profit/loss %

a. Basketball Rs 1000 Rs 1150 ? ?

b. Football Rs 1350 Rs 1050 ? ?

c. School bag Rs 850 ? ? 15% (profit)d. T- shirt Rs 1200 ? ? 10.5% (loss)e. Pair of shoes ? Rs 1700 Rs 50 (Profit) ?

3. a. By selling a pen for Rs 42, a stationer makes a profit of 5%. How much did it cost him? b. By selling a pen for Rs 84, a stationer makes a profit of 5%. How much did it cost him? c. By selling a book at Rs 164, there will be 18% loss, find cost price of the book. d. By selling a book at Rs 320, there will be 20% loss, find cost price of the book.

Profit = 25 %∴SP of whole seller = CP + profit % of CP

= 2016 + 25100 × 2016

= Rs 2520SP of whole seller = CP of retailer∴For retailer, CP = Rs 2520 Profit = 30% SP of retailer = CP + profit % of CP

= 2520 + 30100 × 2520

= Rs 3276CP of customer = SP of retailer

∴ It costs Rs 3276 to the customer.

Exercise 2.1


4. a. A shopkeeper bought a radio set for Rs 7500 and sold it at 8% profit. Find his selling price.

b. A shopkeeper bought a radio set for Rs 8500 and sold it at 10% profit. Find his selling price.

c. A shopkeeper bought a TV set for Rs 7500 and sold it at 6% loss. Find his selling price. d. A shopkeeper bought a TV set for Rs 25500 and sold it at 10% loss. Find his selling

price.5. a. A vegetable seller bought lemons at the rate of Rs 250 per 100 and sold them at Rs 34.50

per dozen. Find the profit percent. b. A vegetable seller bought lemons at the rate of Rs 350 per 200 and sold them at Rs 34.50

per dozen. Find the profit percent.6. a. Find the gain or loss percent if 12 bags are sold at the cost price of 15 bags. b. A dealer gains the selling price of two pen drives by selling 10 pen drives. What is his

profit percent?7. a. A shopkeeper sold a pair of socks for Rs 240 at a loss of 10%. At what price should it be

sold to gain 33%? b. A shopkeeper sold a pair of shoes for Rs 1240 at a loss of 15%. At what price should it

be sold to gain 30%?8. a. A shopkeeper bought 1000 glass items at Rs 20 each. Later on it was found that 100

items were broken and he sold the rest at Rs 25 each. Find his gain or loss percent. b. A fruit seller bought 40 kg of mangoes at Rs 75 per kg. 4 kg of mangoes were bad and

not fit for selling. He sold the remaining mangoes making 12.5% profit. At what rate did he sell the mangoes?

c. A shopkeeper in New road Kathmandu sold a Philip electric iron for Rs 2250 and made a loss of 5%. At what price should he sell it to gain 5%?

d. A shopkeeper bought 75 exercise books for Rs 3000. 5 of which were damaged. He sold 24 exercise books at Rs 50 each and 26 at Rs 45 each. At what price should he sell the remaining exercise books to make a profit of 10% on his total outlay?

9. a. A fan was sold at a loss of 20%. Had it been sold for Rs 336 more then the profit would have been 10%. What is the cost price of the fan?

b. A chair was sold at a loss of 25%. Had it been sold for Rs 1250 more then the profit would have been 10%. What is the cost price of the chair?

10. a. A man bought two table fans for Rs 5000. He made 20 % profit on one and lost 20 % on the other. If he lost 4 % in total transaction, what was the cost price of each fan?

b. Rohit bought two bicycles for Rs 6000. He sold one of them at a profit of 20% and another at a loss of 10%. Calculate his final gain or loss percent if the selling price of both bicycles is the same.

11. a. A sold an article to B with 20% profit. B sold the same article for Rs 504 with 5% profit to C. How much did it cost A?

b. A shopkeeper bought two laptops for Rs 91000. He sold one at a loss of 12% and the


other at a gain of 8 %. If he neither gained nor loss in total transaction, find the cost price of each laptop.

12. Ordinarily a dealer Ram Narayan was making a profit of 12%. When his cost price was decreased by 10%, he lowered his selling price also by 10%. Find his new profit percent.

13. a. A shopkeeper uses a false balance and cheats 5 % while purchasing things and again 5% while selling them. Find his percentage gain by cheating.

b. A shopkeeper uses a false balance and cheats 10 % while purchasing things and again 10 % while selling them. Find his percentage gain by cheating.

14. The manufacturing price of a table fan is Rs 15000. It is sold at a profit of 15 % to a whole seller. The whole seller sold it at a profit of 20% to a retailer. If the retailer sold it to a customer at 25 % profit. How much does it cost to the customer?

15. The manufacturing price of a table fan is Rs 2000. It is sold at a profit of 5 % to a whole seller. The whole seller sold it at a profit of 10% to a retailer. If the retailer sold it to a customer at 15 % profit. How much does it cost to the customer.

16. A reduction of 20% in the price of sugar enables Mrs. Shrestha to buy an extra 5 kg of it for Rs 320. Find: i. the original rate and ii. the reduced rate per kg.

Ans

wer

s 2.1

1a. Profit b. Loss c. Rs 100+xy100

d. Rs x (100-L)100

e. Rs 100SP100+P

f. Rs 100y100-x

2.a.Profit = Rs150, Profit%=15 b. Loss = Rs 300, Loss%=22.22

c. SP = Rs977.5, Profit = Rs127.5 d. SP = Rs1074, Loss = Rs126e. CP = Rs1650, Profit% = 3.03 3. a. Rs 40 b.Rs 80c. Rs200 d. Rs 400 4.a.Rs 8100 b.Rs9350 c.Rs7050 d. Rs 22950 5.a.15% b.64.28% 6a.25% b.25% 7.a.Rs 354.67 b. Rs1896.47 8.a. Profit = 12.5% b. Rs 93.75 c. Rs 2486.84 d. Rs 56.50 9a. Rs 1120 b.Rs3571.42 10.a.Rs2000, Rs3000 b. Gain=2.86% 11.a.Rs400 b.Rs36400, Rs54600,12.12% 13.a.10.25% b.21% 14a.Rs25875 15.Rs2656.5 16. i.Rs16 per kg ii.Rs12.80

2.2 Commission: In many Business activities like selling and buying goods or properties, a third party is involved between manufacturers and customer or between seller or buyer for selling goods or any other properties. The third party or agent gets some money from seller or buyer for providing service. This sum of money is called commission. In general, the commission is paid based on the value of goods or properties sold or bought.Amount of commission = Rate of commission (in %) × total sales.


Example 1 : A book publisher announces the commission to the sales agents as below.i. 5% on annual sales up to 10 lakhsii. 10% on annual sales up to 15 lakhsiii. 20% on annual sales more than 15 lakhs

If an agent sells the books of Rs 1200000 in a year, how much commission does the agent get?Solution: Here, according to the given conditions, the amount of commission received by the sales agent = 10% of Rs1200000 = 1200000 = Rs 120000Hence, the required commission is Rs 120000Example 2 : A salesman receives Rs 12000 in a month including Rs 7000 basic monthly

salary. If total sales of a month is Rs 300000, find the rate of commission.Solution:

Here, total monthly income of a salesman = Rs 12000 Monthly salary = Rs 7000∴Amount of commission received = Rs 12000 – Rs 7000 = Rs 5000 Total sales of the month = Rs 300000

Now, Rate of commission = Amount of commission

Total sales × 100 %

= 5000300000 × 100 %

= 53 %

= 123 %

2.2.1 Bonus :When a factory or company or any other business organization runs into profit, it provides certain percentage of the profit to its employees to encourage them in their work. This sum of money given to the employees as incentive beside their fixed salary is called bonus. It is provided on the basis of the percentage of the profit earned.

Example 3 : A commercial Bank makes a profit of rupees 15000000. It decides to distribute 15% of it equally among its 60 employees. How much bonus does each employ receive?

Solution : Here,Total profit made by the Bank = Rs 15000000 Total bonus amount = 15% of 15000000


= 15100 × 15000000

= Rs 2250000Number of employees = 60∴Bonus received by each employee = Rs 2250000

60

= Rs 37500Thus, each employee receive Rs 37500.

1. a. A sales person sells goods of Rs 144200 and gets a commission of 4.5%. Find his commission.

b. An agent sells a second hand car for Rs 625000 and gets a commission of 3.5%. Find the amount of commission he received.

c. How much commission will be Rs 10000 at the rate of 10%.

2. a. A real estate received a commission of Rs 12000 at the rate of 2.5% by selling a piece of land. At what price did he sell the land?

b. A real estate received a commission of Rs 15000 at the rate of 3.5% by selling a piece of land. At what price did he sell the land?

3. a. The monthly salary of Manohar as a sales man is Rs 9000. He gets 2% commission on monthly total sales. If his total income in a month is Rs 15000. What is the total sale of the month?

b. Monthly salary of Chandra Thapa is Rs 5500 and commission of 3% on the total sales for the month. If his total sales for the month is Rs 15, 50,000. What will be his total income of the month?

c. Find the rate of commission if Rs 3180 is received as commission on the total sales Rs 1,06,000.

d. Find the rate of commission if Rs 4500 is received as commission on the total sales Rs 2,10,000.

4. a. A manufacturing company provides 0.5% commission to a dealer for the sales up to Rs 10 lakh, 1% for the sales from 10 lakh to 20 lakh and 2% for the sales more than 20 lakh. If the dealer purchased the goods worth Rs30, 00,000. How much commission did the dealer get?

Exercise 2.2


b. A person working in a noodle factory gets monthly salary Rs 11000. In a month he got Rs 21000 including commission. If total sales of the month is Rs 5,00,000. Find the rate of commission.

5. a. The monthly salary sales of a person in Kathmandu mall is Rs 5000 and an additional payment of 0.5% on the total monthly sales is provided as commission.

i. Calculate his income if he makes a total sales of Rs 7,50,000 in that month.

ii. What should be his total sale in the next month so that he will receive a total income of Rs 10,000 in the month?

b. A house is sold for Rs 6000000 through an agent offering commission of 1% for the first Rs 1000000, 1.5% for the next Rs 25,00,000 and 2% for the rest. Find total commission received by the agent.

6. a. A brick factory distributes 15% bonus equally to its 25 employees from the net profit of Rs 12,50,000 at the end of a year, find the bonus received by each employee.

b. A wool factory distributes 25 % of the profit as bonus to its 60 employees and each employ gets Rs 8000 as bonus. Find the profit of the company.

c. The total profit of a company is Rs 78,00,000 in a year. If 20 employees of a company received Rs 7800 each as bonus. What percentage of the total profit was distributed as bonus?

d. A department store in Kathmandu earned Rs 15,00,000 as profit and distributed Rs 90,000 as bonus to 20 salesmen equally. What percent of the profit was distributed among them? If it is increased by 2%, what amount will each salesman get as bonus?

e. Profit of a business company increased from 15% to 25% and it becomes Rs 4,70,000. If the business company decided to distribute 30% of the increased profit to its 25 employees, how much bonus will each employee receive?

Ans

wer

s 2.2 1.a.Rs6489 b.Rs21875 c.Rs1000 2.a. Rs480000 b. Rs428571.42

3.a.Rs300000 b.Rs52000 c.3% d.2.14% 4.a.Rs35000 b.2% 5.a. i. Rs8750 ii. Rs1000000 b. Rs97500 6.a.Rs7500 b.Rs1920000 c.2% d.6%, Rs6000 e.Rs2256


Notes: The discount is always reckoned on the marked price. Selling price = marked price – discount‘Selling price' is the amount you actually pay for the thing when you purchase.‘Marked price' is the general price of the thing without any discount. ‘Discount' is a percentage of the marked price. Example 1 : The marked price of a ceiling fan is Rs 1250 and the shopkeeper allows a

discount of 6% on it. Find the selling price of the fan. Solution: Here,

Marked price (M.P.) = 1250 Discount = 6%. Discount = 6% of Marked Price = 6% of Rs 1250

= 6

100 × 1250

= Rs 75 Selling price = Marked Price – discount = Rs (1250 – 75) = Rs 1175. Hence, the selling price of the fan is Rs 1175.

Marked Price:

List Price:

Discount:

In big shops and departmental stores, every article is tagged with a card and its price is written on it. This is called the marked price of that article, abbreviated as MP.

Items which are manufactured in a factory are marked with a price according to the list supplied by the factory, at which the retailer is supposed to sell them. This price is known as the list price of the article.

In order to increase the sale or clear the old stock, sometimes the shopkeepers offer a certain percentage of rebate on the marked price. This rebate is known as discount.

2.3 Discount

Sometimes in order to dispose off old goods or to increase the sales or to attract customers to buy their product, shopkeepers offers goods at reduced prices. You must have seen banners like “Special offer, Huge sale, Discount upto 30 %”. You might have seen advertisement of the type “25% discount on bike on the occasion of Dashain” or “20% discount on shirt”, etc. what does it mean? Let us try to understand: 20 % discount on a shirt mean, if the marked price of a shirt is Rs 1000, then 20 % of Rs 1000 will be deducted from Rs 1000 i.e. we have to pay only

Rs (1000-200) or Rs 800 for that shirt of marked price of Rs 1000.


Example 2 : A trader marks his goods at 40% above the cost price and allows a discount of 25%. What is his gain percent?

Solution: Here,

Let the cost price be Rs 100.

Then, marked price = Rs 140.

Discount = 25% of Marked Price

= 25% of Rs 140

= Rs140 × 25100

= Rs 35.

Selling price = Marked price – discount

= Rs (140 – 35)

= Rs 105.

Gain% = (105 - 100) % = 5%.

Hence, the trader gains 5%.

Example 3: A dealer purchased a washing machine for Rs 7660. He allows a discount of 12% on its marked price and still gains 10%. Find the marked price of the machine.

Solution:

Cost price of the machine = Rs 7660, Gain% = 10%.

Therefore, selling price = 100 + G%

100 × CP

=100 + 10

100 × 7660

=110100 × 7660

= Rs 8426.

Let the marked price be Rs x.

Then, the discount = 12% of Rs x

= Rs x × 12100

= Rs 3x25

Therefore, SP = Marked Price – discount

= Rs (x – 3x25 )


= Rs 22x25

But, the SP = Rs 8426.

Therefore, 22x25 = 8426

or, x = 8426 × 25

22 ∴ x = 9575.

Hence, the marked price of the washing machine is Rs 9575.

Example 4 : How much percent above the cost price should a shopkeeper mark his goods so that after allowing a discount of 25% on the marked price, he gains 20%?

Solution: Here,

Let the cost price be Rs 100.

Gain required = 20%.

Therefore, selling price = Rs 120.

Let the marked price be Rs x.

Then, discount = 25% of Rs x

= Rs (x × 25100 )

= Rs x4

Therefore, selling price = Marked Price – discount

= Rs {x - x4 )

= Rs 3x4

Therefore, 3x4 = 120

or, x = 120 × 4

4Therefore, marked price = Rs 160.

Hence, the marked price is 60% above cost price.

Successive Discounts:

If two or more discounts are allowed one after the other then such discounts are known as successive discounts or discounts in series.

Suppose a discount of 25% is given on an article. Then on the reduced price a discount of 15% is given. In such a case, we say that successive discounts of 25% and 15% are given.


Example 5 : Find the single discount equivalent to two successive discounts of 20% and 10%.

Solution:

Let the marked price of an article be Rs 100.

Then, first discount on it = Rs 20.

Price after first discount = Rs (100 – 20) = Rs 80.

Second discount on it = 10% of Rs 80

= Rs80 × 10100 = Rs 8.

Price after second discount = Rs (80 – 8) = Rs 72.

Net selling price = Rs 72.

Single discount equivalent to given successive discounts = (100 – 72)% = 28%.

Example 6 : A shopkeeper selling an article at a discount of 15% gets loss of Rs 150. If it is sold allowing 10% discount, he gains 120. Find the marked price and the cost price of the article.

Solution :

Let the marked price of an article be Rs x

By question,

M.P. = Rs x , Discount = 15%

Then, S.P. = x – 15% of x

= x – 15100 × x

= 100x – 15x

100

= 85x100

= 17x20

Also, Loss = Rs 150

So, C.P. = S.P. + Loss = 17x20 + 150 =

17x + 300020 …………..(i)

Also,

S.P. = M.P. – Discount

= x – 10 % of x

= x – 10100 × x


= 100x – 10x

100

= 90x100

= 9x10

And Profit = Rs 120

Then, C.P. = S.P. – Profit = 9x10 – 120 =

9x – 120010 ……………..(ii)

From equation (i) and (ii), we get

17x + 300020 =

9x – 120010

Or, 2(9x – 1200) = 17x + 3000

Or, 18x – 2400 = 17x + 3000

Or, 18 x – 17 x = 3000 + 2400

∴x = Rs 5400

Putting the value of x in equation (i), we get

= 17x + 3000

20

= 17 × 5400 + 3000

20

= 91800 + 3000

20

= 94800

10= Rs 4740

Thus, the marked price of the article is Rs 5400 and cost price is Rs 4740.

Exercise 2.3

1. a. Define discount . b. Write the formula of discount percent when marked price is Rs ‘x’ and selling price is

Rs ‘y’. c. What is the selling price of an article if marked price is Rs ‘p’ and discount is Rs ‘q’. d. What is the marked price of an article if selling price is Rs ‘m’ and discount ‘n’.


2. a. The marked price of a rice cooker is Rs 1800. If Rs 600 is discounted , find the price after discount.

b. The marked price of a laptop is Rs 45000. If Rs 6000 is discounted , find the price after discount.

3. a. The marked price of a table is Rs 8000 and selling price Rs 7500 , then find the discount amount.

b. The marked price of a T- shirt is Rs 850 and selling price Rs 750, then find the discount amount.

c. If marked price of a honey bottle is Rs 600 and 10% discount is given, then find the discount amount.

d. If marked price of a gallon of oil is Rs 1000 and 12% discount is given, then find the discount amount.

4. a. After allowing a discount of 8% on a watch, it is sold for Rs 800. Find the marked price of the watch.

b. After allowing a discount of 10% on a toy, it is sold for Rs 1500. Find the marked price of the toy.

c. The price of a sweater was slashed from Rs 960 to Rs 816 by a shopkeeper in the winter season. Find the rate of discount given by him.

d. If one fourth of marked price of a mobile is given as discount then, what is the rate of discount?

5. a. The cost price of an article is Rs 650. If a shopkeeper sells the article by allowing 10% discount, find the actual selling price of the article.

b. The marked price of a ceiling fan is Rs 1560 and the shopkeeper allows a discount of 10% on it. Find the selling price of the fan.

6. a. The price of a T.V. is Rs 30000. If the discount is Rs 1560 , find the percent of discount. b. The price of a refrigerator is Rs 40000. If the discount is Rs 2250 , find the percent of

discount.7. a. A tea set was bought for Rs 528 after getting a discount of 10% on its marked price. Find

the marked price of the tea set. b. A Radio was bought for Rs 725 after getting a discount of 10% on its marked price. Find

the marked price of the tea set.8. a. A trader marks his goods at 30% above the cost price and allows a discount of 22%.

What is his gain percent? b. A trader marks his goods at 20% above the cost price and allows a discount of 15%.

What is his gain percent? 9. a. A dealer purchased a washing machine for Rs 45000. He allows a discount of 15% on

its marked price and still gains 10%. Find the marked price of the machine. b. A dealer purchased a T.V. for Rs 80000. He allows a discount of 15% on its marked price

and still gains 10%. Find the marked price of the T.V.


Ans

wer

s 2.3

1. b. x - y

x % c. P-q d. m+n 2. a.Rs1200 b.Rs 39000

3a.Rs500 b.Rs100 c.Rs60 d.Rs120 4a.Rs869.56b.Rs1666.66 c.15% d.25% 5a.Rs585 b.Rs14046.a.5.2% b.5.62% 7a.Rs586.67 b. Rs 805.55 8a.1.4% b.2% 9a.Rs58235.30 b.Rs103529.4 10.a.43.75% b.52.94% 11.a.28% b.32.85% 12.a.MP = Rs15500 CP = Rs13800 b. MP = Rs48000, CP = Rs 41940

2.4 Taxation :We know that government is providing various public services such as health, education, communication, administration, security etc. Government spends a lot of money in development sectors like construction of roads, bridges, hydroelectricity projects etc. For all these government needs money. Do you know from which sources government collects money? The main source of government’s income is the tax paid by its citizens. Government collects money from the people by taxation. The payment made by an individual or a group of individuals to the government is called tax. Government collects money from the people forming different policies and making rules. There are different types of taxes, examples -Income tax, property tax, vehicle tax, entertainment tax, sales tax, value added tax (VAT) etc.

Income tax

The tax imposed by the government on the income of an individual or a group of individuals is called income tax. When the annual income of an individual or any other business organizations is above a certain specified limit, a certain percent of tax is imposed as decided by the government making rules. The minimum needs of an individual are decided by the government according to his/her personal details, examples: marital status, number of dependents etc. The annual income up to the certain limit as per the minimum need is tax free and it is called allowance and the income above the given limit on which the tax is imposed is

10. a. How much percent above the cost price should a shopkeeper mark his goods so that after allowing a discount of 20% on the marked price, he gains 15%?

b. How much percent above the cost price should a shopkeeper mark his goods so that after allowing a discount of 15% on the marked price, he gains 30%?

11. a. Find the single discount equivalent to two successive discounts of 20% and 10%. b. Find the single discount which is equivalent to two successive discounts of 21% and

15%.12. a. A shopkeeper selling an article at a discount of 12% gets loss of Rs 160. If it is sold

allowing 10% discount, he gains 150. Find the marked price and the cost price of the article.

b. A shopkeeper selling an article at a discount of 13% gets loss of Rs 180. If it is sold allowing 12% discount, he gains 300. Find the marked price and the cost price of the article.


Value Added Tax (VAT):

Tax levied on the cost of goods and services provided by the government is called value added tax (VAT). The VAT rate is decided by the government.

The VAT rate varies from country to country and time to time. At first in our country it was 10% but now it is13%. VAT is always added on the actual selling price.

VAT amount = Rate of VAT (in%) × selling price.

If goods are sold by giving discount on the marked price, first we get the actual selling price deducting discount from the marked price.

i.e. SP after discount = MP – discount amount.

SP with VAT = SP + VAT amount.

Discount: It is the amount which is reduced from the marked price. The rate of discount is decided by the shopkeeper.It is the percentage of marked price.

Discount% = Discount amountMarked price

× 100

Note: 1. VAT is not imposed for all types of goods we purchase and for all types of services. 2. There is no VAT in export.

Taxable income =Total yearly income-yearly allowance along with sum on the above headings (if any)

Income tax = Rate of tax (in%) × Taxable income.

called taxable income. There are some more headings on which tax is not imposed. They are:1. Sum deposited as provident fund of government service holders.2. Sum deposited in citizen investment fund.3. Premium paid for life insurance.4. Expenditure on religious purpose.5. Donation.6. Additional money received for working in remote area.7. Up to 75% of the income which is earned as foreign employment.8. Expenditure in health treatment.

To calculate the income tax, first we have to deduct the tax free allowance along with depositions or expenditure on above heading (if any) to get the taxable income. Then the tax is calculated as per the conditions and rate fixed by the government. The conditions and rate of the tax are changed time to time by the government. Government has imposed 1% social security tax in the income.


Example 1 : Chandra chaudhary purchased a mobile set for Rs12000 after 15% discount. How much money did he pay for the mobile with 13% value added tax?

Solution: Here,Marked price of a mobile set = Rs 12000Discount amount = 13%of Rs 12000

= 13100 × 12000

= Rs 1560SP with discount = Rs 12000-Rs 1560

= Rs 10440VAT amount = 13%of Rs 10440

= 13100 × 10440

= Rs 1357.2SP with VAT = Rs 10440 + Rs 1357.2

= Rs 11797.2 Chandra chaudhary paid Rs11797.2 for the mobile set.

Example 2 : There are 800 students in an institutional school. The average monthly fee of a student is Rs 1500.How much education service tax has to pay by the school at 1% service tax?

Solution: Here,Average monthly fee of a student = Rs 1500Average monthly fee of 800 students = Rs 800 × 1500 = Rs 1200000 Total yearly fee = Rs12 × 1200000 = Rs 14400000Now, Education tax = 1% of 14400000

= 1

100 × 14400000

= Rs 1,44,000Example 3 : The monthly income of an unmarried service holder is Rs 50,000. If 1% social

security tax is levied up to the annual income of 3,50,000. 15% tax is charged above the annual income on Rs 3,50,000 to 4,50,000 and 25% tax is charged above Rs 4,50,000. How much income tax should he pay in a year?

Solution : Here,

Worked Out Examples


Monthly income of a service holder = Rs 50,000∴Yearly income = Rs 12 × 50,000 = Rs 600000 Social security tax = 1 % of Rs 3,50,000

= 1

100 × 350000

= Rs 3500Now, yearly taxable income = Rs 600000 – Rs350000 = Rs250000Income tax for Rs 100000(Rs 350000 to Rs 450000) = 15% of Rs 1,00,000 = Rs 15000Income tax for remaining income more than Rs 450000 = 25% of Rs150000 = Rs 37500∴Total amount of income tax = Rs 3500 + Rs 15000 + Rs 37500 = Rs 56000

Example 4 : Pancha Narayan Maharjan paid Rs 67000 for a laptop with discount of 20% including 13% VAT. What is the marked price of the laptop?

Solution:Let marked price of the laptop = Rs xNow, SP of laptop = marked price – discount% of marked price = x – 20% of x

= x – 20100 × x

= 5x-x

5

= 4x5

SP with VAT = 4x5 + 13% of

4x5

or, 67000 = 4x5 +

13100 ×

4x5

or, 67000 = 4x5 +

13125

or, 67000 = 100x + 13x

125 or, 113x = 67000 × 125

or, x = 67000× 125

113 ∴x = Rs 74115.04∴The required marked price = Rs 74115.04


1. a. What is meant by Income tax ? b. What do you understand by VAT ? c. Write the formula of VAT. d. Write the formula of percent of VAT? e. If M.P. is ‘x’ , discount is ‘y’ , find S.P.2. a. Annual income of an unmarried individual is Rs 2,70,000. How much social security tax

should be paid by him in a year at the rate of 1%. b. The monthly income of a couple is Rs 30,000. If 1% social security tax is levied up to the

annual income of Rs 4,00,000 to a married couple. How much tax should the couple pay in a year?

3. a. There are 650 students in an institutional school. The average monthly fee of a student is Rs 2200. How much education service tax has to be paid by the school in a year at the rate of 1%?

b. The marked price of a bicycle is Rs 5000. If it is sold at Rs 4500, find the discount percent. c. An article is sold on Rs 7500 at a discount of 10%, find the marked price of the article. d. The selling price of an article is Rs 3500. Find its price with 10% VAT. e. A customer pays Rs 6900 for an article with the VAT of Rs1200, find the VAT percent.4. a. The marked price of a radio set is Rs 7520. What will be the price of the radio if 13% VAT is

levied after allowing 10% discount on it? b. A tourist bought a statue at a discount of 20% with 8% VAT. If the VAT amount was

Rs 512, calculate the marked price and the amount of discount. c. A merchant sold a television set for Rs 16950 allowing 20% discount and imposing 13%

VAT, what was the amount of discount? 5. a. After allowing 15% discount on the marked price of a computer, 13% VAT was levied on it.

If its price becomes Rs 23500, what amount was levied on VAT? b. After allowing 15% discount on a marked price of an article and levying 13% VAT, the price

of the article becomes Rs 10,000. Find the VAT amount. c. After allowing 5% discount on the marked price of a watch, 10% VAT is charged on it. If its

price became Rs 1672, find the discount amount.6. a. The annual income of a person is Rs 1,50,000. If the tax allowance is Rs. 90,000 and the tax

rate is 12%. Find the income tax of the person. b. The monthly income of a couple is Rs 50,000. If 1% social security tax is levied up to the

annual income of Rs 4,00,000 and 10% tax should be paid for the remaining income. How much tax should the couple pay in a year?

