202110723-PERFORM-STUDENT-WORKBOOK-MATHEMATICS-G10-FY_Optimized

practice workbook Mathematics Grade 10 Name: Roll No: Section: School Name:

by classklap IMAX is India’s only customised content and educational assessment m. 610+ Published Books Team of IITians & PhDs IMAX Program has authored about 610+ Content development and automation are publications which are used by more than led by a team of IITians, data scientists and 4,50,000+ students. education experts Workshops Lakh Assessments IMAX Program has conducted IMAX Program has conducted more than teacher training workshops for more 85,00,000+ assessments across 17 states in the last 10 years. than 15,000+ teachers. Copyright © 2020 BGM Policy Innovations Pvt Ltd) All rights reserved. No part of this publication, including but not limited to, the content, the presentation layout, session plans, themes, key type of sums, mind maps and illustrations, may be reproduced, stored in a retrieval system or transmitted, in any form or by any means (electronic, mechanical, photocopying, recording or otherwise), without the prior, written permission of the copyright owner of this book. This book is meant for educational and learning purposes. The author(s) of the book has/ have taken all reasonable care to ensure that the contents of the book do not violate any copyright or other intellectual property rights of any person in any manner whatsoever. In the event the author(s) has/have been unable to track any source and if any copyright has been inadvertently infringed, please notify the publisher in writing for any corrective action. Some of the images used in the books have been taken from the following sources www.freepik.com, www.vecteezy.com, www.clipartpanda.com Creative Commons Attribution This book is part of a package of books and is not meant to be sold separately. For MRP, please refer to the package price.

This book is designed to support the teacher in her/his journey of teaching Mathematics for Class 10. The content and topics of this book are in accordance with the latest guidelines issued by CBSE and on the basis of board pattern. The book contains annual academic plan, which shows indicative hours for teaching the class 10 portion. Lesson plans are made available for each lesson listed in the annual plan, practice questions with solutions addressing different skill buckets and different question types are provided for each lesson. Practicing these sheets will help a student gain mastery over the lesson. The practice sheets can be solved by the teacher during the allocated period. There is a self-evaluation sheet at the end of every lesson, this will help the teacher in assessing the learning gap. This sheet can be administered during the allocated period and peer check can be instilled to ensure assessment is completed during the allocated period.



TABLE OF CONTENT • Academic Plan - Teaching and Assessment • Planning Tool • Assessment Pattern: 40 Marks • Assessment Pattern: 80 Marks • Syllabus & Timeline for Assessment Page 1: 1. Real Numbers Page 9: 2. Polynomials Page 21: 3. Pair of Linear Equations in Two Variables Page 31: 4. Quadratic Equations Page 39: 5. Arithmetic Progressions Page 45: 6. Triangles Page 57: 7. Coordinate Geometry Page 63: 8. Introduction to Trigonometry Page 72: 9. Some Applications of Trigonometry Page 78: 10. Circles Page 86: 11. Constructions Page 94: 12. Areas Related to Circles Page 105: 13. Surface Areas and Volumes Page 115: 14. Statistics Page 127: 15. Probability



TEACHING AND ASSESSMENT PLAN (Includes academic plan, pattern of assessment and syllabus of assessment. Assessment pattern for periodic test and term examination are in line with the blue print prescribed by the boards. Pattern of the 40 marks paper is derived by scaling down the 80 marks paper in terms of weightage. Suggestions are included for teachers to aid them while making the students answers the practice sheet questions).



ASSESSMENT PATTERN Marks: 40 Grade 10/Mathematics Max Internal PAPER: BEGINNER PAPER: PROFICIENT Mark Option Q.No Skill Level Difficulty Level Skill Level Difficulty Level Easy Medium Difficult Easy Medium Difficult Section A (Question Type: VSA) 11 Remembering • Remembering • 21 Remembering • Remembering • 31 Understanding • Understanding • 41 Understanding • • Understanding • 51 Applying Applying • 61 Understanding • Understanding • 71 Remembering • Remembering • 81 Analysing • Analysing • 9 1 • Understanding • Understanding • 10 1 • Applying • Applying • Section B (Question Type: SA) 11 2 Understanding • Understanding • 12 2 Understanding • Understanding • 13 2 • Applying • Applying • Section C (Question Type: SA) 14 3 Remembering • Remembering • 15 3 • Remembering • Remembering • 16 3 • Understanding • Understanding • 17 3 Analysing • Analysing • Remembering • Section D (Question Type: LA) 18 4 Remembering • 19 4 • Remembering • Remembering • 20 4 Understanding • Understanding • Beginner Paper: (Easy: 50%, Medium: 40%, Difficult: 10%) Proficient Paper: (Easy: 40%, Medium: 40%, Difficult: 20%) Easy Question: Remembering questions directly from the text or from the given exercises. (Mostly from content of book or end of chapter exercise). Medium Difficulty Question: In-depth understanding of questions, not necessarily from the text. (Slightly modified concepts or end of chapter questions). Difficult Question: Question involving creativity like story writing, analysis question like character analysis, justification of title or extracts (mostly requires creative and thinking skills).

ASSESSMENT PATTERN Marks: 80 Grade 10/Mathematics Max Internal PAPER: BEGINNER PAPER: PROFICIENT Mark Option Q.No Skill Level Difficulty Level Skill Level Difficulty Level Easy Medium Difficult Easy Medium Difficult Section A (Question Type: MCQ) 11 Remembering • Remembering • 21 Understanding • Understanding • 31 Understanding • Understanding • 41 Understanding • Understanding • 51 Applying • Applying • 61 Analysing • Analysing • 71 Analysing • Analysing • 81 Remembering • Remembering • 91 Remembering • Remembering • 10 1 Remembering • Remembering • Section B (Question Type: VSA) 11 1 • Remembering • Remembering • 12 1 Understanding • Understanding • 13 1 • Understanding • Understanding • 14 1 Understanding • Understanding • 15 1 Applying • Applying • 16 1 • Analysing • Analysing • 17 1 Understanding • Understanding • 18 1 Applying • Applying • 19 1 Applying • Applying • 20 1 • Applying • Applying • Section C (Question Type: SA) 21 2 Remembering • Remembering • 22 2 Applying • Applying • 23 2 Remembering • Remembering • 24 2 • Applying • Applying • 25 2 • Understanding • Understanding • 26 2 Analysing • Analysing • Section D (Question Type: SA) 27 3 • Remembering • Remembering • 28 3 • Understanding • Understanding • 29 3 • Understanding • Understanding • 30 3 Applying • Applying • 31 3 Remembering • Remembering • 32 3 Analysing • Analysing • 33 3 • Remembering • Remembering • 34 3 Remembering • Remembering • Section E (Question Type: LA) 35 4 • Remembering • Remembering • 36 4 • Understanding • Understanding • 37 4 Applying • Applying • 38 4 Understanding • Understanding • 39 4 Remembering • Remembering • 40 4 Remembering • Remembering •

SYLLABUS FOR ASSESSMENT Grade 10/Mathematics CHAPTERS PT-1 TE-1 PT-2 MOCK Chapter 1: Real Numbers ✓ ✓ ✓ Chapter 2: Polynomials ✓ ✓ ✓ ✓ Chapter 3: Pair of Linear Equations in Two Variables ✓ ✓ ✓ Chapter 4: Quadratic Equations ✓ ✓ Chapter 5: Arithmetic Progressions ✓ ✓ ✓ ✓ Chapter 6: Triangles ✓ ✓ ✓ Chapter 7: Coordinate Geometry ✓ ✓ Chapter 8: Introduction to Trigonometry ✓ ✓ Chapter 9: Some Applications of Trigonometry ✓ Chapter 10: Circles ✓ ✓ Chapter 11: Constructions ✓ Chapter 12: Areas Related to Circles ✓ Chapter 13: Surface Areas of Volumes ✓ Chapter 14: Statistics ✓ Chapter 15: Probability ✓ Assessment Timeline Periodic Test-1 1st July to 31st July Term Exam 1 23rd September to 21st October Periodic Test-2 16th December to 13th January Mock Test 17th February to 9th March



LESSON WISE PRACTICE (This section has a set of practice questions grouped into different sheets based on different concepts. By solving these questions you will strengthen your subject knowledge. A self-evaluation sheet is provided at the end of every lesson.)



