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Based On NCERT & In Accordance With The Latest Syllabus Prescribed By CBSE

For The Cuttent Session

A COMPLETE

THEORETICAL QUESTIONNAIRE

CONCEPT ILLUSTRATIONS WITH

THE SUITABLE EXAMPLES

WELL CATEGORIZED LISTING OF

OBJECTIVE QUESTIONS

ANSWERS OF ALL QUESTIONS

FOR TEXTUAL PART

MY GRATITUDES CUM PREFACE

Sr. No. TOPICS PAGES

01. Syllabus I

02. Number System & Real Numbers 01 – 07

03. Polynomials & Division Algorithm 08 – 13

04. System Of Linear Equations In Two Variables 14 – 19

05. Euclidian Geometry & Similar Triangles 20 – 23

06. Trigonometric Ratios & Identities 24 – 26

07. Data Analysis & Statistics 27 – 29

08. Chapter 01 - Objective 31 – 32

09. Chapter 02 – Objective 33 – 35

10. Chapter 03 – Objective 36 – 39

11. Chapter 04 – Objective 40 – 42

12. Chapter 05 – Objective 43 – 44

13. Chapter 06 – Objective 45 – 46

14. Answers of Objective Mathematicia 47

TABLE OF CONTENTS

SYLLABUS

Weightage assigned to the different topics of Summative Assessment I:

Sr. No. Topics Marks

01. Number System 11

02. Algebra 23

03. Geometry 17

04. Trigonometry 22

05. Statistics 17

Total Marks 90

Page | 1

INTRODUCTION

We have been ideally made familiar with the Number System in detail in class IX. We perhaps know now how to apply the basic arithmetic operations viz. addition, subtraction, multiplication and division on the natural numbers, integers, rational as well as irrational numbers. We have been taught to locate irrational numbers on the number line as well, we also are aware of Laws of Exponents of Real Numbers. I’m not absolutely right though I hate to admit it. (Pardon me, if you are quite an expert in all these!)

In our current discussion, we shall do a recall of divisibility on integers and shall state some important properties such as, Euclid’s Division Lemma, Euclid’s Division Algorithm and the Fundamental Theorem of Arithmetic. All these shall be used in the remaining part of this chapter to explore and gain more knowledge about integers and real numbers. We shall use Euclid’s Lemma to find HCF (Highest Common Factor) of integers; Fundamental Theorem of Arithmetic shall be used to find the HCF as well as LCM (Lowest Common Multiple) of integers. We shall learn to check irrationality of an irrational number by using contradiction. At the last we shall learn how to decipher without actual division, whether

a rational number p

q, 0q (say) has a terminating or non-terminating decimal expansion.

Before we start, we shall have a quick recap of the Number System for your benefit.

01. Natural numbers: The numbers used in ordinary counting i.e. 1, 2, 3… are called natural numbers (and positive integers as well). The collection (set) of natural numbers is denoted by N. Also if we include 0 to the set of natural numbers, we get set of the whole numbers which is denoted by the symbol W.

02. Integers: The numbers ... 3, 2, 1,0,1,2,3,... are called integers. The set of integers is denoted by the

symbol I or Z. Though now we use Z to symbolize the set of integers.

Also from the above discussion, it is evident that integers are of three types viz.:

a) Positive integers i.e. +Z =1,2,3,...

b) Negative integers i.e. Z = 1, 2, 3,... c) Zero integer i.e. non-positive and non-negative integer.

03. Rational numbers: A number of the form p

q, where p and q are integers and 0q , is called a rational

number. The set of rational numbers is denoted by Q. Zero being an integer, is also a rational number.

04. Irrational numbers: An irrational number has a non-terminating and non-repeating decimal

representation i.e. it can not be expressed in the form of p

q. The set of irrational numbers is denoted by

the letter T. Few examples of irrational numbers are 32, 5 7, 8 3,3 7, 5, , π,... e etc.

Note that π is irrational while 22

7 is rational.

05. Real numbers: The set of all numbers either rational or irrational, is called real number. Set of all the real numbers is denoted by R.

06. Laws of exponents or Indices:

a) a .a a m n m n b) a

aa

m

m n

n c) (a ) am n mn d) (a ) an m mn

e) 0a 1 f) a .b (ab)m m m g) a b

b a

m m

h) 1

aa

n

n.

†The above discussion has been done for the benefit of the reader. I understand you have been taught all

these many a times, yet you tend to forget them. They can be memorized, just start doing this– regular revision for a few days and use them as often you can. And believe me, after some time they will be on

your tips! And obviously, those who are familiar with all these, they may skip this discussion.

Chapter 01. NUMBER SYSTEM & REAL NUMBERS

Page | 2

Now we shall cover our syllabus of class 10. 01. The meaning of Lemma and Algorithm: A lemma is an already proven statement which helps in proving another statement. Also algorithm basically means the steps. It is a series of some rules which are given step wise to solve similar kind of problems.

02. Euclid’s Division Lemma: Given two positive integers a and b, there exist unique integers q and rsatisfying , 0 a bq r r b . Here a, b, q and r are called dividend, divisor, quotient and the remainder

respectively.

03. Finding HCF of two positive integers a and b (such that a > b) using Euclid’s division algorithm:

STEP1- By applying Euclid’s division algorithm, find q and r where , 0 a bq r r b .

STEP2- If 0r , then the HCF of the numbers a and b is “ b ”. If 0r , then apply the division algorithm to b and r taking b as the new dividend and r as the new divisor. STEP3- Continue this process till the remainder comes zero. When the remainder comes zero, the divisor at that stage is the required HCF of the numbers a and b. Here in our syllabus, Euclid’s division lemma is stated for only positive integers. However it can be extended for all the integers except zero (as b 0) .

04. Fundamental Theorem of Arithmetic: Every composite number can be expressed (i.e. factorised) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

05. Finding the HCF and LCM of two integers using Fundamental Theorem of Arithmetic: STEP1- Factorize each of the given integers and then express them as a product of powers of the primes. STEP2- To find the HCF, identify the common prime factors and then find the smallest exponent (power) of these common factors. Now raise these common prime factors to their smallest exponents and multiply each of these to obtain the HCF. STEP3- To find the LCM, list all the prime factors occurring in the prime factorization of the given integers. For each of these factors, find the greatest exponent and raise each prime factor to the greatest exponent and multiply each of these to obtain the LCM.

Do you notice that to find the HCF (Highest Common Factor), we take the smallest exponents of the common prime factors. Whereas in the case of LCM (Lowest Common Multiple), we take those factors which are with the largest exponents.

Remark To find the LCM (or HCF) of two integers a and b, we can use the relation given here if we know already their HCF (or LCM): a b (HCF) (LCM).

06. Theorem: If a prime number p divides a2, then p also divides a where a is a positive integer. That is, if 2 a pq , then a p where q and are positive integers.

07. Condition for a Rational number to have Terminating Decimal Expansion: A rational number

having terminating decimal expansion can always be expressed in the form of p

q where p and q are co-

primes and the prime factorization of denominator i.e., q is of the form 2 5m n where m and n are non- negative integers.

Understanding the meaning of HCF:

The common divisors of 24 and 36 are 1, 2, 3, 4, 6 and 12. The largest among these divisors is 12. This is called Greatest

Common Divisor (GCD) or Highest Common Factor (HCF).

You must remember that a prime number is divisible by only itself and one. Also note that one (i.e. 1) is not considered as a prime number.

Page | 3

If a rational number expressed in the form of p

q is such that q is not of the form 2 5m n , then decimal

expansion of p

q is non- terminating i.e. it has repeating decimal expansion.

Remark If for a rational number p

q, the denominator q is of the form 2 5m n then, it terminates after k

places of decimals where k is the largest of m and n.

WORKED OUT ILLUSTRATIVE EXAMPLES

Ex01. Using Euclid’s division algorithm, find the HCF of 210 and 55. Sol. Clearly 201 > 55. Applying Euclid’s division lemma to 210 and 55, we get

210 55 3+45 ... i

55 45 1+10 ... ii =

45 10 4+5 ... iii =

10 5 2+0 ... iv

The remainder at the (iv) stage is zero. So, the divisor at this stage i.e. 5 is the HCF of 210 and 55.

Ex02. Show that the square of any positive integer is either of the form 3m or 3 1m for some positive integer m. Sol. Let a be any positive integer. Then, it is of the form 3m or 3 1m or 3 2m . Now we have the following cases: Case I When 3a m

2 2 2 2 2(3 ) 9 3(3 ) 3 , 3 a m m m q where q m

Case II When 3 1a m

2 2 2(3 1) 9 6 1 3 (3 2) 1 3 1, (3 2) a m m m m m q where q m m

Case II When 3 2a m

2 2 2 2 2(3 2) 9 12 4 3(3 4 1) 1 3 1, 3 4 1 a m m m m m q where q m m . [H.P.]

Ex03. Prove that there is no natural number n for which 6n ends with the digit zero. Sol. It is evident that any positive integer ending with the digit zero has to be divisible by 5 and 2 so, its prime factorization must contain the primes 5 and 2 both.

We have, 6 (2 3) 2 3 n n n n .

It is observed that the primes in this factorization are 2 and 3. So by uniqueness of the Fundamental

Theorem of Arithmetic, there is no other prime in the factorization of 6n .

5 does not occur in the prime factorization of 6n for any value of n.

6n does not end with the digit zero for any natural number n.

Ex04. Find the HCF and LCM of 90 and 144 by the prime factorization method.

Sol. We have, 290 2 3 5 and 4 2144 2 3 .

1 2HCF(90,144) 2 3 18 and,

4 2 1LCM(90,144) 2 3 5 720 .

Ex05. Prove that 5 is an irrational number.

PROOF: Let us assume that 5 is rational number.