7. a. The monthly income of an unmarried individual is Rs 35500. If 1% social security tax is charged up to the annual income of Rs 3,50,000 and 15% tax is charged above the annual income of Rs 3,50,000 to 4,50,000 and 25% above Rs 4,50,000. How much income tax should be paid by an individual in a year?

b. The monthly income of a married bank manager is Rs 85,000.10% of his monthly income is deducted and deposited as provident fund. If 1% social security tax is levied up to the annual income of Rs 4,00,000, 15% tax is levied on Rs 4,00,000 to Rs 5,00,000 and 25% tax is levied above Rs 5,00,000. How much income tax should he pay in a year.

Exercise 2.4


c. The monthly income of a married civil servant is Rs 32,500. 10% of the salary is deducted as provident fund and another 10% is invested in citizen investment fund. 1% social security tax should be paid up to Rs 4,00,000, 10% tax from Rs 4,00,000 to Rs 5,00,000 and 25% for more than Rs 5,00,000 . How much income tax does she pay in a year?

d. The monthly income of an individual is Rs 75000. 10% of the income of an individual is deducted as provident fund and another 10% is deducted as citizen investment trust. An individual pays Rs 20,000 annually as the premium of life insurance. 1% social security tax is levied up to the yearly income of Rs 3,50,000. 15% tax should be paid from Rs 3,50,000 up to Rs 4,50,000 and 25% for more than 4,50,000. How much tax should be paid by an individual in a year?

8. a. The manufacturing price of two different kinds of articles A and B is Rs 6000 each. The sales tax on these article is not same but a VAT of 10% is added on each after adding the sales tax. If the selling price of article A is Rs 6798 and that of B is Rs 6864, find the rate of sales tax on each kind.

b. The manufacturing price of two different kinds of articles P and Q is Rs 8000 each. The sales tax on these article is not same but a VAT of 13% is added on each after adding the sales tax. If the selling price of article A is Rs 9600 and that of B is Rs 9690, find the rate of sales tax on each kind.

Ans

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s 2.4

1. c.VAT amount = Rate of VAT×SP d. VAT %= VAT amountSP 100%

e. Rs (x – y ) 2.a.Rs 2700 b.Rs 3600 3.a.Rs 171600 b.10% c. Rs 8333.33 d. Rs 3850 e. 21.05% 4.a.Rs 7647.84 b. MP = Rs 8000, Discount = Rs 1600 c.Rs 3750 5a.Rs 2703.54b.Rs 1150.44 c. Rs 80 6.a.Rs 7200 b.Rs 24000 7.a. Rs 14900 b.Rs 123500 c.Rs 3159 d. Rs 83250 8.a. 3% and 4% b. 6.2% and 7.2%

2.5 DividendA huge amount of capital is required to establish a company or to construct any mega project. The required amount of capital is divided into unit value and collected from the public. The small unit of capital is a share and the person who takes the share is called share holder. When the company makes a profit, it divides certain part of profit to share holders in the ratio of number of shares that is called dividend.

Let's study the examples1. A commercial bank has a current capital of Rs20,00,00,000. The value of 1 share is Rs100.

Then the number of shares

= Total capital

Value of one share

= 20,00,00,000

100

= 20,00,000


1. The primary value of the share of a hydro electricity project is Rs 100 and its capital is divided into 50,000 shares. If the project made a profit of Rs 20,00,000 in a year and the Board of director of the project decided to provide 15% of the total profit to the share holders thena. How much dividend does each share get?b. Aakriti has 1500 shares of the company, how much dividend does she get?

Solution :Here number of shares = 50,000Total amount of profit = Rs 20,00,000Rate of dividend = 15%

∴Total dividend = 15100 × 20,00,000

= Rs 300000

a. per share dividend = Total dividend

Total number of shares

= 30000050000

= Rs 6b. Dividend received by Aakriti = Rs 1500 × 6 = Rs 9000

If the bank made a profit of Rs 5,00,00,000 in 2073 and the board of director decided to provide 20% of the total profit as a dividend to each share holder.

Then, total amount of dividend = 20% of 5,00,00,000

= 20100

× 5,00,00,000

= Rs 1,00,00,000

Now Dividend per share = Total amount of dividend


= 1,00,00,00020,00,000

= Rs 5

So, if a person has 500 shares of the bank , then he will get Rs 500 × 5 = Rs 2500 as dividend.

Worked Out Examples


1. Complete the following table.

Net profit Rate of dividend Number of shares Dividend per sharesRs 25,00,000 10% 2500 ?Rs 60,00,000 5% 3000 ?Rs 75,00,000 15% ? Rs 25Rs 1,25,00,000 ? 5000 Rs 50

2. A company has 20,000 shares and made a profit of Rs 75,00,000. The company declared to distribute 5% of net profit to its share holdersThen a. Find per share dividend.

b. How much does a share holder having 250 shares get as dividend?3. A hydropower company has 20,000 shares at the rate of Rs 100 per share. Company

made profit of Rs 5,00,000 in a year and decided to provide 20% of the profit as dividend. If Sony has 300 shares of the company, how much dividend did she receive?

Facts to remember

1. Per share dividend = Total dividend


2. Total amount of dividend = Net profit × Rate of dividend.3. Dividend received by a share holder having ‘n’ shares = per share dividend × n

Exercise 2.5

Answers 2.51. Rs 100, Rs 100, 45000, 2% 2.a.Rs 18.75 b.Rs 4687.50 3.Rs 1500

2.6 Home Arithmetic Introduction:Home arithmetic is the study of regular financial activities in our household. We perform various financial activities in our household daily, weekly or on monthly basis. Among these activities, in this chapter we discuss the expenditure of a house hold for utilizing some facilities provided by the government such as electricity, water and telephone. 2.7.1 Electricity BillHousehold expenses for use of electricityThe consumption of electricity by a house hold is measured in units with the help of electric


meters which is fixed in every household.1 unit of electricity = 1 kilowatt hour or 1000 watt hour

So, the amount of consumption of electricity by an electric appliance of 1000 watt power in 1 hour is 1 unit. The number of units of electricity consumed in a month is given by

Consumed units = present reading – previous reading

Here present reading means reading of the current month and previous reading means the reading of previous month which is in record.

For the payment of electricity bill, the electricity authority has implemented the following rules.

Per unit charge and service charge for the different range of consumed units for different capacity meters are different.

Nepal Electricity Authority has implemented the following rules for the payment of the consumption of electricity.

The table given below shows the capacity of meter of lower voltage limit 220 – 400 volt connected in customer’s house and minimum energy charges.

Meter categories Minimum charge Minimum unitUp to 5A Rs 80 206A – 15A Rs 365 50

16A – 30 A Rs 795 10031A – 60A Rs 1765 200

According to the electricity tariff rates implemented by Nepal Electricity Authority in 2071, 1. Minimum charge up to 20 units Rs 4 per unit.2. Charge from 21 units to 50 units is Rs 7.30 per unit.

I. If the consumed units are up to 30 units, up to 20 units at the rate of Rs 4 per unit and above 20 units at the rate of Rs 7.30 per unit.

II. If the consumed units are more then 30, then the charge is at the rate of Rs 7.30 per unit for whole consumed units.

3. Charge from 51 units to 150 units is Rs 8.60 per unitI. In this case, charge from 0 – 50 units is Rs 7.30 per unit II. Charge from 51 – 150 units is Rs 8.60 per unit.

4. Charge from 150 units to 250 units is rs 9.50 per unit.I. In this case charge from 0- 150 unit is Rs 8.60 per unit.II. Charge from 151 – 250 units is Rs 9.50 per unit

5. Charge above 250 units is Rs 11.0 per unitI. In this case, charge from 0 – 250 units is Rs 9.50 per unitII. Charge above 250 units is Rs 11.0 per unit


Example 1 : The meter reading for the consumption of electricity of a household was 1150 units on 1stBaisakh and 1275 units on 1stJestha. Calculate the amount of money to be paid following the rules given above, if the payment is made within 7 days of meter reading.

Solution: Here,Previous reading = 1150 unitsPresent reading = 1275 unitsConsumed units = 1275 – 1150 units = 125Charge from 0 – 50 units = Rs 7.30 × 50 = Rs 365Number of excessive units = 125 – 50 = 75Charge of 75 units (51- 150 units) = Rs 8.60 × 75 = Rs 645∴Total charge of electricity = Rs 365 + Rs 645 = Rs 1010As the payment is made within 7 days of meter reading,Discount amount = 3 % of Rs 1010

= Rs 3

100× 1010

= Rs 30.30∴ The charge with discount = Rs 1010 – Rs 30.30 = RS 979.90So, the household should pay Rs 979.70 to clear the bill.

The payment schedule after the meter reading are as follows.1. Within 7 days of meter reading 3 % discount is allowed but if minimum bill amounts is

Rs 80, the flat discount of Rs 4 is allowed.2. From 8th day to 22nd day, the payment will be according to the bill.3. From 23rd day to 30th day, 5 % extra fine.4. From 31st day to 40th day, 10 % extra fine.5. From 41st day to 60th day, 25% extra fine.6. If the bill is not paid up to 60th day, the electricity line will

be disconnected.

Worked Out Examples


Example 2 : If a household paid normal charge of Rs 1527.50 for the consumption of electricity in a month. Then how many units of electricity was consumed?

Solution: Here,According to the rules of NEA as given above,Charge up to 150 units = Rs 8.60 × 150 = Rs 1290Extra charge = Rs 1527.50 – Rs 1290 = Rs 237.50Per unit charge for units 151 – 250 is Rs 9.50

∴ Consumed units between 151 – 250 = Rs 237.50Rs 9.50 = 25

∴ Total consumed units of electricity = 150 + 25 = 175

2.7.2 Water Bill

The amount of water used in household is measured in units. It is measured with the help of meter connected to the pipe supplying water in the household.

1000 liter water = 1 unit.

Water tariff rules implemented by Nepal Water Supply Corporation and Kathmandu Upatyaka Khanepani Limited (KUKL) are slightly different.

We follow the following water tariff rules implement by KUKL in this book.

Size of pipe (inch)

Minimum consumption (in litre)

Tap with meter

Minimum charges(Rs)

Additional consumption per 1000 liter(Rs)

12

” 10,000 100 32

34

” 27,000 1910 71

1” 56,000 3960 711.5” 1,55,000 10950 712” 3,20,000 22600 713” 8,81,000 62240 714” 1,810,000 127865 71

Pipes having size more than 12( )” are used in big houses, hotels, hospitals, colony, industry

etc.


Implemented rules and regulations about the schedule of payment of the bill.

payment after the bill issued rebate/fine1.within first and second month 3% rebate

2. within third months No rebate and no fine

3. within the forth month 10% fine

4. within the fifth month 20% fine

5. after fifth month 50% fine

50% of the bill amount is added as sewerage service charge.

Example 1 : The meter reading for the consumption of water of a household was 1550 untils on the 1stMangshir and 1600 units on the 1stPoush. Calculate the charge to be paid including 50% sewerage charge, if the payment of the bill is made within third month after the bill issued.(Take the help of table above)

Solution:Here, previous reading = 1550 unitsCurrent reading = 1600 unitsConsumed units of water = Rs 1600 – 1550 = 50 units

According to the rules given above, considering 1"2 pipe

Charge of minimum 10 units (10, 000 litres) = Rs 100Additional units of water consumption = 50 – 10 = 40 unitsCharge of 40 units = Rs 40 × 32 = Rs 1280Total charge = Rs 100 + Rs 1280 = Rs 1380Sewerage service charge = 50% of bill amount

= Rs 50100 × 1380

= Rs 690∴The charge including 50% sewerage service charge = Rs 1380 + Rs 690

= Rs 2070

Worked Out Examples


Example 2: One 1" pipe is supplying water in a new colony in Budhanilkantha Kathmandu. If 125 units water is consumed in a month, how much bill should be paid by the colony including 50% sewerage charge?

Solution: According to the water tariff rules given above for 1" pipe, charge of minimum 56 units (56000 litres) = Rs 3960.Number of additional units = 125 - 56 = 69 unitscharge of 69 units = Rs 69 × 71 = Rs 4899Total charge = Rs 3960 + Rs 4899 = Rs8859Sewerage service charge = 50% of 8859

= Rs 50100 × 8859

= Rs 4429.50∴ The total amount of bill that should be paid by the colony = Rs 8859 + Rs 4429.50 = Rs 13288.50

2.7.3 Telephone Bill

Nowadays most of the people use recharge card for prepaid and post paid mobile service. But here we discuss about the telephone bill of PSTN/Landline.

The rules of payment of the telephone bills implemented by Nepal Telecome are as follows:

- Minimum charge of first 175 calls is Rs 200.- Rate of charge above 175 calls is Rs 1 per call.- Charge of telephone = minimum charge + charge of additional calls

10% Telephone Service Charge (TSC) and 13% VAT are imposed in telephone bill.Charge with 10% TSC = bill amount + 10% of bill amountCharge with 10% TSC and 13% VAT = charge with TSC + 13% of charge with TSC

Example 1 : The minimum charge up to 175 calls is Rs200. If the charge for each additional call is Re1. How much will be the charge for 315 calls with 10% TSC and 13% VAT?

Solution: Here,Minimum charge for 175 calls = Rs 200 Number of additional calls = 315 – 175 = 140

Worked Out Examples


charge of 140 additional calls = Rs 140 × 1 = Rs 140∴Charge of 315 calls = Rs 200 + Rs 140 = Rs 340Charge with 10% TSC= Rs 340 + 10% of Rs 340

= 340 + 10100 × 340

= Rs 374Charge with 10% TSC and 13% VAT = Rs 374 + 13% of Rs374

= 374 + 13100 × 374

= Rs 422.62So the total charge of 315 calls = Rs 422.62

Example 2 : The minimum charge of telephone call up to 175 calls is Rs 200 and the charge for each additional call is Re 1. If a man paid Rs 675 with 10% TSC and 13% VAT of his one month telephone bill, find number of calls made in the month.

Solution : Here, Charge of 175 calls = Rs 200Let the number of extra calls = x Charge of x extra calls = Rs 1× x = Rs xTotal charge = Rs (x + 200)Charge with 10% TSC = Rs (x + 200) + Rs 10 % of (x + 200)

= (x+200) + 10100 (x + 200)

= Rs 10x + 2000 + x + 200

10 = Rs 11x + 220010

Charge with 10% TSC and 13% VAT

or, 675 = (11x + 2200

10 ) + 13% of (11x + 2200

10 )

or, 675 = 11x + 2200

10 + 13100 (

11x + 220010 )

or, 675 = 11x + 2200

10 + 13 (11x + 2200)

1000

or, 675 = 1100x + 220000 + 143x + 28600

1000

or, 675 = 1243x + 248600

1000or, 1243x + 248600 = 675000


or, 1243x = 426400

or, x = 4264001243

= 343∴Total number of calls made = 175 + 343 = 518

2.7.4 Calculation of taxi fare:

Taxi is one of the most essential means of transportation nowadays. We can hire taxi to go from one place to another in big cities like Kathmandu, Pokhara etc. How much money do we have to pay while using taxi? There is a device fixed in a taxi which records the distance covered and the fare payable. This device is called taximeter.

There is a National Bureau of Standard and Metrology (NBSM) in our country which implements all the rules, regulations, and rate of taxi fare.

For the calculation of fare, NBSM has implemented the following rates.

Time Minimum fare Fare per extra 200 meters6:00 am to 9:00 pm Rs 14 Rs 7.209:00 pm to 6:00 am Rs 21

(1.5 times of the fair of 6:00 am to 9:00 pm)

Rs 10 .80(1.5 times of the fair of 6:00 am to 9:00 pm)

The above table shows the minimum fare which appears on the monitor of the meter, the minimum charge. When it is flagged down between 6:00 am to 9:00 pm it shows Rs 14. Then the fare goes on at the rate of Rs 7.20 per 200 m. but while travelling from 9:00 pm to 6:00 am, the minimum charge is Rs 21 and Rs 10.80 for every 200 m.

If the taxi is asked to wait by the passenger waiting charge of Rs 7.20 per 2 minutes should be paid between 6:00 am to 9:00 pm and the waiting charge between 9:00 pm to 6:00 am is Rs 10.80 per 2 minutes.

Example 1 : If a man hire a taxi at 9:00 am to reach Bir hospital which is at a distance of 7.5 km from his house. How much fair should he pay following the above rates?

Solution: At 9:00amMinimum charge = Rs 14Distance of hospital = 7.5 km = 7.5 × 1000 m = 7500mFare of 200 meters = Rs 7.20

∴Fare of 1 meter = Rs 7.20200

Worked Out Examples


∴Fare of 7500 meter = Rs 7.20200 × 7500

= Rs 270∴Total fare = Rs 14 + Rs 270 = Rs 284

Example 2 : Pradip hired a taxi and travelled a certain distance at 12 pm. He paid the total fare of Rs 915 including additional waiting charge for 10 minutes. If the minimum fare is Rs 21 at first and Rs 10.80 per 200 m and the waiting charge is Rs 10.80 per 2 minutes. Find the distance travelled by him.

Solution: Minimum charge = Rs 21Waiting charge for 2 minutes = Rs 10.80

Waiting charge for 1 minute = Rs 10.80

2

Waiting charge for 10 minute = Rs 10.80

2 × 10 = Rs 54

Remaining fare = Rs 915 – Rs 54 - 21 = Rs 840Rs10.80 is fare of 200 m

∴Re 1 is fare of 200

10.80 m

∴Rs 840 is fare of 200

10.80 × 840

= 15555.55 m (approximately)

= 15555.55

1000 Km

= 15.56 km∴Distance travelled by Pradip = 15.56 km.

1. a. Write the formula to find the electricity consumed units when previous reading and current reading are known.

b. In an electricity meter reading it was found that previous reading is 4762 units and consumed 778 units. Find the present reading.

c. How many units of electricity will be consumed when a water pump of 1000 watts is used for 1 hour?

d. The charge of electricity is Rs 2750 at the rate of Rs 11 per unit find the consumed unit. e. The school tank contains 25,000 liters of water. How many units of water are there? f. Find the water charge of 56 units at the rate of Rs 32 per unit. g. In a telephone bill of a house of a month,previous reading 4272 and present reading

5497 are shown. How many calls are made there?

Exercise 2.6


h. The cost of telephone call in a landline phone is Rs 350. Find the 10% TSC amount. i. What is the taxi fare to travel 16,000m at the rate of Rs 7.20 per 200m? j. How many km can be travelled by Rs3600 in a taxi at the rate of Rs 36 per km?2. a. The charge of electricity is Rs 80 for the first 20 units and Rs 7.30 per unit from 21 units.

Find the charge of 35 units of consumption of electricity. b. The minimum charge of electricity is Rs 80 for the first 20 units and Rs 7.30 per units

from 21 units to 50 units, Rs 8.60 per unit from 51 to 150 units and Rs 9.50 per unit from 151 to 250 units. If 275 units of electricity is consumed in a household in one month, how much charge should be paid by the household if the payment is made within 7 days of meter reading.

c. The electricity meter reading of a household was 1225 units on the first Poush and 1450units on the first Magh. Calculate the charge to be paid for that month using the following rules.Minimum charge up to 20 units = Rs. 80Charge from 21 to 50 units = Rs 7.30 per unitCharge from 51 to 150 units = Rs 8.60 per unitCharge from 151 to 250 units = Rs 9.50 per unitpayment is made on the fifth of next month.

d. The meter reading of the consumption of electricity of a household was 2120 units on first of Jestha and 2265 units on the first of Ashar. Calculate the total charge of electricity following the rules implemented by Nepal Electricity Authority as mentioned in a text box.i. payment was made within 7 days of meter reading.ii. payment was made within 12th days of meter readingiii. payment was made within 25th of meter reading.iv. payment was made within 37th days of meter reading.v. payment was made within 50th days of meter reading.

3. a. A household paid the normal charge of consumption of electricity as Rs 798.50 for the month of Baisakh. Calculate the total consumed units of electricity for that month if the charge from 0 to 50 units is Rs 7.30 per unit and from 51 to 150 units is Rs 8.60 per unit.

b. The charge of consumption of electricity of a household for a month was Rs 1060.75 with 3% rebate. If the per unit charge of electricity is Rs 7.30 for 0 to 50 units and Rs 8.60 for 51 to 150 units. Find the total consumed units.

c. In a house 6 bulbs each of 25watts are glown for 6 hours, an electric heater of 1200watts is used for 3 hours and a rice cooker of 1000 watts is used for 2 hours per day. Calculate the monthly charge for the consumption of electricity if the charge from 0 to 50 units is Rs 7.30 per unit , from 51 to 150 units is Rs 8.60 per unit and from 151 to 250 units is Rs 9.50 per unit.

4. a. The minimum charge of land line telephone up to 175 calls is Rs 200. If the charge for each additional call is Re 1, how much charge should be paid for 375 calls with 10% telephone service charge and 13% VAT?

b. The minimum charge of land line telephone up to 175 calls is Rs 200. The charge for each additional call is Re 1. If the household paid Rs 520.30 with 10% of TSC and 13% VAT to clear the bill of one month. Find the number of additional calls made in the month.


c. The charge of ISD call for UK from Kathmandu is Rs 25 per minute. If Sushmita made the call for 5 minutes to her friend in UK from Kathmandu, how much charge should she pay with 10% TSC and 13% VAT?

d. The charge of ISD from Biratnagar to Tokyo is Rs 125 for the first 3 minutes and Rs 15 for every additional 10 seconds. If a man made a call from Biratnagar to his friend in Tokyo and paid Rs 305, how long did he make a call?

e. The telephone rate up to 6 am is Rs 5 for a call of 3 minutes. From 6 am to 6 pm, the charge is Rs 2 for every 1 minute. If Ram calls his friend from 5:55 am to 6:05 am, find the amount he paid with 10% TSC and 13% VAT.

5. a. A household consumed 35 units of water in a month using 1/2” pipe. Calculate the amount of charge paid by the household including 50% sewerage service charge if the payment is made within the 12th day of the meter reading.

b. A colony consumed 215 units water connecting 3/4” pipe in a month. If the payment of the bill is made within the fourth month after the bill is issued. How much money is required to clear the bill including 50% sewerage service charges.

c. The meter reading of water on 5th day of Asar 2074 was 1523 units and that of 5th day of Shrawan was 1547 units. If the size of the pipe connected is 1/2”, how much money should be paid to clear the bill including 50% sewerage service charge according to the rates implemented by Nepal Water Supply Corporation as mentioned in a book above if the payment is made within the 3rd month of meter reading.

6. a. A man hired a taxi to go to the hospital which is at a distance of 8.5Km from his house at 3:00pm. If the minimum fare is Rs 14 and the fare goes on the rate of Rs 7.20 per 200 meters. Calculate the total taxi fare paid by the man.

b. A taxi passenger travelled 12Km distance by hiring a taxi at 11:00pm. The minimum fare is Rs 21 which appears immediately when the meter was flagged down. Then the fare is Rs 10.80 for every 200 meters. He has to wait on the way for 10 minutes for his friend. If the waiting fare is Rs 10.80 per 2 minutes, calculate the total fare paid by the passenger.

c. Ram Prasad hired a taxi at 10:00 am and paid total fare of Rs 360.75. If minimum fare is Rs 14 and the fare per 200 meters is Rs 7.20, find the distance traveled by him.

d. A woman paid Rs 244 for hiring a taxi whose minimum charge is Rs 14 and then Rs 36 per Km. If the charge includes waiting charge of Rs 50 waiting for 5 minutes, what is the length of the journey?

Ans

wer

s 2.6

1. a. consumed units = current reading - previous reading b. 5540 units c. 1unit d. 250 units e. 25 units f. Rs 1792 g. 1225units h. Rs 35 i. Rs 576 j. 100 KM 2.a. Rs 255.50 b. Rs 2570.50 c. Rs 2131.25 d.i. Rs 1146.54 ii. Rs 1182 iii. Rs1241.1 iv. Rs 1300.2 v. Rs 1477.5 3.a. 101 units b. 135 units c. Rs 1652.54.a. Rs 497.2 b. 219 c. Rs 155.37 d. 5 min e. Rs 22.78 5.a. Rs 1309.5 b. Rs 25175.7 c. Rs 822 6.a. Rs 320 b. Rs723 c. 9.63Km d. 5Km


Group - A ( 3 × 1 = 3 )

1. a. The cost price of an article be Rs 4000 and its loss be Rs 80. Find the selling price of that article.

b. A commission agent sells some article for Rs 300. What commission does he get at 5.6%

c. In a electricity meter reading, previous reading 1543 and present reading 2225 were shown. Find the consumed units.

Group – B ( 2 × 2 = 4 )

2. a. A company gains Rs 570000 in a year. If 5% of profit is divided among 150 employees as bonus, how much will each employee get?

c. The water meter reading on 1st Chaitra and 1st Baisakh are 05925 and 05938 respectively. Calculate the total amount to be paid for the month of Chaitra under the following rules.

Minimum charge up to 10 units = Rs 100

Per unit charge above 10 units = Rs 30

Sewerage charge = 50% of the cost of water consumption charge

Water meter rental charge = Rs 5

Group – C ( 3 × 4 = 12 )

3. a. A business firm has a monthly income of Rs 56,000. The business firm is free of tax up to the income of Rs 2,00,000. Out of the taxable income, Rs 2,50,000 was taxed at the rate of 15% and the rest at 25%. Find the amount of the tax paid by the business firm.

b. A company runs with Rs 18500000 investment. The company has profit of 8 % of the investment and 6 % of the profit is distributed among 250 employee as bonus. How much bonus will each employee receive?

c. Sudan travelled 8 km from Kathmandu New Bus park to Kalanki by taxi. The minimum fare of taxi is Rs 15 at the first and the fare goes on the rate of Rs 7.20 per 200 m. The waiting charge of Rs 7.30 per a 2 minute was charged for the waiting of 20 minutes during the journey. How much total fare including 25% tips is paid by Sudan?

Group- D (1 × 5 = 5)

4. Marked price of an article is 20 % above its purchased price. If the shopkeeper sold it by allowing 15% discount, then

i. Calculate the profit percentage.

ii. Calculate the marked price and profit amount if the shopkeeper paid Rs2500 for the article.

Model Question F.M. - 24


Test yourself F.M. - 30

1. a. What is the formula for S.P. if C.P. and loss percent are given.

b. If M.P. is ‘a’ and discount is ‘b’ then find S.P.

c. Define : Commission with example.

d. The profit of one share is Rs 25. Find the profit of 600 shares.

e. In an electricity meter reading, previous reading of 1620 and present reading of 2130 were shown. Find the consumed unit.

f. What do you understand by seven units water ?

2. a. The marked price of an article is Rs 2500. If it is sold for Rs 2000, find the rate of discount.

b. What will be total sales to get commission of Rs 1200 at the rate of 3 13 % ?

c. The net profit of a life insurance company is Rs 75 crore. If 30 % of profit is distributed as dividend, find the total dividend.

d. A man hired a taxi and travelled 11 km. If minimum fare is Rs 14 and the fare goes on at the rate of Rs 7.20 per 200 m , calculate total fare paid by the man.

3. a. Anil bought two watches for Rs 500. He sold one at a loss of 12% and other at a gain of 8%. He neither gained nor lost in his transaction. Find the cost of each watch.

b. The marked price of a radio was 40% above the cost price. When it was sold allowing 30% discount on it there was loss of Rs 100, what was the marked price?

c. The monthly income of an unmarried individual is Rs 55500. If 1% social security tax is charged up to the annual income of Rs 3,50,000 and 15% tax is charged above the annual income of Rs3,50,000 to 4,50,000 and 25% above Rs 4,50,000. How much income tax should be paid by an individual in a year?

d. The minimum charge of land line telephone up to 175 calls is Rs 200. If the charge for each additional call is Re 1, how much charge should be paid for 375 calls with 10% telephone service charge and 13% VAT?