1. Real Numbers Learning Outcome By the end of this chapter, a student will be able to: • Analyze the properties of a given number. • Explain the concept of Euclid’s Division Lemma and • Determine the irrationality of a given number. • Find the decimal expansion of a rational number. Fundamental Theorem of Arithmetic. • Calculate the HCF and LCM of integers. Concept Map Key Points and q ≠ 0. Example: 2, 5 , 0.01011011101110…., • A number ‘r’ is called rational number, if it can be 0.34334333433334….etc. written in the form p , where ‘p’ and ‘q’ are integers A number whose decimal expansion is non- q terminating and non-recurring is called an Irrational Number. and q ≠ 0. Rational numbers includes natural • The collection of all rational and irrational numbers numbers, whole numbers and integers. When p is makes up the real numbers, denoted by R. • Composite number is a positive integer with q exception of ‘1’ and has at least one factor other than the number itself and 1. represented on a number line, we assume that q ≠ Example: 4 has factors 1, 4 and 2, Hence it is a 0 and that ‘p’ and ‘q’ have no common factors other composite number. than 1 (that is p and q are co-primes). 7 has factors only 1,7 and hence it is a prime • A group of rational numbers whose value is same number. but have different representation in p form are • Co-primes are sets of numbers which do not have a common factor other than 1. q Example: 2, 15 can be called co-primes as the common factor is 1. called Equivalent Rational Numbers or Fractions. 4, 9 can be called co-primes as the common factor is 1. • 1 ,w24h, o84s,e135000de, 1c98im, eatlc. expansion is either 3, 15 are not co-primes since 3, 5 are also common AExanmumplbe:er2 factors apart from 1. terminating or non-terminating recurring is called • Highest Common Factor (HCF) of two positive a Rational Number. integers ‘a’, ‘b’ is the largest positive integer ‘d’ that Example: 0.34, 0.333333.., 0.25, 0.881881881…, etc. divides both ‘a’ and ‘b’. • A number ‘s’ is called irrational, if it cannot be written in the form p , where ‘p’ and ‘q’ are integers q 1

1. Real Numbers Example: HCF(50,10) = 10 is zero. The divisor at this stage will be the HCF(20,150) = 10 required HCF. HCF(4,9) = 1 (Remainder = 0, hence the divisor (10) is the • Least Common Multiple (LCM) of two numbers is sHCF.) the smallest number that they both divide evenly o This algorithm works because of HCF(c,d) = into. HCF(d,r) where the symbol HCF(c,d) denotes Example: LCM(50,10) = 50 HCF of c and d. LCM(20,150) = 300 o The algorithm can be extended for all integers LCM(4,9) = 36 except zero. • HCF of two numbers is the product of the smallest o The algorithm can be used to find the power of each common prime factor in the properties of numbers. numbers. LCM is the product of the greatest power • Fundamental Theorem of Arithmetic states that of each prime factor involved in the numbers. “every composite number can be expressed as a Example: 20 = 2251 150 = 213152 product of primes and this factorisation is unique, apart from the order in which the prime factors HCF(20,150) = 2151 = 10 occur.” The theorem can also be stated as “the Prime LCM(20,150) = 223152 = 300 factorization of a natural number is unique except for the order of its factors.” • Algorithm is a series of well defined steps which Example: 125 = 5× 5× 5 gives a procedure for solving a type of problem. • For any two positive integers a, b, HCF(a,b) × A lemma is a proven statement which is used for LCM(a,b) = a × b. This formula can be used to find proving another statement. the LCM of two positive integers if their HCF is known as the HCF can be same for different sets of • Euclid’s Division Lemma states that “Given positive numbers. integers a and b, there exist unique integers ‘q’ and ‘r’ satisfying, Example: HCF (150, 20)× LCM (150, 20) 150 X 20 a = bq + r, 0 ≤ r < b The lemma is a restatement of the long division 10 X 300 = 3000 process and in general ‘q’ is called quotient, ‘b’ is divisor and ‘r’ is remainder. Here ‘q’ or ‘r’ can also 3000 = 3000 be zero. In other words, “Any positive integer ‘a’ can • Proof by Contradiction: When proving something be divided by another positive integer ‘b’ in such a way that it leaves a remainder ‘r’ that is smaller is true using proof by contradiction, you assume a than ‘b’. statement to be false and try to prove it. This leads Example: 150 = 20× 7 +10 to the assumption to be not false and hence the initial statement must be true. Here a = 150, b = 20, r = 10 • If a prime number ‘p’ divides a2 then p divides ‘a’ 150 = 30× 5 + 0 Here a = 150, b = 30, r = 0 where ‘a’ is a positive integer. • Euclid’s Division Algorithm is used to obtain the Example: 5 divides 102 = 100, 5 divides 10 HCF of two positive integers say c and d with c > d as follows: • A number x= p , where p and q are co-prime and q o Step 1: Apply Euclid’s division lemma to c and q d. So we find whole numbers q and r such has 2,5 as prime factors, i.e.,q = 2n5m where m and that c = dq + r, n are non-negative integers then ‘x’ will have a 0≤r<d terminating decimal expansion. o St(hLteeenpt 1c25=:0I1=f5r02=0a×n0d7, d+d1i=s0,t2hh0ee,reHqCF= Example: 0=.25 =25 25 100 2251 7, r = 10) d. If r ≠ 0, • A number x= p , where p and q are co-prime and q of c and q apply the division lemma to d and r. has prime factors other than 2 and 5 i.e. q ≠ 2n5m o Step 3: Continue the process till the remainder where m and n are non-negative integers, ‘x’ will be a rational number whose decimal expansion which is non-terminating (recurring). Example: 0.333…. = 1 3 2

1. Real Numbers Work Plan CONCEPT COVERAGE COVERAGE DETAILS PRACTICE SHEET PS – 1 Pre-requisites • Rational, Irrational and Real Numbers PS – 2 • Co-primes, Factorisation PS – 3 Real Numbers – Euclid’s • Euclid’ Division Lemma PS-4 Division Algorithm • Euclid’s Division Algorithm Self Evaluation Sheet • Computing HCF of Two Positive Numbers • Properties of Numbers. Real Numbers – Fundamental • Statement Theorem of Arithmetic • HCF and LCM • Properties of Numbers Real Numbers – Revisiting • Proof for irrationality of a number using Irrational Number the concept of Fundamental theorem of - Revisiting Rational Numbers arithmetic. and their Decimal Expansions • Decimal expansion of rational numbers using factorisation. Worksheet for \"Real Numbers\" Evaluation with Self Check or ---- Peer Check* 3

PRACTICE SHEET - 1 (PS-1) 1. Which among the following numbers are co-primes of 8? i)  6  ii) 7 2. State whether the following statement is true or false. Give reason. All rational numbers are natural numbers.” 3. Show that 4.3689 is a rational number. 4. Factorise 66150. 5. Which of the following numbers are rational numbers and irrational numbers: i) 1 ii) 2 iii) 4 iv) 4 32 6. Why are all the following numbers rational numbers? i) 0.34  ii) 0.3434345  iii) 0.3434343434….  iv) 34.343434 PRACTICE SHEET - 2 (PS-2) 1. Find the HCF of 100 and 80 using Euclid’s algorithm. 2. Find the HCF of 284 and 4096. 3. Prove that every positive even integer can be of the form 2q, where q is a whole number. 4. Show that the square of any positive integer is of the form 3m or 3m+1. 5. Find the HCF and LCM of 25, 120, 80 using prime factorisation method. 6. Without doing the actual multiplication check if 7 × 5 × 3 × 2 + 7 is a composite number or not? 7. Find the HCF and LCM of the following numbers i) 24, 45, 60 ii) 15, 81, 99 8. Given HCF (24,45) = 3. Find the LCM (24,45). 9. If HCF (150,81) is given by m+1, where m is a integer, find the value of m. Also find the LCM of 150 and 81 without prime factorising the numbers. 10. If a, b, 5 are the three distinct prime factors of 900, find the value of a, b. 11. Find a, b, c and d. 12. Rahul brings 180 candies and 70 biscuits to the class to distribute among his friends. He distributes the candies and biscuits equally among his friends. What is the maximum number of friends Rahul can have? How many candies and biscuits each friend get? 4