As 5 is rational, so there exists two coprime integers a and b where 0b such that

5 a

b

So, 5a b

Squaring on both the sides, we get: 2 25 ...(i)a b

Page | 4

Therefore, 5 divides 2a and hence 5 divides a as well. So, we can write: 5a c .

Substituting the value of a in (i), we get: 2 2(5 ) 5c b

2 225 5 c b

2 25 b c .

This means that 5 divides 2b and hence 5 divides b as well. Therefore, a and b have at least 5 as a common factor i.e. they are not coprimes.

This leads to the contradiction which has arisen due to our incorrect assumption that 5 is rational.

Hence 5 is irrational number. [H.P.]

Ex06. Prove that 5 2 3 is an irrational.

PROOF: Let us assume that 5 2 3 is rational.

Then there exists two coprime integers a and b where 0b such that

5 2 3 a

b

So, 5

32 2

a

b

5

32

b a

b

Since a and b are integers, so 5

2

b a

b is rational, and so 3 is rational as well.

But this contradicts the fact that 3 is irrational.

This contradiction has arisen because of our incorrect assumption that 5 2 3 is rational.

Hence 5 2 3 is irrational. [H.P.]

Ex07. State whether 3 2

23

2 .5 will have a terminating decimal expansion or a non-terminating repeating

decimal expansion? Also, find the number of places of decimals after which the decimal expansion terminates if it has a terminating decimal expansion.

Sol. It is clearly seen that the prime factorization of denominator of 3 2

23

2 .5 is of the form 2 .5m n . So, it has

terminating decimal expansion which terminates after 3 places of decimal.

Ex08. Find the largest number which divides 245 and 1029 leaving remainder 5 in each case. Sol. It is said that the required number leaves remainder 5 when it divides 245 and 1029. This means 245 5 240 and 1029 5 1024 are completely divisible by the required number. This implies that the required number is a common factor of 240 and 1024. It is also said that this required number is the largest number satisfying the given conditions. Therefore, it follows that the required number is HCF of 240 and 1024.

We have, 4240 2 3 5 and 101024 2 .

So, 4HCF(240, 1024) 2 16 .

Hence, required number is 16.

EXERCISE FOR PRACTICE

Q01. Fill in the blanks in the following: a) The sequence of well defined steps to solve any problem is known as___________________.

b) Numbers having non-terminating, non-repeating decimal expansion are known as_____________.

c) A proven statement used as a stepping stone towards the proof of another statement is known as_________.

d) Fundamental Theorem of Arithmetic states ___________________________________.

Page | 5

e) The probable factors of the denominator of rational number having terminating decimal expansion is of the form___________.

f) The prime factorization of composite numbers is_________________.

g) The algorithm which is used to find the HCF of two positive numbers is__________________.

h) For any two numbers, HCF LCM _____________ of numbers.

i) If a and b be two prime numbers then, HCF(a, b) = _________ and LCM(a, b) = _________.

j) If x is a rational number and y is irrational then, xy is _______________ number.

Q02. Write a rational number between 2 and 3 . .3/2Ans

Q03. What will be the value of 0.3 0.4 ? .7/9Ans

Q04. Evaluate 18 50 . What type of number is it, rational or irrational? .30;RationalAns

Q05. If 4 a q r then what are the conditions for a and r? . : Positive integer s.t. 4; 0 4 Ans a a r

Q06. What is the smallest number by which 5 3 be multiplied to make it a rational number? Also find

the number so obtained. . 5 3;2 Ans

Q07. Write the digit at the unit’s place of 9n where Zn . .For even power 1;for odd power 9 Ans

Q08. Find one rational and one irrational number between 3 and 5 . 1

.2; 3etc.2

Ans

Q09. If the number np were to end with the digit 0 then, what are the possible value(s) of p?

.10's multipleAns

Q10. There is a circular path around a sports field. Preeti takes 18minutes to complete one round of the field, while Renu takes 12minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the same point?

.36minutesAns

Q11. State Euclid’s Division Lemma and hence find the HCF of 28 and 16.

Q12. Using Euclid’s Division Algorithm, find the HCF of 9828 and 14742.

Q13. Find the HCF of the followings: a) 72 and 120 b) 52 and 130 c) 960 and 432 d) 4052 and 12576

Q14. Find the greatest common factor of 2730 and 9350.

Q15. Find HCF of 56, 96 and 324 by Euclid’s division algorithm. .4Ans

Q16. State Fundamental Theorem of Arithmetic and hence find the LCM of 867 and 255.

Q17. a) If LCM(480, 672) = 3360, find HCF(480, 672). [Ans.96 b) If HCF(306, 657) = 9, find LCM(306, 657). [Ans.22338 Q18. Find HCF LCM for the numbers 100 and 190. [Ans.19000

Q19. State Fundamental Theorem of Arithmetic and hence find the unique factorization of 120.

3.2 3 5 Ans

Q20. Find the LCM and HCF of 120 and 144 by using Fundamental Theorem of Arithmetic.

.720; 24Ans

Q21. Write the HCF of smallest composite number and the smallest prime number. .2Ans

Q22. The LCM of two numbers is thrice its HCF. The sum of LCM and HCF is 64. If one of the numbers is

48, find the other number. .16Ans

Q23. Find HCF of 81 and 237. Express it as a linear combination of 81 and 237.

.3 237 81 ; such that 13, 38 Ans x y x y

Q24. If the HCF of 210 and 55 is expressible in the form of 210 5 55 y , find the value of y. . 19Ans

Q25. Can two numbers have 12 as their HCF and 340 as their LCM? Explain.

:No, LCM is always multiple of HCFHint

Page | 6

Q26. Check whether there is any natural number n for which 4n can end with the digit zero. .NoAns

Q27. Check whether there is any natural number n for which 34n can end with the digit zero. .NoAns

Q28. Show that 9n can never end with the digit zero.

Q29. If p

q is a rational number ( 0)q , what is the condition on q so that the decimal representation of

p

q is

terminating? .Primefactorization of should beof theform 2 5m nAns q

Q30. Find whether the decimal representation of the numbers (a) 51

120 (b)

637

7280 is terminating or non-

terminating. Q31. The decimal expansion of following numbers will terminate after how many places:

a)4 3

47

2 5 b)

3

359

2 5 c)

2 2

503

2 5 .Four, three & two placesAns

Q32. Without actually performing the long division, state whether 543

225 has a terminating decimal expansion

or non-terminating recurring decimal expansion.

Q33. What type of decimal expansion will 3 represent? After how many places will the decimal expansion

terminate?

Q34. Without actual division, find whether the rational number 3 2

1323

6 35 has a terminating or a non-

terminating decimal expansion?

Q35. Write whether the rational number 51

1500 will have a terminating decimal expansion or non-terminating

repeating decimal expansion.

Q36. After how many places of decimals, the decimal expansion of the rational number 43

2000 will terminate?

Q37. Prove that 13 is irrational. Q38. Prove that 1

2 is irrational.

Q39. Prove that 2 3 7 is irrational. Q40. Prove that 7 5 is an irrational number.

Q41. Prove that 5 2 3 is an irrational number. Q42. Show that 5 3 is irrational.

Q43. Show that 1

2 5 is not rational number. Q44. Prove that

25 3

7 is irrational.

Q45. Prove that for any prime positive integer p , p is an irrational number.

Q46. Complete the missing entries in the adjacent factor tree:

Q52. With the help of Euclid’s division lemma, show that the cube of any positive integer is either of the form 9m , 9 1m or 9 8m for some positive integer m.

Q53. Show that every positive even integer is of the form 2m and that every positive odd integer is of the form 2m+1 for some positive integer m.

Q47. Show that only one out of , 2a a and 4a is divisible by 3.

Q48. Sow that square of any positive odd integer is of the form 8q+1 for

some positive integer q .

Q49. Show that any positive even integer is of the form 8p, 8p+2, 8p+4 and 8p+6, where p is some integer.

Q50. Show that the product of three consecutive natural numbers is divisible by 6.

Q51. Show that any positive odd integer is of the form 6p +1 , 6p +3 or

6 5p where p is some integer.

Page | 7

Q54. Show that any positive odd integer is of the form 4q+1 or 4q+3, where q is some integer.

Q55. Show that 2 1n is divisible by 8, if n is an odd positive integer.

Q56. Prove that the product of two consecutive positive integers is divisible by 2.

Q57. Prove that the product of three consecutive positive integer is divisible by 6.

Q58. Show that the square of any positive integer is of the form 4q or 4q+1 for some integer q.

Q59. Show that one and only one out of n, n+4, n+8, n+12 and n+16 is divisible by 5, where n is any integer.

Q60. Show that one out of n, n+3, n+6 or n+9 is divisible by 4, where n is any positive integer.

Q61. Show that the number of the form 7n , where n is any natural number, can not have unit digit zero.

Q62. Find a largest number that divides 2053 and 967 leaving a remainder of 5 and 7 respectively. .64Ans

Q63. Write two irrational numbers between 1 and 2. . 2 & 3Ans

Q64. Show that the sum and product of two irrational numbers 7 5 and 7 5 are rational numbers.

Q65. Explain why 3 5 7 11 11 is a composite number?

Q66. The HCF and LCM of 207 and x are 23 and 1449 respectively. Determine the value of x.

Q67. State division algorithm for the positive integers.

Page | 8

INTRODUCTION

Almost a year back, we had heard about this amusing word ‘polynomials’ and let me reveal – you have to bear this all long in many forms at almost all the places in mathematics and science! Do you remember we were taught about polynomials in one variable, their factors and degrees? We learnt about linear, quadratic and cubic polynomials. Also we got through monomial, binomial and trinomial etc. We studied about zeroes of a polynomial and factorization of algebraic expressions using various identities as well.