4. a. The meter reading for the consumption of electricity of a household was 1280 units on 1stBaisakh and 1570 units on 1stJestha. Calculate the amount of money to be paid, if the payment is made within 7 days of meter reading.

b. A shopkeeper selling an article at a discount of 13% gets loss of Rs 180. If it is sold allowing 12% discount, he gains 300. Find the marked price and the cost price of the article.


3Chapter

MensurationEstimated period : 25

Specific ObjectivesAt the end of this unit, the students will be able to

• solve the problems related to area of rectangle, square, triangle, parallelogram, rhombus, kite, quadrilateral, circular figures and paths.

• calculate the area of four walls, ceiling, floor and find the cost of plastering, carpeting, painting

• calculate the area of solids like cuboids, cube, prism, cylinder .• calculate the volume of solids like cuboids, cube, prism, cylinder etc• estimate the number of bricks required for wall and finding the cost

Teaching materialsChart paper, scissors, pencil, eraser and different solid models, net of different solids.

Note to the teacher1. Review the basic concept of area, perimeter and volume of different plane figures and

solids with formulae.2. Ask the students to measure the different dimensions of the things available in the

class like book, white board, floor and walls of the class room.3. Ask the students to make different models of solids like cube, cuboids, cylinder, prism

etc and to measure the different dimensions of them.4. Ask them to find their area, volume etc using formulae.5. Give them concept of estimation of required number of tiles, marbles, stones, bricks,

etc and cost estimation with the help of daily life examples.

Contents• Simple problems on area(Carpeting, stone paving, path construction, coloring,

plastering, etc) and its cost estimation on the related field.• Problems on surface area, cross section area and volume of prisms, cubes and cuboids• The practical project work on Mensuration and problem collection and solution.


3.1 Area of plane figuresReview1. Area of triangle

a. When base and height of triangle are given,

Area of triangle (A) = 12 × base × height

= 12 × b × h

b. Area of scalene triangle: supposing three sides of triangle as a, b, c.

Then, s = a + b + c

2

Now area of triangle = s (s - a) (s - b) (s - c) Perimeter of triangle = sum of three sides.

c. Area of right angled triangle

A = 12 × product of sides containing right angle

= 12 × p × b

d. Area of an equilateral triangle:PQR is an equilateral triangle in which PQ = PR = QR = a, let PD ⊥QR.∴ Here ∆ PQD ≅ ∆PRD

∴QD = DR = a2

In right angled ∆ PRD

PD2 = PR2 - DR2

or, h = a2 -( a2 )2

= a2 - a2

4

= 3a2

4 = a 3

2

Now area of triangle ∆PQR = 12 × base × height

= 12 × a × a 3

2

= 34 a

2

Perimeter = a + a + a = 3a

A

c b

aB C


3.1 Area of plane figuresReview1. Area of triangle

a. When base and height of triangle are given,

Area of triangle (A) = 12 × base × height

= 12 × b × h

b. Area of scalene triangle: supposing three sides of triangle as a, b, c.

Then, s = a + b + c

2

Now area of triangle = s (s - a) (s - b) (s - c) Perimeter of triangle = sum of three sides.

c. Area of right angled triangle

A = 12 × product of sides containing right angle

= 12 × p × b

d. Area of an equilateral triangle:PQR is an equilateral triangle in which PQ = PR = QR = a, let PD ⊥QR.∴ Here ∆ PQD ≅ ∆PRD

∴QD = DR = a2

In right angled ∆ PRD

PD2 = PR2 - DR2

or, h = a2 -( a2 )2

= a2 - a2

4

= 3a2

4 = a 3

2

Now area of triangle ∆PQR = 12 × base × height

= 12 × a × a 3

2

= 34 a

2

Perimeter = a + a + a = 3a

2. Area of rectangle:Length of rectangle = lBreadth of rectangle = bDiagonal of rectangle = d

Area (A) = l × bPerimeter (P) = l + l + b + b = 2l + 2b = 2 (l + b)

Diagonal (d) = l2+ b2 3. Area of square:

In square, l = b

Area of square (A) = l2 = d2

2 Perimeter (P) = 4lDiagonal (d) = 2 l

4. Area of parallelogram Let b be base and h be height of parallelogram. Then area of parallelogram (A) = b × hPerimeter (p) = 2a + 2b = 2(a + b)

5. Rhombus Rhombus is a parallelogram having all sides equal

Area of rhombus = 12 × d1 × d2

Where d1 and d2 are diagonals.Perimeter sum of four sides = 4a

6. Area of trapezium: Let a, b be the parallel sides and h be height of trapezium.

Area of trapezium (A) = 12 × height (sum of parallel sides)

= 12 × h (a+b)

Perimeter = sum of 4 sides7. Area of kite:

Let d1 and d2 be diagonals of a kite

Area of kite (A) = 12 × product of diagonals

= 12 × d1 × d2

h

h

d1


8. Area of quadrilateral:

Area of quadrilateral (A) = 12 × diagonal × sum of

perpendiculars drawn to the diagonal from opposite vertices.

= 12 × d (p1 + p2)

9. Area of circle Let r be the radius of circle having center o. area of circle (A) = pr2

Circumference = 2pr

Area of semi circle = πr2

2

Circumference of semicircle = 2πr2 + d

= πr + 2r = r (π + 2)

3.1.1 Area of pathways1. Area of path running outside a rectangular field:

Let us consider a rectangular field ABCD having length l and breadth b. There is a path of uniform width d running outside the field. PQRS is the field with path.

Length (PQ) = l+d+d= l+2dBreadth (RQ) = b+d+d= b+2dArea of path (the shaded region):

= area of PQRS – area of ABCD= (l+2d)(b+2d) – lb= lb+2ld+2bd+4d2-lb= 2ld+2bd+4d2

= 2d(l+b+2d)If the field is square, l=b.Area of path = 2d(l+l+2d) = 2d(2l+2d) = 2d x 2( l+d) = 4d(l+d)

2. Area of path running inside a rectangular fieldLet us consider a rectangular field ABCD having length l and breadth b. There is a path of uniform width d running inside the field.Length (SR) of rectangular PQRS = l-d-d= l-2d

S R

Ql + 2d

b + 2d

C

BA

D

bd

d

P

od

A

D

B

C

l - 2d

b - 2d

S

d

dR

QP


Breadth (QR) of rectangular PQRS = b-d-d = b-2dArea of path (the shaded region):

= area of ABCD – area of PQRS= lb-(l-2d) (b-2d)= lb-(lb-2ld-2bd+4d2)= lb -lb + 2ld+2bd-4d2

= 2d (l + b - 2d)3. Area of two paths crossing each other perpendicularly:

Let us consider a rectangular field ABCD having length l and breadth b. PQRS and WXYZ are two paths having uniform width 'd' running from the middle of every side of the field and crossing each other at right angle.Here, area of path PQRS = l x dArea of path WXYZ = b x dArea of square IJKL = d x d =d2

Now, area of two crossing path= area of PQRS + area of WXYZ- area of IJKL= ld+bd- d2

= d(l+b-d)If the field is square l = b.Then area of two paths = d (l + l - d)

= d(2l-d)4. Area of circular pathway/Annulus:

In the figure there are two circles having the same center o.Radius of the outer circle = R and radius of inner circle = rArea of outer circle = pR2

Area of inner circle = pr2

Area of the path (shaded region) = Area of outer circle – Area of inner circle = pR2- pr2

= p(R2-r2)Area, cost and quantities.

Let A= area of pathR = rate (per unit cost)T = total costN = number of bricks, tiles required to pave the patha = area of each brick or tiles

Then, T = A x R

or, A = TR

or, R = TA

T= N x R

R O

r

W

I

L

J

K

S

Rb

YI

Z

d

dP

Q

XA

D

B

C

If the field is square, l = bArea of path = 2d(l+b-2d) = 2d (l + b - 2d)

= 2d(l+l-2d)= 2d(2l-2d)= 2d x 2 (l-d)= 4d(l-d)


or, N = TR

or, R = TA

T = N × R

or, N = TR

or, R = TN

A = N × a

or, N = Aa

or, a = AN

Example 1 : If area of square garden is 225 m2, find its perimeter.Solution: Here,

Area of square = 225 m2 Or, l2 =225 m2

Or, l = 225 Or, l = 15 m

Now perimeter of square garden = 4 l = 4 × 15m = 60m∴The required perimeter of the square garden is 60m.

Example 2 : Find area of triangle ABC alongside whose sides are given.Solution:

Here, AB = 7 cm (c) BC = 9cm (a) AC = 8cm (b)Perimeter = 7cm + 9cm + 8cm = 24cm

S = 242 cm = 12cm

Worked Out Examples


Now, Area of ABC = s (s - a) (s - b) (s - c) = 12 (12 - 9) (12 - 8) (12 - 7) = 12 × 3 × 4 × 5 = 720

= 26.83 cm2

Example 3 : Square flowerbed is surrounded by a path 10 m wide around it. If the area of the path is 2000 m², find the area of the square flower-bed.

Solution: Here,In the adjoining figure, ABCD is the square flowerbed. EFGH is the outer boundary of the path. Let each side of the flowerbed = x m Then, the area of the square flowerbed ABCD (x × x) m² = x² m²Now, the side of the square EFGH = (x + 10 + 10) m = (x + 20) m So, the area of square EFGH = (x + 20) (x + 20) m² = (x + 20) m²Therefore, area of the path = Area of EFGH - Area of ABCD = [(x + 20)² - x²] m² = [x² + 400 + 40x - x²] m² = (40x + 400) m²But the area of path given = 2000 m²Therefore, 40x + 400 = 2000

Or, 40x = 2000 - 400 Or, 40x = 1600

Or, x = 160040 = 40

Therefore, side of square flowerbed = 40 m Therefore, the area of the square flowerbed = 40 × 40 m² = 1600 m²

Example 4 : Find the area of shaded part in the given figure.Solution:

Sides of square ABCD (l) = 21 cmDiameter of two semi circles (d) = 21cm

Now area of two semi circle = pd2

4

= × 21 × 21

4

227

= 6932 cm2

= 346.5 cm2

H

E

G

F

D

A

C

B

A D

B C←21 cm →


Now, area of shaded portion = Area of square ABCD- Area of two semi circles = l2- 346.5 = (21)2 – 346.5 = 441 – 346.5 = 94.5 cm2

Example 5 : Find the area of circular path in the given figure. How much money will be required to pave marble in the path, if the rate of paving marble is Rs 250 per cm2 ?

Solution: Here,Radius of outer circle (R) = 10 cmRadius of inner circle (r) = 5 cmArea of path (A) = area of outer circle – area of inner circle = pR2 - pr2

= p(R2– r2)

= 227 x (102 – 52)

= 227 x (100 – 25)

= 227 x 75

= 235.71 cm2

Rate of paving the marbles ( R) = Rs 250 per cm2

Cost of paving marble (T) = A x R = 235.71 x Rs 250 = Rs 5892.85

Example 6 : A rectangular field is 80 m by 50 m. A path of width 2m runs immediately inside its boundary. If the path is to be paved by bricks, find the cost of paving it at the rate of Rs 60 per m2.

Solution: HereLength of rectangular field (l)= 80mBreadth rectangular field (b)= 50mWidth of the path (d) = 2mAs the path is inside the boundary of the fieldArea of path(A) = 2d (l + b - 2d) = 2 x 2(80 + 50 - 2 x 2) = 512 m2

Cost of paving bricks ( C) = A × R = 512 × Rs 60 = Rs 30720

10 cm5 cm


Example 7 : A rectangular garden is 60 m long and 52 m wide. Two paths each of 2m wide are running across the middle of the garden at right angle.i. Calculate the cost of paving the bricks of size 16 cm by 12 cm on the path

at 10.50 per brickii. Find the cost of growing grass in the remaining space of the garden at Rs

50 per m2.Solution: Here,

Length of garden (l) = 60m Breadth of garden (b) = 52 mWidth of paths (d) = 2m Now, Area of the crossing paths (A) = d(l + b - d) = 2(60 + 52-2) = 2 × 110 = 220m2

Area of a brick (a) = 16cm × 12 cm

= 16100 m ×

12100 m

= 12625 m

2

i. Number of bricks required to pave the paths

(N) = Area of paths

Area of a brick

= 22012625

= 220 × 62512

= 11458 approx.

Cost of a brick ( R) = Rs 10.5

Cost of paving bricks = N × R

=11458 × Rs. 10.5

= Rs 120309

Again, a area of garden = 60m × 52 m

= 3120 m2

Area of remaining space = Area of garden – Area of path

= 3120 m2 -220m2

= 2900m2

ii. Cost of growing grass in

the remaining space

= Area × Rate

= 2900 × Rs. 50

= Rs 145000


1. a. Write the formula of area of right angled triangle. b. Write the formula of perimeter of a triangle having sides a cm, b cm and c cm c. What is the formula of area of equilateral triangle ? d. Write the formula of area of circle whose radius is ‘x’ cm.2. Find the area of the following: a. A triangle having sides 6 cm, 7cm and 5cm. b. A triangle having sides 7 cm, 8 cm and 9 cm c. An equilateral triangle having side 6cm. d. An equilateral triangle having side 7 cm. e. A rhombus with diagonals 6.4 cm and 8.6 cm. f. A rhombus with diagonals 6.4 cm and 11.1 cm. g. A trapezium with parallel sides 10 cm and 6 cm and height 5.4 cm. h. A quadrilateral having a diagonal 8.4 cm and length of perpendiculars to the diagonal

from opposite vertices are 3.4 cm and 3.8 cm. i. A circle having diameter 14cm.3. a. Find the area of square whose one side is 3 cm. b. Find the area of square whose one side is 5 cm. c. Find the perimeter of a square whose area is 64 cm2

d. Find the perimeter of a square whose area is 100 cm2

e. If area of an equilateral triangle is 15 cm2, find its perimeter. f. If area of an equilateral triangle is 16 cm2, find its perimeter. g. The perimeter of an equilateral triangle is 15 cm, find its area. h. The perimeter of an equilateral triangle is 18 cm, find its area. i. Find the area of circular field whose perimeter is 44 cm. j. Find the area of circular field whose perimeter is 176 cm.4. Find the area of the shaded region.

a. b. c.

Exercise 3.1

3 cm15 cm

3 cm8 cm 3 cm

7 cm2 cm

d. e. f.


g. The given figure is one quarter of a circle having radius 14cm. findi. length of arc PQii. perimeter of the given figureiii. area of the figureiv. area of triangle POQv. area of the shaded part

5. a. If the cost of constructing a path running inside a rectangular field at Rs 150 per sq. meter is Rs 1680, find the area of the path.

b. If the cost of constructing a path running inside a rectangular field at Rs 300 per sq. meter is Rs 3300, find the area of the path

6. a. 6000 bricks each of 20 cm x 15 cm are required to pave on the path. Find the area of the path.

b. 8000 bricks each of 25 cm x 20 cm are required to pave on the path. Find the area of the path.

7. a. A 50m long and 40m broad park is surrounded by 3.5 m wide path.i. Find the area of the path.ii. Calculate the cost of paving blocks on the path at Rs 360 per square meter.

b. A 60 m long and 44 m broad park is surrounded by 2 m wide path.i. Find the area of the path.ii. Calculate the cost of paving blocks on the path at Rs 360 per square meter.

c. A path of 2.5 m wide is to be constructed surrounding a park 35m long and 20m broad. Find the cost of covering the path with square marbles of size 324 cm2 at the rate of Rs 25 per piece.

d. A rectangular ground 28m long and 20m broad is surrounded by a path whose area is 324 m2. Find the width of the path.

8. a. A rectangular garden is 60 m long and 45 m broad. A 2.5 m wide path is running inside the garden. Calculatei. Cost of gravelling the path at Rs 150 per square meterii. Cost of growing Dubo in the remaining space at Rs 175 per square meter.

b. A rectangular garden is 70 m long and 50 m broad. A 3.5 m wide path is running inside the garden. Calculatei. Cost of gravelling the path at Rs 250 per square meterii. Cost of growing Dubo in the remaining space at Rs 180 per square meter.

9. a. A rectangular ground is 45 m long and 30 m wide. Two crossing paths at right angle to each other 3 cm wide are running across the middle of the ground. Findi. The cost of paving stone of size 25 cm by 15cm on the path at Rs 75 per stone.ii. The cost of growing grass in the remaining space on the ground at Rs 145 per sq.

meter. b. Two crossroads, each of width 8 m, run at right angles through the centre of a rectangular

garden of length 500 m and breadth 250 m and parallel to its sides. Find : i. the area of the garden excluding crossroads. ii. the area of the crossroads. iii. the cost of paving the road with stones if rate of paving is 150 per square meter.

P Q

14cm

O


c. The circumference of a circular field is 512m. A path 2.1 m wide runs around it. Find the cost of paving the path with bricks at Rs 12 per sq. metre.

d. The cost of constructing a path 3.5 m wide inside the boundary of a square garden at Rs 75.5 per sq. m is Rs 105700. What is the cost of covering the empty space with turfs at the rate of Rs 60 per square meter?

10. A path 5 m wide runs along inside a rectangular field. The length of the rectangular field is three times the breadth of the field. If the area of the path is 500 m², then find the length and breadth of the field.

Ans

wer

s 3.1

1.a. 12

Product of sides containing right angle b. P=a+b+c

c. 34 a2 d. px2 cm2 2.a.6 6 cm2 b.12 5 cm2 c.9 3 cm2

d. 3494

cm2 e.27.52cm2 f.35.52cm2 g.43.2cm2 h.30.24cm2 i.154cm2 3.a.9cm2 b.25cm2 c.32cm d.40cme.17.65cm f.18.23cm g.10.82cm2 h.9 3 cm2 i.154cm2

j. 2464cm2 4.a.102cm2 b.56cm2 c.60cm2 d.13.5cm2 e.1092.86cm2 f.45.86cm2 g i.22cm ii.50cm iii.154cm2 iv.98cm2 v.56cm2 5a.11.2m2 b.11m2 6a.180m2 b.400m2 7a.679m2, Rs244440 b.432m2, Rs155520 c. Rs231481.48 d.3m 8.a. i .Rs75000ii.Rs385000 b. i.Rs197750 ii.Rs487620 9.a.i. Rs432000 ii.Rs164430 b.i.119064m2 ii.5936m2 iii. Rs890400 c.Rs13068 d.Rs558735 10.45m, 15m

3.2 Area of 4 walls, ceiling and floor.The figure alongside shows a room consisting of 4 walls ABCD, EFGH, AGFD, BCEH. Let l,b and h be the length, breadth and height of the room respectively.Now Area of wall ABCD = l × hArea of wall EFGH = l × hArea of wall BCEH = b × hArea of wall AGFD = b × hArea of 4 walls = l × h + l × h + b × h + b × h = 2 l h + 2 b h = 2 h ( l + b )Similarly, Area of floor = l × b = l b∴Area of ceiling = l b∴Area of 4 walls, ceiling and floor = 2 h ( l + b ) + lb + l b


Example 1 : A room is 25m long and 15m broad. Find the length of carpet with breadth 5 m required to cover the floor of a room. Also find the cost of carpeting at Rs 250 per meter.

Solution:Length of room (l) = 25mBreadth of room (b) = 15mArea of room (A) = l × b = 25m × 15m = 375m2 We know,Area of carpet = Area of roomOr, lc × bc = 375m2

Or, lc × 5m = 375m2

= 2 l h + 2 b h + 2 l b = 2 (lh+lb+bh), which is also the total surface area of cuboid.Again area of 4 walls of a room = 2 h (l + b) =2 (l + b) × h = Perimeter × height.If l1 and h1 be length and height of a window. And l2 and h2 be length and height of a door. Then, Area of window = l1 × h1

Area of door = l2 × h2

Now, area of 4 walls excluding the area of window and door.= 2h(l + b) – l1h1- l2h2

We have to subtract area of the windows and doors from the area of 4 walls to get area of smalls excluding area of windows and doors.

Carpeting a room:

Let lr and br be length and breadth of a room. Again let lc and bc be length and breadth of carpet.We know that, Area of room = area of carpet

Or, lr × br = lc × bc

Worked Out Examples


Or, lc = 3755

Or, lc = 75mCost of 1m carpet (R)= Rs 250Total cost of carpeting the room = Rate × length = 250 × 75 = Rs 18750

Example 2 : A room is 12 m long , 8 m broad and 5m high. Find the cost of plastering its walls and ceiling at Rs 50 per sq. meter.

Solution: Here,Length of room (l) =12mBreadth of room (b) = 8mHeight of room (h) = 5mArea of 4 walls and ceiling (A) = 2h (l+b)+lb = 2 × 5 (12 + 8) + 12 × 8 = 10 × 20 + 96 = 296m2 Rate of plastering (R ) = Rs 50 per m2 Cost of plastering 4 walls and ceiling = A × R = 296 × 50 = Rs 14800

Example 3 : A square room is 5m high. The cost of carpeting its floor at Rs 25.sq meter is Rs 8000. Find the cost of carpeting its walls at Rs 75 per square meter.

Solution: Here,For carpeting the floor of the square room,Rate of carpeting (R ) = Rs 125 per square meterCost of carpeting = Rs 8000

Area of floor (A) = Total cost of carpeting

Rate of carpeting

= 8000125

= 64m2

Side of the room (l) = 64 = 8m Now area of four walls: = 2h (l + b) = 2 × 5 (8 + 8) = 10 × 16 = 160m2


Cost of painting the walls = Area × Rate = 160 × 75 = Rs 12000

Example 4 : A room is 20m long, 15m broad, and 5m high. It contains two windows of size 2 × 1.5m each and a door of size 2 × 3m. Calculate a. The area of 4 walls excluding the windows and door.b. Cost of painting the walls at Rs 60 per m2.

Solution:Length of room (l) = 20m Breadth of room (b) = 15mHeight of room (h) = 5m Area of 4 walls including windows and door = 2h ( l + b) = 2 × 5 (20 + 15) = 10 × 35 = 350 m2 Area of two windows = 2(2 × 1.5) = 6 m2 Area of a door = 2 × 3 = 6m2 a. Area of 4 walls excluding windows and door (A) = area of 4 walls including

windows and door - area of windows and door = 350-6-6 = 338 m2 b. Rate of painting the walls = Rs 60 per m2 Cost of painting the walls = area × rate = 338 × Rs 60 = Rs 20280

Example 5 : The length of room is two times its breadth and breadth is two times its height. If the cost of plastering its walls at Rs 60 per square meter is Rs 18000. Find the cost of paving marbles on its floor at Rs 350 per square meter.

Solution: Here,Let, height of the room (h) = x cmThen, breadth of the room (b) = 2x cmAnd length of the room (l) = 4x cmFor plastering the walls,Rate of plastering (R) = Rs 60 per m2

Cost of plastering the room = Rs 18720

Area of 4 walls of a room = cost of plasteringRate of plastering


= 18000125

= 300m2 Or, 2h (l + b) = 300m2 Or, 2x (4x + 2x) = 300Or, 2x × 6x = 300

Or, x2 = 30012

Or, x2 = 25Or, x = 25

Or, x = 5ml = 4x = 4 × 5 = 20mb = 2 x = 2 × 5 = 10mNow area of floor = l × b = 20m × 10m = 200 m2

Rate of paving marbles on the floor = Rs 350 per sq metreCost of paving marbles on the floor = Area × Rate = 200 × Rs 350 = Rs 70000

1. a. Find the length of carpet 3m wide which cover the floor of the room 24 m × 18 m. b. A square room is 12m long. How many meter of carpet of width 3 meter is needed for

carpeting its floor? Also find the cost of carpet at Rs 105 per meter. c. The length of the room is 15m and breadth of the room is 10m. Find the length of

carpet with 5 m width. Also find the cost of carpet at Rs 300 per meter.2. a. A room is 10m long, 10m broad and 5m high.

i. Find the area of its floor.ii. Find the area of 4 walls.iii. Find the area of its ceiling.iv. Find the area of 4 walls, ceiling and floor.

b. A square room is 8m long and 4m high.i. Find the area of its floor.ii. Find the area of its ceiling.iii. Find the area of its 4 wallsiv. Find the area of 4 walls ceiling and floor.

Exercise 3.2


c. A room is 12m long, 8m wide and 4 m high. It contains two windows each of 2.5 m long and 1.5 m high and two doors each of 1.5m long and 3m high.i. Find the area of 4 walls.ii. Find the area of 2 windows.iii. Find the area of 2 doors.iv. Find the area of 4 walls excluding windows and doors.

3. a. A room is 10m long, 6.5m wide and 4.5 m high. Find the cost of plastering walls and ceiling of the room at Rs 250/sq. meter.

b. A room is 8m long, 6m wide and 4m high. Find the number of wall papers of size 1m by 0.5m required to paste on the wall. Also find the cost of papering the walls at Rs 125 per piece.

c. A room is 10 m long and 6m wide. How many pieces of marbles 1.5 m long and 1m wide are required to pave on the floor of the room. Also find the cost of marbles at Rs 850 per piece.

4. a. How many pieces of carpet 3.5m long and 2m wide are required to cover the floor of a square room of length 14m? Also find the cost of carpet at Rs 350 per piece.

b. 300 square papers of same size are required to cover a rectangular wall of 24m long and 30m wide. Find the area of the piece of paper.

5. a. The cost of painting a wall of a room 15m long and 4m high at Rs 20 per sq meter is Rs4000. Find the width of the room.

b. The cost of plastering the walls and ceiling of a room at Rs 45 per square meter is Rs 12250. Find the cost of colouring the walls and ceiling at Rs 40 per sq. meter.

6 a. A square room is 9m long and 5m high. It contains three windows of size 2.5 m x 3 m and a door of size 1.5 m x 4 m. Find the cost of plastering the walls and ceiling at Rs 75 per square meter.

b. A room is three times long as it is broad and height is 4.5m. If the cost of carpeting the floor at Rs 70 per sq. meter is 5250, find the cost of colouring the walls at Rs 25 per sq meter.

c. The cost of plastering the walls of a room which is 5m high and whose breadth is one third of its length is Rs 2400 at the rate of Rs30 per sq meter. Find the cost of carpeting its floor at Rs 350per sq. meter.

7. a. A room is 10 m long and 5m high. If the cost of paving marbles on the floor at Rs190 per sq meter is Rs 11400, find the cost of papering its walls at Rs 85 per sq. meter.

b. The cost of carpeting a square room at Rs72 per sq meter is Rs 3528. If the cost of plastering its walls at Rs 16 per sq meter is Rs 2240, find the height of the room.

c. The cost of plastering a floor of a room, which is 10m long at Rs 65 per sq meter is Rs 5200. If the cost of painting its walls at Rs 26 per sq meter is Rs 5616, find the height of the room.

8. a. The length of a hall is two times its breadth and the breadth is two times its height. If the cost of painting its walls at Rs 75 per sq meter is Rs 32400, find the cost of paving marbles on the floor at Rs450 per sq meter.