PRACTICE SHEET - 3 (PS-3) 1. Find whether 11 is rational or irrational number. 2. Prove that 6 – 2 is a irrational number. 3. Which of the following numbers are irrational numbers, explain? i) 7 ii) 9 − 25 iii) 9 + 2 4. Express the following rational numbers in p form and factorize p and q. q i) 4.9  ii) 3.14  iii) 0.3885 5. Find the nature of the numbers given below. (Rational – terminating, non- terminating, irrational). i) 12 ii) 12 iii) 12 iv) 12 15 17 2 4 6. If 0.16666666….. is to be written in the form of p discuss the prime factors of q. q 5

PRACTICE SHEET - 4 (PS-4) I. Choose the correct option. (D) 225 1. Which of these numbers is an irrational number? (D) 2q (A) 11 (B) 16 (C) 196 (D) 600 2. Identify the general form of a positive even integer. (A) 3q + 1 (B) 4q + 3 (C) 4q + 1 3. Choose the number with 2² × 3² × 5² as prime factorisation. (A) 400 (B) 900 (C) 1200 4. Which of the following is a rational number between and ? (A) (B) (C) (D) 5. The sum of a rational number and an irrational number is ________. (A) an irrational number (B) a rational number (C) an integer (D) a whole number 6. The product of two irrational numbers is ________. (A) always an irrational number (B) always a rational number (C) always an integer (D) either rational or irrational 7. The factors in the denominator of a rational number that can be expressed as a terminating decimal are __________. (A) powers of 3 and 5 (B) powers of 5 and 7 (C) powers of 2 and 5 (D) powers of 3 and 7 8. How many rational numbers are there between and ? (A) 1 (B) 16 (C) 24 (D) Infinitely many 9. The rational number 0.0452 can be expressed in the form of as ____________. (A) (B) (C) (D) 10. Which of these rational numbers has the decimal expansion of 0.148? (A) (B) (C) (D) II. Short answer questions. 1. Prime factorise 1144. 2. Write the decimal expansions of the following: (i) (ii) 6

PRACTICE SHEET - 4 (PS-4) 3. Using Euclid’s division algorithm, find the HCF of 250 and 400. III. Long answer questions. 1. Show that is irrational. 2. Prove that is irrational, where p and q are primes. 7

SELF-EVALUATION SHEET Marks: 15 Time: 30 Mins 1. A book store wants to arrange 650 short note books and 350 long note books in the racks in the form of stack (one book over another). Find the number of books a stack should contain if the books are not to be mixed? (1 Mark) 2. Find whether 3n ends with digit 6 for any positive integer value of n. (1 Marks) 3. Which of the following numbers is a composite number and find the factors of the numbers? Is it possible to check without doing actual multiplication? (3 Marks) i) 3 × 5 + 3 × 3 × 3 + 2 × 3 ii)  3 × 5 + 7 × 11 4. A father and son go for walking in a park. The son walks faster and moves ahead after starting and completes one round in 15 minutes. The father completes one round of walk in 24 minutes. After how much time will the father and son meet at the starting point in the park? (2 Marks) 5. Ravi and Rahul arrange seats for function. There are two designs of chairs and their quantities are 350, 200. In a given row, only one type of chairs should be present. Determine the minimum number of rows the chairs can be placed and what is the number of each type of chair per row. (2 Marks) 8

SELF-EVALUATION SHEET Marks: 15 Time: 30 Mins (3 Marks) 6. Prove 3 is an irrational number. 7. Determine the prime factors in the denominator of the rational number 123.456789123.  (1 Mark) 8. Write the decimal expansion of i) 7 ii) 3  (2 Marks) 16 215171 9

2. Polynomials Learning Outcome By the end of this chapter, a student will be able to: • Determine quadratic polynomials if zeroes are • Calculate the Zero of polynomials. known. • Determine the number of zeroes of the polynomial • Determine zeroes of higher degree polynomials if from graphs. one or two of its zeroes are given. Concept Map Key Points Degree of Name Standard Example Polynomial Form • Expressions are formed from variables and constants. The value of an expression changes with 1 Linear p(x) = ax + b p(x) = 3x – 5, the value of the variables. Terms are added to form Polynomial a, b are real p(y) = 2 – 8y, expressions. Terms themselves can be formed as numbers p(z) = 19z – 26 product of factors. The numerical factor of a term is a≠0 called its coefficient or numerical coefficient. Example: 3x + 4y – 3xy + y2 + 11, 2 Quadratic p(x) = ax2 + p(x) = 3x2 – 5x 3x, 4y, –3xy, y2, 11 are the terms and the numbers in Polynomial bx + c +2 each term is called coefficient. +11 is the constant. a, b, c are q(y) = 18y2 + real numbers 91y – 21 • A monomial is having only one term, a binomial a≠0 q(t) = –t2 has two terms, trinomial has three terms and a polynomial has one or more terms. 3 Cubic p(x) = ax3 + p(x) = –x3 – 5x Polynomial bx2 + cx + d + 25 • An expression containing one or more terms with a, b, c, d are non-zero coefficients is called polynomial. Also real numbers q (t ) = 5 t3 + 25t2 in a polynomial, the exponent of the variables a≠0 should only be whole numbers. In a polynomial, 3 the highest power of the variable is called Degree of polynomial. The degree of a non-zero constant + 9t − 95 polynomial is zero. The degree of a Zero polynomial is not defined. Expressions such as 1 , x3 + 25 7x are Example: x2 + 3x − 5 10

2. Polynomials not polynomials as the degree of the polynomial is not a whole number. • Value of Polynomial In any polynomial p(x) in variable x, if k is any real number, then the value obtained by replacing x with k in p(x) is called value of p(x) at x = k and denoted by p(k). Example: If p(x) = x2 – 5, then p(2) = (2)2 – 5 = –1 • Zero of Polynomial A real number k, is said to be zero of a polynomial p(x), if p(k) = 0. Example: Find zero of p(x) = 5x – 3 If k is zero of p(x), then p(k) = 0 ⇒ 5(k) – 3 = 0 ⇒k= 3 We know that zero of the polynomial ax + b is −b . 5 Here a = 1 b = –1 a Zero of a linear polynomial p(x) = ax + b is Zero of the polynomial = 1 = x-coordinate of the −b = −constant term point of intersection. a coefficient of x A linear equation y = constant is parallel to x-axis • Geometrical Meaning of Zeroes of Polynomial and does not intersect the x-axis. The polynomial 1) Linear Polynomial: in this case will not have any zero. The standard form of a linear polynomial p(x) is ax + b where a ≠ 0. 2) Quadratic Polynomial The graph of the linear polynomial is a straight line. The standard form of a quadratic polynomial p(x) is In order to plot the polynomial in a graph, (x, y) ax2 + bx + c where a ≠ 0. coordinates are needed. The graph of a quadratic equation y = ax2 + bx + Let y = p(x). The coordinate of a point will be (x, c has one of the two shapes, either open upwards p(x)). Two coordinate points are needed to draw a like ‘∪’ or open downwards like ‘∩’ depending straight line. on whether a > 0 or a < 0. These curves are called If y = 0, then the value of x is known as the Zero of parabolas. the polynomial. By substituting for values of x, the values of y can In a graph, y = 0 represents the x-axis. Hence the be obtained and joining the points, the graph of the point of intersection of y = p(x) and x- axis will be quadratic polynomial can be obtained. the Zero of the polynomial. Figure shows different quadratic polynomials. From Linear equation, y = ax +b intersects the x axis at the figure it is clear the values of a, b, c determine y = 0 ⇒ 0 = ax + b ⇒ x = −b the shape and position of the graph. a The zero of the polynomial will be −b . a A linear equation has at utmost zero, namely, the x-coordinate of the point where the graph of y = ax +b intersects the x-axis. Example: Find the zero of the polynomial y = x – 1 and verify the same graphically. To find points on the line At x = 2, y = (2) – 1 = 1 At x = 4, y = (4) – 1 = 3 The two points are (2,1) and (4,3) and a straight line can be drawn through these points. From the figure, the straight line y = x – 1 intersects the x-axis at (1,0) 11