Now in our current venture we shall study about geometrical interpretation of zeroes of a polynomial. We shall confine our exploration around quadratic polynomials mainly, though I shall be telling you about various aspects of cubic polynomials too in the class! We shall learn how to find zeroes of a quadratic polynomial and we shall also establish relationship between its coefficients and the zeroes. We shall extend our study to learn further about Division Algorithm for Polynomials with real coefficients.

And if that is all not enough, hang on! We shall enjoy a bit too, trust me!!

01. Polynomial: An algebraic expression involving some constants and the variable terms is known as a

polynomial. If x be a variable, n be a positive integer and 0 1 2, , ,..., na a a a be the constants, then 1 2

1 2 1 0... n n

n na x a x a x a x a is known as a polynomial in variable x with degree n.

A polynomial of degree n can have at the most (n+1) terms.

02. Degree of a polynomial: It is the exponent (power) of the highest degree term in a polynomial. That is

the highest power of variable x (say) in a polynomial is called its degree. e.g. 3 22 4 7 5 x x x is a degree 3 polynomial.

03. Linear polynomial: It is a polynomial of degree one. The general notation for a linear polynomial is given as ( ) p x ax b , 0a , where a and b are constants.

04. Quadratic polynomial: A polynomial of degree two is called as quadratic polynomial. The general

notation for a quadratic polynomial is given as 2( ) , p x ax bx c 0a , where a, b and c are constants.

05. Cubic Polynomial: It is a polynomial of degree three. The general notation for a cubic polynomial is

given as 3 2( ) p x ax bx cx d , 0a , where a, b, c and d are constants.

06. Bi- quadratic polynomial: It is a polynomial of degree four. The general notation for a bi-quadratic

polynomial is given as 4 3 2( ) p x ax bx cx dx e , 0a , where a, b, c, d and e are constants.

07. Zeroes of a polynomial: The value(s) of x for which the polynomial ( )p x becomes zero is (are) called

zero(es) of the polynomial. Geometrically, zeroes of ( )p x are the x-coordinates of the points where the

graph of ( )y p x intersects the X- axis.

A polynomial of degree n has at the most n zeroes.

A quadratic polynomial has at the most two zeroes.

If a polynomial is of the form ( )x p y then geometrically, its zeroes are y-coordinates of the

points where the graph of ( )x p y intersects the Y-axis.

08. If α and β are the zeroes of quadratic polynomial 2( ) , p x ax bx c 0a then,

Sum of zeroes ( . . α β) b

i ea

and Product of zeroes ( . . αβ) c

i ea

i.e., 2

coefficient ofα β

coefficient of

x

x and

2

constant termαβ

coefficient of x.

09. If α , β and γ are the zeroes of cubic polynomial 3 2( ) p x ax bx cx d , 0a , then,

(By OP Gupta – 9650 350 480)

Chapter 02. POLYNOMIALS & DIVISION ALGORITHM

Page | 9

α β γ , αβ βγ γα b c

a a and αβ γ

d

a.

10. To find a Quadratic polynomial if its zeroes are given: If α and β are the zeroes of quadratic

polynomial say ( )p x then the polynomial is given as,

2 2( ) S P or ( S P) p x x x k x x where S α β, P αβ and k is any real number.

Note that here S and P represent the sum of zeroes and product of zeroes of ( )p x respectively.

11. Division Algorithm for Polynomials: If ( )p x and ( )g x are any two polynomials with ( ) 0g x , then

we can find polynomials ( )q x and ( )r x such that

( ) ( ) ( ) ( ) p x g x q x r x , where ( ) 0 or degree of ( ) degree of ( ) r x r x g x .

If ( )r x = 0 then the polynomial ( )g x is a factor of polynomial ( )p x .

12. Algebraic Identities:

a) 2 2 2( ) 2 a b a ab b b) 2 2 2( ) 2 a b a ab b

c) 3 3 2 2 3( ) 3 3 a b a a b ab b d) 3 3 2 2 3( ) 3 3 a b a a b ab b

e) 2 2 ( )( ) a b a b a b f) 3 3 2 2( )( ) a b a b a ab b

g) 3 3 2 2( )( ) a b a b a ab b h) 2 2 2 2( ) 2 2 2 a b c a b c ab bc ca

i) 3 3 3 2 2 23 ( )( ) a b c abc a b c a b c ab bc ca .

WORKED OUT ILLUSTRATIVE EXAMPLES

Ex01. Find the zeroes of 23 8 4 3 x x and verify the relationship between its zeroes and coefficients.

Sol. We have, 2 23 8 4 3 3 6 2 4 3 x x x x x

3 ( 2 3) 2( 2 3) ( 2 3)( 3 2) x x x x x

So, the value of 23 8 4 3 x x is zero when ( 2 3) 0 x or ( 3 2) 0 x , i.e., when 2 3x or 2

3x .

Therefore, the zeroes of 23 8 4 3 x x are 2 3x and 2

3x .

Now, sum of zeroes2

2 8 8 Coefficient of2 3

Coefficient of3 3 3

x

x

and, product of zeroes2

2 4 3 Constant term(2 3) 4

Coefficient of3 3

x.

Ex02. Find a quadratic polynomial whose zeroes are 1/4 and –1.

Sol. Let the required quadratic polynomial be 2( ) , 0 p x ax bx c a , and its zeroes be denoted by α and β .

We have, sum of zeroes1 3

S α β ( 1)4 4

and, product of zeroes1 1

P αβ ( 1)4 4

.

So, the required quadratic polynomial is 2 2( ) S P or ( S P) p x x x k x x ,

i.e., 2 23 1 3 1( )

4 4 4 4

p x x x x x or 2(4 3 1) k x x where k is any real number.

Ex03. Obtain all zeroes of 4 3 2( ) 3 6 2 10 5 f x x x x x if two of its zeroes are 5

3 and –

5

3.

Page | 10

Sol. Since two zeroes are 5

3 and –

5

3, so 25 5 5

3 3 3

x x x is a factor of the given polynomial.

Now, we shall divide the polynomial 4 3 2( ) 3 6 2 10 5 f x x x x x by 2 5

3x .

2

2 4 3 2

4 2

3 2

3

2

2

3 6 35

3 6 2 10 53

3 5

6 3 10 5

6 10

3 5

3 5

0

x x

x x x x x

x x

x x x

x x

x

x

So, 4 3 2 2 253 6 2 10 5 (3 6 3)

3

x x x x x x x .

Now by splitting the middle term of quotient 23 6 3 x x we factorize it as 23( 1)x . So, its zeroes are given

by 1 x and 1 x . Therefore, the zeroes of given polynomial ( )f x are 5 5

, , 13 3

and 1 .

EXERCISE FOR PRACTICE

Q01. Which of the followings are not polynomials?

a) 3 2 23 7 x x x b) 2 x px q c) 2

2

18 x

x d) 3 22 3 5 6 x x x .

Q02. What do you understand by the value of a polynomial at a given point?

Q03. If 3 2( ) 3 2 6 5 p x x x x , find (2)p .

Q04. Find the zeroes of the polynomial 2 ( ) mx m n x n .

Q05. Show that the quadratic polynomial 2 4 5 x x has no zeroes. [In fact, instead of saying ‘no zeroes’, it is better to say ‘no real zeroes’. I shall tell you about this aspect in the chapter on Quadratic Equations though this is not included in our syllabus of class X.]

Q06. Find the zeroes of 2 2x and verify the relationship between the zeroes and the coefficient.

Q07. Find the zeroes of 2( ) 4 8 f m m m and hence verify the relationship between the zeroes and its

coefficient.

Q08. Find the zeroes of 2( ) 6 15 f x x x and establish a relationship between the zeroes and its

coefficient.

Q09. Find the zeroes of 2( ) 4 3 5 2 3 f x x x and verify the relationship between the zeroes and its

coefficient.

Q10. The sum of the zeroes of the quadratic polynomial 2( ) 2 3 f y ky y k is same as their product,

determine the value of k.

Q11. If 1 is a zero of the polynomial 2( ) 3( 1) 1 f u au a u then, find the value of a. . 1Ans a

Q12. For what value of k, –4 is a zero of the polynomial 2 (2 2) x x k ? .9Ans

Q13. If one zero of polynomial 2 2( 9) 13 6 k x x k is reciprocal of the other, find the value of k. . 3Ans k

Page | 11

Q14. If one zero of 24 9 8 x k x is negative of the other, determine the value of k. . 0Ans k

Q15. Find the value of k such that the polynomial 23 2 5 x k x x k has the sum of its zeroes as half of

their product. .1Ans

Q16. If –5 is one of the zeroes of 22 15 x px and quadratic polynomial 2( ) p x x m has both of its

zeroes equal to each other then, find the value of m. . 140Ans

Q17. The difference between the squares of the zeroes of 2 45 x px is 216, find the value of p. . 18Ans

Q18. Form a quadratic polynomial whose zeroes are the squares of the zeroes of 2 2 5x x .

Q19. Find the zeroes of the polynomial 22 7 3 x x and hence find the sum and product of its zeroes. Q20. Form a quadratic polynomial whose zeroes are 2 and –6. Verify the relation between the coefficients and zeroes of the polynomial.

Q21. Form a quadratic polynomial whose sum and product of zeroes are 2 and 1

3.

Q22. Find a quadratic polynomial whose sum and product of zeroes are 2 and 3 respectively.

Q23. Find the quadratic polynomial whose one zero is 2 3 .

Q24. If a and b are zeroes of the polynomial 2 6 x x then, find a quadratic polynomial whose zeroes are 3 2a b and 2 3a b .

Q25. If α and β are the zeroes of polynomial 2 , 0 ax bx c a then, evaluate:

a) 2 2α β b) α β

β α c)

2 2α β

β α

d) 3 3

1 1

α β .