Ans

wer

s 3.2

1.a.144m b.48m, Rs 5040 c. 30m , Rs 9000 2.a.i. 100m2 ii.200m2 iii.100m2 iv.400m2 b i.64m2 ii.64m2 iii. 128m2 iv.256m2 2.c.i.160m2 ii.7.5m2 iii.9m2 iv.143.5m2 3. a.Rs 53375 b.224 ,Rs 28000 c.40,Rs 34000 4.a.28Rs 9800 b.2.4m2 5.a.10m b.Rs 10888.88 6a.Rs 17437.5 b. Rs 4500 c.Rs 4200 7.a.Rs 13600 b.5m c.6m 8a.Rs 129600 b. Rs 35437.5

3.3 Area and Volume of solidSolids have definite shape and they occupy space. Solids have three dimensions length, breadth and height. Cube, cuboid, cylinder, cone, pyramid, sphere are solid figures. Different types of prism are also solid figures.Area and volume of some solids:1. Cuboid

A cuboid is a solid object having length, breadth and height (thickness). It has six rectangular faces. In the adjoining figure length, breadth and height of the cuboid are l, b, h respectively. Area of lateral surfaces = 2h ( l+ b)Total surface area = 2 × Area of bases section+ lateral surface = 2lb + 2h(l + b) = 2{lb +h (l + b)} = 2{ l b +l h + b h}Volume of cuboid = l × b × h

2. CubeIn cube length, breadth and height are equal.So area of lateral surfaces = 2l(l+l) = 2l × 2l = 4l2 Total surface area = area of bases + area of lateral surfaces = 2(l × l) +4l2 = 2l2+ 4l2

= 6l2

Volume of cube = l × l × l = l3

b. The length of a room is twice its breadth and thrice its height. If the cost of carpeting the floor at Rs160 per square meter is Rs 18000, find the cost of plastering its walls and ceiling at Rs 105 per sq meter.


In U shaped prism, cross section is in U shape

In any prism, Lateral surface area = Perimeter of cross sections × lengthTotal surface area = Lateral surface area + 2 area of cross section.Volume = Area of cross sections × height.Estimation of number of bricks and cost of building wall:Let L,B,H be length, breadth and height of a wall.

Triangular prism Circular prism

walls h

roof

floor

height

breadthlength

When a plane parallel to the base of the prism intersects the prism, the intersection is the crosssection of that prism.

cross-section is a circle

axis

cylinder

Base

Rectangular prism

In L shaped prism, cross section is in L shape

In a Cube, cross section is a square.

In a cylinder, cross section is a circle.

In a Cuboid, cross section is a rectangle.

3. PrismIt is a solid having uniform cross section. Prism is any solid which has any two opposite faces congruent and parallel. Prism has any kind of polygonal bases and all other faces are rectangular. Only circular prism(cylinder) has circular crosssection. Given below are some examples of different prisms.


Let l, b, h be the length, breadth and height of a brick.Then volume of the wall (V) = L × B × HVolume of the brick (v) = l × b × hIf N be the number of bricks required to build the wall

N = volume of wall

volume of bricks = Vv

If T be the total cost for building the wall and R be the cost of 1 brick, then T = N × R

Or, N = TR

Or, R = TN

If the wall contains windows and doors, we have to subtract the volume of the space occupied by windows and doors to get the volume of the wall excluding the volume occupied by windows and doors.

Example 1 : Find i. the area of cross section ii. lateral surface area iii. total surface area iv. volume of the following prism.

i. area of cross section = 12 × (4+4)-6 × 4 = 12 × 8-24 = 96-24 = 72cm2

Or, Area of cross- section= 12 × 4 + 6 × 4= 48 + 24 = 72 cm2

ii. Lateral surface area = perimeter of cross section x length = (12 + 4 + 6 + 4 + 6 + 8) × 10 = 40×10 = 400 cm2 iii. Total surface area = lateral surface area + 2 area of cross sections = 400+2×72 = 400 + 144 = 544cm2

6cm

4 + 4 = 8 cm

12cm4cm6cm

4cm

Worked Out Examples


iv. Volume of prism = area of cross sections × length = 72×10 = 720cm3

Example 2 : How many bricks of size 20cm × 10cm × 5cm will be required to build a wall of size 5m × 2m × 50cm? Also find the total cost of bricks at Rs 12000 for 1000 bricks.

Solution: Here,Volume of a brick (v) = 20cm × 10cm × 5cm

= 20100 m ×

10100 m ×

5100 m

= 15 ×

110 ×

120 m

3

= 1

1000 m3

Volume of wall (V) = 5 m × 2m × 50100 m

= 5m3 Number of bricks required to build the wall

(N) = volume of wall

volume of a brick

= 51

1000 = 5000Now, cost of 1000 bricks = Rs 12000

Cost of 1 brick = 120001000

= Rs 12Cost of 5000 bricks = Rs 12 × 12000 = Rs 144000So, the required number of bricks is 5000 and the cost is Rs 144000.Example 3 : A wall is 15m long, 5m high and 60cm wide. It contains 3 square windows of

length 2m each and two gateways of size 3m × 1.5m.i. How many bricks of size 15cm × 10cm × 5cm are required to build the

wall?ii. Find the cost of bricks at the rate Rs 10000 for 1000 bricks.

Solutions:Volume of wall = 15 m × 5m ×

60100 m

= 45 m3

Volume of wall occupied by 3 square windows = 3 × 22 × 0.6m3

= 7.2 m3


Volume of wall occupied by two gateways = 2(3×1.5) × 0.6 = 32.4 m3 Volume of wall occupied by bricks (V) = 45m3-7.2m3-5.4m3

= 24m3

Volume of a bricks (v) = 15100 m ×

10100 m ×

5100 m

= 3

4000 m3

i. Number of bricks required to build the wall (N) = vυ

= 243

4000 = 43200ii. Cost of 1000 bricks = Rs 10000

Cost of 1 brick = Rs. 100001000

= Rs 10 Cost of 43200 bricks = Rs 10 × 43200 = Rs 432000

Example 4 : A square room contains 384 m3 of air. The cost of carpeting the room at Rs 55 per sq meter is Rs3250. Find the cost of painting its walls at Rs 95 per sq meter.

Solution: Here,Volume of a room = 384 m3 For carpeting the room,Cost of 1m2 (R) = Rs 55Total cost (T) = 3250

Area of the room(floor) = TR

Or, l2 = 325055

Or, l2 = 64m2

∴ l = 8m Now height of the room (h) =

VolumeArea

= 38464

= 6mArea of 4 walls = 2h(l+b) = 2 × 6(8+8) = 12 × 16 = 192m2 Rate of painting the wall = Rs 95 per sq meterCost of painting the wall = area × rate = 192 × 95 = Rs 18240


1. a. How many faces are there in a cube? b. If length of the cube is ‘b’ then find the area of cross section. c. A prism has perimeter ‘p’ and height ‘h’ then write the formula of its curved surface

area. d. If area of cross-section of a prism is ‘A’ cm2 and height ‘h’ cm then find its volume. e. The volume of the wall (V) and volume of a brick is (v) then find the number of

bricks. f. The cost of N bricks is Rs 16000. If the cost of a brick is Rs 20, find the value of N. g. If cost of a brick is Rs ‘c’ and number of bricks is ‘N’ then write the formula to

calculate the total cost of bricks.2. a. The base area of prism is 10 cm2 and height 4 cm then, find its volume. b. The lateral surface area and area of cross section of a prism are 72 cm2 and 35 cm2

respectively. Find the total surface area of the prism. c. Find the volume of cube whose edge is 20 cm. d. The volume of cube is 343 cm3, find its total surface area. e. The volume of cube is 512 cm3, find its total surface area. f. The total surface area of the cube is 96 cm2 . Find its volume g. If the total surface area of lidless cubical box is 90 cm2, find the length of edge of

the box. h. A prism has base area 200 square meter and volume 1460 cubic meter, find its

height. i. A prism has height 6 cm and volume 240 cm3. Find the cross-section. j. A prism whose base perimeter is 20 cm and height 10 cm. Find its curved surface

area. k. Find the perimeter of a prism whose lateral surface area is 320 cm2 and height 10 cm. l. The perimeter of cross section of a prism of length 6cm is 32 cm. find its lateral

surface area. m. A prism has cross section area 25 cm2 and volume 250cm2. What will be its height? n. The area of cross section of a prism is 35 cm2 and its height is 12cm. Find its volume. o. The length, breadth and height of cuboid are respectively 9 cm, 7 cm and 5 cm then

find its surface area. p. What is the total cost for consructing a wall having measure 300 m3 at the rate of

Rs 160 per cubic- meter.

Exercise 3.3


4. Calculate the area of cross section, lateral surface area, total surface area and volume of the following prism.

3. Find the lateral surface area and total surface area of the following solids.

a. b.

c. d.

a. b.

c. d.

e. 2cm

2cm

2cm

2cm

2cm6cm

8cm

5cm

5cm

9cm 15cm

18cm

10cm

5cm

3cm

5cm 6.5cm

4cm 3cm

4cm

4cm

6cm

6cm

12cm

3cm

4cm

4cm

8cm

3cm

10cm6cm

7cm5cm

4cm4cm

4cm

4cm

4cm8cm

12cm

4cm

2cm


5. a. The length of a rectangular box is double of its breadth. If its height is 25cm and volume is 45000cm3, find the length and breadth of the box.

b. A cubical water tank is filled in 5103 seconds at the rate of 1 liter in 7 seconds. Calculate i. the internal volume and length of side of tank. ii. total internal surface area of the

tank. c. The internal dimension of a box is 72 cm × 60 cm × 66 cm and dimension of a piece of

soap is 22 cm × 6 cm × 15 cm. How many pieces of soap can be kept in the box ?6. a. How many bricks of size 20 cm × 10 cm × 15 cm will be required to construct a wall of

length 60m, width 50cm and height 4m with 4 windows of dimension 1.5m × 4m b. A wall contains two windows of size 2m × 1.5m each. The wall is 20m long and 40cm

wide. 4700 bricks of size 25 cm × 16 cm × 10 cm are required to construct the wall. Find the height of the wall.

c. A wall is 30m long, 5m high and 50 cm wide. It contains two windows each of 2m × 1.5m and a door of size 4m × 1.5mi. Find the number of bricks required to construct the wall having size of 15 cm × 10

cm × 5 cmii. Find the cost of bricks at the rate of Rs9.50 per piece.

d. The dimension of a wall is 15 m × 7.8 m × 75 cm. How many bricks of size 15 cm × 12 cm × 6 cm will be required to build the wall leaving 10% of the space for the cement

work. Also, find the cost of bricks at the rate of Rs 750 per 100 brick.7. a. The volume of a square room is 512 m3. Find the height of the room if cost of painting

the walls at the rate of Rs 40 per square meter is Rs 10240. b. The volume of a room is 384 m3, whose length is twice its breath. If the cost of painting

its four walls at Rs 150 per square meter is Rs 43200, find the height of the room. c. The length of a room is double of its breadth and it contains 396 m3 of air. If the cost

of plastering its floor at Rs 15 per square meter is 1080, find the cost of plastering its walls at Rs 30 per square meter.

Ans

wer

s 3.3

1.a.6 b.b2 c. ph cm2 d. Ah cm3 e. N= Vv f.800

g. Total cost = N×c 2a.40cm3 b.142cm2 c.8000cm3

d.294cm2 e.384cm2 f.64cm3 g.3 2 cm h. 7.3cm

i.40cm2 j.200cm2 k.32cm l.192cm2 m.10cm

n.420cm3 o.286cm2 p .Rs48000

3.a.400cm2, 600cm2 b. 304cm2, 424cm2 c. 840cm2, 1110cm2

d.288cm2, 400cm2 4.a.47cm2, 221cm2, 315cm2, 305.5cm3

b.60cm2, 252cm2, 372cm2, 360cm3 c. 56cm2, 288cm2, 400cm2, 448cm3

d. 37.9cm2, 168cm2, 243.8cm2, 265.3cm3 e. 36cm2, 168cm2, 240cm2, 216cm3

5.a.60cm, 30cm bi. 729000cm3, 90cm ii. 48600cm2

c. 144 6.a. 36000 b. 2.65m c.i.92000 ii. Rs874000

d. 73125, Rs 548437.5 7.a.8m b.12m c. Rs 5940


Group- A (1 × 1 = 1)

1. If the area of a room is 36 m2 and its height is 3.5m, find its volume. Group- B (3 × 2 = 6)

2. a. A uniform path of width 2 m is running outside the rectangular park of length 40m and breadth 32m. Find the area of the path.

b. Find total surface area of given solid prism.

c. If the volume of a cube is 125 cm3, find its total surface area.Group- C (1 × 4 = 4)

3. A room is 15 m long and 9 m wide. If the cost for plastering its 4 walls and ceiling at Rs 20 per square meter is Rs 2504. Find the height of the room.

Group- B (1 × 5 = 5)

4. The cost of carpeting the floor of a room, whose breadth is twice the height and breadth is half of its length at the rate of Rs 600 per square meter is Rs 84,000. What will be the cost of plastering its 4 walls at the rate of Rs 30 per square meter ?



1. a. A rectangle has length ‘a’ and breadth ‘b’, then write the formula to find perimeter.

b. If 1m3 = 1000 l, then how much water is there in a tank of capacity 30 m3?

2. a. Find the area of an equilateral triangle having perimeter 36 cm.

b. Find the area of path 6 m wide which is surrounding the square pond of length 50 m.

c. A wall is constructed using 2000 bricks where the volume of each brick is 1600 cm3. Find the volume of wall.

3. a. The perimeter of both a square and a rectangle are each equal to 30 m and difference of their area is 6 m2. Find the breadth of the rectangle.

b. A room is 12 m long and 9 m wide. If the total cost for plastering it’s 4 walls and celling at Rs 8 per square meter is Rs 3008 . Find the height of room.

c. The length of the room is two times of its breadth and five times its height. If volume of the room is 1600 m3, find the cost of plastering its walls at Rs 10 per square meter.

4. a. A room is three times as long as it is broad and its height is 4.6m. If the cost of carpeting its floor at Rs 240 per m2 be Rs 1800. Find the cost of colouring the walls at Rs 54per m2.

b. A wall 3 m long, 0.5 m wide and 2 m high has two windows each of 1.5m × 1m. If the size of each brick is 15 cm × 5cm × 4cm, find the number of bricks required to construct the wall. Also find the total cost of brick if each bricks costs Rs 18.

4cm5cm

5cm

5cm3cm

6.5cm


Specific objectives : At the end of this unit , the students will be able to

• factorize the algebraic expression.• simplify the algebraic expression based on indices with negative and fractional power. • solve the problems based on exponential equation(without quadratic form).• solve simultaneous linear equation by using elimination method, substitution method

and graphical method.• solve the quadratic equation by using factorization, by completing square and by

using formula.• prove the identities related to proportion.• solve problems related to ratio and proportion.

Teaching Materials :Tiles, chart, graph, balance, blocks representing the concept of a2– b2, (a + b)2, (a + b)3, a3 + b3, a3 – b3.

Note to the teacher :1. Give the concept of factorization.2. Give the clear concept to solve exponential expression and equation.3. Solve some problems to give clear concept how to solve simultaneous linear equations

of two variables.

Curriculum • Factorization.• Indices.• Simple problem related to ratio and proportion.• Simple problem related to continued proportion.• Linear equations.• Quadratic equation.

Specification GridCongnitive Domain/

TopicKnowledge (K) Comprehensive(C) Application(A) Higher ability

(HA)Total no. of questions

Total marks

Each 1 mark Each 2 marks Each of 4 marks Each of 5 marks

Algebraic expressionIndicesRatio and proportionLinear equationQuadratic equation

1

5

2

1

9 24


4Chapter

AlgebraEstimated period : 30


4.1. Factorization: Algebra is the branch of mathematics in which symbols are used to represent numbers or variables in arithmetical operations.

Example : {2 + (3 x 5) } ÷5 is an arithmetical expression.

Whereas {a + ( b x c ) ÷ d} is an expression in algebra.

Algebraic symbols: Letter representing number and various operational symbols indicate algebraic operations.

Algebraic terms: Parts of an expression separated by signs + or – are called the terms.

Example : 2ab, 3a, - 5b, and - c are the terms of expression 2ab + 3a – 5b – c.

In another words, the product of constant and variable is called a term.

Algebraic Expression: An expression containing or using algebraic symbols and operations such as 2x, 3a + 5b – c, 3a ÷ 4x + 5, x – 3 + y are known as algebraic expressions.

An algebraic expression having one term, two terms, three terms, four terms and more than four terms are respectively called monomial, binomial, trinomial and polynomial.

Factorization of algebraic expression:

Meaning of factors: We know that a(b + c) = ab + ac (Distributive law). Interchanging the sides, we can write it as ab + ac = a(b + c).

Here a and (b + c) are called the factors of ab + ac.

When an expression is written as a product of two or more factors, we say that it is resolved into factors. There are several types of expressions which can be factorized. We should learn some special devices to factorize them. We will revise what we had done in the previous class and consider some more below.

Factorization of each term having common factor. In xa + xb, we see that x is a factor of whole expression. The other factor (a + b) is obtained by dividing the expression by x.

Example 1 : Resolve ab – ac + ad into factors.Solution: Here,

‘a’ is factor of each term in the expression. ‘a’ is therefore one factor. We divide ab – ac + ad by ‘a, and get another value b – c + d.∴ ab – ac + ad = a(b – c + d)

Example 2 : Factorize: 15x3 + 25x2y – 35xy2

Solution: Here, 5x is common in all terms.∴ 15x3 + 25x2y – 35xy2 = 5x (3x2 + 5xy – 7 y2)

Worked Out Examples


Example 3 : Factorize: a (x + 2y) – 2b (x + 2y) + 3c(x + 2y)Solution: Here,

we see that ( x + 2y) is a common factor. For convenience, we may write p for x + 2y.Then,The expression = ap – 2bp + 3cp = p(a – 2b + 3c) =(x + 2y)(a -2b + 3c) Replacing p by x + 2y.After a little practice we may take common factor (x + 2y) directly without substituting a single letter p for it.

Example 4 : Factorize: x(a -5) + y( 5 – a )Solution: Here,

At first sight it appears that there is no common factor, But5 – a = - (a – 5)x(a – 5) + y(5 – a) can be written as x(a – 5) – y(a – 5) = (a – 5)(x – y)

Note: The relation (a – b) = - (b – a) and (b – a) = -( a – b) are frequently used in fctorization.

Resolve into factors1. 3a2 + 15b2 2. 20a2b2 – 35b4

3. 7x4 + 14x3 – 28x 4. a2x2 – abxy – acxy5. 2a3bc – 4abc2 + 8ab2c 6. 2m3n + 3mn3 – 4m2n2

7. 2x(a – b) – y(a – b) 8. 2ab(x + y) – 3cd(x + y)9. 3(x2 + y2) + 6m(x2 + y2) 10. a(x – y) + b(y – x)11. 3a(x2 + 2x + 7) – 4b(x2 + 2x + 7) 12. 12x(a – 2b) – 18y(2b – a)13. 2a(3x2 – 2x + 4) – 5b(3x2 – 2x + 4) 14. p(a + b + c) + 1q(a + b + c) + r(a + b + c)

Exercise 4.1

Ans

wer

s 4.1 1.3(a2+5b2) 2. 5b2(4a2-7b2) 3. 7x(x3+2x2-4) 4. ax(ax-by-cy)

5. 2abc(a2-2c+4b) 6.mn(2m2+3n2-4mn) 7. (a-b)(2x-y) 8. (x+y)(2ab-3cd) 9. 3(x2+y2)(1+2m) 10. (x-y)(a-b) 11. (x2+2x+7)(3a-4b) 12. 6(a-2b)(2x+3y) 13. (3x2-2x+4) (2a-5b) 14. (a+b+c)(p+q+r)

Factors by grouping: In some cases, terms are to be arranged in suitable group so that each group has a common factor.


Example 1 : Factorize: a2 + ab + ac + bcSolution: Here,

There is no factor common to all the terms. But we can take terms in pair so that each pair has a common factor.Thus, a2 + ab + ac + bc = a(a + b) + c (a +b) = (a + b)(a + c) This relation is also clear from the adjoining diagram.

Example 2 : Factorize : ab + cd – bd – acSolution: Here, we have to rearrange the terms before grouping them in pairs.

Thus, ab + cd – bd – ac = ab – ac – bd + cd= a(b – c ) – d( b – c )= (b – c)(a –d)

Example 3 : Factorize: a3 – a2 – ab + a + b – 1 Solution: Here,

= a3 – a2 – ab + a + b – 1= a3 – a2 – ab + b + a – 1 = a2(a – 1 ) – b(a – 1) +1(a -1)= (a -1)(a2 –b +1)

Example 4 : Factorize: x – x(3z – 2y) + 6yzSolution: Removing the bracket before grouping the terms

x2 – x(3z + 2y) + 6yz= x2 – 3xz – 2xy + 6yz= x(x – 3z) – 2y(x – 3z)= (x -3z)(x – 2y)

Resolve the factors:1. a2 + ax + ab + bx 2. 2ax – 2ay – 3by + 3bx3. x2 – xy + xz – yz 4. 2r – 2s – rt + st5. p3 – p2 q + p q2 – q3 6. 12a3 – 15a + 16a2 – 20

Exercise 4.2

Worked Out Examples


Example 1 : Resolve x2 + 7x + 12 into factors.Solution: Here,

We have to find two numbers whose sum is 7 and product is 12.We notice that 3 + 4 = 7 and 3 × 4 = 12Replacing 7 by (3 + 4),

x2 + 7x + 12= x2 + (4 + 3) x + 12= x2 +4x + 3x + 12= x(x + 4) + 3 (x + 4 )= (x +4 ) (x + 3 )

Example 2 : Resolve: x2 – 13x + 40Solution: Here,

We have to find two numbers whose sum is 13 and product is 40.We see that 5 + 8 = 13 and 5 × 8 = 40Replacing 13 by (5+ 8)

= x2 – 13x + 40= x2 – (8 + 5)x + 40

Factors of 1212 = 1 × 12, 2 × 6, 3 × 4

Factors of 4040 = 1 × 40, 2 × 20, 4 × 10, 8 × 5

Ans

wer

s 4.2 1. (a+x)(a+b) 2. (x-y)(2a+3b) 3. (x-y)(x+z) 4. (r-s) (2-t)

5. (p-q)(p2+q2) 6. (4a2- 5)(3a+4) 7. (a-b)(a2+3b2) 8. (a+2b) (3a-4c) 9. (a-1)(a3+3) 10. (x+7)(3x2-5) 11. (a+b)(ax+by+c) 12. (x+y)(a-b+c) 13. (b-1)(ab+c) 14. (2x-5z)(x-y+3) 15. (p-r)(pr+q2)

7. a3 – 6b3 – 2a2b + 3ab2 8. 3a2 – 4ac + 6ab – 8bc9. a4 – a3 + 3a – 3 10. 3x3 + 21x2 – 5x – 3511. a2x + abx + ac + aby + b2y + bc 12. ax + cx + ay – by –bx + cy13. ab2 – (a – c)b – c 14. 2x2 + 6x -15z – 2xy +5yz – 5xz15. (p2 – q2)r + (q2 – r2)P

Factorization of the form x2 + px + q : Here, we have to find the two factors of ‘q’ whose sum is ‘p’ Let’s see the following examples to clear the concept.

Worked Out Examples


1. x2 + 3x + 2 2. x2 + 4x + 33. p2 + 10x +9 4. x2 + 11x + 285. a2 + 7a +10 6. m2 + 9m + 207. p2 + 12p + 20 8. k2 + 17k + 309. 12x2 + 12x + 27 10. c2 – 14c + 4811. x2 - 25x + 150 12. m2 – 31m + 24013. a2 + 22ab + 72b2 14. a2b2 – 25abc + 156c2

15. (m + n)2 – 16(m + n) + 63 16. (x + y)2 -15(x + y) + 3617. (2x – y)2 + 19( 2x – y)z + 70z2

= x2 – 8x – 5x + 40= x(x – 8) – 5(x – 8)= (x -8)(x -5)

Exercise 4.3

Ans

wer

s 4.3 1. (x+1)(x+2) 2. (x+1)(x+3) 3. (p+1)(p+9) 4. (x+4)(x+7)

5. (a+2)(a+5) 6. (m+4)(m+5) 7. (p+2)(p+10) 8. (k+2)(k+15) 9. (x+3)(x+9) 10. (c-6)(c-8) 11. (x-10)(x-15) 12.(m-15)(m-16) 13. (a+4b)(a+18b) 14. (ab-12c)(ab-13c) 15. (m+n-7)(m+n-9) 16. (x+y-3)(x+y-12) 17. (2x-y+5z)(2x-y+14z)

Expression of the form : x2 + px –q2 and x2 – px – q2 i.e. with last term negative.Suppose there are two numbers a and b such that a – b = p and ab = qthen, x2 +px – q = x2 + (a – b)x – ab = x2 +ax – bx –ab = x(x + a) – b(x + a) = (x +a)(x – b)and x2 – px – q = x2 - (a – b)x – ab = x2 - ax + bx –ab = x(x – a) +b(x – a) = (x –a)(x +b)Thus, both x2 + px –q2 and x2 – px – q2 can be factorized if we can find a and b such that ab = q and a – b = p


Example 1 : Resolve: x2 + 7x –18 into factors.Solution: Here,

= x2 + 7x –18 = x2 + (9 – 2)x – 18= x2 +9x – 2x – 18= x(x + 9) – 2(x + 9)= (x +9)(x -2)

Example 2 : Factorize: a2 – 8a - 48Solution: Here,

a2 – 8a – 48 = a2 – (12-4)a - 48 = a2 – 12a + 4a – 48 = a(a-12) +4(a-12) = (a - 12)(a + 4)

Example 3 : Factorize: (x – y)2 + 18(x - y)(x + 2y) – 144(x +2y)2

Solution: Here, Put x – y = a and x + 2y = bthen, (x – y)2 + 18(x - y)(x + 2y) – 144(x +2y)2

= a2 + 18ab – 144b2

= a2+(24 -6)ab – 144b2

= a2 +24ab – 6ab – 144b2

= a(a +24b) – 6b(a + 24b)= (a + 24b)(a – 6b)

Replacing a by x – y and b by x + 2yThen = {(x – y) + 24(x + 2y)}{(x – y) – 6(x + 2y)} = (x – y + 24x + 48y)(x – y – 6x – 12y) = (25x + 47y)(-5x – 13y) = - (25x + 47y)(5x + 13y)

Factorization of the expression of the form ax2 + bx + c and ax2 - bx + c Here, we have to find two numbers whose product is ac and sum is b.Let’s learn the following examples to clear the concept.Example 4 : Factorize: 5x2y2 – 41xy + 84Solution : Here,

5 × 84 = 420. Hence we have to find two numbers whose product is 420 and sum is 41. They are 20 and 21.∴5x2y2 – 41xy + 84= 5x2y2 –(20 + 21)xy + 84

We have to find two numbers whose product is 48 and difference is 8. They are 12 and 4.

We have to find two numbers whose product is 18 and difference is 7. They are 9 and 2.

Worked Out Examples


=5x2y2 -20xy – 21xy + 84=5xy(xy – 4) – 21(xy – 4)= (xy – 4)(5xy – 21)

Resolve the factors:1. x2 + x – 20 2. x2 + 3x – 283. 3x2 + 8x + 4 4. 3a2 - 11a + 105. p2 + p – 72 6. m2 – 4m - 457. 6m2 – 13m + 5 8. m2 + 2m – 249. p2 – 7pq – 30q2 10. 2a2 + 11a + 1511. 5x2 – 19x + 12 12. a2b2 + 7ab – 12013. 3x2 – 13ax + 12a2 14. 1 + 13xy – 378x2y2

15. 15x2y2 – 17xy + 4 16. x2y2 – 13xyz – 168z2

17. (a +b)2 – 9(a + b) – 112 18. (x + y)2 + 5(x +y) - 36 19. (c +d)2 – 10(c + d) – 144 20. (3a – b)2 – 3(3a – b)c – 40c2

21. 2(x +y)2 – 17(x + y) + 15 22. 6(x +y)2 + 11(x + y)z + 3z2

Example 1 : Resolve 5x2 + 11x – 12 into factors.Solution: Here,

5x2 + 11x – 12= 5x2 + (15 – 4)x – 12= 5x2 + 15x – 4x – 12

420 = 2 × 2 × 3 × 5 × 7 = 7 × 60 or 35 × 12 or 30 × 14 and 20 × 21Here, 20 + 21 = 41

5 × 12 = 60, hence we have to find two numbers whose product is 60 and difference is 11. They are 15 and 4

Exercise 4.4

Ans

wer

s 4.4 1.(x+5)(x-4) 2.(x+7)(x-4) 3.(x+2)(3x+2) 4. (a-2)(3a-5)

5. (p+9)(p-8) 6. (m-9)(m+5) 7. (3m-5)(2m-1) 8. (m+6)(m-4) 9. (p-10q)(p+3q) 10. (a+3)(2a+5) 11. (x-3)(x-4) 12. (ab+15)(ab-8) 13. (x-3a)(3x-4a) 14. (1+27xy)(1-14xy) 15. (5xy-4)(3xy-1) 16. (xy-21z)(xy+8z) 17. (a+b-16)(a+b+7) 18. (x+y+9)(x+y-4) 19. (c+d-18)(c+d+8) 20. (3a-b-8c) (3a-b+5c) 21. (2x+2y-15)(x+y-1) 22. (2x+2y+3z)(3x+3y+z)

Factorization of the expression of the form : ax2 +bx +c and ax2 – bx – cHere, we have to find two numbers whose product is ac and difference is b.Let’s learn the following examples to clear the concept.