2. Polynomials The coordinates of A and A’ are the two zeroes of the quadratic polynomial ax2 + bx + c. o Case II: The graph touches the x-axis at exactly one point, i.e., at two coincident points. So the two points A and A’ coincide to become point A. The graph does not cross the x-axis and remains on one side of the axis. The x-coordinate of A is the only zero for the quadratic polynomial. o Case III: Here the graph is either completely above the x-axis or completely below the x-axis, so it does not intersect the x-point at any point. So the quadratic polynomial in this case has no zero. The zeroes of a quadratic polynomial s A quadratic polynomial can have either two distinct ax2 + bx + c, a ≠ 0 are precisely the x-coordinates zeroes or two equal zeroes (i.e., one zero) or no of the points where the parabola y = ax2 + bx + c zero. A quadratic polynomial of degree 2 can have a maximum of two zeroes. intersects the x-axis. 3) Cubic Polynomial o Case I: The graph crosses the x-axis at two The standard form of cubic polynomial is points A and A’. ax3 + bx2 + cx + d where a ≠ 0. A cubic polynomial has degree 3 and it can have a maximum of three zeroes. • In general a polynomial of degree n can have a maximum of n zeroes, i.e., the graph of y = p(x), p(x) is a polynomial of degree n, can intersect the x-axis at a maximum of n points. • Relationship between Zeroes and Coefficients of 12

2. Polynomials a Polynomial is the remainder. Select the next term from and add it to the remainder and this is the new A quadratic polynomial, p (x) = ax2 + bx + c , a ≠ 0 dividend. o Repeat the calculation of quotient and can have two factors (x − α) and (x − β) . Here α, β subtraction until the remainder is either zero or the degree of remainder is less than the degree are the zeroes of the polynomial. of divisor. We can write We see that ax2 + bx + c = k (x − α)(x − β) , (k is a constant) Dividend = Divisor × Quotient + Remainder. Example: Divide 6x4 + 3x3 + 2x + 5 by x +1 = kx2 − k (α + β) x + k (αβ) Writing the given polynomial in the standard form we get 6x4 + 3x3 + 0x2 + 2x + 5 ∴ a = k b∴=a−=kk(α b+=β−)k (αc+=β)kαβc = kαβ α + β =  − b  αβ= c  a  a = α+β = −b = -Coefficient of x  6x3 − 3x2 + 3x − 1 a Coefficient of x2 = αβ = c = Constant term x +  6x4 + 3x3 + 0x2 + 2x Quotient = a Coefficient of x 2 1 +5 6x4 = 6x3 A cubic polynomial, p(x) = ax3 +bx2 + cx + d can  6x4 + 6x3 x have three factors (x – α), (x – β) and (x – γ). Here α,   –  – β and γ are the zeroes of the cubic polynomial. We can write     − 3x3  + 0x2 Add 0x2 to the result     − 3x3  − 3x2 of subtraction ax3 + bx2 + cx + d = k (x − α)(x − β)(x − γ)    +   + Quotient = Sum of zeroes = α + β + γ = −b       3x2 + 2x − 3x3 = −3x2 a       3x2 + 3x x      −    − Sum of product of two zeroes = αβ + βγ + γα = c Add 2x to the result a         − x + 5 of subtraction         − x − 1 Quotient = Product of the zeroes = αβγ = −d            + a             6 3x2 = 3x x • If α, β are zeroes of a polynomials then ( x − α) and ( x − β) are factors of the polynomial. Add 5 to the result of subtraction • Division Algorithm for Polynomials If p(x) and g(x) are any two polynomials with g(x) ≠ Quotient = −x = −1 0, then we can find polynomials q(x) and r(x) such x that p(x) = g(x) × q(x) + r(x) where r(x) = 0 or degree of r(x) < degree of g(x). Further division This is known as the Division Algorithm for is not possible polynomials. as degree of the remainder (0) is less • Procedure for division of polynomial p(x) by g(x) than the degree of o First arrange the terms of p(x) (Dividend) and divisor (1) g(x) (divisor) in the decreasing order of their degrees. ( )∴6x4 + 3x3 + 2x + 5 = (x +1)× 6x3 − 3x2 + 3x −1 + 9 o Divide the highest degree term of the dividend Dividend = Divisor × Quotient+ Remainder by the highest degree of the divisor and the Note: In the division of polynomials, if the result is quotient. remainder is zero, then the divisor or quotient can Example: If 5x5 is highest degree of dividend be considered as factors of the polynomial. and x3 is the highest degree term of divisor then the quotient term will be 5x5 = 5x2 . x3 o Subtract the terms of the dividend and the product of quotient and divisor and the result 13

2. Polynomials Work Plan CONCEPT COVERAGE COVERAGE DETAILS PRACTICE SHEET PS – 1 Pre-requisites • Linear polynomial PS – 2 • Zeroes of Linear polynomial PS – 3 PS – 4 • Geometrical Meaning of the Zeroes of a Polynomial Polynomials • Relationship between Zeroes and Coefficients of a Polynomial • Division algorithm for Polynomials Worksheet for “Polynomials” PS-5 Evaluation with Self ---- Self Evaluation Sheet Check or Peer Check* 14

PRACTICE SHEET - 1 (PS-1) 1. State whether following statements are True / False. Give reasons or examples. i) All monomials are polynomials but all polynomials are not monomials. ii) x3 + 3 is a binomial in x. iii) ax3 + bx + 2 is a cubic polynomial if a = 0 and b = 5. 2. Find the value of p(x) at x = 0, 1, 2 i) p (x) =3x2 -5x ii) p (x) =x2 - x 5 3. If a graph of a linear equation is given, is it possible to find the zero of the polynomial without knowing the equation? Give reasons. 4. Find the linear polynomial which has zero at the given values. 5. A graph of a polynomial is as shown. Find the zero of the polynomial. 15

PRACTICE SHEET - 2 (PS-2) 1. What is the minimum and maximum numbers of zeroes can a polynomial of degree 5 have? 2. If a graph of a polynomial intersects y axis at y = 5 and passes through origin. State the number of zeroes the polynomial can have and value of zero of the polynomial? 3. From the given graphs of y = p(x), find the number of zeroes of the polynomial and the degree of the polynomial. 16

PRACTICE SHEET - 3 (PS-3) 1. Find the sum and product of the zeroes of the given polynomials: i) 3x2 – 5x + 12 ii)  5x2 − 1 x + 10 5 ( )2. −1 For the polynomial a2 − b2 x2 +(a −b) x +(a + b) show that sum of zeroes is a+b and product of zeroes is 1 . a−b 3. Find the ratio of sum of zeroes and product of zeroes for the polynomial: -x2 + 3 . 4. Find the zeroes of the quadratic polynomial 2x2 − 7x − 4 and verify the relationship between the zeroes and coefficients. 5. Verify the relationship between the zeroes and coefficients of the polynomial x2 – 5. 6. Verify the relation between the zeroes and coefficients of the polynomial: p (x) = 4x2 +16x +16 . 7. A quadratic polynomial has sum of zeroes as 5 and product of zeroes as –2. Find the polynomial. 8. A quadratic polynomial has two equal zeroes and the sum of the zeroes is 1. Determine the polynomial. 9. A polynomial px2 + qx − 5 has the –5 and 5 as its sum and product of its zeroes respectively. Find the value of the polynomial at x = 2. 10. A quadratic polynomial intersects the x axis only once at x = 5. Determine the polynomial. 11. A quadratic polynomial has its sum of zeroes equal to the product of its zeroes. Find the polynomial. 17

PRACTICE SHEET - 4 (PS-4) 1. Divide 4x3 - 3x2 + 2 by 2x+1. State the divisor, dividend, quotient and remainder. 2. Divide x3 − 5x2 − 8x −14 by 3x −1 3. Divide 5x5 − 4x3 − 3x2 + 2x − 4 by x +1 and state the quotient and remainder. 4. Divide x4 − 7 x3 − 2x2 by x − 2 and verify using the division algorithm. Writing the Dividend in standard form we get x4 − 7x3 − 2x2 + 0x + 0. 5. Find all the zeroes of the polynomial x4 + 4x2 − 7 x2 − 22x + 24 if 1, 2 are two Zeroes of the polynomial. 18