Q26. If sum of the squares of zeroes of 2( ) 8 f x x x is 40, find the value of . . 12Ans

Q27. If α and β are the zeroes of polynomial 2( ) 2 5 f x x x m satisfying the relation 2 2 21α β αβ

4

then, find the value of m. .2Ans

Q28. If α and β are the zeroes of polynomial 2( ) 2 f x x x , find a polynomial whose zeroes are 2α 1

and 2β 1 .

Q29. If α and β are the zeroes of polynomial 2( ) 5 f x x x k such that α β 1 then, find the value of k.

Q30. If α and β are the zeroes of 22 7 3 x x then, find the sum of the reciprocal of its zeroes. .7/3Ans

Q31. If m and n are zeroes of 23 11 4 x x then, find the value of m n

n m.

Q32. If p and q are zeroes of polynomial 2t 4t 3 , show that 1 1 14

2 03

pqp q

.

Q33. If ( 6)x is a factor of 3 2 0 x ax bx and 7 a b , find the values of a and b.

Q34. Find the divisor of 2 1x which gives the quotient as 1x and leaves 2 as remainder.

Q35. Divide the polynomial 3 2( ) 14 5 9 1 g x x x x by polynomial ( ) 2 1 h x x . Find its quotient and the

remainder.

Q36. Divide 4 3 22 9 5 3 8 x x x x by 2 4 1x x and hence verify the division algorithm.

Q37. State the division algorithm for polynomials. Divide the polynomial 2 3( ) 3 3 5 f x x x x by the

polynomial 2( ) 1 g x x x and hence verify the division algorithm.

Q38. Divide 2 36 19 6 x x x by 22 5 3 x x and hence verify the division algorithm.

Q39. Apply the division algorithm to find the quotient and remainder on dividing 3 2( ) 6 11 6 f x x x x

by the polynomial ( ) 2 g x x .

Q40. Apply the division algorithm to find the quotient and remainder on dividing 4 2( ) 3 4 5 p x x x x by

the polynomial 2( ) 1 m x x x .

Page | 12

Q41. By division algorithm, check whether or not 2 3 1x x is a factor of 4 3 23 5 7 2 2 x x x x . .YesAns

Q42. By division algorithm, check whether or not 2 5 4x x is a factor of 4 3 25 3 4 x x x x . .NoAns

Q43. Find all zeroes of polynomial 4 33 6 4 x x x if it is given that 2 are its known zeroes. .1,2Ans

Q44. Find all zeroes of the polynomial 4 3 2( ) 2 3 3 6 2 f x x x x x if two of its known zeroes are given as

2 . .1,1/2Ans

Q45. Find all zeroes of polynomial 4 3 25 4 10 12 x x x x if it is given that 2 are its known zeroes.

Q46. It is given that 1 is one of the zeroes of 37 6 x x , find its other zeroes.

Q47. Find all zeroes of polynomial 4 3 25 9 15 18 x x x x if its known zeroes are 3 .

Q48. Find all zeroes of polynomial 4 3 26 26 138 35 x x x x if its known zeroes are 2 3 . . 5,7Ans

Q49. Find all zeroes of polynomial 4 3 24 2 12 3 x x x x if its known zeroes are 2 3 .

Q50. Find the remaining zeroes of the polynomial 4 3 2( ) 4 20 23 5 6 p x x x x x if it is given that

(2) 0 (3) p p . . 1/2Ans

Q51. Obtain all zeroes of 3 25 15 3 9 x x x if two of its zeroes are 3/5 .

Q52. Find all zeroes of 4 3 2( ) 2 2 7 3 6 p x x x x x if two of its factors are given as 3/2x . . 1,2Ans

Q53. Find all the zeroes of the polynomial 4 3 234 4 120 x x x x if its known zeroes are 2 . . 6,5Ans

Q54. Obtain all the zeroes of 4 3 27 17 17 6 x x x x if two of its zeroes are 1 and 2.

Q55. Find k so that 2 2 x x k is a factor of 4 3 22 14 5 6 x x x x . Also find all the zeroes of the two

polynomials. 2. 3;zeroesof 2 3 are 1, 3&of otherare 1,2, 3, 1/2 Ans k x x

Q56. For what value of k, the polynomial 4 3 210 25 15 x x x x k is exactly divisible by x + 7. . 91Ans

Q57. On dividing the polynomial 3 2( ) 3 2 f x x x x by another polynomial ( )g x , the quotient ( )q x and

remainder ( )r x are obtained as ( ) 2 q x x and ( ) 4 2 r x x respectively. Determine the polynomial

( )g x . 2. 1 Ans x x

Q58. On dividing 3 23 2 5 5 x x x by a polynomial ( )p x , the quotient and remainder were found to be

2 2 x x and –7 respectively. Find the polynomial ( )p x . .3 1Ans x

Q59. What must be subtracted from 4 3 28 14 2 7 8 x x x x so that the resulting polynomial is exactly

divisible by 24 3 2 x x . .14 10Ans x

Q60. What must be subtracted from the polynomial 4 3 24 2 8 3 7 x x x x so that the resulting polynomial

is exactly divisible by 22 2 x x . .5 11Ans x

Q61. What must be added to the polynomial 4 3 24 2 2 1 x x x x so that the resulting polynomial is

divisible by 2 2 3x x ?

Q62. What must be added to or subtracted from 4 3 2( ) 8 14 2 8 12 p x x x x x so that the polynomial

24 3 2 x x becomes a factor of ( )p x ? . 15 14,15 14 Ans x x

Q63. Find the values of a and b so that 4 3 28 x x x ax b is divisible by 2 1x . . 1, 7 Ans a b

Q64. If the polynomial 4 3 2( ) 6 16 25 10 p x x x x x is divided by 2 2 x x k , the remainder comes out

to be x a , find the value of k and a. . 5 Ans k a

Q65. If the polynomial 4 3 26 8 17 21 7 x x x x is divided by another polynomial 23 4 1 x x , the

remainder comes out to be ax b , find a and b. . 1, 2 Ans a b

Q66. If the zeroes of the polynomial 3 2( ) 3 1 f x x x x are given as ,a b a and a b then, find the

values of a and b. . 1, 2 Ans a b

Page | 13

Q67. The coefficient of x in the quadratic polynomial 2( ) f x x px q was wrongly written as 17 in place

of 13 and the zeroes thus found were –2 and –15. Find the zeroes of correct polynomial.

Q68. Find the sum and product of the zeroes of 2 2 2 2 2( ) p x p q x q , also cite the relationship between the

zeroes and the coefficients.

Q69. Given a linear polynomial in x, state how many zeroes it can have and why? Illustrate with an example.

Q70. Write a quadratic polynomial whose zeroes are given as α and β .

PREDICTING THE ZEROES OF A POLYNOMIAL BY OBSERVING ITS GRAPH

Q71. Find the number of zero(es) and zero(es) in the following graphs: Y ( )y = f x (0,3) ( )y = g x

X O X X O X

Y Y Y Y ( )y = f x

Y

7-

6-

5-

4-

3-

2-

1-

-

–1-

–2-

–3-

–2

-

–1-

1

-

2-

3

-

4-

5

-

(1,2)

(0, 3)

O X

(–1, 0) ( )y = g x

(0,–3)

X O X

Y (0, 4)

(–4, 0) (–2, 0) O (2, 0) X

( )y = p x

O (1, 0) X

Y (0, 4) (0, 5/2) (0, 1)

(0,–1)

Y

( )x = f y

(a)

(b)

(c)

(d)

(e) (f)

Page | 14

INTRODUCTION

In our previous classes, we have been taught about linear equations in single variable and their applications in solving simple real life based word problems. In class IX, we learnt about linear equations in two variables of the form 0 a x by c where 0a , 0b ; a, b and c are real constants. We were

made familiar with the graphs of these linear equations in two variables as well. Remember chapter 3 on Coordinate Geometry?

In our current study, we shall learn about systems of linear equations in two variables, their solution by graphical and various algebraic methods namely Substitution Method, Elimination Method and Cross-Multiplication Method. We shall also study about consistency and inconsistency of the system of linear equations. At last, we shall discuss some applications of linear equations in solving real life based problems from different areas.

01. Linear equation in two variables: It is an equation of the form 0 a x by c ; 0, 0a b where a,

b, c are constants corresponding to real numbers.

02. System of linear equations: It is a pair of linear equations in two variables of the form 0 a x by c , 0 lx my n where a, b, c; l, m, n are constants corresponding to real numbers.

03. Graphical Method of Solving System of Linear Equations:

STEP1- Obtain the system of linear equations in x and y in the following form: 0 ...( ) ax by c i

0 ...( ) lx my n ii

STEP2- Draw the graphs of equations (i) and (ii) in STEP1. Say L1 and L2 represent the lines of equations (i) and (ii) respectively. STEP3- The point of intersection say (α,β ) of lines L1 and L2 corresponds to the solution of the

given system of equations. That is the solution of these equation is given as say αx and βy . This is case of consistent system having the unique solution.

STEP4- If the lines L1 and L2 are coincident, then the given system of equations has infinitely many solutions. This is case of consistent system having infinitely many solutions. STEP5- If the lines L1 and L2 are parallel, then the given system of equations has no solution. This is the case of inconsistent system having no solution.

04. Consistent and inconsistent systems: A system of simultaneous linear equations having at least one solution is said to be consistent system. Whereas if a system of simultaneous linear equations has no solution then, it is said to be an inconsistent system.

05. Table depicting conditions of Consistency and inconsistency of the pair of lines:

Sr. No. Pair of Equations Comparing the Ratio Graphical Meaning Algebraic Interpretations

1.

1 1 1 0 a x b y c

1 1

2 2

a b

a b

Intersecting Lines

Unique Solution (Consistent system)

2.

and

2 2 2 0 a x b y c

1 1 1

2 2 2

a b c

a b c

Coincident Lines

Infinitely Many Solutions (Consistent System)

3.