Worked Out Examples


= 5x( x + 3) – 4(x + 3)= (x + 3)(5x – 4)

Example 2 : Factorize: 20x2 – 9xy – 20y2.Solution: Here,

20 × 20 = 400, hence we have to find two numbers whose product is 400 and difference is 9. They are 25 and 16.Replacing 9 by (25 – 16)20x2 – 9xy – 20y2 = 20x2 – (25 – 16)xy – 20y2

= 20x2 –25xy +16xy – 20y2

= 5x(4x –5y) + 4y(4x – 5y) = (4x – 5y)(5x + 4y)

Example 3 : Factorize: 30a2 – 85ab + 35b2

Solution:Here, we notice that 5 is a factor of each term ∴30a2 - 85ab + 35b2

= 5(6a2 – 17ab + 7b2) = 5[ 6a2 – (14 + 3) ab + 7b2] = 5[6a2 – 14ab – 3ab + 7b2] = 5[2a(3a – 7b) –b(3a – 7b)] = 5(3a – 7b)(2a – b)

Note: If each term has a common factor, it should be always be taken out first. Example 4 : Factorize: –12x2 + 10 – 7xSolution: Here,

The expression must be rearranged in descending (or ascending) powers of x.Then the expression = –12x2 – 7x + 10Since–12 × 10 = –120,we have to find two numbers whose product is 120 and difference is 7. They are 15 and 8.–12x2 – 7x + 10 = –12x2 – (15 – 8)x + 10 = –12x2 –15x + 8x + 10 = –3x(4x + 5) + 2(4x + 5) = (4x + 5)( –3x + 2)

Example 5 : Factorize: (x + 1)( 2x – 1) – 20Solution: Here,

(x + 1) ( 2x – 1) – 20 = 2x2 + 2x – x – 1 – 20= 2x2 + x – 21= 2x2 + (7 – 6)x – 21= 2x2 + 7x – 6x – 21= x(2x + 7) – 3(2x + 7)= (2x + 7)(x – 3)

400 = 2 × 2 × 2 × 2 × 5 × 5 = 200 × 2 or 100 × 4 or 50 × 8 or 25 × 16 etc. And 25 – 16 = 9

∵ 6 × 7 = 42, two numbers whose product is 42 and sum is 17 are 14 and 3


Resolve into factors1. 2x2 + x – 3 2. 2x2 + 3x – 23. 3p2 – 7p – 6 4. 4m2 - 7m – 155. 5a2 - 13a – 6 6. 3x2 - 2x – 87. 27 – 3x – 2x2 8. 3 + 11x – 4x2

9. 8x2 -14x + 3 10. 49b2 + 21bc + 2c2

11. 3x2 – 10xy – 8y2 12. 15a2 – a – 6 13. 12p2 – 17pq – 7q2 14. 22m2 + mn – 6n2 15. 24 + 42x – 45x2 16. 10 – 58x – 12x2

17. 2(x + y)2 – 5(x + y) – 25 18. 42x2 – 77xy – 9y2

19. 5(3a + b)2 + 6(3a + b) – 8 20. 15 + 16xy – 15x2y2

21. 6(a + b)2 + 5(a + b) – 4 22. 3(x + y)2 + 7(x + y) + 223. 12(p + q)2 + (p + q) – 35 24. 4(2x - 3y)2 – (2x - 3y) – 14

Exercise 4.5

A a-b

a-b

E b B

G

C

b

b

F

D

b

J I

H

Ans

wer

s 4.5

1. (2x+3)(x-1) 2. (2x-1)(x+2) 3.(p-3)(3p+2) 4.(m-3)(4m+5) 5.(a-3)(5a+2) 6. (3x+4)(x-2) 7. (9+2x)(3-x) 8. (3-x)(4x+1) 9. (2x-3)(4x-1) 10. (7b+2c)(7b+c) 11. (3x+2y)(x-4y) 12. (3a-2)(5a+3) 13. (4p-7q)(3p+q) 14. (2m-n)(11m+6n) 15. 3(4x-3)(2x+5) 16. 2(1-6x)(5+x) 17. (x+y-5)(2x+2y+5) 18. 7(3x-7y)(2x+y) 19. (3a+b+2)(15a+5b-4) 20. (3+5xy)(5-3xy) 21. (2a+2b-1)(3a+3b+4)22. (x+y+2)(3x+3y+1) 23. (4p+4q+7)(3p+3q-5) 24. (2x-3y-2)(x-12y+7)

Factorization of algebraic expression of types a2 – b2 and a4 – b4.

Factorization is the process of changing algebraic expression into a product form. Generally after factorization we get all factors as the prime factors.

Geometrical proof of a2 – b2

a2 – b2 is the difference of two squares.

Here, ABCD is a square with side ‘a’ and

EFGB is another square with side ‘b’.

In the figure, the area of the shaded portion is a2 – b2.

Again, When we produce EF to H we get a rectangle FGCH,

where FH = a – b and FG = b.


Example 1: Factorize the following expression.a. 16a2 – b2

Solution: Here,= 16a2 – b2

= (4a)2 – b2

= (4a + b)(4a – b)b. 64x4 – 81y2

Solution: Here, = 64x4 – 81y2

= (8x2)2 – (9y)2

= (8x2 + 9y)(8x2 – 9y)c. 36m4 – 64n2

Solution: Here, = 36m4 – 64n2

= (6m2)2 – (8n)2

= (6m2 + 8n)(6m2 – 8n)

d. 1 - a2

b2

Solution: Here,

= 1 - a2

b2

= (1)2 – ( ab

)2

= (1+ ab ) (1 - a

b)

Now, Produce AD to J so that DJ = b, Draw rectangle JDHI so that rectangle, FGCH = rectangle JDHI. From the figure, area of shaded region is equal to area of the rectangle AEIJ, where

AE = a –b and AJ = a + b

So, a2 – b2 = (a – b) (a + b)

For the factorization of a4 – b4

a4 – b4 = (a2)2 – (b2)2 = (a2 + b2) (a2 – b2) = (a2 + b2) (a + b) (a – b)Note: a2 – b2 = (a – b)(a + b) a4 – b4 = (a2 + b2)(a + b) (a – b)

Worked Out Examples


e. a2

b2 - b2

a2 Solution: Here,

= a2

b2 - b2

a2

= ab( )2

– ba( )2

= ( ab

+ ba

) ( ab

- ba

)

Example 2: Resolve into factors.a. m4 – 16n4

Solution: Here,

= m4 – 16n4

= (m2)2 – (4n2)2

= (m2+4n2)(m2 – 4n2)

= (m2+4n2){(m)2 – (2n)2}

= (m2+4n2)(m + 2n)(m – 2n)

b. 256a8 – 1

Solution: Here,

= (16a4)2 – (1)2

= (16a4 + 1)(16a4 – 1)

= (16a4 + 1){(4a2)2 – (1)2}

= (16a4 + 1)(4a2 + 1)(4a2 – 1)

= (16a4 + 1)(4a2 + 1){(2a)2 – (1)2}

= (16a4 + 1)(4a2 + 1)(2a + 1)(2a – 1)

c. 405p4 – 3125q4

Solution: Here,

= 405p4 – 3125q4

= 5(81p4 – 625q4)

= 5{(9p2)2 – (25q2)2}

= 5(9p2 + 25q2) (9p2 - 25q2)

= 5(9p2 + 25q2) {(3p)2 – (5q)2}

= 5 (9p2 + 25q2)(3p + 5q)(3p – 5q)Example 3: Factorize the followings.

a. a2 – b2 – 4bc – 4c2


Solution: Here,= a2 – b2 – 4bc – 4c2

= a2 –( b2 + 4bc + 4c2)= a2 – (b +2c)2

= (a +b +2c)(a-b – 2c)b. m2 + n2 – 8pq – p2 – 2mn – 16q2

Solution: Here,= m2 + n2 + 8pq – p2 – 2mn – 16q2

= m2 – 2mn + n2 – p2 – 8pq – 16q2

= m2 – 2mn + n2 – (p2 + 8pq +16q2)= (m – n)2 – (p + 4q)2

= {(m – n) + (p + 4q)}{(m –n) – (p + 4q)}= (m – n + p + 4q)(m –n – p – 4q)

1. a2 – 9 2. 144 – a2 3. 16x2 – 36

4. 4x2

49y2 – 25y2

16x2 5. a8 – b4 6. 16x2

625 - 81y4

7. 32 - 14x4 8. a8 – b8 9. x6y8 – 16 z2

10. a10 – b2c4 11. x2 – y2 +z2 + 2xz 12. a2 + b2 – c2 – d2 + 2ab – 2cd13. x2 – 10zx + 25z2 – y2 – 2y – 1 14. a2 – 4ab - 5b2 – c2 + 6bc15. x2 -6xy +9y2 – 4z2 16. 9x4 – 24x2y2 + 4y4

Exercise 4.6

Ans

wer

s 4.6

1.(a+3)(a-3) 2. (a+12)(12-a) 3. (4x+6)(4x-6)

4. ( 2x7y

+ 5y4x

) ( 2x7y

- 5y4x

) 5. (a4+b2)(a2+b)(a2-b) 6. ( 4x25

+ 9y2) ( 4x25

- 9y2)

7. (4 2 + 12x2 ) (4 2 - 1

2x2 ) 8. (a4+b4)(a2+b2)(a+b)(a-b)9.(x3y4+4z) (x3y4-4z) 10. (a5+bc2) (a5 -bc2) 11. (x+y+z) (x-y+z) 12. (a+b+c+d)(a+b-c-d) 13. (x-5z+y+1)(x-5z-y-1) 14. (a+b-c)(a-5b+c) 15. (x-3y+2z)(x-3y-2z) 16. (3x2+6xy+2y2)(3x2-6xy+2y2)

Factorization of sum or difference of two cubes

To find the factors of a3 + b3 and a3 – b3 , we use the following formulaea3 + b3 = (a + b) (a2 – ab + b2)a3 – b3 = (a – b) (a2 + ab + b2)


Example 1: Resolve into factors.a. a3 – 8

Solution: Here,= a3 – 8= (a)3 – (2)3

= (a –2)(a2 + a.2 + 22)= (a –2)(a2 + 2a + 4)

b. x3 + 1 Solution: Here,= x3 + 1= (x)3 + (1)3

= (x + 1)(x2 – x .1 + 12)= (x + 1) )(x2 – x + 12)

c. a6 – 1Solution: Here,= (a2)3 – (1)3

= (a2 -1){(a2)2 + a2.1+ 12}= (a2 -1)(a4 + a2+ 12)= (a2 -1){(a2)2 + 2.a2.1+ 12 -a2}= (a2 -1){(a2+ 1)2 - a2}= (a + 1)(a -1)(a2 +1 +a)(a2 +1 –a)= (a + 1)(a -1)(a2 +a +1)(a2 –a +1)

d. a3 + 1a3

Solution: Here,= a3 + 1

a3 = a3 + ( 1

a)3

= (a + 1a

){a2 – a. 1a

+ ( 1a

) 2}

= (a + 1a

){a2 – 1 + ( 1a

) 2}

= (a + 1a

)(a2 – 1 + 1a2 )

Example 2: Resolve into factors:a. 64a3 – (2b – 3c)3

Solution: Here,

Worked Out Examples


= 64a3 – (2b – 3c)3

= (4a)3 – (2b – 3c)3

= {4a – (2b – 3c)}{(4a)2 + 4a. (2b – 3c) + (2b – 3c)2}= (4a – 2b + 3c)(16a2 + 8ab – 12ac + 4b2 – 12bc + 9c2)= (4a – 2b + 3c)(16a2+ 4b2 + 9c2+ 8ab – 12ac – 12bc)

b. (x – y)3 – (y – x)3

Solution: Here,= (x – y)3 – (y – x)3

= {(x – y) – (y –x )}{(x – y )2 + (x – y)(y – x) + (y –x )2

= (x – y – y + x){x2 -2xy + y2 + xy – x2 – y2 + xy + y2 – 2xy + x2)= (2x – 2y)(x2 – 2xy + y2)= 2(x –y)(x2 – 2xy + y2)= 2(x –y)(x – y)2

= 2(x – y)3

c. a3 + b3 + a + bSolution: Here,= a3 + b3 + a + b= (a + b)(a2 – ab + b2) + 1(a + b)(a + b)(a2 – ab + b2 + 1)

Example 3: Find the value of x3 – y3 if x – y = 4 and xy = 3Solution: Here,

x3 – y3 =(x – y)(x2 + xy + y2) = (x – y){(x – y)2 + 2xy + xy} = (x – y){(x – y)2 + 3xy} = 4(42 + 3 ×3) = 4 (16 + 9) = 4 × 25 = 100

Example 4: Find the value of a3 + 1a3 , if a + 1

a = 6

Solution: Here,a3 + 1

a3 = a3 + ( 1a

)3

= (a + 1a

) (a2 – a. 1a

+ 1a2 )

= (a + 1a

)(a2 + 1a2 - 1)

= x3 – y3 = (x - y)3 + 3xy (x - y) = 43 + 3 × 3 × 4= 64 +36= 100

or,


= (a + 1a

){(a + 1a

)2 - 2a. 1a

- 1}

= (a + 1a

){(a + 1a

)2 - 2 - 1}

= (a + 1a

){(a + 1a

)2 - 3}

= 6(62 – 3) = 6 (36 – 3) = 6 × 33 = 198

Factorize the following.1. x3 – 8y3 2. a3b3 -1 3. 216a4 – ab3 4. 512x9 - y9 5. a6 – y6 6. 27p3 + 8q3

7. 343a3 + b3 8. 1 + 64x6 9. 8a3 - 18

10. 375x3 + 1029y3 11. 8a3 + 18 12. 16x3 + 1

32b3

13. 1 – (a – b)6 14. (m + n)3 – (m – n)3 15. (4x – 3y)3 + (3x – 4y)3 16. a3 – 27x3 + 2ay – 6xy 17. x3 + x + y3 + y 18. (x + y )3 + z3

19. (x + a )3 + 8a3x3 20. If a + b = 7 and ab = 8, find the value of a3 + b3 21. If x – y = 7 and xy = 9, find the value of x3 – y3.

22. If m – 1m = 9, find the value of m3 - 1

m3 . 23. If 3a – 13a= 4, find the value of 27a3- 1

27a3

24. If x+ 1x

= a, prove that x3 + 1x3 = a3 – 3a

Exercise 4.7

Ans

wer

s 4.7

1.(x-2y)(x2+2xy+4y2) 2. (ab-1)(a2b2+ab+1) 3. a (6a-b)(36a2+6ab+b2)

4. (2x-y)(4x2+2xy+y2)(64x6+8x3y3+y6) 5. (a+y)(a-y)(a2+y2+ay)(a2+y2-ay)

6. (3p+2q)(9p2-6pq+4q2) 7. (7a+b)(49a2-7ab+b2) 8. (4x2+1)(4x2+2x+1)(4x2-2x+1)

9. (2a - 12

) (4a2 + a+ 12

) 10. 3(5x+7y)(25x2-35xy+49y2) 11. (2a+ 12

)(4a2-a+ 14

)

12. 2(2x+ 14b

) (4x2 - x2b

- 116b2)

13. (1-(a-b)2)(1+(a-b)2+(a-b)4)

14. 2n(3m2+mn+n2) 15. (7x-7y)(13x2-23xy+13y2) 16. (a-3x)(a2+9x2+3ax+2y)

17. (x+y) (x2-xy+y2+1) 18. (x+y+z)(x2+y2+z2+2xy-xz-yz) 19. (x+a+2ax)((x+a)2-2ax2+2a2x+4a2x2)

20. 175 21. 532 22. 756 23. 76


Example 1: Factorize: a4 + a2 + 1Solution: Here,

= a4 + a2 + 1= (a2)2 + 2a2.1 – a2 + 1= (a2 + 1)2 – a2

= (a2 + 1 + a)(a2 + 1 –a)= (a2+ a +1)(a2 –a + 1)

Example 2: Factorize: a4 + 4Solution: Here,

= a4 + 4= (a2)2 + 2.a2.2 +(2)2 –2a2.2= (a2+ 2)2 –(2a)2

= (a2 + 2 + 2a)(a2 + 2 – 2a)= (a2+ 2a + 2)(a2 – 2a + 2)

Example 3: Factorize: x4 – 3x2 + 1Solution: Here,

= x4 – 3x2 + 1= x4 – 2x2 – x2 + 1= (x2)2 – 2x2.1 + 1 - x2

= (x2 -1)2 – (x)2

=(x2 -1 + x)(x2 -1 –x)= (x2 + x -1)(x2 –x -1)

Example 4: Factorize: x2

y2 + 1 + y2

x2 Solution: Here,

= x2

y2 + 1 + y2

x2

Factorization of the expression in the form of a4 + a2b2 + b4

The expression in the form of a4 + a2b2 + b4 can be expressed in the form of a2 – b2 by adding and subtracting a2b2

a4 + a2b2 + b4 = (a2)2 + 2a2b 2 – a2b2 + (b2)2

= (a2 + b2)2 – (ab)2 [∴a2 + 2ab +b2 =(a + b)2] = (a2 + b2 + ab)(a2 + b2 –ab) [∴a2 – b2 =(a – b)(a + b)] =(a2+ ab + b2)(a2–ab + b2) ∴ a4 + a2b2 + b4 =(a2+ ab + b2)(a2–ab + b2)

Worked Out Examples


= ( xy

)2 + 2 xy

× yx

– xy

× yx

+ ( yx

)2

= ( xy

)2 + 2 xy

× yx

+ ( yx

)2 – xy

× yx

= ( xy

+ yx

)2 – 12

= ( xy

+ yx

+ 1) ( xy

+ yx

– 1)

= ( xy

+ 1 + yx

) ( xy

– 1 + yx

)

Example 5: Factorize: x4

y4 + 1 - 7x2

y2 Solution:

= x4

y4 + 1 - 7x2

y2

= ( x2

y2 + 1)2 – 2 x2

y2 1 - 7x2

y2 [∵a2 + b2 = (a + b)2 – 2ab]

= ( x2

y2 + 1)2 – 2x2

y2 - 7x2

y2

= ( x2

y2 + 1)2 – 9x2

y2

= ( x2

y2 + 1)2 – ( 3xy

)2

= ( x2

y2 + 1 + 3xy

) ( x2

y2 + 1 – 3xy

) [∵a2 – b2 = (a + b) (a – b)]

= ( x2

y2 + 3xy

+ 1) ( x2

y2 – 3xy

+ 1)

Example 6: Factorize: x2 -10x + 24 + 6y – 9y2

Solution: Here,= x2 -10x + 24 + 6y – 9y2

= x2 – 2x .5 + 25 – 1 + 2.1 .3y – (3y)2

= (x -5)2 – {1 – 2.1 .3y + (3y)2}= (x -5)2 – (1 -3y)2

= (x – 5 + 1 – 3y){(x- 5) – (1 -3y)} [∵ a2 – b2 = (a + b)(a – b)]= (x – 4 – 3y)(x – 5 – 1 + 3y)= (x – 4-3y)(x – 6 + 3y)= (x – 3y-4)(x + 3y - 6)


1. y4 + y2 + 1 2. x8 + x4 + 1 3. x4 + x2y2 + y4 4. x4 + 4y4 5. 81x4 + 64y4 6. 64x4 + y4

7. x4 - 7x2 + 1 8. x4 - 5x2y2 + 4y4 9. 49x4 – 154x2y2 + 9y4 10. 25a4 – 34a2x2 + 9x4 11. 256x4 – x2y2 +49y4 12. 1225x4 + 31x2y2 + 64y4

13. 2025x4 + 185x2y2 + 2401y4 14. 4a4 + 35a2b2 + 121b4 15. x4 + 1 + 1x4

16. x4

y4 + x2

y2 + 1 17. a4

b4 + a2

b2 + 1 18. x4 – 8x2 – 33 – 14y – y2

19. x4 – 10x2 + 24 + 6y – 9y2 20. x2 + 50xy + 609y2 – 8yz – z2

21. (1 – a2) (1 – b2) + 4ab 22. x2 – a2 +4xy + 2ab – b2 + 4y2

Exercise 4.8A

nsw

ers 4

.8

1.(y2+y+1)(y2-y+1) 2. (x2+x+1)(x2-x+1) 3. (x2+xy+y2)(x2-xy+y2)

4. (x2+2xy+2y2)(x2-2xy+2y2) 5. (9x2+12xy+8y2) (9x2-12xy+8y2)

6. (8x2+4xy+y2)(8x2-4xy+y2) 7. (x2+3x+1)(x2-3x+1)

8. (x2+3xy+2y2)(x2-3xy+2y2) 9. (7x2+14xy+3y2)(7x2-14xy+3y2)

10. (5a2+7ax+3x2)(5a2-7ax+3x2) 11. (16x2+15xy+7y2)(16x2-15xy+7y2)

12. (35x2+23xy+8y2)(35x2-23xy+8y2) 13. (45x2+65xy+49y2)(45x2-65xy+49y2)

14. (2a2+3ab+11b2)(2a2-3ab+11b2) 15. (x2+ 1+ 1x2 ) (x2-1+ 1

x2 ) 16. ( x2

y2 + xy

+1) ( x2

y2 - xy

+1)

17. ( a2

b2 + ab

+1) ( a2

b2 - ab

+1) 18. (x2+y+3)(x2-y-11) 19. (x2+3y-7) (x2-3y-3)

20. (x+29y+z)(x+21y-z) 21. (ab+a-b+1)(ab-a+b+1) 22. (x+2y+a-b)(x+2y-a+b)

4.2. IndicesIndices is the plural term of index. We can write a × a = a2, a × a × a × a = a4 and so on. In the same way a × a × a × a ....... to m factors is expressed as am and read as ‘a’ to the power m. Here, a is called base and m is called the index or power or exponent.

We have already learnt about the laws of indices in previous class. Let’s recall here

1. am × an = am+n 2. am ÷ an = am – n 3. a0 = 1 4. (am)n = amn

Complete the following

1. (am)(an) = 2. (ab)m = 3. (abc)m = 4. ( a

b )m =

5. a-m = 6. amn =


Example 1: Simplify:

a. 81( )34

b. 127( )-2

3 c.

2431024( )-2

3 d. 34 × 9-3 × 816

914 × 33 × 27-2 Solution:

a. 81( )34

= 34( )34

= 34 × 34 = 33 = 27

b. 127( )-2

3= -2

3

127

= -23

133×

= 1

3-2 = 32 = 9

c. 2431024( )-2

3=

-3

-35

51024

243 = -3

-35

545 ×

35 × =

3-3

4-3 = 133 × 43 =

6427

d. 34 × 9-3 × 816

914 × 33 × 27-2 = 34 × (32)-3 × (34)6

(32)14 × 33 × (33)-2 = 34- 6+24-28-3+6 = 3-3 = 133 =

127

Example 2: Express each of the following without the radical sign and negative indices.

a. x103 =

13(x10) =

13x10 × = x10/3

b. a-341

= 1

4

1(a-3)

= -34

1(a)

= a 34

c. a-44 × ( a10 )

- 14 = a- 44 × a

-140 = a

-1 - 140 = a-1 × a

-140 = a

-4140 = 1

a41/40

d. 27x63× 16x124

× (27x6)13 × (16x12)

14

= (3x2)33× (2x3)44

× (3x2)3 ×

13 × (2x3)

4 ×

14

= 3x2 × 2x3 × 3x2 × 2x3

= 36 x10

e. x3 y4z

= {(z 12 y)

12 x3}

14

= (z 1

16y 18 x3)

14

= z 1

16 y 18 . x

34

= x3/4. y1/8. z1/6

Worked Out Examples


Example 3: Simplify:(xa-b)a+b × (xb-c)b+c × (xc-a)c+a

Solution: Here,= x(a-b) (a+b) × x(b-c) (b+c) × x(c-a) (c+a)

= xa2-b2 × xb2-c2 × xc2-a2

= xa2-b2 + b2-c2+c2-a2

= x0

= 1

Example 5: Prove that : ax

ay( )x2 + xy + y2 ×

ay

az( )y2 + yz + z2

× az

ax( )z2 + zx + z2

= 1Solution:

= ax

ay( )x2 + xy + y2 ×

ay

az( )y2 + yz + z2

× az

ax( )z2 + zx + z2

= a(x-y) x2 + xy + y2 × a(y-z)y2 + yz + z2

× a(z-x) (z2 + yz+x2)

= ax3 - y3 + y3 - z3 + z3- x3

= a°= 1 = R.H.S proved

Example 6: Prove that : (a + 1

b)m (a – 1

b)m

(b + 1a

)m (b – 1a

)m = a

b( )2m

Solution:

L.H.S = (a + 1

b)m (a – 1

b)m

(b + 1a

)m (b – 1a

)m

= ( ab + 1

b)m× ( ab - 1

b)m

( ab + 1a

)m× ( ab - 1a

)m

= ab + 1

bab - 1

ba

ab + 1a

ab - 1× ××( () )m m

= ab( )m a

b( )m×

= ab( )m + m

= ab( )2m

= R.H.S proved.

Example 4: Simplify: xa

xbab

× xb

xcbc

× xc

xaca

Solution : Here,

= xa

xbab

× xb

xcbc

× xc

xaca

= xa

xb( )1

ab × xb

xc( )1bc

× xc

xa( )1ac

= x a-bab + b-c

bc + c-aca

= x c (a-b) + a (b-c) + b (c-a)abc

= x a (c-bc + ab- ac + be - acabc

= x °abc = x° = 1


Example 7: If a = x 13 - x

-13 and prove that a3 + 3a = x - 1

x.

Solution: L.H.S = a3 + 3a

= (x13 - x

– 13 )3 + 3 (x

13 - x

– 13 )

= (x13 )3 – (x

– 13 )3 –3 . x

13 . x

– 13 (x

13 - x

– 13 ) + 3 (x

13 - x

13 )

= x - x -1 - 3 (x13 - x

– 13 + 3 (x

13 - x

13 )

= x – x-1

= x - 1x

= R.H.S proved

Example 8 : Simplify :1

1+xa-b + xc-b + 1

1+xb-c + xa-c + 1

1+xc-a + xb-a

Solution:

= 1

1+xa-b + xc-b + 1

1+xb-c + xa-c + 1

1+xc-a + xb-a

= 11 + xa

xb + xc

xb

+

11 + xb

xc + xa

xc

+ 11 + xc

xa + xb

xa

= 1

xb + xa + xc

xb

+ 1xc + xb + xa

xc

+ 1xa + xc + xb

xa

= xb

xb + xa + xc + xc

xc + xb + xa +xa

xa + xc + xb

= xb+xc + xa

xa + xb + xc

= 1

Exercise 4.9

1. a. What power of x yields for unit value? b. What power of (a + b) has the value 1? c. If x = (x + y)0 + 6, what is the value of x?

d. What is the power of 7 so that its value becomes 149 ?