PRACTICE SHEET - 5 (PS-5) I. Choose the correct option. 1. The highest power of the variable in a polynomial is called its _____. (A) coefficient (B) zero (C) degree (D) All of these 2. The value of the variable that satisfies a given linear equation is called its _______. (A) root (B) evaluation (C) solution (D) Both (A) and (B) 3. Which of these is a linear equation? (A) 19 – 15t = 45 (B) 4t² + 34 = 236 (C) 4t² + 15t = -25 (D) 234t + 146t³ = 0 4. Identify the zero of the equation 5m – 12 = -47. −11 (A) 7 (B) -7 (C) 35 (D) 5 5. Identify the cubic polynomial from among the following. 2 – x³, x³, x³ – x² + x³, 3x³ – 2x² + x – 1 (A) 27 + 45x³ (B) 14x³ – 36x² (C) -13x³ – 23x² + 15x – 112 (D) All of these 6. The graph of a quadratic polynomial is a _____________. (A) parabola (B) line parallel to x – axis (C) line parallel to y – axis (D) line passing through origin 7. The roots of the quadratic polynomial x² + 2x + 1 = 0 are __________. (A) 1, 1 (B) -1, -1 (C) 0, 1 (D) 1, -1 8. Identify the equation of the line in the given graph. (A) x = 3 (B) x = -3 (C) y = 4 (D) y = -4 9. The sum of the zeroes of a quadratic polynomial in ‘x’ is given by (A) − Coefficient of x2 (B) Coefficient of x2 Coefficien t of x Coefficien t of x (C) Coefficient of x (D) − Coefficient of x Coefficien t of x2 Coefficien t of x2 10. The solutions of 2m² – 3m + 1 = 0 are __________. (A) m = −1, m = 1 (B)=m 1,=m 1 (C) m = 1, m = − 1 (D) m = −1, m = − 1 2 2 2 2 II. Short answer questions. 1. Find the zeroes of the quadratic polynomial p² + 7p + 10. 2. Find the quadratic polynomial in ‘p’, the sum of whose roots is -5 and product of roots is 6. 3. Verify that -4, 2 and -3 are the roots of the cubic polynomial p(x) = x³ + 5x² – 2x – 24 = 0. Also, verify the relationship between its zeroes and the coefficients. III. Long answer questions. 1. What are the roots of the quadratic polynomial y² + 4y + 4 = 0? Verify the relation between its roots and coefficients. 2. Verify the division algorithm for the given division: 19

SELF-EVALUATION SHEET Marks: 15 Time: 30 Mins 1. The graph of a polynomial of degree n is as shown in the figure. Find the number of zeroes of the polynomial and the degree of the polynomial. (2 Marks) 2. Find the zeroes of the polynomial 3x2 – 4x –4 and verify the relationship between the zeroes and coefficients. (3 Marks) 3. For the polynomial 3x2 + mx + n if the sum of zeroes is 5 and product of zeroes is –12, find the values of m and n. (2 Marks) 4. Determine the polynomials if the numbers provided are the sum and product of zeroes respectively: (2 Marks) i)  0, –5 ii) –5, 0 5. Divide 4x4 + 6x3 − 2x2 − 6x − 2 by x − 2 and verify using division algorithm. (4 Marks) 6. A polynomial of degree 5 is to be divided by a polynomial of degree 3. What will be the degree of the quotient if the remainder of division is (2 Marks) i) 0  ii) degree 1 State whether the divisor is a factor of the dividend. 20

3. Pair of Linear Equations in Two Variables Learning Outcome By the end of this chapter, a student will be able to: • Convert given pair of equations into pair of linear • Solve a pair of linear equations in two variables equations in two variables and find the solution using graphical method. • Solve a pair of linear equations in two variables using algebraic method. Concept Map Key Points • Every solution of the equation is a point on the line representing it. • Any equation which can be put in the form: ax + by + c = 0, where a, b and c are real numbers • Graphical Method of Solution of a Pair of Linear Equations and a, b are both non zero is called is a linear The solution to a pair of linear equations in two equation in two variables. variables can be obtained by plotting the graphs x and y are the two variables in the equation. It is of the two equations. The solution to the linear customary to denote the variables by x and y but equations is nothing but the coordinates of a point other letters may also be used. (x, y) which would satisfy both the equations. Example: 3x + y = 22 , 5t − 4x = 5 , p − 8q = −2 , etc. When two lines in are present in a plane, following • A solution to a linear equation in two variables three possibilities can happen means a pair of values, one for x and one for y o  The two lines intersect at one point. o  The two lines will not intersect, i.e., they are which satisfies the given equation. The solution is parallel. written as an ordered pair. A linear equation in two o  The two lines will be coincident. variables has infinitely many solutions. • The graphical representation of ax + by + c = Parallel Lines Intersecting Lines Coincident Lines 0 will be a straight line in Cartesian plane. A linear equation in two variables is represented o Parallel Lines: When the graphs of the two geometrically by a line whose points make up equations are parallel, there is no common the collection of solutions of the equation. This point between them (no intersection point). geometrical representation is called graph of the Since there is no common point, the pair of linear equation. 21

3. Pair of Linear Equations in Two Variables linear equations will have no solution i.e., no • Algebraic Methods of Solving a Pair of Linear value of x, y will satisfy both the equations. A Equations: Substitution Method pair of linear equations which has no solution Step 1: Identify the variables and write the is called an inconsistent pair. equations in standard form. Rearrange the terms of o Intersecting lines: When the graphs of the any one of the equation such that LHS has either of two equations intersect, the coordinates the variables. corresponding to the point of intersection is Step 2: In the second equation, substitute for the the solution to the equations. The equations variable in the LHS of step 1 to convert the second have one and only one solution. A pair of equation in two variables into linear equation in linear equations which has a solution is called one variable. consistent pair of linear equations. Step 3: Solve the equation in one variable. While o Coincident lines: When the graphs of the two solving the equations, the following three cases equations are coincident, then all the points of can arise: the graph representing one equation lie on the - If the variable gets eliminated then the given graph of the representing the second equation. pair of equations in two variables is dependent Hence we can say that they have infinitely and has infinite solutions i.e., any value of many solutions. A pair of linear equations in variables which satisfy one of the equations will two variables which are equivalent (coincident) satisfy the other equation. has infinitely many distinct common solutions Example: 5��=5, −1� � �= −1, etc. are called a dependent pair. A dependent pair - If the variable gets eliminated and a false of linear equations is always consistent. statement is created, then the given pair of equations in two variables has no solution. • A pair of linear equations represented by Example: −5 = −3, 1 = 10, etc. a1x + b1 y + c1 = 0 - If the variable gets a value then the given pair a2 x + b2 y + c2 = 0 of linear equations is consistent and hence has i) If the pair of equations are intersecting, then one solution. a1 ↑ b1 Example: x = −5, y = 3, t = 25, etc. a2 b2 Step 4: Substitute the value of the variable if any of ii) If the pair of equations are coincident, then the equations and solve for the second variable to a=1 b=1 c1 have a solution. a2 b2 c2 Example: Solve x + y = 2 and 2x − 3y = 4. iii) If the pair of equations are parallel, then Step 1: x , y are the variables in the given a1 = b1 ≠ c1 equations. a2 b2 c2 Rearranging the term of one equation, say x + y = 2 Example: we get y = 2 − x Step 2: Substituting for y in the other equation, i.e., 2x +3y −5 = 0 3x − 5y +10 = 0 3x − 5y +10 = 0 2x − 3y = 4 we get 5x − 7y −5 = 0 6x −10 y + 20 = 0 6x −10 y − 20 = 0 2x −3(2− x) = 4 2≠ 3 3 = −5 = 10 3 = −5 ≠ 10 5 −7 6 −10 20 6 −10 −20 Step 3: Solving the equation for x we get Equations are 2x − 6 + 3x = 4 Equations are Equations are inconsistent. ⇒ 5x = 10 consistent. dependent and The graphs will ⇒ x = 10 = 2 The graphs consistent. be are parallel. will have an The graphs will 5 intersection point. be coincident. Step 4: Substituting x = 2 in any one equation, say x + y = 2 we get In the graphical method of finding solutions, the 2+ y =2 accuracy of the solution of a pair of linear equations ⇒ y=0 depends on the how well the coordinates of the ∴ x = 2 , y = 0 is the solution of the given pair intersecting points are read from the graph. To of linear equations. avoid errors due to reading from the graphs, algebraic methods for solving a pair of linear • Algebraic Methods of Solving a Pair of Linear equations are used. Equations: Elimination Method Step 1: Multiply both the equations by some 22