1 1 1

2 2 2

a b c

a b c

Parallel Lines

No Solution (Inconsistent System)

(By OP Gupta – 9650 350 480)

Chapter 03. SYSTEM OF LINEAR EQUATIONS IN TWO VARIABLES

Page | 15

06. Algebraic Methods of Solving System of Linear Equations:

a) Substitution Method b) Elimination Method c) Cross- Multiplication Method

Here we shall discuss these methods one by one with the help of examples!

Substitution Method:

Consider system of equations

3 5 1 x y …(i)

and 1 x y …(ii)

From equation (ii), we get

1 x y …(iii)

Substituting 1 x y in equation (i), we get

3( 1) 5 1 y y

3 3 5 1 y y

2 1 3 y

2 2 y

1 y .

Putting 1 y in (iii), we get

1 1 x

2 x .

Hence the solution of the given system of equations is 2, 1 x y .

Isn’t it interesting? We shall now proceed to the next method i.e., elimination method of solving the pair of linear equations.

Elimination Method:

Consider system of equations

8 5 9 x y …(i)

and 3 2 4 x y …(ii)

Let us eliminate y. To do so multiply both the sides of equation (i) and (ii) by 2 and 5 respectively i.e.,

i.e. 8 5 9 2 x y

and 3 2 4 5 x y

So, 16 10 18 x y …(iii)

and 15 10 20 x y …(iv)

Subtracting (iv) from (iii), we get

16 10 15 10 18 20 x y x y

2 x .

Putting 2 x in (i), we get

8 2 5 9 y

16 5 9 y

5 9 16 y

5 25y

5 y .

Hence the solution of the given system of equations is 2, 5 x y .

Now the last one i.e., cross- multiplication method which is a bit tricky! So you are supposed to give it more attention than you can offer. (How do we do so, well, even I am not aware of this!) Note that for your benefit I have provided here the discussion in two forms– firstly the variable format and then, secondly I have added an example too!

Cross- Multiplication Method: Consider system of equations of the form 1 1 1 0 a x b y c and 2 2 2 0 a x b y c .

Then, their solution can be obtained by the following expression,

1 2 2 1 1 2 2 1 1 2 2 1

1

x y

b c b c c a c a a b a b

Also, you can memorize the above with the help of following diagram,

1 1

2 2

x

b c

b c

1 1

2 2

y

a c

a c

1 1

2 2

1

a b

a b

Let us take an example now!

Page | 16

Consider system of equations 2 3 0 x y

and 4 3 0 x y .

By cross-multiplication, we have

1

1 3 1 3 4 3 2 3 2 1 4 1

x y

1

3 3 12 6 2 4

x y

1

6 6 6

x y

Consider 1

6 6

x and

1

6 6

y

6

16

x and 6

16

y .

Hence the solution of the given system of equations is 1, 1 x y .

In any case, if you are not able to make head-tail of anything in this cross-multiplication method, just wait till we do it in our class. I shall make it stick to your brain-chips (I mean the brain cells)!

07. A Special Case when Coefficient of x and y are interchanged in the two equations:

Consider the two equations as ax by m and bx ay n .

STEP1- Add the two equations to obtain a b x b a y m n

i.e. a b x y m n ...

m nx y i

a b

STEP2- Subtract the second equation from the first equation to obtain a b x b a y m n

i.e. a b x y m n ...

m nx y ii

a b

STEP3- Add (i) and (ii) to obtain the value of x and then subtract (ii) from (i) to obtain value of y.

Following example will illustrate the above discussion:

Consider the system of equations 37 41 70 x y …(i)

and 41 37 86 x y …(ii)

Adding the equations (i) and (ii), we get

78 78 156 2 x y x y …(iii)

Subtracting equation (ii) from (i), we get

4 4 16 4 x y x y …(iv)

Now on adding equations (iii) and (iv), we obtain

2 6 3 x x .

Also on subtracting equation (iv) from (iii), we obtain

2 2 1 y y .

Hence the solution of the given system of equations is 3, 1 x y .

Page | 17

EXERCISE FOR PRACTICE

Q01. Without actually drawing the graph, can you comment on the type of graph made by a given pair of linear equations in two variables? Q02. If ratio of coefficients of x is not equal to the ratio of coefficients of y in a given pair of linear equations in two variables then, what will be the type of graph? Q03. Comment on the type of solution and type of graph for the following pair of linear equations: 2 5 9, 5 6 8 x y x y .

Q04. For what value of k the equations 2 3, 3 1 x y x ky has a unique solution?

Q05. Comment on the consistency or inconsistency of a pair of linear equations in two variables having intersecting lines on the graph. Q06. Check graphically whether the pair of equations 3 2 4 0 x y and 2 2 0 x y is consistent.

Also find the coordinates of the points where the graphs of the lines of equations meet the y-axis.

Q07. For what value of k the equations 1 5 and 1 9 8 1 x k y k x y k has infinitely many

solutions?

Q08. Find value of k for which pair of equations 2 3 7 and 1 2 3 x y k x k y k has infinitely

many solutions.

Q09. Find value of k for which pair of equations 2 10, 3 3 12 x ky x k y represent parallel lines.

Q10. Find the value of m and n for which pair of linear equations 2 3 7 0 x y and

2 28 mx m n y has infinitely many solutions.

Q11. For what value of k the system of equations 3 3, 12 kx y k x ky k has infinitely many

solutions?

Q12. Find the value of k for which the pair of equations 3 1 and 2 1 1 2 1 x y k x k y k has

no solutions. Q13. Find the value of k so that the pair of equations 2 7 0 and 2 14 x y x ky will represent

coincident lines.

Q14. For what value of p the system of equations 2 2 1 3 2 1 and 2 3 7 p x p y p x y has

unique solution.

Q15. Find the value of m so that the pair of linear equations given as: 3 1 3 2 m x y and

2 1 2 5 0 m x m y is consistent.

Q16. Check whether x = 4 and y = 5 is not a solution of the pair of equations 3 1, 2 11 x y x y .

Q17. Write equation of a line intersecting with the line whose equation is 2 3 7 x y .

Q18. Give a linear equation which may be coincident with 2 3 4 x y . .4 6 8 Ans x y

Q19. Solve the followings:

a) 6 8

6, 3 5 x xy y

b) 28 17 63, 17 28 62 x y x y

c) 7

0.8, 102

2

xy

yx

d) 2 4 0,2 4 0 ax by a b bx ay b a

e) 5 1 6 3

2, 11 2 1 2

x y x y f) 1, 5

x y x y

xy xy

g)

2

2 21, 1

a bax by bx ay

a b h)

1 1 1 18, 9

2 3 3 2

x y x y

i) 2 12 3 17, 2 3 5 x y x y j) 139 56 641, 56 139 724 x y x y

k) 37 43 123, 43 37 117 x y x y l) 5 1 15 5

2, 2 x y x y x y x y

m) 2 3 4 9

2, 1 x y x y

n) 2, 6; 0, 0

p q p q

p qpq pq

Page | 18

o) 2 3 17 5 1

, 23 2 3 2 5 3 2 3 2

x y x y x y x y

p)

6 7 1 13, ; 0, 0

2 3

x y x y

x y x y x y x y

q)

1 12 1 7 4, 2

2 2 3 7 3 2 2 2 3 3 2

x y x y x y x y

r)

1 1 3 1 1 1,

3 3 4 2 3 2 3 8

x y x y x y x y

. )3,2; ( )2/5,3/5; ( )7,13; ( )3,4; ( )1,2Ans a b h i k

Q20. Show that the pair of equations 3 4 7,12 16 28 x y x y have infinitely many solutions.

Q21. Show that the system of equations 3 4 8 x y and 6 8 10 x y is inconsistent.

Q22. Solve the pair of linear eqs. 2, 4 x y x y . Also find the value of p if 2 3p x .

Q23. Riya is walking along the line joining (1,4) and (0,6) while Ritu is walking along the line joining (3,4) and (1,0). Represent this situation on the graph and find the point where both of them cross each other. Q24. Form a pair of linear equations for: The sum of the numerator and denominator of fraction is 3 less than twice the denominator. If the numerator and denominator both are decreased by 1, the numerator becomes half the denominator. . 3,2 1 if & represent Nr and Dr respectively Ans x y x y x y

Q25. Anuj gives `9000 to some athletes of a school as scholarship every month. Had there been 20 more athletes each would have got `160 less. Form a pair of linear equations for this situation.Q26. The sum of numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction. .5/7Ans

Q27. A two digit number is four times the sum of the digits and twice the product of the digits. Find the

number. .36Ans

Q28. The sum of the digits of a two digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. Q29. If from twice the greater of two numbers, 20 is subtracted, the result is the other number. Also, if from twice the smaller number, 5 is subtracted, the result is the greater number. Find the two

numbers. .15,10Ans

Q30. If two digit number is obtained by either multiplying the sum of the digits by 8 and adding 1; or by multiplying the difference of the digits by 13 and adding 2. Find the numbers. How many such

numbers are there? .41Ans

Q31. A number consists of three digits whose sum is 17. The middle one exceeds the sum of other two by 1. If the digits are reversed, the number is diminished by 396. Find the number. .692Ans

Q32. In a competitive examination, one mark is awarded for each correct answer while 1/2 mark is deducted for each wrong answer. Payal answered 120 questions and got 90 marks. How many questions did she answer correctly? Q33. In a unit test, the number of students passed and the number of those who failed were in the ratio 3:1. Had 8 more students appeared and 6 less passed, the ratio of passes to failures would have been

2:1. Find the number of students actually appeared in the test. .136Ans

Q34. A person can row a boat at the rate of 5km/hr in still water. He takes thrice as much time in going 40km upstream as in going 40km downstream. Find the speed of the stream. Q35. A boatman rows his boat 35km upstream and 55km downstream in 12hours. He can row 30km upstream and 44km downstream in 10hours. Find the speed of stream and that of the boat in still water. Hence find the total time taken by him to row 50km upstream and 77km downstream.