2. Find the value of :

a. 23 × 2-3 b. 63 × 163 c. 23 × 3-2 × 62

d. 4-8 × 16-6 × 323 e. 5x2 ÷ (2x)7 f. (4x3)2÷(2x)7.3. Find the value of :

a. (32)-32 b. 1

64( )23 c. (27)

-13 d. 81

49( )-32

e. (128)47 ÷(81)

34 f. 125

27[( )23 ÷ 5

3( )2 ] g. 7291

6 h. (534 )

-43

i. 827( )

-23 j.

6255

4

4. Remove the radical signs and negative indices. a. 65a63 b. x-64 c. x-24

÷ x-68

d. a-31

4 e. a3 × b23

× 9104b-26 f. 363 × 36 -14

g. x-4 × x-2x3 h. 54 3

5. Simplify:

a. 312 (3

12 + 9

-14 ) b. 4

12 (4

12 + 16

-14 )

c. 3x7y11z-13 × 72x-1 yz43

d. 64a4b4 ÷ 16a2b 2 × 256a6b6

e. (16a3b3)24 ÷ 8a6b63 f. (x+y)-83

. (x+y)23

g. a2x+3y × a3x-6y

ax+2y × a4x - 4y h. 210+x × 43x-4

22(2x+1) × 23x-1

i. 3x+1 × 27(x+1) (x+1)

3x(x+1) × 9x+1 j. 95 × 274 × 32

816 × 3 × 9-5

6. Simplify :

a. 2x+1 + 2x

2x × 3 b. 3x+1 + 3x

3x × 5 c. 33x+2 - 33x+1

6 ×27x

d. 11n+2 - 55 × 11n-1

11n × 116 e. 7n+2 - 35.7n-1

7n × 11 f. 5.3k - 9.3k-2

3k-3k-1

7. Simplify:

a. (xl+m)3 × (xm+n)3 × (xn+l)3

(xl × xm × xn)6


b. x

xcb

bc

bc

× x

xac

ca

ca

× x

xba

ab

ab

c. ax

a-y( )x2-xy+y2× ay

a-z( )y2-yz+z2× az

a-x( )z2-zx+x2

d. ap

x-q( )p-q× xq

x-r( )q-r× xr

x-p( )r-p

8. Simplify:

a. xa

xb( )c× xb

xc( )a× xc

xa( )b b. xa

xb( )a+b× xb

xc( )b+c× xc

xa( )c+a

c. xa

xb( )a2+ab+b2× xb

xc( )b2+bc+c2× xc

xa( )c2+ca+a2

d. ax

ay( )x2+xy+y2× ay

az( )y2+yz+z2× az

aa( )x2+zx+z2

e. xa2 - b2a+b × xb2 - c2b+c

× xc2 - a2c+a f. x

bc

cbx

1bc

×

xca

acx

1ac

× xab

bax

1ab

g. xa+b

xa-b( )c-a

× xb+c

xb-c( )a-b

×xc+a

xc-a( )b-c

h. xa+b

xc( )a-b

× xb+c

xa( )b-c

×xc+a

xb( )c-a

9. Prove that:

a. 1

a-b( )x

1a-c

× 1

b-c( )x

1b-a

× 1

c-a( )x

1c-b

= 1

b. xa

xb( )c

×

xb

xc( )a

×

xc

x-a( )b

= x2ab

c.

m ×1

b2( )a2 -

n 1a2( )b2 -

n-m1b( )a -

m - n1a( )b +

=

ab( )m+n

d. 11+ax-y + az-y

+ 1

1+ay-z + ax-z + 1

1 + az-x + ay-x = 1

e. 11+ap-q + ar-q +

11+aq-r + ap-r +

11 + ar-p + aq-p = 1


Example 1: Solve: 3x+6 = 81x

Solution: Here, 3x+6 = 81x

or, 3x+6 = 34x or, x + 6 = 4x (Since bases are same )

10.a. If a + b + c = 0, prove that : 1x a + x-b +1

+ 1

xb+x-c+1+

1xc+x-a+1 = 1

b. If x = aq+rap, y = ar+paq and z = ap+q ar show that xb-c yc-a za-b = 1

c. If lmn = 1, then prove that : 11 + l+ m-1 +

1l + m + n-1 +

11 + n + l-1 = 1

11. a. Ifx = 313 + 3

-13 , then prove that 3x3 – 9x = 10

b. If x = 213 + 2

-13 then prove that 2x3 – 6x = 5

c. If x2 + 2 = 323 + 3

-23 ,then prove that 3x3 + 9x = 8

4.3. Exponential Equation

In the equation when a and b are constant and x is the variable. Such equation with variable as the exponent is called the exponential equation or an equation based on index.For example 3x = 27, 24-x = 32 are the exponential equations.

To solve the exponential equation, we have to use the axiom ax = ab, (a ≠ 0) then x = b, it means while solving exponential equation we need to make base of both side same and compare the power.

Ans

wer

s 4.9

1. a. 0 b. 0 c. 7 d. -2 2.a. 1 b. 1 c. 32 d. 1225 e. 5

27x5

f. 18x

3.a. 11522

b. 116

c. 13

d. 343729

e. 1627

f.1 g. 13

h. 15

i. 94

j. 1 4.a. a2 653 b.

132x

c. x1/4 d. a3/4 e. a32 b

243

f. 7776 g. 1x2 ×

12x3x

h. 51/24 5.a. 4 b. 5 c. 6x2y4z d. 32a4b4 e. 2ab

f. 1(x+y)2 g. 1

ay h. 2 i. 32x2+4x + 3 j. 39 6. a. 1 b. 45

c. 1

d. 1 e. 4 f. 6 7. a. 1 b. 1 c. a2(x3+y3+z3) d. 1

8.a. 1 b. 1 c. 1 d. 1 e. 1 f. 1 g. 1 h. 1

Worked Out Examples


or, 4x - x = 6or, 3x = 6

or, x = 63

∴ x = 2Example 2: Solve: 33x+1 = 92x-1

Solution: Hence, 33x+1 = 92x+1

or, 33x+1 = 32(2x-1)

or, 3x + 1 = 2 (2x - 1) ( Since bases are same )or, 3x + 1 = 4x - 2or, 4x - 3x = 1 + 2or, x = 3∴ x = 3

Example: 3 Solve: 25x+2 = 52x +1 + 620Solution: Here,

or, 25x+2 = 52x+1 + 620 or, 25x × 252 = 52x × 51 + 620or, 52x × 625 - 52x × 5 = 620or, 52x (625 - 5) = 620or, 52x = 1or, 52x = 50

or, 2x = 0∴ x = 0

Example 5: Solve: 32x+3

3 + 32(x+1)

9 - 90 = 0

Solution: Here,32x+3

3 + 32(x+1)

9 - 90 = 0

or, 32x +3 -1 + 32(x+1)

32 - 90 = 0

or, 32x +2 + 32x+2

32 - 90 = 0

or, 32x +2 + 32x + 2 - 2 = 90

or, 32x × 32 + 32x = 90or, 32x (32 + 1) = 90or, 32x (9 + 1) = 90

or, 32x = 9010

or, 32x = 9or, 32x = 32

By equating,or, 2x = 2

Example 4: Solve: 3x+3 × 5x+2 = 3Solution: Here,

3x+3 × 5x+2 = 3or, 3x × 33 × 5x × 52 = 3

or, 3x × 5x = 3

33 × 52

or, (3 × 5)x = 31-3 × 5-2 or, (3 × 5)x = 3-2 × 5-2

or, (3 × 5)x = (3 × 5)-2

∴ x = -2

Example 6: Solve: 35x - 4 × a3x - 2 = 33x - 2 × a2x - 1

Solution: Here,

35x - 4 × a3x - 2 = 33x - 2 × a2x - 1

35x - 4 × a3x - 2

33x - 2 × a2x - 1 = 1

or, 35x - 4 - 3x + 2 × a 3x - 2 - 2x + 1 = 1

or, 32x - 2 × ax - 1 = 1

or, 32(x-1) × ax-1 = 1

or, 9x-1 × ax-1 = 1

or, (9a)x - 1 = (9a)0

or, x - 1 = 0

or, x = 1

∴ x = 1.


or, x = 22

∴ x = 1Example 7: 4x - 3 × 2x + 1 + 8 = 0Solution:

4x - 3 × 2x + 1 + 8 = 0or, 22x - 3 × 2x × 21 + 8 = 0or, (2x)2 - 3 × 2x × 21 + 8 = 0

Let 2x = a a2 - 6a + 8 = 0or, a2 - 4a - 2a + 8 = 0or, a(a - 4) - 2(a - 4) + 8 = 0or, (a - 4) (a - 2) = 0

Either, a - 4 = 0∴ a = 4or, 2x = 4or, 2x = 22

or, ∴ x = 2or, a - 2 = 0or, a = 2or, 2x = 2or, 2x = 21

∴ x = 1∴ x = 1 or 2

Example 8: If 4x + 14x = 16 1

16 then show that x = ± 2

Solution:

Given, 4x + 14x = 16 1

16

or, 42x + 14x

= 25716

let 4x = a then

or, a2 + 1a

= 25716

or, 16a2 + 16 = 257aor, 16a2 - 257a + 16 = 0or, 16a2 - 256a - a + 16 = 0or, 16a (a - 16) - 1(a - 16) = 0or, (a - 16) (16a - 1) = 0

Either a - 16 = 0or, a = 16


∴ 4x = 16or, 4x = 42

∴ x = 2or, 16a - 1 = 0or, 16a = 1

∴ a = 116

∴ 4x = 142

or, 4x = 4-2

∴ x = - 2So, x = ± 2 proved.

Example 9: If ax = by = cz and b2 = ac, then

Prove that 1x

+

1z

= 2y

Solution:

Given, b2 = acor, (b2)xz = (ac)xz

or, b2xz = axz cxz

or, b2xz = (ax)z × (cz)x

or, b2xz = byz × byx

or, b2xz = byz + yx

or, 2xz = yz + yx

(Dividing all terms by xyz)

or, 2xzxyz

=

yzxyz

+

yxxyz

or, 2y

=

1x

+

1z

proved.

1. a. If ax = ac then what is the value of x ? b. If ax = 1 then what is the value of x ? c. If 32x = 1 then what is the value of x ?2. Solve the following equation. a. 2x= 32 b. 8x = 64 c. 4x = 128 d. ax+5 = 1 e. 125x = 5x-2 f. 9x - 10 × 3x + 9=0

g. 4x + 128 = 3.2x+3 h. 5x -1 + 5-x = 1 15 i. 5x + 1

5x = 25 1

25 j. 4x - 4x-1 = 192 k. 2x+2 + 2x+4

4 = 1 l. 4.3x + 1 = 27 + 9x

m. 3x+5 = 3x+3 + 83

n. 4x + 14x = 16

116 o. 9x + 27 = 4 × 3x+1

3. If ax = y, by = z and zc = x then prove that xyz = 14. If am = bn and b = a2 , then prove that m – 2n = 0

5. It ax = by = cz and b2 = ac prove that: 1x

+ 1z

= 2y

6. It a = 10m, b= 10n and anbm = 100 prove that mn = 1 7. If xyp-1 = a, xy q-1 = b, xy r-1 = c then prove that aq-r br-p cp-q = 1

Exercise 4.10


Ans

wer

s 4.1

01.a. c b. 0 c. 0 2.a. 5 b. 2 c. 7

2 d. -5 e. -1 f. 0 or 2

g. 3 , 4 h. 0 or 1 i. ±2 j.4 k.-3 l. 2 , 1 m. -4 n. 2 , -2 o. 1or 2

Example 1: Find the ratio of : a. 40 minutes to 60 minutes b. 1 year to 9 months

= 40 minutes60 minutes

= 12 months9 months

= 23

= 43

= 2:3 = 4:3

c. 4 days to 2 weeks d. 1 kilogram to 750 gram

= 4 days14 days

= 1000 gm750 gm

= 27

= 43

= 2:7 = 4:3

4.4. Ratio and ProportionRatio: The ratio of two number or quantities of same kind is defined in terms of fraction is used to compare two known quantities. If a and b are two numbers or quantities of same kind, then ratio of a and b is denoted by a:b (read as ‘a is to b) defined by a

b ,where a is called first term or antecedent and b is called second term or consequent. It should be noted that the quantities of different kind cannot be compared.

Example: Ratio of 10 to 15 is 10:15 or 1015 or 2

3 or 2:3

Ratio of 30 min to 1 hrs is 30 min :60 min = 3060 or 1

2 or 1:2

If x and y are two number such that x:y =3:4 then x can be expressed in term of y and y can be expressed in term of x.

i.e. xy

=

34 or x = 3

4 y and y

x =

43 or y = 4

3 x

Ratio of two or more numbers or quantities is always free from units.

Worked Out Examples


Example 2: Express a in the term of b and b in term of a if a:b = 25:6.Solution: Here,

a: b = 5:6

So, ab

=

56

or, a = 5b6

Again,

ab

=

56

or, ba

=

65

or, b = 6a5

or, b = 6a5

Types of Ratio:

1. Duplicate ratio: If the ratio ab

is multiplied by same ratio ab

to get a2

b2 is called duplicate ratio of a

b .

Example :

Duplicate ratio of 3:4 is 32

42 = 916 = 9 : 16

2. Sub-duplicate ratio: Sub duplicate is the ratio of square root of each antecedent and

consequent of given ratio . Sub- duplicate ratio of xy

can be written as xy

.Example :

Sub duplicate ratio of 64:729 is 64729

= 827 = 8 : 27

3. Triplicate ratio

If the ratio ab

is multiplied three times of itself to get a3

b3 is called triplicate ratio of ab

.Example:

Triplicate ratio of the ratio 3:4 = 33

43 = 2764 = 27 : 64

4. Sub-triplicate ratio: Sub triplicate ratio is the ratio of the cube root of antecedent and

consequent of given ratio only. Sub-triplicate ratio of can be written as x3

y3

Example:

Sub- triplicate of ratio 64:729 is 643

7293

= (64)13

(729)13

= 49

= 4: 9


Example 3 : If x:y = 3:4, find the ratio of 3x + y2x+3y

Solution: Here,

Given x:y = 3:4

Let k be any number such that x = 3k and y = 4k.

Now 3x+2y2x+3y

= 3 ×3k + 2 × 4k2 × 3k + 3×4k

= 9k+8k6k+12k

= 17 k18 k

= 1718

= 17 : 18

Example 4 : If x:y:z =2:3:4 then find the value of the ratio x2 + y2

y2 +z2

Solution: Here,

Given x: y:z = 2 : 3 : 4

Let k be any number such that x:y:z = 2:3:4 = k then x = 2k, y = 3k and z = 4k

Now, x2 + y2

y2 + z2 =

(2k)2 + (3k)2

(3k)2 + (4k)2

= 4k2 + 9k2

9k2 + 16k2

= 13k2

25k2

= 1325

Example 5 : Find x:y, 3x - y2x +y

=

12

Solution: Here,

Let x:y = xy

= k, then x = yk

Now,

3x - y2x +y

=

12

or, 3×yk-y2×yk-y

=

12

or, y(3k-1)y(2k-1)

=

12


or, 3k - 12k - 1

=

12

or, 6k - 2 = 2k + 1

or, 6k - 2k = 1 + 2

or, 4k = 3

or, k = 34

or, x:y = 3:4 Example 6 : Two numbers are in the ratio of 4:7. When 6 is subtracted from each of them, new ratio is 10 : 19, find the numbers.

Solution : Here,

Given, two numbers are in the ratio of 4:7

Let k be any constant number. Then the required numbers are 4k and 7k.

According to question,

(4k-6)(7k-6)

= 1019

or, 76k - 114 = 70k - 60

or, 76k - 70k = - 60 + 114

= - 60 + 114

or, 6k = 54

or, k = 546

∴ k = 9

Hence, the required number are 7 × 9 = 63, 4 × 9 = 36

Example 7: If two number are in the ratio of 3:4 and their sum is 49, find the numbers.

Solution: Here,

Given, two numbers are in the ratio of 3:4

Let k be any constant number. then the required numbers are 3k and 4k.

According to questions,

3k + 4k = 49

or, 7k = 49

or, k = 497


∴ k = 7

The required number are 3 × 7 = 21 and 4 × 7 = 28

Example 8: What number must be added to each other of the ratio 9:11 so that sum may be in the ratio 5:6?

Solution: Here,

Let k be added to each term of 9:11.

According to question,

(9 + k): (11 + k) = 5:6

or, 9+k11+k

=

56

or, 54 + 6k = 55 + 5k

or, 6k - 5k = 55 - 54

k = 1

∴ 1 is to be added.

Example 9 : If the two numbers are in the ratio 5:6 and their sum is 33,find the numbers.

Solution: Here,

Let x and y be two numbers.

∴ xy

=

56

and x + y = 33

or, 6x = 5y and y = 33 - x

∴ 6x = 5 (33 - x) [put y = 33 - x]

6x = 165 - 5x

6x + 5x = 165

11x = 165

11x = 11 × 15

x = 15 [while negleting the value of 11 from both]

or, y = 33 - 15

or, y = 18

∴ The two required number are 15 and 18.


1. a. What is ratio?

b. Find the ratio of 50 paisa and Re 1.

c. If there are 15 girls and 20 boys in a class , what is the ratio of number of girls to number of boys in the class.

d. If the antecedent of a coublet be 130 and the ratio is 1:13, find the consequent.

e. If the consequent of a coublet be 7 and the ratio be 1: 18, find the antecedent.

f. If x : 15 = 12 : 45 , find the value of x.

g. 6 : 24 = x : 18 , find the value of x.

h. If x7

= 314

, find the value of x.

2. a. Express x in the term of y, if x:y=2:3.

b. Express b in the term of a if a:b = 3:4.

3. Find duplicate and triplicate of ratios.

a. 2:3 b. 3:4 c. 4:5

4. Find the sub-duplicate of the ratios.

a.1:64 b. 729: 4096 c. 64 : 125

5. Find the sub-triplicate of the ratios.

a.1:64 b. 729: 4096 c. 64 : 125

6. a. If a:b = 2:3 find the ratio

i. 3a+4b5a-2b

ii. a2+b2

a2-b2

b. If a:b =4:5 find (6a-3b):(5a+2b)

c. If p:q =7:8 find (5p-4q):(7p+7q)

7. If a:b:c= 3:4:7 find the ratio of

a. 3a+2b-c

6a-2c

b. a2 - 2bc + c2

b2 -2ac+2ab

8. Find a:b if

a. 3a - 7b4a - b

=

43

b. 2a+b3a+2b

=

58

c. a + 3b3a +2b

=

57

Exercise 4.11


9. a. Two numbers are in the ratio 2:5 and their sum is 77. Find the numbers.

b. Two numbers are in the ratio 3:7 and their sum is 200. Find the numbers.

10. a. Two numbers are in ratio 5:7 and their difference is 16. Find the numbers.

b. Two numbers are in ratio 6:5 and their difference is 10. Find the numbers.

11. a. Two numbers are in the ratio 11:7 and if 7 is added on each number the resulting numbers will be in the ratio of 3:2, find the numbers.

b. Two numbers are in the ratio 7:11 and if 5 is added on each number the resulting numbers will be in the ratio of 2:3, find the numbers.

12. a. What number must be added to each term of the ratio 3:5 to get sum which is in the ratio of 13:5?

b. What number must be added to each term of the ratio 4:5 to get sum which is in the ratio of 11:12?

c. What number must be subtracted from each term of the ratio 4:9 to get the difference which is in the ratio of 9:4?

d. What number must be subtracted from each term of the ratio 7:9 to get the difference which is in the ratio of 3:4?

13. Two brother’s ages are the ratio of 6:5. 15 years ago, the ratio of their ages was 3:2.Findtheir present ages?

14. If the ratio of the ages of mother and her daughter is 5:7. 5 years ago the ratio of their ages was 3:2.Findtheir present ages?

15. a. Divide Rs 870 in the ratio of 7:8.

b. A basket contains 144 oranges, divide the oranges in the ratio 7:5.

c. Divide 30 into three parts in the ratio of 2:3:5.

1 a. consult your teacher b. 1:2 c. 3:4 d. 1690 e. 711

f. 4 g. 92

h. 32

2.a. 2y3

b. 4a3

3.a. 49

, 827

b. 916

, 2764

4.a. 18

b. 916

c. 45

5.a. 14

b. 916

c. 4:5 6.a. 9:2, 13:-5 b. 3:10 c. 1:45 7.a. 5:2 b. 1:2 8.a. -17:7

b. 2:1 c. 11:8 9.a. 22 ,55 b. 60 , 140 10.a. 40, 56 b. 60 , 50 11.a. 77 , 49

b. 35 , 55 12.a. -254

b. 7 c. 13 d. 1

13. 30, 25 14. 3511

,2511

, 15.a. 406, 464 b.84 , 60 c. 6 , 9 and 15

Exer

cise

4.1

1


Proportion:Proportion is the equality of two ratios. Four numbers a, b, c and d are in proportion when the ratio of first pair is equal to the ratio of the second pair. It is denoted by a:b:: c:d. In the above proportion a:b::c:d , a and d are extremes, where a is the first proportional and d is fourth proportional. Again b and c are mean proportional. If four members are in the proportion then product of extremes is equal to product of means. i.e ad = bc.Since, 2:3 equal to 4:6 ; 2, 3, 4 and 6 are in the proportion. 2 and 6 are extremes and 3 and 4 are means. Continued proportion : A continued proportion is an order set of three or more numbers or quantities such that the ratio between any two successive ones is the same.If three numbers a, b, c are in continued proportion then it is denoted by a:b::b:c and defined

by ab

= bc

.

In the above proportion a:b::b:c , a and c are extremes and b is mean. Then square of mean is equal to the product of extremes i.e. since 1:2 equal to 2:4 then 1, 2, 4 are in continued proportion.Discontinued proportion: The proportion of the form

ab

= cd

is called discontinued proportion. Here two ratios are equal but b ≠ c

For example 23

= 46

is an example of discontinued proportion.

Some important results.

1. If , ab

= cd

we can multiply both side by bd. Then ad=bc, i.e we can cross multiply.

2. If ab

= cd

then 1 ÷ ab

= 1 ÷ cd

or ba

= dc

or this result is called “Invertendo”.

3. If ab

= cd

multiplying each side by bc

We have,

ab

× bc

=

cd

×bc

Or, abb2

= cbdc

This result is called “Alternando”4. If

ab

=

cd

We have,

ab

+ 1 =

cd

+ 1

Or, a + bb

=

c +dd

This result is called “componendo”.

5. If ab

=

cd

,

ab

- 1 =

cd

- 1

Or, a - bb

=

c -dd


Example 1 : Find ab

, if 8a + 5b8a - 5b

=

73

Solution : Here,

8a + 5b8a - 5b

=

73

First method.By componendo and Dividendo,

(8a + 5b) + (8a - 5b)(8a + 5b) - (8a - 5b)

=

7 + 37 - 3

16a10b

= 104

This result is called “Dividendo”.

6. If ab

=

cd

then,

∴ a + b

a - b=

c + dc - d

(Dividing componendo by dividendo)

This result is called “Componendo and Dividendo”.The k-method.Many results in the ratio and proportion can be obtained by putting each equal ratio to k (or any other symbol)Thus if

ab

=

cd

= ef

, we may put each ratio equal to k.

ab

=

cd

= ef

= k

Then a = bk, c = dk, e = fkIf a, b, c, d are in the continued proportion, we may put

ab

=

bc

= cd

= kThen c = dk, b = ck = dk2, a = bk = dk3.The Addition TheoremWe will now use k-method to prove this useful result.If a:b = c:d = e:f = ....Then each of these ratio =

a+c+e+...b+d+f+...

Let ab

=

cd

= ef

= k

Then a = bk , c = dk , e = fk

Thus each of the ratio a+c+e+...b+d+f+...

= bk + dk + fk + ...b + d + f + ...

= k

This result is called “Addendo”.

Worked Out Examples


or,

ab

=10 ×1016×4

or,

ab

= 2516

Second method Or, 24a + 15 b = 56a - 35 b

Or, 24a - 56a = - 35b - 15b

Or, - 31a = -50 b

∴ ab

=

-50-32

=

2516

Example 2 : If a+3b+2c+6da+3b-2c+6d

= a-3b+2c-6da-3b-2c+6d

, prove that ab

=

cd

Solution:

First method

a+3b+2c+6da+3b-2c+6d

= a-3b+2c-6da-3b-2c+6d

∴2a + 6b

4c + 12d =

2a - 6b4c - 12a

(by componendo and dividendo)

or, 2(a+3b)4(c+d)

=

2(a-3b)4(a-3d)

or,

a + 3bc + 3d

=

a - 3bc - 3d

or, a + 3ba - 3b

=

c + 3dc - 3d

(by componendo and dividendo)

ab

=

cd

Hence, proved.

Second method:

Let ab

=

cd

= k

Then a = bk, c = dk

LHS = a+3b + 2c + 6da + 3b - 2c - 6d

= bk + 3b + 2dk + 6dbk + 3b - 2dk - 2d

= b(k+3) + 2d(k+3)b(k+3) - 2d(k+3)


= (k + 3) (b + 2d)(k + 3) (b - 2d)

=

b + 2db - 2d

Again,

RHS = a - 3b + 2c - 6da - 3b - 2c + 6d

= bk - 3b + 2dk - 6dbk - 3b - 2dk + 6d

=

b(k-3)+2d(k-3)b(k-3) -2d(k-3)

= (k - 3) (b + 2d)(k - 3) (b - 2d)

= b + 2db - 2d

Hence,

ab

=

cd

Proved.

Example 3 : If a : b = b : c = c : d prove that (a + b)(c + d) = (b + c)2

Solution : Here,

Let ab

=

bc

=

cd

= k

Then, c = dk

b = ck = dk2

a = bk = dk3

∴ (a + b) (c + d) = (dk3 + dk2) (dk + d)

= dk2 (k + 1)d (k + 1)

= d2k2 (k + 1)2

And (b +c)2 = (dk2 + dk)2

= {dk (k + 1)}2 = d2k2 (k + 1)2

∴ (a + b) (c + d) = (b + c)2

Example 4 : If ab

=

cd

=

ef

prove that 4a - 9d2

4b-9d( )2

= 3c2+4e2

3d2 + 4f2

Solution: Here,

Let ab

=

cd

=

ef

= k Then a = bk, c = dk, e = fk

∴ L.H.S = 4a - 9d2

4b-9d( )2


= 4bk-9dk4b-9d{ }2

= k(4b-9d)

4b-9d{ }2

= k 2

R.H.S = 3d2k2 + 4f2 k2

3d2 + 4f2

= k2 (3d2 + 4f2)(3d2 + 4f2)

= k2

∴ L.H.S = R. H. S

Example 5 : If xa

=

yb

=

zc

, prove that x3 - y3

a3 - b3 = y - zb - c( )3

Solution:

Let xa

=

yb

=

zc

= k

Then x = ak, y = bk, z = ck

∴ L.H.S = x3 - y3

a3 - b3

= a3 k3 - b3 k3

a3 - b3

= k3 (a3 - b3)a3 - b3

= k3

R.H.S. = y - zb - c( )3

= bk - ckb - c( )3

k(b-c)b - c[ ]3

= k3

∴ L.H.S. = R. H. S

Example 6 : If ab

=

bc

=

cd

, show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2

Solution:

Let ab

=

bc

=

cd

= k


Then c = dk b = ck = dk2

a = bk = dk3

∴ (a2 + b2 + c2) (b2 + c2 + d2)

= {(dk3)2 + (dk2)2 + (dk)2} {(dk2)2 + (dk)2 + d2}

= (d2k6 + d2k4 + d2k2) (d 2k4 + d2k2 + d2)

= d2k2 (k4 + k2 + 1)d2 (k4 + k2 + 1)

d4k2 (k4 + k2 + 1)2

and (ab + bc + cd)2

= (dk3 dk2 + dk2dk + dkd)2

= {d2k (k4 + k2 + 1)}2

= d4k2 (k4k2 + 1)2

∴ (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2

The work will be shortened if we simplify substituting a = bk, b = ck, c = dk

Thus, (a2 + b2 + c2 ) (b2 + c2 + d2)

= (b2k2 + c2k2 + d2k2 ) (b2 + c2 + d2) = k2 (b2 + c2 + d2)2

and (ab + bc + cd)2 = (b2k + c2k + d2k)2 = k2(b2 + c2 + d2)2

∴ (a2 + b2 + c2 ) (b2 + c2 + d2 ) = (ab + bc + cd)2

1. a. What is proportion ? b. What is continued proportion? c. If 6,10,x,40 are in proportion, find the value of x. d. If 1,a,5,15 are in proportion, find the value of a. e. What constant number must be subtracted from each of 8,10,17,22 to make them in

proportion? f. What number should be subtracted from each of the terms 11,10,29 and 26 so that they

are in proportion?