3. Pair of Linear Equations in Two Variables suitable non-zero constants to make coefficients of x      y     1 one of the variable numerically equal in both the equations. b1 c1 a1 b1 Step 2: Add or subtract one equation from another so that one variable gets eliminated. This can lead b2 c2 a2 b2 to the following conditions: - If a true statement is obtained involving no Cross multiplying the coefficients and writing the equation as shown we get variable, then the original pair of equations has infinitely many solutions i.e., dependent pair x= y=1 and consistent pair. b1c2 − b2c1 c1a2 − c2a1 a1b2 − a1b1 Example: 5 = 5, −1 = −1, 0 = 0, etc. - If a false statement is obtained involving no If the denominator of the third portion terms variable, then the original pair of equations has become zero, then the values of x and y cannot no solution, i.e., inconsistent pair of equations. Example: 0 = −18, 1 = −5, 2 = 3, etc. be found. Hence - If an equation in one variable is obtained. Solving this equation will result in a numerical - When a1 ↑ b1 i.e., a1b2 − a1b1 ≠ 0, we get a unique value for the variable i.e., consistent pair of a2 b2 equations. Example: x = −1, y = 5, t = 20, etc. solution. Step 3: Use the numerical value of the variable solved in step 2 to find the value of the second - When a=1 b=1 c1 , i.e., a1b2 − a1b1 = 0, variable. a2 b2 c2 Example: Solve x + y = 2 and 2x − 3y = 4. Step 1: To make the coefficients of x equal in both b1c2 − b2c1 = 0 and c1a2 − c2a1 = 0. There are the given equations, infinitely many solutions. Any solution which satisfies one of the equations 2×(x + y) = 2×2 ⇒ 2x + 2y = 4 − − − −(i) will satisfy the second equation also. 2x − 3y = 4 − − − − − (ii) - When a1 = b1 ≠ c1 , i.e., a1b2 − a1b1 = 0, a2 b2 c2 Step 2: Subtracting (i) from (ii) we get 2x −3y = 4 b1c2 − b2c1 ≠ 0 and c1a2 − c2a1 ≠ 0. There are no 2x + 2y = 4 solutions to the given pair of linear equations. Step 3: Find the values of x and y using the above −− − equation. (Using first and third portions find the value of x. Using second and third portions find the −5y = 0 value of y .) ⇒ y=0 In this method, the coefficients are cross multiplied Step 3: Using y = 0 in (i) we get and their difference is used to find the solution of the equations and thus the name Cross Multiplication 2x + 2(0) = 4 ⇒ x = 2 Method. Example: Solve x + y = 2 and 2x − 3y = 4. ∴ x = 2 , y = 0 is the solution of the given pair of Wgertiting the given equations in standard form we linear equations. x + y − 2 = 0 − − −(i) • Algebraic Methods of Solving a Pair of Linear 2x − 3y − 4 = 0 − − − (ii) Equations: Cross Multiplication Method Step 1: Writing the given pair of linear equations in To solve the equations by cross multiplication standard form we get, method, we draw the diagram as shown below. a1x + b1 y + c1 = 0 − − − (i) x      y     1 a2 x + b2 y + c2 = 0 − − − (ii) 1 –2 1 1 Step 2: Writing the coefficients and constants in –3 –4 2 –3 the form of a diagram as shown: Cross multiplying the coefficients we get, (1) ( −4 ) x = ( −2 ) ( 2 ) y (1) ( −4 ) = 1 (1) ( 2 ) − − (−2)(−3) (1)(−3) − ⇒x=y=1 −4 − 6 −4 + 4 −3 − 2 23

3. Pair of Linear Equations in Two Variables ⇒ x =y= 1 −10 0 −5 ⇒ x = −10 = 2 and y = 0× 1 = 0 −5 −5 The solution to the given pair of linear equations is at x = 2 and y = 0. • Equations Reducible to a Pair of Linear Equations in Two Variables Step 1: Rearrange the terms of the equations by using appropriate methods to bring the variables to the numerator. The following methods can be used to convert a given equation to a linear equation: - Adding or subtracting, multiplying, dividing, squaring a given equation on both sides by the same amount does not alter the equation. - Substitute another variable for expression in x and y, etc. The substitution in a pair of linear equations should be same. Step 2: Ensure that the powers of the variables is equal to 1 so that the equations become Linear equations. Step 3: If the equations get converted into the form ax + by + c = 0 they can be solved using any of the methods discussed earlier. Not all equations can be converted into Linear equations in two variables. Example: a) y − 4 = 5 x ⇒ y − 4x = 20 ⇒ −4x + y − 20 = 0. This is a linear equation in two variables. b) x − y = 1 ( )⇒ x − y 2 = 12 ⇒ x + y − 2 x y −1 = 0. This is not linear equation. But if x = p and y = q the given equation becomes p − q =1⇒ p − q −1= 0. This is a linear equation in two variables (p, q). Solve for p and q and then find the values of x and y. c) 1 + 1 = 5 and 1 − 1 = 2 x−3 y−4 x−3 y−3 Here if we substitute 1 =p and 1 = q then the first equation will be a linear equation in two variables x−3 y−4 i.e., p, q but second equation will not be a linear equation in two variables. 24

3. Pair of Linear Equations in Two Variables Work Plan CONCEPT COVERAGE COVERAGE DETAILS PRACTICE SHEET Pre-requisites • Linear equations in Two Variables PS – 1 • Graph of Linear equations Pair of Linear Equations in • Graphical Method of Solution of a Pair of PS – 2 Two Variables Linear Equations. • Algebraic methods of Solving a Pair of PS – 3 Linear Equations o Substitution Method o Elimination Method o Cross Multiplication Method • Equations Reducible to a Pair of Linear PS – 4 equations in Two variables Worksheet for “Pair of Linear Equations in Two Variables” PS-5 Evaluation with Self Check ---- Self Evaluation Sheet or Peer Check* 25

PRACTICE SHEET - 1 (PS-1) 1. Plot the graph of the equation x + 2 y = 5. 2. Draw the graph of the following equations: i)  y = x  ii)  y = 3  iii)  x = 4 Are any graphs parallel to the axes? PRACTICE SHEET - 2 (PS-2) 1. Ravi goes to a shop and buys a few pens and with the remaining money he buys chocolates. With Rs. 30, Ravi purchased 5 pens and 5 chocolates and he purchased 4 pens and 5 chocolates with Rs. 25. Find the cost of the pens and chocolates and represent the situation algebraically and graphically. 2. Check the nature of the given pair of linear equations and find the solution graphically only if the equations are consistent. 2y− x =8 4y −2 = 9x 3. Check whether equations are consistent, inconsistent or dependent. i) 3x − 5 y + 2 = 0; 6x −10 y + 4 = 0 =ii)  y 4=x; x 4 y iii)  x = 5; y = 5 =iv)  x y=; 5x 5 y 4. Ravi and Raju are close friends and go together to a bakery. In the bakery there are cakes which cost Rs. 5, sweets which cost Rs. 3. They pick 5 items and their bill amount was Rs. 21. Find how many cakes and sweets they buy using graphical method. 26

PRACTICE SHEET - 3 (PS-3) 1. Solve the pair of linear equations by were present during her birthday. Substitution Method. 10. Solve the given pair of linear equations using x − 2 y = 15 and 2x − y = 4 Cross multiplication Method: 4x + 3 y = 15 2. Solve the pair of linear equations by 3x − 5 y = 10 Substitution Method. 11. Solve the given pair of linear equations using 3x − 2 5 y = 2 3 and x − y = 5 Cross multiplication method: 3. Solve the pair of linear equations by 7 x − 3 y = 10 Substitution Method. 2x + 5 y = 10 5x + 3 y = 10 and 10x + 6 y = 5 12. Find the value of p in the given pair of linear If the equations are plotted, what will the equations such that they have unique solution. nature of the lines? i) px + 3 y = −10 and 5x − y = 15 4. Solve the pair of linear equations by ii) 2x + 4 y = −3 and x − py = 20 Substitution Method. x − 4 y = 2 and 2x − 8 y = 4 iii) 3x − y = p and 6x − 2 y = 10 5. A farmer has 5 acres of land. He can grow 13. Find the values of p and q so that the given paddy, sugarcane, corn, etc. in his field. The money he earns for 1 acre of paddy is Rs. 1 lakh pair of linear equations have and for sugarcane he gets Rs. 2 lakh per acre. i)  unique solution  ii)  Infinite solution  The cost of growing paddy per acre is Rs. 0.2 iii)  No solution lakh and for growing sugarcane is Rs. 0.6 lakh per one acre. In year 2018 the farmer spent Rs. px + qy = ( p + q) 1 lakh on his farm and earned Rs. 4 lakh after selling paddy and sugarcane. How many acres 6x + 4 y = 9. did he use to grow paddy and sugarcane? (Use substitution Method) 6. By Elimination method solve the following equations 10x + 5 y = 20 and 5x +15 y = 10. 7. A two digit number has the sum of digits as 12. If the position of the two digits is interchanged, then the value of the number increases by 54. Find the digits of the number. Find the numbers using elimination method. 8. A 52 year old woman had a elder daughter and a younger son. Her age is equal to twice the sum of the ages of her children. The difference in the age of her children was 6 years. Find the current ages of her children. (Use elimination Method) 9. Reshma’s father and brother worked in military. During her birthday one of them was present. If Reshma’s father was there for her birthday he gave her Rs. 500 and if her brother was there for her birthday he would give her Rs. 300. She had been putting the money in her piggy bank from her 10th birthday until her 18th birthday. After her 18th birthday party, she counts the money in the piggy bank and found Rs. 3500. How many times her father, brother 27