.3km/hr, 8km/hr, 17hrs.Ans

Page | 19

Q36. Neeraj covers 60km in 1

12

hours with the wind and 2hours against the wind. Find the speed of

Neeraj and also the speed of wind. .3km/hr, 5km/hrAns

Q37. The distance between Kritika’s school and metro station is 300m. She starts running from school towards metro station. While her classmate Renu starts running from metro station to their school. They meet after 4minutes. Had Kritika doubled her speed and Renu reduced her speed to third of her original speed, they would have met one minute earlier. Find their original speeds. .45km/hr, 30km/hrAns

Q38. Pawan chases Vineet who is 5km ahead. Vineet is travelling at a speed of 80km/hr and Pawan

chases at an average speed of 90km/hr. After how much time will they meet? .30minutesAns

Q39. A chemist has one solution which is 50% acid and a second solution which is 25% acid. How much of each should be mixed to make 10 litres of a 40% acid solution? Q40. 27 pencils and 31 rubbers together costs `85 while 31 pencils and 27 rubbers together costs `89. Find the cost of 2 pencils and 1 rubber. .Ans Pencil: `2, Rubber: `1

Q41. In a function, if 10 guests are sent from room A to room B, the number of guests in rooms A and B are same. If 20 guests are sent from room B to room A, the number of guests in room A gets doubled the number of guests in room B. Find the number of guests in both the rooms in the beginning. .100, 80Ans

Q42. In a function, Sandeep wished to give `21 to each person present and found that he fell short of `4 so he distributed `20 to each and found that `1 were left over. How much money did he gave and how many persons were there? .Ans `101, 5

Q43. The area of a rectangle remains the same if its length is increased by 7cm and the breadth is decreased by 3cm. The area remains unaffected if length is decreased by 7cm and the breadth is increased by 5cm. Find the dimensions of the rectangle. .28 ,15Ans m m

Q44. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Prerna paid `27 for a book kept for 7 days while Sunayna paid `21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. .Ans `15, `3

Q45. A mobile company charges a fixed amount as monthly rental which includes 100 minutes free per month. And it charges a fixed amount thereafter for every additional minute. Abhilasha paid `433 for 370 minutes and Priyanka had to pay `398 for 300 minutes. Find the bill amount under the same plan, if Shivam uses 400minutes of calling. .Ans `448

Page | 20

Important facts and theorems:

01. If two figures are given, having the same shape but not necessarily the same size then, they are called the similar figures.

02. All the congruent figures are similar but the converse is not necessarily true.

03. Two polygons of the same number of sides are similar, if a) their corresponding angles are equal and b) their corresponding sides are in the same ratio.

04. Similar triangles: Two triangles are said to be similar, if their corresponding angles are equal and their corresponding sides are proportional i.e., the corresponding sides are in the same ratio.

05. Criterion for Similarity:

a) . . .A A A Similarity

,ABC PQR when A P B Q and C R

b) . .A A Similarity

ABC PQR when A P and B Q

c) . . .S A S Similarity

AB CA

ABC PQR when A P andPQ RP

d) . . .S S S Similarity

AB BC CA

ABC PQR whenPQ QR RP

.

06. #Basic Proportionality Theorem (B.P.T. or Thales Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

GIVEN: A ABC in which DE BC , and DE intersects AB in D and AC in E.

TO PROVE: AD AE

DB EC .

CONSTRUCTION: Join ,BE CD and draw EN BA and DM CA .

PROOF: As EN BA , therefore EN is the height of the triangles ADE and DBE.

Now, 1

2ar ADE AD EN

1

2Area of Base Height

and, 1

2ar DBE DB EN

1

2 ...( )1

2

AD ENar ADE ADi

ar DBE DBDB EN

Similarly, 1

2ar ADE AE DM and

1

2ar DEC EC DM

1

2 ...( )1

2

AE DMar ADE AEii

ar DEC ECEC DM

Since DBE and DEC are on the same base DE and between the same parallels DE and BC.

...( )ar DEC ar DBE iii

By (i), (ii) and (iii), we get

(By OP Gupta – 9650 350 480)

Chapter 04. EUCLIDIAN GEOMETRY & SIMILAR TRIANGLES

Page | 21

AD AE

DB EC . H.P.

Remarks: a) 1 1AD AE AD AE AB AC

DB EC DB EC DB EC

b) 1 1AD AE DB EC DB EC AB AC

DB EC AD AE AD AE AD AE .

07. Converse of Thales Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

08. #The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. GIVEN: Two triangles ABC and PQR such that ABC PQR .

TO PROVE:

2 2 2ar ABC AB BC CA

ar PQR PQ QR RP

.

CONSTRUCTION: Draw AL BC and PM QR .

PROOF: In triangles ABL and PQM, B Q ABC PQR

o90 L M

So, ABL PQM AA similarity criterion

...( )AB AL

iPQ PM

Also as ABC PQR Given

...( )AB BC CA

iiPQ QR RP

Now 1

2ar ABC BC AL and

1

2ar PQR QR PM

So,

1

2 ...( )1

2

BC ALar ABC BC ALiii

ar PQR QR PMQR PM

By (i), (ii) and (iii), we get

2ar ABC AB

ar PQR PQ

.

Hence,

2 2 2ar ABC AB BC CA

ar PQR PQ QR RP

. H.P.

09. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other. Here o, 90 ABC B and BD AC .

We have, ,ADB ABC BDC ABC and ADB BDC .

Page | 22

10. #Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides. GIVEN: A right angled triangle ABC in which o90 B .

TO PROVE: 2 2 2AC AB BC . CONSTRUCTION: From the vertex B draw BD AC . PROOF: In triangles ABC and ADB, we have o90 ABC ADB

and A A Common

So, ABC ADB AA similarity criterion

AB AC

AD AB

2 ...( )AB AD AC i

Now in triangles ABC and BDC, we have o90 ABC CDB

and C C Common

So, ABC BDC AA similarity criterion

BC AC

DC BC

2 ...( )BC DC AC ii

Adding eqs.(i) and (ii), we get 2 2AB BC AD AC DC AC

2 2AB BC AC AD DC

2 2 .AB BC AC AC 2 2 2AB BC AC

Hence, 2 2 2AC AB BC . H.P.

11. #Converse of Pythagoras Theorem: In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. GIVEN: A triangle ABC in which 2 2 2AC AB BC .

TO PROVE: o90 B . CONSTRUCTION: Construct a triangle PQR right angled at Q such that PQ = AB and QR = BC. PROOF: Since triangle PQR is right angled at Q so, we have

2 2 2PR PQ QR Pythagoras Theorem

2 2 2 ...( )PR AB BC i By construction

But 2 2 2 ...( )AC AB BC ii Given

2 2 ( ) ( )AC PR Using i and ii

...( )AC PR iii

Now in ABC and PQR , we have

AB PQ By construction

Page | 23

BC QR By construction

and ( )AC PR From iii

ABC PQR SSS Congruence

Therefore, . . . .B Q C P C T

But o90 Q By construction

So, o90 B H.P.

12. Statement of Baudhayan Theorem: The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e. length and breadth).

The theorems marked with “ # ” are supposed to be learnt properly by you since their proof is supposed to be asked in the examinations!

Important questions from the NCERT book: Examples 2, 3, 6, 7, 8, 9, 10, 11, 12, 13, 14. Also from exercises:

Ex6.2 Q1,Q2, Q4 to Q10; Ex6.3 All the questions; Ex6.4 All the questions; Ex6.5 Q2 t o Q16.

Page | 24

01. Introduction with the Trigonometric ratios: Consider a ABC right angled at B , such that θ ACB . We know AB = Perpendicular (p), BC = Base (b), CA= Hypotenuse (h).

Then the trigonometric ratios are expressed as:

sin θ AB p

CA h, cosθ

BC b

CA h, tanθ

AB p

BC b,

cosecθ CA h

AB p, secθ

CA h

BC b, cotθ

BC b

AB p.

B b C

Please note that the angle (i.e. the angle in consideration) is always made by the base and the hypotenuse.

02. Trigonometric ratios for Complimentary angles: Consider a right angle ABC such that o o90 , θ 90 θ B C A ,

A

B C

• osin 90 θ cosθ , ocos 90 θ sinθ

• otan 90 θ cot θ , ocot 90 θ tanθ

• osec 90 θ cosec θ , ocosec 90 θ secθ .

03. Basic Trigonometric Identities: In ABC which is right angled at B , we have

sin θ p

h and cosθ

b

h.

p h

B C b

2

2 2

2sin θ cos θ

hBy Pythagoras Theorem

h

2 2sin θ cos θ 1 .

Similarly, other trigonometric identities which have been given below, can be obtained:

• 2 2 2 2sin θ cos θ 1 sin θ 1 cos θ and 2 2cos θ 1 sin θ

• 2 2 2 2sec θ 1+tan θ tan θ sec θ 1 and 2 2sec θ tan θ 1

• 2 2 2 2cosec θ 1 cot θ cot θ cosec θ 1 and 2 2cosec θ cot θ 1

A

h p

(By OP Gupta – 9650 350 480)

Chapter 05. INTRODUCTION TO TRIGONOMETRY (Trigonometric Ratios & Identities)

θ

θ

(90o

– θ)

so that, we have:

o osin sin 90 θ cosθ cos sin 90 θ cosθ BC

A CCA

.