2. a. Find xy

when 4x + 3y4x - 3y

= 119

Exercise 4.12


b. Find ab

when 7a - 3b7a + 3b

= 59

3. If a:b =c:d, prove that

a. 3a + 4c: 3b + 4d = 5a - c:5b - d

b. a:a + b = c:c + d

c. a2 + c2 :b2 + d2 = ac:bd

d. a + 2b:c + 2d = 2a - b:2c - d

e. ma + nb:mc + nd = b2c:ad2

f. a2 + b2:a2 - b2 = ac + bd:ac - bd

g. a2d - bc2 = ac(b - d)

h. a2

b2 +

c2

d2 = 2acbd

i. a2 - ac + c2:b2 - bd + d2 = (a + c)2 : (b + d)2

j. (a + c)3 : (b + d)3 = a(a - c)2: b(b - d)2

k. a + b:c + d = a2 + b2 : c2 + d2

l. a2 + c2 : b2 + d2 = ma + nc:mb + nd

m. (a2 + b2) (b2 + d2) = (ab + cd)2

n. a2 + 3b2:a2 - ab + b2 = c2 + 3d2: c2 - cd + d2

o. a: (a+ b) (a + c) = 1:a + b + c + d

p. (a + c)3 : (b + d)3 = b (a - c)4 : a (b - d)4

q. a2 (b + d) (c + d) = bc (a + b) (a + c)

r. 4 (a + b) (c + d) = bd 2

+[ ]a+bb

c +dd

4. If a:b =b:c prove that.

a. a:a + b = a - b:a - c

b. (a - b)2: a = ( b - c)2:c

c. a2 + ab + b2:b2 + bc + c2 = a:c

d. (a + b + c) (b - c) = ab - c2

e. abc (a + b + c)3 = (ab + bc + ca)3

f. a3 + b3

b3 + c3 = a(a-b)c(b-c)

5. If a, b, c, d are in continued proportion prove that


a. a:d = a3:b3

b. a:d = pa3 + qb3 + rc3:pb3 + qc3 + rd3

c. a2 + ab + b2:c2 + cd + d2 = ab:cd

d. a:d = a3 + b3 + c3:b3 + c3 + d3

e. a2+abb

= b2 +bcc

= c2 +cdd

f. (a+b)2

ab = (b+c)2

bc = (c+d)2

cd

g. (b +c ) (b + d) = (c + a) (c + d)

h. (a + d) ( b + c) = (a + c) (b + d) + (b - c)2

i. (b - c)2 + (c - a)2 + (d - b)2 = (a - d)2

j. (a + d)2 + (b - c)2 = (b + c)2 + (a - d)2

6. If ab

=

cd

=

ef

= k, show that each of the following given ratios is equal to k.

a. pq - qc - repb - qd - rd

b. xa - yc - 2exb - yd -2f

c. a + mc + ncb + md + nf

7. a. If a:b = c:d, prove that (a+b) (a+c)

abc =

1a

+

1b

+

1c

+

1d

b. If ab = cd = ef, show that (a2 + c2 + e2 ) (b2 + d2 + f2) = (ab + cd + ef)2

c. If a: b = b:c = c:d, prove that ab - bc + cd = (a-b+c) (b-c+d)

1.a . consult your teacher b. consult your teacher c. 24 d. 3 e.2 f. 2 2.a. 152

b. 32

Exercise 4.12

4.5. Algebraic fractions/ Rational fractionFraction having algebraic expression in numerator and denominator are called algebraic fraction. To simplify such algebraic fraction we have to follow same process as numeric fraction i.e rational number. The process of addition and subtraction of two or more algebraic fraction consist of following steps.a. Factorize the denominator and numerator if required.b. Find the L.C.M of denominators.c. Divide the L.C.M by denominator of first fraction and multiply the quotient with

numerator of the fraction. Follow same process for other fractions. d. Simplify the numerator and reduce the last fraction into lowest terms.


Example 1: Simplify : a. 1x+1

+ 1x+1

b. aa + b

+

aa - b

c. 2a2 - b2

+

3a2 + 2ab + b2

a. Solution : Here, b. Solution : Here c. Solution: Here,

1

x+1 + 1

x+1 a

a + b +

aa - b 2

a2 - b2 +

3a2 + 2ab + b2

= 1+1x+1

= a(a-b) + a(a + b)

(a+b) (a - b) =

2(a+b) (a-b) -

3(a+b) (a+b)

= 2x+1

= a2 - ab +a2 + ab

a2 - b2 = 2(a+b) - 3(a-b)

(a+b) (a-b) (a+b)

= 2a2

a2 - b2 =

2a + 2b - 3a + 3b(a + b)2 ( a - b)

= 5b - a

(a+b)2 (a-b)Example 2: Simplify : 2

(a - 2) (a - 3) - 2

(a - 1) (a - 3) Solution : Here

= 2(a - 2) (a - 3)

- 2(a - 1) (a - 3)

= 2(a-1) -2 (a -2)(a -1) (a - 2) (a - 3)

= 2a - 2 - 2a + 4(a -1) (a - 2) (a - 3)

=

2(a -1) (a - 2) (a - 3)

Example 4: Simplify : xx - y

- yx + y

+ 2xy x2 - y2

Solution : Here

= x2 + xy - xy + y2

(x -y) (x + y) + 2xy

x2 - y2

= x2 + y2

(x -y) (x + y) + 2xy

(x-y) (x+y)

= x2 + y2 + 2xy(x-y) (x+y)

= (x + y)2

(x-y) (x+y)= x + y

x - y

Example 3: Simplify : 1x2 - 2x - 8

- 1x2 + 5x + 4

Solution : Here

= 1x2 – 2x - 8

- 1x2 + 5x + 4

= 1(x -2) (x + 4)

- 1(x + 1) (x + 4)

= x + 1 - x + 2(x +1) (x - 2) (x + 4)

=

3(x +1) (x - 2) (x + 4)

Worked Out Examples


Example 5: Simplify : a-2a2 - 3a + 2

+ a-3a2 - 5a + 6

+ a-5a2 - 8a + 5

Solution : Here

= a - 2a2 - 3a + 2

+ a - 3a2 - 5a + 2

+

a - 5a2 - 8a + 5

= a - 2(a -1) (a-2)

+ a - 3(a -2) (a-3)

+ a - 5(a -3) (a-5)

= 1a - 1

+

1a - 2

+

1a - 3

= (a - 2) (a - 3) + (a - 1) – (a - 3) + (a - 1) (a - 2) (a-1) (a - 2) (a - 3)

= a2 – 5a + 6 + a2 - 4a + 3 + a2 - 3a + 2(a -1) (a - 2) (a - 3)

= 3a2 - 12a + 11(a-1) (a - 2) (a - 3)

Example 7: Simplify : 1 +x1 - x

+ 8x1+x4

-

1-x1+x

+

4x1+x2

Solution : Here

= 1 + x1 - x

+ 8x1 + x4

– 1 - x1 + x

+ 4x1 + x2

= 1 + x1 - x

– 1 -x1 + x

+ 4x1 + x2 + 8x

1 + x4

= 1+2x+x2-1+2x - x2 + 4x(1 - x2) (1 + x2)

+ 8x1 + x4

= 8x1 - x4 + 8x

1 + x4

= 8x (1 + x4)+ 8x (1-x4)(1 - x4) (1 + x4)

= 8x + 8x 5 + 8x - 8x5

1 - x8

= 16x1 - x8

Example 8: Simplify : 1a + b

+ ba2 - ab + b2

+

a2

a3 + b3

Solution : Here

1a + b

+ ba2 - ab + b2

+

a2

a3 + b3

= 1a + b

+ ba2 - ab + b2

+

a2

(a+ b) (a2 – ab + b2)


Exercise 4.13

1. Simplify :

a. 3

xy +

5xy

b. 3aa-b

- 3ba-b

c. 1

x(x+a) +

1x(x-a)

d. x + yx - y

- x - yx + y

e. a2 + b2

a2 - b2 -

a - b2 (a - b)

f. 8

(x + 3)2 +

8x2 - 9

g.

12m - 3n

- m + n

4m2 - 9n2 h.

8a2

b2-4a2 +

2ab + 2a

2. a. 1

a - 2b +

1a + 2b

- 4a

a2-4b2 b. 4

1 + x -

31 - x

+ 7

x2 - 1 c.

1p + q

+ 1

p - q-

2pp2 - q2

d.

1m - n

+ 1

m + n-

2nm2-n2

e. a + ba - b

+ a - ba + b

- a2+b2

a2-b2 f.

xx+2y

- x

2y -x-

8y2

x2-4y2

g. 1

a+2b +

12b-a

+ 8b 2

4ab2-a3 h.

11 + x

+ 1

1 - x+

21 +x2 +

41 +x4

i. 1

m2 + 2mn + n2 +

2n(m+n)2 (m-n)

+1

m2 - n2

3. a. 3a

+ 2

a + 2 -

5a + 1

b. 1

x - 1 -

3x

+3

x +1-

1x +2

c. 1

x - 6 -

1x-4

+1

x +6-

1x +4

4. a. 2

a2-1 -

1a + 1

- 1

(a+1)2 b. 6

a2-9 -

1a -3

- 1

2(a+3)2 c. 3

a+3 +

4a - 3

+ 9a

2(9-a2)

5. a. a + 5

a2 + 11a+ 30 +

a - 318 - 3a - a2

b. x2 - 5x

x2 - 8x +15 +

10 - 2xx2 - 8x + 15

c. 1

a2 + 4a + 3 +

1a2 + 8a + 15

+1

a2 + 6a + 5 d.

17 - 5x - 2x2

-2

x2 - 5x + 4 +

328 + x - 2x2

e. 3

(x + 2) (4 - x) +

1(x - 3) (x - 4)

+1

(x - 3) (x + 2)

6. 1a + x

+ 2aa2+x2

+ 4a3

a4 +x4 +

8a7

x8–a8

= a2 - ab + b2 + ab + b2 + a2

(a + b) (a2 -ab + b2)

= 2a2 + 2b2

(a + b) (a 2 - ab + b2)

= 2(a2 + b2)a3 + b3


1. a. 8xy b. 3 c. x

x2-a2 d. 4xyx+y e. a2+3b2

2 (a2-b2) f. 16x

(x-3) (x+3)2 g. m + 2n4m2 - 9m2 h. 2a

b-2a 2a. a-b4b2 -a2 b. 7x+6

x2-1 c. 0

d. 2m+n e. a

2+b2

a2-b2 f. 2 g. 4ba (2b -a) h. 8

1 -x8 3.a. a + 6

a (a+1) (a+2) b. 6x (x+2) (x2-1) c. 20x2

(x2 - 16) (x2 - 36) 4.a 4 + a - a2

(a + 1)2 (a -1)

b. -(2a+7)2 (a+3)2 c. 5a - 6

2 (a2 -9) 5a. 0 b. x+2x - 3 c. 3

(a+1) (a+5) d. - (7 + 8x)

(x-1) (x-4) (2x+7) e. 7 - x(x+2) (x-3) (x-4) 6. 1

x - a

Exer

cise

4.1

3

4.6. Equation in two variables:Linear equations in one variableReview : Before going to equation in two variable we should review the linear equation in one variable.Linear equation in one variableAn expression of the form ax+b, where a and b are constant and a ≠ 0, is called linear expression or a first degree expression in the variable x. Therefore, an equation of the type ax + b = 0, where a and b are constant and a≠0 is called a linear equation in the variable x.Following are some examples of linear equation in one variable:a. 2x = 8 b. 4y = 9 c. 3z = 7 d. 2x + 4= 7All the above mentioned examples have only one variable and are linear in nature. So, they are known as linear equation in one variable. A linear equation may have one variable or two variables.Now moving into the actual topic, linear equation is an equation that gives a straight line when plotted on a graph. Linear equation in one variable is an equation with one unknown quantity which when plotted on the graph gives a straight line.Any value of the variable which satisfies the equation is called the solution or the root of the equation. The equation x + 5 = 7 is true when x = 2, so 2 is the solution(or root of the equation x + 5 = 7) .To solve an equation is to find its solution and process of finding the solution is called solving equation.Example 1 : Solve : a. 3x + 18 = 21 b. 2 ( 4 – x ) + 3 ( 7 – x ) – 1 = 51x a. Solution : Here, b. Solution : Here

3x + 18 = 21 2 ( 4 – x ) + 3 ( 7 – x ) –1 = 51xOr, 3x = 21 – 18 Or, 8 – 2x + 21 – 3x – 1 = 51x

Or, x = 33

= 1 Or, – 5x +28 = 51xThus, the value of x is 1 Or, 28 = 56 x ∴ x = 2 Hence, the value of x is 2


c. x -15

+ x - 92

= 3

Solution : Here,

x - 15

+ x - 92

= 3

Or, 2(x - 1) + 5 (x - 9)10

= 3

Or, 2x – 2 +5x – 45 = 30Or, 7x = 30 + 47

Or, x = 777

= 11

Hence, the value of x is 11Example 2 : The sum of two numbers is 24, one of the numbers is twice the other. Find the numbers.Solution: Here,Let one of the numbers be ‘x’.It is given that 2nd number is two times the first number. So, 2nd number = 2x.By question, x + 2x = 24. Or, 3x = 24

Or, x = 243

= 8∴One number is 8and other number is 2x = 2 × 8 = 16

Example 3 : The difference between two numbers is 70. If the numbers are in ratio 3:5, find the numbers.Solution: Here,Let the common number be ‘x’.Then,The 1st number = 3x and 2nd number = 5x.Now, 5x – 3x = 70 Or, 2x = 70

Or, x = 702

= 35The numbers are 3x = 3 × 35 = 105 and 5x = 5 × 35 = 175.Hence, the required numbers are 105 and 175.


4.7. Simultaneous equation Consider the equation x-y = 7 , here we see thatWhen x= 3, y = - 4When x= 8, y = 1When x= 9, y = 2 and so on.In this way, there is a value of y for every value of x and the equation has an infinite number of solutions.Again, consider the equation x + y = 15.It has also infinite number of solutions such as x = 7, y = 8, x = 8, y = 7, x = 12, y = 3 and so on.Thus x and y have infinite number of values when the two equations are taken separately. But if we take both equations simultaneously, we see that they are satisfied by only one pair value x = 11, y = 4.Two such equations which are satisfied by the same values of unknown variables are called simultaneous equations in two variables.Method of solving simultaneous Equations.We have three methods of solving simultaneous equations of two unknown values which are given below. Substitution method :i. Take one equation and get the value of x or y in terms of the other.

Example; if 2x+3y = 5 write y = 5 - 2x3 or x =

5 - 3y2

ii. Substitute this value in the other equation.iii. Solve the simple equation so obtained.iv. Substitute this value in the equation obtained from step i and get the value of the other

unknown.Note: Number the equation so that the explanation of your work may be clear.

Example 1 : Solve : 2x + y = 8 and 3x - 2y = 5 Solution : Here, 2x + y = 8 ...................(i) 3x - 2y = 5 ..................(ii)

From (i), if we write 2x = 8 - y and x = 8 - y2

we will get a fraction. But the coefficient of y is 1 so, we can write the values of y more conveniently as, y = 8 - 2x ...................(iii)Substituting this value of y in (ii) we have, 3x - 2 (8 - 2x) = 5

Worked Out Examples


or, 3x - 16 + 4x = 5 or, 7x = 21

or, x = 217

∴ x = 3

Substituting this value of x in (iii) we have

y = 8 - 2 × 3 = 8 - 6 = 2

∴ x = 3,y = 2

Checking: Putting the value of x and y in equation (iii)

2 = 8 - 2 × 3

2 = 2 (True)Example 2 : Solve: 2x + 5y = 9 and 3x + 5y = 11 Solution : Here, 2x + 5y = 9 …………….(i) 3x + 5y = 11 ……………(ii)Here, y has the same coefficient 5 in both the equations. So we can easily Substitute the value from one equation to the other.Thus from (i) 5y = 9 - 2x 5y = 9 - 2x ..................(iii)Substituting it in (ii) we get, 3x + 9 - 2x = 11 3x - 2x = 11 - 9 x = 2Substituting this value of x in (iii), we get

y = 9 - 2 × 25

= 9 - 45

= 55

= 1∴ x is 2 and y is 1Example 3 : Solve: 2x + 3y = 17 and 3x - 5y = - 3 Solution : Here, 2x + 3y = 17 …………..(i) 3x - 5y = - 3 ..………..(ii)


From (i), 2x = 17 - 3y

x = 17 - 3y2

(iii)

Solving equation (ii) and (iii)

3x - 5y = -3

3 (17 - 3y)2

- 5y = - 3

51 - 9y - 10y = - 6

or, -10y - 9y = - 6 - 51

or, - 19y = - 57

or, y = 5719

∴ y = 3

Substituting y = 3 in (iii), x = 17-3×32

= 4

∴ x = 4, y = 3

Example 4 : Solve: 7x = 3y and 13x = 10y - 31Solution : Here, 7x = 3y ………………….(i) 13x = 10y - 31 .......……..(ii)

From (i) y = 7x3

………(iii)

Substituting it in (ii), we have,

13x = 10 × 7x3

- 31

or, 13x = 70 - 933

or, 39x = 70x - 93or, 39x - 70x = - 93or, -31x = - 93

or, x = -93- 31

or, x = 3Substituting 3 for in (iii),we have

y = 7 × 33

y = 7 ∴ x = 3, y = 7


Elimination methodIf the coefficient of one of the variable is equal, we can eliminate that variable by addition or subtraction.Example 5: Solve by elimination method: 3x + 5y = 11 and 4x + 5y = 13.Solution : Here, 3x + 5y = 11 ………….(i) 4x + 5y = 13 ……………(ii)Here the coefficient of y are equalSubtracting (ii) from (i), we have 3x + 5y = 11 4x + 5y = 13

(-) (-) (-)- x = - 2

∴ x = 2Substituting x = 2 in (i) we get, 3 × 2 + 5y = 11 5y = 11 - 6 5y = 5 ∴ y = 1Example 6 : Solve by elimination method. 4x + 3y = 20 and 6x - 3y = 0Solution : Here, 4x + 3y = 20 ..........(i) 6x - 3y = 0 .............(ii)Here, we have 3y in (i) and - 3y in (ii)Adding (i) and (ii) we have

4x + 3y = 206x - 3y = 010x = 20

∴ x = 2010

= 2

Substituting x = 2 in (i) We get, 4 × 2 + 3y = 20 8 + 3y = 20 or, 3y = 12 or, y = 4 ∴x = 2 y = 4


The equation are not always given in such simple forms. But, we can multiply one or both the equations by a suitable number and make the coefficient of one variable equal. The method is illustrated below.

Example 7 : Solve x + 2y = 7 and 2x + 3y = 12Solution : Here, x + 2y = 7 ………….(i) 2x + 3y = 12 ………..(ii)Here, the coefficient of x in equation (i) and (ii) is not same. So,we can multiply (i) by 2 to make the coefficient of equal.Then, 2x + 4y = 14 …….(iii)Subtracting (iii) from (ii) we have

2x + 3y = 122x - 4y =14

-y = -2

or, y = 2

Putting y = 2 in (i) we have x + 4 = 7or, x = 3∴ x = 3, y = 2Note: Here, we could also have multiplied (i) by 3 and (ii) by 2 to eliminate y.

Example 8 : Solve : 2x + 4y = 10 and 5x - 6y = 9Solution : Here, 2x + 4y = 10 ………….(i) 5x - 6y = 9 …………..(ii)Here, we can multiply (i) by 3 and (ii) by 2 and eliminate y. 6x + 12y = 30 …………(iii) 10 x - 12y = 18……….. (iv)Adding (iii) and (iv) we get,

6x + 12y=3010x-12y=18

16x = 48

∴x = 3Putting x = 3 in (i), 6 + 4y = 10or, 4y = 4


or, y = 1∴ x = 3, y = 1

Example 9 : Solve - 3x5

+

2y3

= 1 and

2x5

+

y3

= 4Solution: Here,

-3x5

+

2y3

= 1

or, - 9x + 10 y 15

= 1

or, - 9x + 10 y = 15 .........(i)

and 2x5

+

y3

= 4

or, 6x + 5y15

= 4

or, 6x + 5y = 60 ...........(ii)Multiply equation (i) by 2 and (ii) by 3 and adding

- 18x + 20 y = 3018 x + 15 y = 180

35y = 210

or, y = 21035

∴ y = 6Now put y = 6 in equation (ii) we get 6x + 5y = 60or, 6x + 5 × 6 = 60or, 6x = 60 - 30or, 6x = 30or, x = 30

6∴ x = 5So, x = 5 and y = 6

Example 10 : Solve: 2x

+

3y

= 2 and 8x

+

9y

= 7Solution: Here,

2x

+

3y

= 2 .............(i)

8x

+

9y

= 7 .............(ii)

Multiplying equation (i) by 4 and subtracting from equation (ii)

8x

+

9y

= 7


8x

+ 12y

= 8

- - -

9y

- 12y

= -1

or, 9 - 12y

= -1

or, -3y

= -1

or, y = 3

Now put y = 3 in equation (i) we get

2x

+

3y

= 2

2x

+

33

= 2

2x

= 2 - 1

2x

= 1

x = 2 ∴ x = 2 and y = 3Graphical method To solve the pair of simultaneous equation, we draw the graph of two linear equations.When we draw the graph of two linear equations in two variables on the same axes, we get two lines in same plane. Then there arise three conditions:Conditions:i. They may intersect at one and only one point on the plane.ii. They may not intersect at all (i.e. the lines are parallel).iii. They may intersect infinite number of points (i.e. the lines may coincide). To solve the

equation by the method of graph we need follow these steps.Steps to follow.a. Make table for each equation.b. Draw the line graph of the both equations on the same plane.c. The lines drawn meets at a point.d. Read the coordinates of the point of intersection, which is the solution of the equations.


Example 11 : Draw the graph of x + 2y = 6 and x - 2y + 2 = 0 and find the solution.Given equation are x + 2y = 6 -------------(i) x - 2y + 2 = 0 ----------(ii)Changing the equation (i) and (ii) in the form y = mx + c from equation (i), x + 2y = 6or, x = 6 - 2y

or, y = 6 - x2

x 4 2 0y 1 2 3

∴ (4,1), (2,2), (0,3)From equation (ii), x - 2y + 2 = 0or, x + 2 = 2y

or, y = x + 22

x -2 0 2y 0 1 2

∴ (-2,0), (0,1),(2,2)

Since the two lines intersect at a point (2,2). Solution of the equations is (2,2) i.e. x = 2 and y = 2.

Example 12: Solve graphically 2x + 3y = 12 and x + 2y = 7 2x + 3y = 12 ----------------(i) x + 2y = 7 ---------------(ii)From equation (i) 2x + 3y = 12 3y = 12 - 2x

y = -2x + 123

Table for equation (i)

x 0 3 6y 4 2 0

∴ (0,4), (3,2), (6,0)From equation (ii) x + 2y = 7 2y = 7 - x

Y

O X

•

•

•

•

•


y = - x + 72

Table for equation(ii)

x 1 3 5y 3 2 1

∴ (1,3), (3,2), (5,1)

a

Xd

Y

Since two line intersect at point (3,2) so the value of x = 3 and y = 2Verbal Problems:Verbal problems are the word problems, where one or more unknown quantities are given and we have to form mathematical statement using basic fundamental operations of mathematics like addition, subtraction, multiplication and division. Here we are concerned with two unknown quantities.In order to solve problems related to unknown quantities, we have follow the following general procedure.1. Read the question carefully so that you clearly understand what is given to you and what

are asked to find out.2. Choose two letters (say x and y) to represent two unknown numbers which you have to

find.3. Write the statement of problem in algebraic symbols so that two equations are formed.4. Solve the equations and find out the values of x and y .5. Verify the answer by restoring two values in both of the equations.

Example 13 : Find the two numbers whose sum is 36 and difference is 12.Solution:Let the two numbers be x and y.According to the first condition x + y = 36 -------------------(i)According to the second condition x - y = 12 .......................(ii)


Adding equation (i) and (ii)

x + y = 36x - y = 122x = 48

or, x = 482

∴ x = 24Now, putting the value of x in equation (i) we get x + y = 36or, 24 + y = 36or, y = 36 - 24∴ y = 12Therefore, the required number are 29 and 12.For VerificationRestoring x = 24, y = 12 in equation (i) x - y = 12 24 - 12 = 12 ∴ 12 = 12Here, required numbers are true.

Example 14 : Two times a number is greater than 3 times smaller by 6. 6 times the smaller number exceeds 3 times the greater by 6. Find the numbers.Solution : Here: Let the greater number be x and smaller number by y.According to the 1st condition 2x - 3y = 6 --------------------(i) [two times x is greater that 3 timrs y by 6] According to the 2nd condition 6y - 3x = 6 (the difference of 6 times y and 3 times x is 6)or, 3(2y - x) = 6

or, 2y - x = 63

or, 2y - x = 2 ...................(ii) Now, From equation (ii) x = 2y - 2 -------------(iii)Solving equation (i) and (iii) 2x - 3y = 6


2(2y - 2) - 3y = 6 4y - 4 - 3y = 6 y = 6 + 4 y = 10Putting the value of y in equation (iii) x = 2y - 2 = 2 × 10 - 2 = 2 × 10 - 2 = 20 - 2 = 18Therefore, the required numbers are 18 and 10.

Example 15: The sum of the ages of mother and her daughter is 37. After 10 years the mother will be twice as old as the daughter. Find their present ages.

Solution : Here,Let the present age of mother be x years and present age of daughter be y years According to the 1st condition x + y = 37 ------------(i)According to the second condition x + 10 = 2(y + 10) or, x + 10 = 2y + 20or, x = 2y + 20 - 10or, x = 2y + 10 ............................(ii)Solving equation (i) and (ii) x + y = 37or, 2y + 10 + y = 37or, 3y = 37 - 10or, 3y = 27

or, y = 273

or, y = 9Putting the value of y in equation (ii) x = 2y + 10or, x = 2 × 9 + 10or, x = 18 + 10∴ x = 28

∴The present age of mother is 28 years and the present age of daughter is 9 years.


Example 16 : 4 kg of tomatoes and 6 kg of potatoes cost Rs 208, 6kg of tomatoes and 4kg of potatoes cost Rs 192. Find the cost per kg of each.

Solution : Here,

Let the price of tomatoes be Rs. x per kg and price of potatoes be Rs. y per kg.