PRACTICE SHEET - 4 (PS-4) 1. Solve the given pair of linear equations by reducing them to pair of linear equations by substitution method. 5+ 6 =5 and x −1 + 4 = 15 x −1 y −1 y −1 2. Solve the given pair of linear equations by reducing them to pair of linear equations by Cross multiplication method. 5+ y = 20 and x + y =2 x 3. Are the given equations a pair of linear equations in two variables? i) x + y = 1 and 1+ 1 = 4  ii) 6x − 4 y =3 and x − 2 y = 5xy x y 8 xy 4. 5 Boys and 2 girls of Class 10 can complete a project in 3 days. 4 boys and 4 girls can complete the same project in 2 days. Find the time taken to build the model if i)  1 boy works on the project ii)  1 girl works on the project 5. An inverter has capacity to power 2 bulbs and 3 fans for 3 hours. It can also supply power to 4 bulbs and 1 fan for 6 hours. If one bulb is to be powered by the inverter how many hours can it be powered? 28

PRACTICE SHEET - 5 (PS-5) I. Choose the correct option. 1. Which of these points lies on the x- axis? (A) (-3, 0) (B) (5, 0) (C) (0, 0) (D)All of these 2. The equation of y – axis is ______. (C) x = 0 (D) y = 0 (A) y = a (B) x = a 3. An inconsistent system of equations has ___________ solution (s). (A) 1 (B) 0 (C) 2 (D) infinitely many 4. A pair of linear equations has infinitely many solutions. What are they named as? (A) Dependent or consistent pair of linear equations (B) Inconsistent pair of linear equations (C) Coincident pair of linear equations (D) Intersecting pair of linear equations 5. Identify the ratio comparison that describes a pair of parallel lines. (A) a=1 b=1 c1 (B) a1 ↑ b1 (C) a1 = b1 ≠ c1 (D) a1 ↑ b1 ↑ c1 a2 b2 c2 a2 b2 a2 b2 c2 a2 b2 c2 6. The graph of a system of linear equations has only one solution. What type of lines are they? (A) Intersecting lines (B) Parallel lines (C) Coincident lines (D) Distinct lines 7. (x, y) is the solution of the pair of linear equations ax + by = a – b and bx – ay = a + b. What are the re- spective values of ‘x’ and ‘y’? (A) 2, 1 (B) -1 , 2 (C) -1, -1 (D) 1, -1 8. The solution of the pair of linear equations in two variables x + y = 4 and x – y = 2 is (a, b). What are the respective values of ‘a’ and ‘b’? (A) 3, 1 (B) 3, -1 (C) 1, 3 (D) -3, 1 9. a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are a pair of linear equations in ‘x’ and ‘y’. Using cross multiplica- tion method, what is the ratio that gives the value of ‘x’? (A) b1c2 + b2c1 (B) b1c2 − b2c1 (C) c1a − a2c1 (D) b1c2 − b2c1 a1b2 + a2b1 a1b2 − a2b1 a1b2 − a2b1 c1a − a2c1 10. A pair of linear equations is inconsistent. The lines are ____________. (A) always intersecting (B) either coincident or parallel (C)always coincident (D) always parallel II. Short answer questions. 1. Find the value of ‘k’ for which the pair of linear equations 5m + kn = -7 and m + 2n = 3 has no solution. 2. How many solutions does the system of equations 2p + 3q = 1 and 4p + 6q = 4 have? 3. The difference of two numbers is 1000 and the difference of their squares is 256000. Find the numbers. III. Long answer questions. 1. The angles of a ∆PQR are such that ∠R = 3∠Q = 2(∠P + ∠Q) . Find their measures. 2. If the digits of a two-digit number are interchanged, the number so obtained is greater than the origi- nal number by 27. If the sum of the two digits of the number is 11, what is the original number? 29

SELF-EVALUATION SHEET Marks: 15 Time: 30 Mins 1. 2x + y = 5 and x + 2 y = 5 form a pair of linear equations in two variables. Check if the equations are consistent. (2 Marks) 2. Find the value of p so that the given equations have infinite solutions: px + 3 y = 15 and 4x +12 y = 60.  (1 Marks) 3. Find the solution of the equations using graphical method if they are consistent. (4 Marks) x + 3 y = 6    x + y = 4. 4. One fine morning Sehwag decides to make runs by hitting sixes and fours. He played 60 balls in the match and did not score any runs in 15 balls only and ended with a score of 210. How many fours and sixes did Sehwag hit during the match. Solve the problem using Elimination and Cross Multiplication methods and compare the answers. (4 Marks) 5. Solve the given pair of equations by reducing them to pair of linear equations using Substitution Method. 1+ 1 = 10 and 5− 4= y 3 + 4  (4 Marks) x y −1 x y −1 −1 30

4. Quadratic Equations Learning Outcome By the end of this chapter, a student will be able to: • Determine the roots of quadratic equation by • Determine whether a given equation is a quadratic quadratic formula. equation. • Determine the nature of the roots of a quadratic • Determine the roots of a quadratic equation by equation. factorisation. • Determine the roots of a quadratic equation by method of completing the square. Concept Map Key Points • Solution to a quadratic equation by factorization Let the given quadratic equation be ax2 + bx + c = • ax2 + bx + c and a ≠ 0 is a general form of a quadratic 0 and a ≠ 0. polynomial and when this polynomial is equated Step 1: Find the product a × c. to zero, we get a equation. The general form of Step 2: Determine two numbers m, n which are quadratic equation is ax2 + bx + c = 0 factors of the value (a × c) such that m + n = b. Step 3: The given equation can be written as • A quadratic equation in variable x is an equation ax2 + mx + nx + c = 0 of the form ax2 + bx + c = 0, where a, b and c are real Step 4: Take out common factors among the numbers and a ≠ 0. This is also called the standard elements and write the equation in the form form of a quadratic equation. Step 5: The root of the quadratic equation will be −q • A given equation has to be simplified in order x = p and x = −s to decide whether the equation is a quadratic r equation. The values of p, q, r and s can be either positive or Example: negative. i) x(x+1) - 2x = 0 Example: Find the root of x2 + 8x +12 =0 ⇒ x2 + x - 2x = 0 ⇒ x2 - x = 0 Step 1: The product of 12 × 1 = 12 Step 2: 12 = 4 × 3 = 6 × 2 This is a quadratic equation But 4 + 3 ≠ 8 and 6 + 2 =8 ii) x(x+1) = x2 Step 3: x2 + 8x +12 = x2 + 2x + 6x +12 ⇒ x2 + x - x2 = 0 Step 4: = x (x + 2) + 6(x + 2) ⇒x=0    =( x + 2)( x + 6) This is not a quadratic equation. • A real number a is called a root of the quadratic Step 5: Given x2 + 8x +12 =0 equation ax2 + bx + c = 0, a ≠ 0 if aa2 + ba + c = 0. ⇒ ( x + 2)( x + 6) =0 x = a is the solution of the quadratic equation i.e., a ⇒ x + 2 =0 x + 6 =0 satisfies the equation. ⇒ x =−2 x = −6 The zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same. 31

4. Quadratic Equations The roots of the given equation are -2 and -6. • Solution of a quadratic equation using Quadratic • Solution of a Quadratic equation by Completing formula the square Let ax2 + bx + c =0 and a ≠ 0 is the given quadratic Let ax2 + bx + c =0 and a ≠ 0 is the given quadratic equation, then if b2 − 4ac > 0 , then the roots or solution of the quadratic equation are given by equation         x = −b ± b2 − 4ac Step 1: Write ax2 = (px )2 , where p is a real number 2a Step 2: Write bx =2 × (px ) × q , where q is a real number Step 3: Add and subtract q2. • Nature of the roots For the quadratic equation ax2 + bx + c =0 and Step 4: The given quadratic equation becomes a≠0 ( )(px + q)2 + −q2 + c =0 i) If b2 − 4ac > 0 , then the two roots are real and Step 5: The value of x will be px + q =± −q2 + c distinct.     ⇒ x ± −q2 + c − q ii) If b2 − 4ac =0 , then the two roots are real and = p equal. Example: Find the root of x2 + 8x +12 =0 iii) If b2 − 4ac < 0 , then there are no real roots or roots are imaginary. x2 + 8x +12 =0 ( )⇒ ( x)2 + 2( x)(4) + 42 + −42 +12 =0 ⇒ ( x + 4)2 − 4 =0 ⇒ ( x + 4)2 =4 ⇒ x + 4 =±2 ⇒ x + 4 =2 x + 4 =−2 ⇒ x =−2 x = −6 32

4. Quadratic Equations Work Plan CONCEPT COVERAGE COVERAGE DETAILS PRACTICE SHEET PS – 1 Pre-requisites • Introduction to Quadratic equations PS – 2 PS - 3 • Solution of a Quadratic equation by Factorisation PS - 4 PS-5 Quadratic Equation • Solution of a Quadratic equation by Self Evaluation Sheet Completing the square • Solution of a Quadratic equation using Quadratic formula • Nature of Roots Worksheet for “Quadratic Equations” Evaluation with Self Check ---- or Peer Check* 33

PRACTICE SHEET - 1 (PS-1) 1) Determine if any of the equations given below are quadratic equations: i) 2x3 + 3x2 + 2x + 5 ii) 3x2 + 4 = 0 iii) 4y2 = 5 iv) 4x + 5y - 5 = 0 v) 3x + 4 = 0 vi) 4r3 + 3r - 8 = 0 2) The length of a rectangle is 5 units more than the breadth of the rectangle. If the area of the rectangle is 50 sq. units, then express the problem mathematically. 3) The sum of square of a number and 5 times the number will be equal to 100. Express the statement in the standard form of quadratic equation. 4) The product of the two numbers whose difference is 5 was 300. Write a quadratic equation to express the problem. 5) Sum of the value of the area of the square and the value of the perimeter is 100. Express the problem mathematically. 6) Ramu makes a trip each year and spends Rs 15000 during the trip. If he spends Rs. 500 more in a day, then the duration of his trip will reduce by 5 days. Determine the money he spent on each day of the trip. 7) A metro is planned between two points in a city which are 10 km apart. The number of passengers travelling through the metro increased tremendously and the operator decided to make more trips per day. By increasing the speed of metro by 1 km/min, the time per trip reduced by 2 min and the train made more trips in a day. Express the problem mathematically to determine the speed of the train. (One trip means going from a station and returning back to the starting station 2 × 15 = 30 km) 8) Which of the following equations are quadratic equations? i) ( x + y)2 − y2 − 2xy + 2x + 8 =0 ii) (2x + 5)(3x − 6) = ( x + 1)(2x + 4) iii) ( x + 1)3 − ( x − 1)3 =x iv) x( x + 1)( x + 2) = x3 + 4 v) x( x −1)= x( x + 5) vi) ( x + 5)(2x − 3) =x − 2 PRACTICE SHEET - 2 (PS-2) 1) Find the roots of the given quadratic equations by factorisation method. i) 2x2 + 2x - 12 = 0 ii) 9x2 - 6x + 9 = 0 iii) 4x(x - 2) + 5 = 4(x - 1) iv) 5x2 + 6x - 8 = 0 v) x2 − x + 4 =0 vi) x2 − 7 x − 1 =0 25 12 ( )  vii) 15 + x 3 − 5 − x2 =0 viii) 2x2 + 7 x + 5 =0 ix) 25x2 + 3 − 10 3x =0   2 2 2 +x ( ) ( )2) Find the roots of 3 x+2 2 14 − 4 6 =0 3) The sum of two numbers is 12 and the product of numbers is 35. Determine the numbers. 4) If the length of the side of a square was increased by two 4 units then the ratio of the area of the final square to area of initial square is 2.25. Determine initial and final length of the side of the square. 34

PRACTICE SHEET - 3 (PS-3) 1) Find the roots of the following quadratic equation by Method of Completing the square: i) 12x2 + 2x - 2 = 0 ii) x2 - 3 = 0 iii) 2x2 + 13x + 20 = 0 iv) 2x2 - 6x + 4 = 0 22 v) x 2 + 3x =0 vi) 3x 2 − 5 x − 1 =0 vii) x 2 + 3x − 6 =0 viii) 5x 2 − 3 5x − 4 =0 ix) x 2 − 7 x + 1 =0 10 2) The sum of a number and its reciprocal is 7 more than the product of the number and its reciprocal. 6 Determine the numbers using method of completing the squares. 3) A lorry full of vegetables starts from a farmer’s place and reaches a market which is 60 km away and empties the vegetables at the market. It takes 3 hr to unload contents of the lorry and while returning the driver increases the speed by 10 km/hr. Determine the speed of the lorry during onward and return travel if the total time taken by the lorry to complete the journey is 6 hr (including the waiting time). Solve using method of completing the square. PRACTICE SHEET - 4 (PS-4) 1) Find the nature of the root of the following equation. Determine the root if the they are real. i) x2 + 12x + 1 = 0    ii) x2 - 4x - 25 = 0   iii) x2 − 1 x + 1 =0 iv) 3x 2 − 4 5x + 15 =0 3 5   v) x 2 − 2x + 4 3 =0 vi) 5x 2 − 5x =0 3   vii) 3x 2 + 5x − 1 =0 viii) x 1 1 + x 1 3 =5 + −      ix) x 2 + 7 − 2 7x =0 2) Determine the value of m in each of the following equation so that the quadratic equation has real and equal roots: i) x2 + mx + 12 = 0   ii) m + 6x - m = 0 iii) 3x2 + 2x + m = 0 3) The cost of making 1 sq. ft. of lawn in a garden is Rs. 50, the cost of putting a fence around the lawn is Rs. 10 for 1 ft. of perimeter and all other charges are Rs. 60. If the lawn is in the shape of a square and the total cost of the work as Rs. 2100 then find the size of the lawn. 4) A triangle with an altitude 6 cm more than the length of it base has an area of 100 cm2. Determine the length of the base and altitude of the triangle. 35

PRACTICE SHEET - 5 (PS-5) I. Choose the correct option. 1. Which of these is a quadratic equation? (A) (p + 2)² = 3(p + 4) (B) 15d² +12d = 3d(5d – 7) (C) 19m + 23 = 24m – 25 (D) (12k – 3)² = (12k +5)² 2. The general form of a quadratic equation is _______________. (A) bx + c = 0 (B) ax + b = 0 (C) ax³ + bx + c = 0 (D) ax² + bx + c = 0 3. The quadratic formula is ________. (A) −b + b2 + 4ac (B) −b ± b2 − 4ac (C) −b − b2 + 4ac (D) All of these 2a 2a 2a 4. Which of the following are the roots of the equation (m – 3)² – 144 = 0 ? (D) (m – 9) (m – 15) (A) (m – 9) (m + 15) (B) (m + 9) (m + 15) (C) (m + 9) (m – 15) 5. The condition when a quadratic equation has no real roots is _____. (A) b² – 4ac < 0 (B) b² – 4ac > 0 (C) b² – 4ac = 0 (D) Both (B) and (C) 6. The quadratic expression x2 – 5x + 6 can be factorised as ________. (D) (x + 3) (x + 2) (A) (x – 3) (x + 2) (B) (x – 3) (x – 2) (C) (x + 3) (x – 2) (D) real and distinct 7. The nature of roots of a quadratic equation depends on __________. (A) the sign of the coefficient of the square term (B) the sign of the coefficient of the constant term (C) the discriminant (D) All of these 8. What is the nature of roots of a quadratic equation if b² – 4ac > 0 ? (A) no real roots (B) real and equal (C) no roots 9. p = 2 is a root of the equation 2p² – mp + 6 = 0. The value of ‘m’ is: (A) -7 (B) -5 (C) 7 (D) 5 (D) n² + 3n + 10 = 0 10. Identify the quadratic equation whose roots are -2 and 5. (A) y² – 3y – 10 = 0 (B) x² + 3x – 10 = 0 (C) s² – 3s + 10 = 0 36