According to the 1st condition,

4x + 6y = 208

2x + 3y = 104 ………………(i)

According to the 2nd condition

6x + 4y = 192

3x + 2y = 91 .................(ii)

Multiply equation (i) by 2 and (ii) by 3, and subtracting equation (i) from equation (ii) we get,

4x + 6y = 208 + 9x +6y= + 273

- 5x = - 65

or, x = 13

Putting the value of x in equation (i)

2x + 3y = 104

or, 2 × 13 + 3y = 104

or, 26 + 3y = 104

or, 3y = 104 - 26

or, 3y = 78

or, y = 783

∴ y = 26

The cost per kg of tomatoes is Rs 13 and the cost per kg of potatoes is 26.

Example 17 : The perimeter of a rectangular field is 102m.If the length is decreased by 3 meters and breadth is decreased by 2 meters it becomes a square find the length and breadth of rectangular field.

Solution : Here,

Let the length of rectangular filed be x meters and breadth be y meters.

According to the 1st condition,

2(x + y) = 102 m

x + y = 51 m --------------(i)


According to the 2nd condition,

x - 3 = y - 2

x - 3 + 2 = y

x - 1 = y

y = x - 1 ---------------(ii)

Solving equation (i) and (ii)

x + y = 51

x + x - 1 = 51

2x = 51 + 1

2x = 52

x = 522

x = 26Again,

Putting the value of x in equation (ii) we get

y = x - 1

y = 26 - 1

y = 25

∴ The required length and breadth of rectangular field are 26m and 25m respectively.

Example 18 : In a two digit number. The unit digit is thrice the tens digit. If 36 is added to the number, the digits interchange their place. Find the numbers.

Solution:

Let the digit in the units place of two digit number is x and the digit in the tens place be y the the number is 10y+x.

According to 1st condition : x = 3y .............(i)

The number obtained by reversing the digits is 10x + y.

If 36 is added to the number, digits interchange their places,

According to 2nd condition. 10 y + x + 36 = 10 x + y

or, 10y + x + 36 - 10x - y = 0

or, 9y - 9x + 36 = 0

or, 9y - 9x = - 36

or, y - x = - 4........... (ii)

∴ x - y = 4


Substituting the value of x = 3y in equation (i), we get

3y - y = 4

or, 2y = 4

or, y = 42

∴ y = 2

Substituting the value of y = 2 in equation (i),we get

x - 2 = 4

or, x = 4 + 2

∴ x = 6

Therefore, the required two digit number

= 10y + x

= 10 × 2 + 6

= 26

Exercise 4.14

1. a. What is linear equation ?

b. What are simultaneous equations?

c. Find a point on the line whose equation is y = 4x.

d. Write any point lies on X- axis.

e. Write any point on Y- axis.

2. Solve :

a. x+ 4 = 9 b. 3x -6 = 10 c. 5(x+5) = 30

d. x5

- 8 = 12 e. 2x - 4

3 = 6 f. 3(y – 2) = 2(2 – y)

g. 2(x – 3) = 3(3 – x) h. 5x - 51x - 15

= 2 i. 4y + 45y - 7

= 4

3. a. In equation x = 2y + 33

if y = 3 then what is the value of x?

b. In an equation 2x + 3y = 10 if x = 2 then what is the value of y?

c. In the equation 3x + 2y = 12 and x + 3y = 11 to solve by using elimination method to eliminate x which equation should be multiply by what number?


4. Solve the following equations using substitution method.

a. x + y = 6 and 3x + y = 15 b. 3x - 2y = 11 and x + 3y = 11

c. 3x - 2y = 3 and 3x + 5y = 24 d. 3x - 2y = 8 and x + 2y = 8

e. 4x - 3y = 2 and 3x + 4y = 39 f. 5x - 2y = 20 and 3x + 5y = 43

g. 3x + 7y = 13 and 9x + y = 19 h. 4x - 3y = - 1 and 3x + 2y = 12

i. y = 5x + 1 and 2x - 5y = 51 j. 3x + 4y = 27 and 7x - 2y = 29

k. x + 1y - 1

=

12

and x - 1y + 1

=

27

l. x5

+ y8

= 2 and x4

+ y10

= 2 120

m. 1x

+ 1y

= 5 and 1

3x +

12y

= 2 n. 3x

+ 4y

= 24 and 4x

- 3y

= 2

5. Solve the following equations using elimination method.

a. 7x + 9y = 41 and 2x + 3y = 13 b. 5x + 6y = 27 and 3x + 4y = 17

c. 2x - y = 1 and y = 3x d. 3x - 2y = 11 and x - 4y = - 3

e. 3x + 2y = 100 and x + 4y = 120 f. 9x - 8y = 12 and 2x + 3y = 12

g. x - 6y + 36 = 0 and 3x - 4y = 4 h. 3x + 2y - 340 = 0 and 2x + 3y = 360

i. 2x + y - 12 = 0 and y = - 0.5x + 6 j. x2

+ y3

= 4 and 2x - y = 4

k. 1x

+ 1y

= 34

and 2x

+ 3y

= 52

l. 1

5x +

13y

- 2 = 0 and 5x

= 12y

+ 7

6. Solve the following equations graphically :

a. x + y = 16 and x - y = - 4 b. 3x + y = 15 and 2x + 3y = 17

c. 3x - 2y = 4 and 5x - y = 23 d. 4x - 3y = 6 and 3x - 4y = 1

e. x + y = 25 and x - y = 5 f. x + y = 9 and x + 2y = 12

g. 2x + y = 6 and x + 2y = 6 h. x + 2y = 10 and 2x - y = 0

i. x + y6

= 5 and x2

+ y3

= 4

j. x2

+ y3

= 4 and x2

- y3

= 4

7. a. Find two numbers whose sum is 160 and difference is 32.

b. Find two numbers whose sum is 80 and difference is 16.

c. The sum of two numbers is 50 and their difference is 22. Find the numbers.

d. One number is three times the other number. The difference between the two numbers is 12. Find the two numbers.

e. The sum of two numbers is 9. The difference in their squares is also 9. Find the numbers.


8. a. Two times a number is greater than 3 times another number by 6. 6 times the smaller number exceeds 3 times the greater by 6. Find the number.

b. 4 times the sum of two numbers is 52 and 3 times their difference is 9. Find the numbers.

c. 3 times the sum of two numbers is 48.If 5 times their difference is 30,find the numbers.

9. a. Sonika is 5 years older than Pramisha. 5 year ago Sonika was twice as old as Pramisha.Find their present ages.

b. 20 years ago, the age of father was four times the age of his son. In four years time his age will be twice that of his son. Find their present ages.

c. A mother is 3 times as old as her daughter. 12 years hence, she will be twice as old as her daughter. Find their present ages.

d. Ten years ago, mother was 12 times as old as her daughter and ten years, hence she will be twice as old as her daughter will be. Find the present ages.

10. a. The cost of 2kg of chicken and 1kg of mutton is Rs 1500.If the cost of 1kg of chicken and 2kg of mutton is Rs 1950. Find the cost per kg of chicken and mutton separately.

b. 9 kg of oranges and 10 kg of apples cost Rs 825 and 6 kg of oranges and 15 kg of apples cost Rs 960. Find the cost of each per kg.

11. a. Four times a breadth of the room is equal to 3 times the length. If the breadth had been 1 meter more and the length 1 meter less the room would have been a square. Find the dimension of room.

b. The perimeter of a rectangular ground is 100 m. If two times of its breadth is 10 m more than its length, find the length and breadth of the ground.

12. a. A number consists of two digits whose sum is 5. When the digits are reversed, the number becomes greater by 9. Find the number.

b. The sum of a two digit number and the number obtained by reversing the digit is 110. The difference between the digits is 4. Find the number.

c. Seven times a 2 – digit number is equal to 4 times the number obtained by reversing the digits. The difference between the digits is 1. Find the number.

13. a. If 2 is added to the numerator and denominator of the fraction it becomes 910

and if 3

is subtracted from the numerator and denominator it become 45

. Find the fraction.

b. The sum of the numerator and denominator of a fraction is 12. If the denominator is

increased by 1, the fraction becomes 76

. Find the fraction.


1 Consult to your teacher

2 a. 5 b. 163 c. 1 d. 100 e. 11 f. 2 g. 3 h. 7 i. 2

3 a. 3 b . 2 c .2nd equation must be muntiply by 3

4 a. 92

and 32

b. 5, 2 c. 3,3 d. 4, 2 e. 5, 6 f. 6, 5 g. 2,1 h. 2, 3 i. -5623 , -257

23

j. 5, 3 k. 5, 13 l. 5, 8 m. 13

and 12

n. 516, 5

18

5a. 2, 3 b. 3, 2 c. -1, -3 d. 5, 2 e. 16, 26 f. x = 13243 , y = 84

43 g. 12, 8

h. 60, 80 i. 4, 4 j. 327 , 36

7 k. -4, 1 l. x = 61395 , y = 61

129

6 a. 6, 10 b. 4 , 3 c. 6, 7 d. 3 , 2 e. 15 , 10 f. 6 , 3 g. 2 h. 2, 4 i. 4 , 6 j. 8,0

7. a. 96, 64 b. 48, 32 c. 36 , 14 d. 18, 6 e. 5, 4 8. a. 18, 10 b. 8 , 5

c. 11, 5 9a. 15, 10 b. 68, 32 c. 36, 12 d. 34, 12

10 a. Rs. 350, 800 b. Rs. 37, 2465

11a. l = 8m, b = 6m b. l = 30m, b = 20m 12a. 32 b. 73 c. -12 13 a. 7

8 b. 75

Exer

cise

4.1

4

4.8. Quadratic Equation:Consider the equation

3x = 7……..(i) 7x2 - 9 = 0 ………..(i) and 2x2 - 9x + 10 = 0 ………..(ii)

The equations (i) , (ii) and (iii) all have one variable. The equation (i) is of degree 1. It is known as linear equation. Again the equations (ii) and (iii) the highest power of x is 2. Such equations of the second degree are called quadratic equations.

The polynomial equation of degree two is known as quadratic equation. The general form of quadratic equation is ax2 + bx + c = 0 where a ≠ 0

A quadratic equation in which the term containing x is absent i.e. b=0 is called pure quadratic equation. x2 – 36 = 0 is a pure quadratic equation.

A quadratic equation containing terms x2 and x as well is called adfected quadratic equation.

2x2 – 4x = 0 and 3x2 +5x – 2 = 0 are adfected quadratic equation.

Method of solving pure quadratic equations.

1. Remove the brackets fraction ,if any.

2. Solve the equation in the same way as you solve a simple equation.

3. Then take the square root of both side to get the two values (one positive and other negative)


Example 1 : Solve : 4x2 - 64 = 0Solution: Here,

x2 = 644

or, x2 = 16or, x = ± 16 (Taking square root of both sides)or, x = ± 4Thus, the required values of x are 4 or – 4.Example 2: Solve (x + 5 (x - 2) = 3 (x + 18)Solution : Here, (x + 5 ) (x - 2) = 3(x + 18)or, x2 - 2x + 5x - 10 = 3x + 54or, x2 + 3x - 10 = 3x + 54or, x2 = 64or, x = ± 64∴x = ± 8Thus, the required values of x are 8 or – 8

Example 3: Solve 4x2 +510

- 2x2 - 5

15 =

7x2 - 25

20

Solution : Here,The L.C.M of 10,15, and 20 is 60.Multiplying each side by 60, we haveor, 6(4x2 + 5) - 4 (2x2 - 5) = 3 (7x2 - 25)or, 24x2 + 30 - 8x2 + 20 = 21x2 - 75or, 24x2 - 8x2 - 21x2 = - 75 - 30 - 20or, -5x2 = - 125or, x2 = 25or, x = ± 25∴ x = ± 5Thus, the required values of x are 5 or – 5.

Example 4 : 2 + x2 - x

+

2 - x2 + x

= 4 14

Solution : Here

Worked Out Examples


(2 + x)2 + (2 - x)2

(2 - x) (2 + x) =

174

or, 4 + 4x + x2 + 4 - 4x + x2

4 - x2 =

174

or, 8 + 2x2

4 - x2 =

174

or, 32 + 8x2 = 68 - 17x2 (by cross multiplication)or, 25x2 = 36

or, x2 = 3625

or, x = ± 3625

or, x = ± 65

∴ x = ± 1 15

Thus, the required values of x are 1 15

or - 1 15

Method of solving adfected quadratic equation:The general form of an adfected quadratic equation is ax2 + bx + c = 0 (a ≠ 0). Here a is the coefficient of x2, b is the coefficient of x and c is constant term. The method of solving an adfected quadratic equation is based on following principle.If A × B = 0Then either A = 0 or, B = 0Both A and B may be zero . But at least one term must be equal to zero.Example, if (x - 2) ( x - 3) = 0Either x - 2 = 0 ……….(i) or, x - 3 = 0 ……………(ii)From (i) x = 2 and from (ii) x = 3Thus, the required values of x are 2 or 3.Solving a quadratic equation by factorization methodIn this method the polynomial of the form ax2 + bx+ c = 0 is factorized and expressed as the product of two linear factors.Following points should be remembered during factorization1. Remove brackets or fraction ( if any)2. Simplify and reduce equation to the form ax2 + bx + c = 0.3. Resolve the left hand side into factors.4. Apply the above principle so that each factor is equal to zero.5. Solve these equations and get two values of x.


Example 5 : Solve: x2 - 11x + 30 = 0Solution : Here, x2 - 11x + 30 = 0or, x2 - 5x - 6x + 30 = 0or, x(x - 5) - 6(x - 5) = 0or, (x - 5) (x - 6) = 0Either, x - 5 = 0 ----------(i)or, x - 6 = 0 ----------(ii)From i, x = 5From ii, x = 6Thus, the required value of x are 5 or 6.

Example 6 : Solve :

52x -

x - 2 = 4 - 10x

Solution : Here,

52x -

x - 2 = 4 - 10x

or, 2x - 52(x -2)

= 4x - 10x

or, 2x - 52(x -2)

= 2(2x-5)x

or, x (2x - 5) = 2 × 2 (x - 2) (2x - 5)or, x (2x - 5) - 4 (x-2) (2x - 5) = 0or, (2x - 5) (x - 4 (x-2)} = 0or, (2x - 5) (x - 4x + 8) = 0or, (2x - 5) (-3x + 8) = 0

Either, 2x - 5 = 0 ----------- (i)or, -3x + 8 = 0 ------------- (ii)From (i), 2x - 5 = 0 or, 2x = 5

or, x = 52

from (ii), -3x + 8 = 0

or, 3x = 8

or, x = 83


Thus, the value of x are

52

or

83

.

Example 7 : Solve x + 3x + 2

+

x - 3x - 2

=

2x - 3x - 1

Solution : Here,

x + 3x + 2

+

x - 3x - 2

=

2x - 3x - 1

To remove x from the numerators, we may write the equation as.

or, x +2+1x + 2

+

x-2-1x - 2

= 2(x-1)-1

x - 1

or, 1 + 1x + 2

+ 1 - 1x - 2

= 2 -

- 1x - 1

or, 1x + 2

- 1x - 2

=

- 1x - 1

or, (x - 2) - (x + 2)x2 - 4

= - 1x - 1

or, - 4x2 - 4

=

- 1x - 1

or, x2 - 4 = 4x - 4

or, x2 - 4x = 0

or, x(x - 4) = 0

Either x = 0 .......................(i)

or, x - 4 = 0 ......................(ii)

From (ii)

x = 4Thus, the value of x are 0 or 4.Solving quadratic equation by completing the square.A quadratic equation can also be solved by completing the squares. For the solution of quadratic equation by this method, we use the following steps.

a. Reduce the given equation in the form of x2 + 2bx = c or, x2 - 2ax = cb. Add b2 on both side of the equation ( 1 being coefficient of x)c. Make perfect squares root on both sides.d. Take square roots on both sides which will give two values of the variable by taking

positive and negative sign alternatively.Examples 8 : Solve: 2x2 - 7x + 3 = 0Solution : Here,or, 2x2 - 7x + 3 = 0or, 2x2 - 7x = - 3 (Transpose 3 to R.H.S )


or, x2 - 72

x = -32

(dividing both sides by 2)

or, x2 - 2. 74

x = -32

(2 × 74

= 72

)

Thus L.H.S is in the form x2 – 2ax

Now, add ( 74

)2 to each side in order to make the left hand side a perfect square.

Then x2 - 2. 74

x + ( 74

)2 = -32

+ ( 7

4 )2

or, (x -

74

)2 = 2516

or, or, (x -

74

)2 = (+ 54

)2

Taking the square root of each side,

x - 74

= ± 54

Taking +ve sign , x = 54

+

74

= 123

= 3 Taking – ve sign, x = -54

+

74

= -5 + 74

= 24

= 12

Thus, the required values of x are 3 or 1

2.

Example 9 : Solve: 2x - 92x - 7

-

2x - 72x - 9

= 712

Solution : Here,

2x - 92x - 7

-

2x - 72x - 9

= 712

or, 2x - 7 - 22x - 7

- 2x - 9 + 2

2x - 9 = 7

12

or, 1 - 22x - 7

- 1 -

22x - 9

= 712

or, - 22x - 7

-

22x - 9

= 712

or, 12x - 7

+

12x - 9

= -724

or, 4x - 16(2x - 7) (2x - 9)

= -724

or, 4x - 163x2 - 32 x + 63

= -724


or, 96x - 384 = - 28x2 + 224x - 441or, 28x2 - 128 x = - 57

or, x2 - 12828

x = -5758

or, x2 -

327

x = -5758

or, x2 - 216

7 x +

162( )2

= -578

+

25649

or, (x - 167

)2 = - 399 + 1024196

=

625196

or, (x - 167

)2

= (± 25

14)2

or, x - 167

= ± 2514

or, x = ± 2514

+

167

Taking +ve sign x =

2514

+

167

Taking +ve sign, x = - 25

14 +

167

or, x = 25 + 3214

or, x = -25 + 3214

or, x = 5714

or, x = 714

or,

x = 12

Thus, the required values of x are 5714

or 12

.

Formula for solving a quadratic equationEvery quadratic equation can be written in the form ax2 + bx + c = 0. The expression ax2 + bx + c has no simple factors. We will therefore use the method of completing the square to solve it. ax2 + bx + c = 0or, ax2 + bx = - c

or, x2 + bxa

=

-ca

(dividing both sides by a)

or, x2 + b2a

. 2x + b2a( )2

=

b2

4a2 - c

a

or, b2a( )x + 2 = b2-4ac

4a2

or, (x +

b2a

)2

= (±

2ab2 - 4ac )2

or, x = 2a

- b ± b2 - 4ac

This is a formula to solve any quadratic equation.


Examples 10 : Solve: 2x2 - 6x + 3 = 0Solution : Here,Comparing it with the equation ax2 + bx + c = 0,a = 2, b = - 6 and c = 3

x = 2a- b ± b2 - 4ac

x = 2 × 2

-(-6)± (-6)2 - 4 × 2 × 3

or, x = 46 ± 12

or, x = 4

6±2 3

or, x = 2

3 ± 3

Thus, the required value of x are 2

3 + 3 or 2

3 - 3

Example 11 : Solve: x+3x+2

= 3x-72x-3

Solution : Here,Multiplying cross wise (x + 3) (2x - 3) = (x + 2) (3x - 7)or, 2x2 + 3x - 9 = 3x2 - x - 14or, - x2 + 4x + 5 = 0or, x2 - 4x - 5 = 0Comparing this equation with ax2 + bx + c = 0 a = 1, b = - 4 and c = - 5

x = 2a- b ± b2 - 4ac

x = 2 ×1-(-4) ± (-4)2 - 4×1× (-5)

x = 2 × 24 ± 16 + 20

x = 24 ± 36

Taking +ve sign

x = 4 + 6

2

x = 102

x = 5


Taking -ve sign

x = 4 - 6

2

x = -22

x = - 1

Thus, the value of x are 5 or -1.

1. a. What is quadratic equation ?

b. State any two methods for solving quadratic equation.

c. Write the roots of quadratic equation ax2 + bx + c = 0. (a ≠ 0)

d. Write down the roots of of a quadratic equation lx2 + mx + n = 0

e. Write the general form of quadratic equation.

2. Find the value of x :

a. x ( x – 2 ) = 5 b. x ( x – 6 ) = 0 c. x2 = 81 d. x2 – 100 = 125 e. 6x2 = 24

3. a. What are the roots of quadratic equation x2 = 64?

b. What are the roots of quadratic equation x2 = 81?

4. Substitute the values and evaluate

a. x2 - 7 = 29 b. 3(x + 3) (x - 3) = x2 + 5 c. 2(x2 - 8) + x(4 - x) = 4(x + 12)

d. 32x2

-

35x2

=

19

e. 8x +

7x

= 65x7

f. 2x2 + 1015

= 7 - 50 + x2

25

g. x+1x+3

=

x+72x+8

h. x2

+ 2x

= x8

+ 8x

i. (x - 3a)2 - 3a (a - 2x) = 55a2

j. 4x-3

-

4x+3

= 13

k. 21+x

+ 31-x

= 8

l. 11+x

- 1x-1

= 23x2

m. x+4x-4

+

x-4x+4

= 3 13

n. x+3x-3

+ x-3x+3

= 2 12

o. 1x+9

+ 1x+4

= 1x-9

+ 1x-4

5. Solve the following equation:

a. x2 + 3x - 10 = 0 b. x2 - 5x - 36 = 0 c. x (x - 9) = 36

d. 6x2 + 13x = 5 e. 4x2 + 20x + 25 = 0 f. 30x = 8x2 + 27

g. (x - 3) (x + 5) = 48 h. (x + 4( 2x - 3) = 6 i. x2 + 6x + 9 = (2x + 1)2

Exercise 4.15


j. (2x - 3) (x - 5) = (x - 3)2 k. 11 (x +1) (x + 2) = 38(x +1) + 9x

l. 13

(x - 2) (x - 3) = 14

(x - 4) (x + 2 m. x - 6x

= 5

n. x + 2 - 6x+2

= 1 o. 3a2x2 - 4abx + b2 = 0 p. pqx2 +x(p+q) + 1 = 0

q. a2x2 - a(b + c)x + bc = 0 r. x2 - 3ax + 2bx - 6ab = 0 s. x2 - 2ax + a2 - b2 = 0

t. (x - 3)2 = (x - 3) (2x + 1) u. (x - 3)2 - 3(x - 3) + 2 = 0 v. 3(x +1)2 + 2(x + 1) - 1= 0

w. x+3x+2

- 3x-72x-3

= 0 x. xx - 1

+ x +1

x = 2

12

y. x+1x

+

xx+1

= 2

16

z. x + 1x - 1

= 3x - 72x - 5

6. Solve the following equations by using factorization method.

a. x2 + 8x + 12 = 0 b. x2 - 7x - 9x = 0 c. x2 - 2x - 1 = 0

d. 3x2 - x - 2 = 0 e. 3x2 + 2x - 5 = 0 f. x2 + 9x - 52 = 0

g. 6x2 + 5x - 4 = 0 h. 7x2 = 8 - 10x i. 3x2 - 7x - 2 = 0

j. x2 - 2x = 2 k. 2x2 + x = 5 l. 4x - 4x2 = - 7

m. 3x2 - 5x = 3 n. x2 = 1 + 4x o. 2x2 + 7x = 3

p. -x2 + 4x + 77 = 0 q. x2 + 117x = 2x r. x2 + x12

=

12

s. x2 - 2x + 34

= 0 t. x2 + 2x3

=

359

u. x2 - 14 = 5x

v. x+1x-1

+

x-1x+1

= 2 12

w. 2x-12x+1

+ 2x+12x-1

= 3415

7. Use the formula to solve the following equations:

a. x2 + 6x = 40 b. 2x2 - 3x - 2 = 0 c. 7x2 - 9x + 2 = 0

d. 4x2 - 3x = 6 e. 2x2 - 3x = 3 f. 2x2 = 3x + 7

g. 24x2 + 12x + 1 = 0 h. x2 + 7ax + a2 = 0 i. 41x2 = 5a2 + ax

j. 5x2 - 9x - 4 = 0 k. x(x + 27

) = 349

l. x2 + 2x = 323

m. 1x-2

+

1x-3

=

110

n.

1x+2

+

1x+3

=

110


1. a. b. c. - b + b2 - 4ac2a

, - b - b2 - 4ac2a

d. -m+ m2-4nl2l

, -m- m2-4nl2l

e. ax2 + bx + c = 0

2. a. 5 or 7 b. 0 or 6 c. +9 d. +15 e. +2

3. a. +8 b. +9

4. a. +6 b. +4 c. +8 d. +8110 e. +

73

f. +32513 g. + 13 h. + 4 i. + 7a j. + 9 k. x = -1 + 97

13

l. + 12

m. 2,3 n. + 9 o. + 6

5. a. 2,-5 b. -4,9 c. 12,-3 d. 13

, -52

e. -52

f. 94

, 32 g. 7, -9 h. 2,

-92

i. -43

,2 j. 1,6 k. 2, -811

l. 6,8 m. 6, -1 n. 1, -4 o. ba ,

b3a

p. -1p ,

-1q

q. ba

, ca r. 3a, -2b s. (a+b), (a-b) t. 3, -4 u. 4, 5 v. -2

3 , -2

w. 5,-1 x. 5 + 172

y. 2, -3 z. 3,4

6. a. -2 , -6 b. 0 , 16 c. + 2 d. -23

,1 e. 1, -53

f. 4, -13

g. 13

,-43

h. 47

, -2 i. 7 + 736

j. 2+ 2 32

k. -1+ 414

l. 1+ 2 22

m. 5 + 616

n. 2 + 2 o. -7+ 734

p. -7, 11 q. 0 , -115 r. 23

, -34

s. 12

, 32 t.

-73

, 53

u. 7,-2 v. + 3 w. + 2

7. a. -10 or 4 b. -12

or 2 c. 1 or 27

d. 3 + 1058

e. 3 + 334

f. 3 + 654

g. -12 + 4 348

h. -7a + 3a 52

i. a (1+ 821)82

j. 9 + 18110

k. 17

or -37

l. 17 or -19

m. 25+ 4012

n. 15+ 4012

Exer

cise

4.1

5




1. a. If 5x = 1, what is the value of x?

b. If x:15 = 12: 45 , find the value of x.

c. Solve : x2 – 64 x = 0.

2. a. Factorize : 25m4 – 9m2n2 + 16n4

b. Simplify : 7x+2 × 8 - 3 × 7x

7x+2 - 4 × 7x

c. If a = bc, b = ca and c = ab , prove that : abc = 1

d. Solve by using substitution method : x +2y =9 and x= 3y+2

e. Solve : 4x2 – 12x = 0

3. a. Simplify : ax

ay( )x2 + xy + y2 × ay

az( )y2 + yz + z2 × az

aa( )x2 + zx + z2

b. If a:b=b:c=c:d prove that: (a+b)(c+d)=(b+c)2

c. The sum of the ages of mother and her daughter is 37. After 10 years the mother will be twice as old as the daughter. Find their present ages.

4. a. Solve by graphical method : 3x+2y=9, 2x+y=5

b. Solve: x+3x+2

= 3x-72x-5

Group - A (1 × 1 = 1 )

1. a. If x7

= 314

, find the value of x. Group - B (5 × 2 = 10 )2. a. Factorize : a4 – 14 a2 b2 + b4

b. Solve : 2x+2 + 2x+3

2 = 1

c. Two numbers are in the ratio 3 : 5. If 8 is added to each the ratio becomes 7 : 9. Find the numbers

d. Solve : x = 2y and x + 4y = 6 e. Solve : x2 – 49 x = 0

Group - C (4 × 2 = 8 )

3. a. If ax = by = cz and b2 = ac , show that 1x

+

1z

=

2y

b. Solve : x - 44

+

6x+ 1

= 2Group - D (5 × 1 = 5 )

4. If xb +c

=

yc+a

= za +b

then prove that : x(y-z)b2 -c2

= y(z -x)c2 -a2

= z(x-y)a2- b2

school mathematics 9

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