Similarly, other complimentary angle- trigonometric relations which have been given below, can be obtained:

A

θ

Squaring and then adding these Trigonometric ratios, we get

2 2

2 2sin θ cos θ

p b

h h

2 2

2 2

2sin θ cos θ

p b

h

Page | 25

04. Relationship between trigonometric ratios:

• sinθ

tanθcosθ

= • 1

tanθcotθ

= • 1

cotθtanθ

=

• cosθ

cotθsinθ

= • 1

cosecθsinθ

= • 1

secθcosθ

=

05. Trigonometric Ratios of Standard angles:

Angle in Degree

T – Ratios

o0 o30 o45 o60 o90

sin 0 1

2

1

2 3

2 1

cos 1 3

2

1

2

1

2 0

tan 0 1

3 1

3 Not Defined

cosec Not Defined 2 2 2

3 1

sec 1 2

3 2 2 Not Defined

cot Not Defined 3 1 1

3 0

06. Geometrical Method to find the value of Trigonometric Ratios of Standard Angles:

a) Trigonometric Ratios of o45

Consider a right triangle ABC with right angle at B such that o45 A . C

a

Assume AB BC a . Then by Pythagoras Theorem, B a A

2 2 2AC AB BC

2 2 2AC a a

2 22AC a

2AC a

Therefore in triangle ABC, we have o45 A , Base AB a , Perpendicular BC a and

Hypotenuse 2 AC a .

o 1sin 45

2 2

BC a

AC a, o 1

cos 452 2

AB a

AC a,

otan 45 1 BC a

AB a, o

o

1cot 45 1

tan 45 ,

o

o

1cosec45 2

sin 45 , o

o

1sec45 2

cos 45 .

Then, o180 A B C

o o o45 90 180 C

o45 C

A C

AB BC

o45

Page | 26

b) Trigonometric Ratios of o30 and o60 Consider an equilateral triangle ABC with each side of length 2a . We know that each angle in

equilateral triangle is of o60 . So all the angles in ABC are also of measure o60 . Let AD be the perpendicular from vertex A on BC . As ABC is equilateral, so AD is the bisector of A and D is the mid-point of BC .

BD DC a and o30 BAD .

A

2a 2a

B a D a C I. Trigonometric Ratios of 30o

In the right triangle ABD , we have

Base 3AD a , Perpendicular BD a , Hypotenuse 2AB a and o30 BAD .

o 1sin 30

2 2

BD a

AB a, o 3 3

cos302 2

AD a

AB a,

o 1tan30

3 3

BD a

AD a, o

o

1cot 30 3

tan 30 ,

o

o

1cosec30 2

sin30 , o

o

1 2sec30

cos30 3 .

II. Trigonometric Ratios of 60o

In the right triangle ABD , we have

Base BD a , Perpendicular 3AD a , Hypotenuse 2AB a and o60 ABD .

o 3 3sin 60

2 2

AD a

AB a, o 1

cos602 2

BD a

AB a,

o 3tan 60 3

AD a

BD a, o

o

1 1cot 60

tan 60 3 ,

o

o

1 2cosec60

sin60 3 , o

o

1sec60 2

cos60 .

Important questions from the NCERT book: Examples 2, 5, 8, 10, 11, 12, 15. Also from exercises: Ex8.1 Q2, Q6, Q10;

Ex8.2 Q1(iii), (v), Q3; Ex8.3 Q2 to Q7; Ex8.4 Q4(ii), Q5(iii), (v), (vii), (viii),

(ix), (x).

BRAIN TEASERS

Q01. If sin θ + cosθ = x then, prove that sin6 θ + cos6 θ = 2 24 3(x 1)

4

.

Q02. If sinα a sinβ= and tanα b tanβ= , then prove that 2

2

2

a 1cos α

b 1

= .

Q03. If 8 tanθ = 3 cos θ , 0 < θ < 180o, then find the value of sin θ . Q04. If cosec θ – sin θ = l and sec θ – cos θ = m, show that l2m2 (l2 + m2 + 3) = 1.

Q05. If a cos θ – b sin θ = c, then prove that a sin θ + b cosθ = 2 2 2a b c .

Q06. If tanθ secθ l +m n and tanθ secθ l m n , then show that

2 2

1

nl ln m n n m

ml lm lm ml.

Q07. If cosec θ – sin θ = a3 and sec θ – cos θ = b3 then, prove that 2 2 2 2a b (a b ) 1 .

o60 o60

o30

Thus in ABD , D is a right angle, hypotenuse 2AB a and BD a . So by Pythagoras Theorem, we have

2 2 2AB AD BD

2 2 22a AD a

3AD a

Page | 27

IMPORTANT TERMS & FORMULAE

01. Mean of Grouped Data

a) Direct Method:

i i

i

f xx

f

where ix represents the variates (class- marks) and if their corresponding frequencies.

b) Assumed Mean Method:

i i

i

f dx a

f

where a is the assumed mean and i id x a is the deviation of a from each of the variates ix .

c) Step Deviation Method:

i i

i

f ux a h

f

where a is the assumed mean and h is the class-width. Also ii

x au

h

.

Assumed mean and step deviation methods are known as Short Cut Methods too. 02. Mode of Grouped Data

1 0

1 0 22

f fMode l h

f f f

where l lower limit of the modal-class,

h size of the class interval,

1f frequency of the modal-class,

0f frequency of the class preceding the modal-class,

2f frequency of the class succeeding the modal-class.

Note that the Modal-class is that class which has got the maximum frequency. 03. Median of Discrete Frequency Distribution

STEP1- Find the cumulative frequency.

STEP2- Find N if and then N

2.

STEP3- Observe the cumulative frequency which is just greater than the 2

N value and find out the

correspo corresponding value of the variable x .

STEP4- The value of the variable obtained in the STEP3 is the required Median.

The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.

(By OP Gupta – 9650 350 480)

Chapter 06. DATA ANALYSIS & STATISTICS

Page | 28

04. Median of Grouped Data or Continuous Frequency Distribution

NC

2

Median l hf

where l lower limit of the median- class,

N if number of observations,

C cumulative frequency of the class preceding the median class, f frequency of the median- class,

h class- size. The median class is obtained by marking that class- interval which corresponds to the

cumulative frequency which is just greater than the 2

N value.

05. Empirical Relationship between the three measures of Central Tendency

3 2 Median Mode Mean .

06. Graphical Representation of Cumulative Frequency Distribution

A curve that represents the cumulative frequency distribution of grouped data is called an ogive or cumulative frequency curve. The two types of Ogives are more than type ogive and less than type ogive.

An ogive representing a cumulative frequency distribution of ‘more than’ type is called a more than ogive. An ogive representing a cumulative frequency distribution of ‘less than’ type is called a less than ogive. Ogives can be used to find the median of a grouped data. The median of grouped data can be obtained graphically by plotting the Ogives of the less than type and more than type and locate the point of intersection of both the Ogives. The x-coordinate of the point of intersection of two Ogives gives the median of the grouped data. X LOWER LIMITS

An Ogive representing a Cumulative Frequency distribution of ‘more than’ type

CU

MU

LA

TIV

E F

RE

QU

EN

CY

5

Y

60-

55-

50-

45-

40-

35-

30-

25-

20-

15-

10-

5 -

O

5

-

15-

25

-

35-

45

-

55-

65

-

75-

85

-

95-

105-

Scale On x-axis: 2cm = 10units On y-axis: 1cm = 5units

Page | 29

X UPPER LIMITS

An Ogive representing a Cumulative Frequency distribution of ‘less than’ type X

CLASS LIMITS

An Ogive representing a ‘more than’ and ‘less than’ type ogives

to be used to find Median

CU

MU

LA

TIV

E F

RE

QU

EN

CY

5

Y

60-

55-

50-

45-

40-

35-

30-

25-

20-

15-

10-

5 -

O

5

-

15-

25

-

35-

45

-

55-

65

-

75-

85

-

95-

105-

Median = 62

Scale On x-axis: 2cm = 10units On y-axis: 1cm = 5units

CU

MU

LA

TIV

E F

RE

QU

EN

CY

5

Y

60-

55-

50-

45-

40-

35-

30-

25-

20-

15-

10-

5 -

O

5

-

15-

25

-

35-

45

-

55-

65

-

75-

85

-

95-

105-

Scale On x-axis: 2cm = 10units On y-axis: 1cm = 5units

(Exclusively for Summative Assessment I of class X)

Page | 31

01.NUMBER SYSTEM & REAL NUMBERS

Mathematics is incredibly wonderful, but to get to the point where you can see how great it is, you must learn your basics well!

Q01. The smallest prime number is: (a) 0 (b) 1 (c) 2 (d) 3 Q02. The sum of first five prime numbers is: (a) 26 (b) 15 (c) 39 (d) 28 Q03. Total prime numbers between 1 and 100 are: (a) 31 (b) 25 (c) 22 (d) 20 Q04. The unit’s digit obtained on simplifying 207×781×39×94 is: (a) 9 (b) 1 (c) 7 (d) 2

Q05. The number 3 is a/an: (a) integer (b) rational number (c) irrational number (d) None of these

Q06. The HCF and LCM of 6, 72 and 120 is: (a) 8, 360 (b) 6, 340 (c) 6, 360 (d) None of these Q07. The total number of even prime numbers is: (a) 0 (b) 1 (c) 2 (d) None of these Q08. 22/7 is a: (a) prime number (b) an integer (c) a rational number (d) an irrational number

Q09. The sum of two numbers is 37 and their product is 342. The numbers are: (a) 18, 19 (b) 23, 14 (c) 24, 13 (d) 28, 9

Q10. A number is as bigger than 22 as much it is smaller than 72. The number is: (a) 92 (b) 47 (c) 24 (d) None of these

Q11. If HCF and LCM of two numbers are 4 and 9696, then the product of two numbers is: (a) 9696 (b) 24242 (c) 38784 (d) 4848

Q12. 5 2 3 is: (a) a natural number (b) an integer (c) a rational number (d) an irrational number

Q13. If 3 2 6 9

9 49 7

7 81 9

x

then, the value of x is:

(a) 12 (b) 9 (c) 8 (d) 6

Q14. The number .211 2111 21111 211111… is a: (a) terminating decimal (b) non-terminating repeating decimal (c) non-terminating decimal which is non-repeating (d) None of these

Q15. If 32nm , where m and n are positive integers, then the value of mnn is:

(a) 32 (b) 25 (c) 105 (d) 255

Q16. Any one of the numbers , 2a a and 4a is a multiple of:

(a) 2 (b) 3 (c) 5 (d) 7

Q17. If p is a prime number and p divides 2k , then p divides:

(a) 22k (b) k (c) 3k (d) None of these

Q18. If the HCF of 85 and 153 is expressible in the form of 85 153n then, the value of n is: (a) 3 (b) 2 (c) 4 (d) 1

Q19. Given that LCM (91, 26) = 182 then, HCF (91, 26) is: (a) 13 (b) 26 (c) 7 (d) 9

Q20. Out of the four numbers (i)3

15

5

(ii) 2.123123 (iii) 2.123123... (iv) 2 3 2 2 3 2 , the

rational number is: (a) i (b) ii (c) iii (d) iv

Q21. 7 11 13 6 is: (a) a prime number (b) a composite number (c) an even number (d) None

Q22. If 5nnp a , for np to end with the digit zero _______a for any natural number n:

Page | 32

(a) any natural number (b) an odd number (c) any even number (d) None

Q23. HCF is always: (a) multiple of LCM (b) factor of LCM (c) divisible by LCM (d) Option a and c both

Q24. In Euclid’s division lemma where a bq r and a, b are positive integers, which one is correct:

(a) 0 r b (b) 0 r b (c) 0 r b (d) 0 r b

Q25. If p is a positive rational number which is not a perfect square then, 3 p is:

(a) an integer (b) rational number (c) irrational number (d) Option (a) and (c) both

Q26. 2 5 is: (a) a rational number (b) a natural number (c) equal to zero (d) an irrational number

Q27. The number given below which always ends with the digit 6 for all natural nos. n is:

(a) 4n (b) 2n (c) 6n (d) 8n

Q28. 2 5 5 2 is:

(a) a rational number (b) a natural number (c) equal to zero (d) an irrational number

Q29. Let 2 32 5

7x

be a rational number. Then x has decimal expansion which terminates:

(a) after four places of decimal (b) after three places of decimal (c) after two places of decimal (d) after five places of decimal

Q30. The decimal expansion of 63

72 175 is:

(a) terminating (b) non-terminating (c) non-terminating and repeating (d) None of these Q31. 2.35 is: (a) an integer (b) a rational number (c) an irrational no. (d) None of these Q32. Which one is not a natural number:

(a) 25 (b) 121 (c) 100 (d) 7

Q33. (35 – 7 3)2 is equal to:

(a) (2115 –192) (b) (2115–192) (c) (192+2115) (d) (192 – 4215) Q34. Which number is natural? (a) –8 (b) 2 (c) 0 (d) 1/2 Q35. Zero is: (a) natural no. (b) whole no. (c) non divisible no. (d) divisible no. Q36. Which of the following is a rational number having terminating decimal expansion: (a) 36/100 (b) 41/8 (c)329/400 (d) All these Q37. Which one of the following statement is true? (a) Every natural number is a whole number (b) Every integer is a whole number (c) Every rational number is a whole number (d) None of these

Q38. 6 5 2 5 is equal to: (a) 70 (b) 50 (c) 60 (d) 55

Q39. LCM of 25, 35 and 105 is: (a) 555 (b) 565 (c) 575 (d) None of these

Page | 33

02.POLYNOMIALS & ALGEBRAIC EXPRESSIONS

A mathematician who is not also a poet will never be a complete mathematician. In fact, pure mathematics is, in its way, the poetry of logical ideas!

Q01. The quadratic polynomials with the sum and the products of its zeroes as 1/4 and –1 respectively, is: (a) 4x2 + x + 1 (b) 4x2 + x + 4 (c) 4x2 +x –1 (d) 4x2 – x –4

Q02. If 2

2

1102x

x , then the value of

1xx

is:

(a) 8 (b) 10 (c) 12 (d) 13 Q03. If p(x)= 3x3+x2+2x+5 is divided by g(x)=x2+2x+1, then the remainder will be: (a) 8x+10 (b) 9x+10 (c)10x+10 (d)11x+10

Q04. The quadratic polynomial, the sum and product of whose zeroes are –3 and 2 respectively, is: (a) x2 + 3x +2 (b) x2 –3x +2 (c) x2 + 3x–2 (d) –x2 + 3x +2

Q05. The zeroes of quadratic polynomial t2–15 are:

(a) 15, 15 (b) 15, 12 (c) 15, 12 (d) 15, 15

Q06. A quadratic polynomials, the sum and product of whose zeroes are –1/4 and 1/4 respectively, is: (a) 4x2 +x+1 (b) x2–3x+2 (c) x2 +3x–2 (d) None of these

Q07. If 1

3xx

, then 2

2

1x

x is equal to:

(a) 82/9 (b) 10/3 (c) 7 (d) 11

Q08. If x1/3 + y1/3+ z1/3 =0, then: (a) x + y + z = 0 (b) x + y + z =3xyz (c) (x + y + z)3 =27xyz (d) x3+y3+z3 =0

Q09. If p(x) = 3x2–5x, then p(2) =_______:

(a) 2 (b) 3 (c) 0 (d) None of these Q10. The quadratic polynomials whose zeroes are 3/5 and –1/2, is: (a) 10x2–x–3 (b)10x2+x–3 (c) 10x2–x+3 (d) None of these Q11. If α and β are the zeroes of 2x2 + 5x – 10, then the value of αβ is: (a) –5/2 (b) 5 (c) –5 (d) 2/5

Q12. A real number is a zero of the polynomial f x if:

(a) 0f (b) 0f (c) 0f (d) 0f

Q13. The zeroes of a polynomial f x are the coordinates of the points where the graph of y f x

intersects: (a) X-axis (b) Y-axis (c) Origin (d) None

Q14. If is a zero of f x then, _________ is one of the factors of f x :

(a) 2x (b) x (c) x (d) 2x

Q15. If y a is factor of f y then, _______ is a zero of f y :

(a) y (b) a (c) 2 y (d) a

Q16. Out of the followings, the incorrect statement for a quadratic polynomial is: (a) no real zeroes (b) two equal real zeroes (c) two distinct zeroes (d) three real zeroes

Q17. A cubic polynomial x f y cuts Y-axis at atmost:

(a) one point (b) two points (c) three points (d) four points

Q18. Graph of 2ax bx c intersects X-axis at two distinct points if:

(a) 2 4 0b ac (b) 2 4 0b ac (c) 2 4 0b ac (d) 2 4 0b ac

Q19. Polynomial 2 1f x x has ___________________ zeroes:

(a) only one real (b) no real (c) only two real (d) one real and one non-real Q20. If P is the sum of zeroes and S is product then, the corresponding quadratic polynomial may be:

(a) 2 S Px x (b) 2 S Px x

(c) 2 P Sx x (d) 2 +S Px x

Q21. If zeroes of the quadratic polynomial 2ax bx c are reciprocal of each other, then:

Page | 34

(a) a=c (b) a=b (c) b=c (d) a+c=0

Q22. If the sum of the zeroes of quadratic polynomial 23 6 x kx is 3, the value of k is:

(a) 3 (b) –3 (c) 6 (d) None

Q23. The other two zeroes of 3 28 19 12 x x x if one of its zero is unity, is: (a) –3, 4 (b) –3, –4 (c) 3, –4 (d) None Q24. Quadratic polynomial, the sum and product of whose zeroes are –3 and 2, is:

(a) 2 3 2 x x (b) 2 3 2 x x (c) 2 3 2 x x (d) 2 3 2 x x

Q25. The third zero of the polynomial 3 27 2 14 x x x , if two of its zeroes are 2 , is: (a) 7 (b) –7 (c) 14 (d) –14

Q26. If 5

3 and

5

3 are two zeroes of 4 3 23 6 2 10 5 x x x x , then its other two zeroes are:

(a) –1, –1 (b) 1, –1 (c) 1, 1 (d) 3,–3

Q27. If 3 and 3 are two zeroes of 4 3 23 9 6x x x x , then its other two zeroes are: (a) –1, –2 (b) 1, 2 (c) –1, 2 (d) None

Q28. If ,a b a and a b are zeroes of the polynomial 3 23 1 x x x then, the value of a b is:

(a) 1 2 (b) 1 2 (c) 1 2 (d) 1 3

Q29. If a real number a is zero of the polynomial f x , then:

(a) 1 f a (b) 1 f a (c) 0f a (d) 1f a

Q30. If the product of two of the zeroes of polynomial 3 22 9 13 6 x x x is 2, the third zero of the polynomial is: (a) –1 (b) –2 (c) 3/2 (d) –3/2

Q31. If –4 is a zero of the polynomial 2 2 2 x x k then, the value of k is:

(a) 3 (b) 9 (c) 6 (d) –9

Q32. If one zero of 2 24 13 4 k x x k is reciprocal of the other, then k is:

(a) 2 (b) –2 (c) 1 (d) –1

Q33. If the sum of the zeroes of polynomial 3 22 3 4 5 p x x kx x is 6 then, the value of k is:

(a) 2 (b) 4 (c) –2 (d) –4

Q34. If one of the zero of polynomial 25 13 p x x x m is reciprocal of the other then, value of m is:

(a) 2 (b) 4 (c) –2 (d) None of these

Q35. For the polynomial 4 3 22 5 8 x x x x , the maximum number of zeroes it has are: (a) 2 (b) 4 (c) 3 (d) 0

Q36. For the polynomial 22 5 4x x , the maximum number of zeroes (real or imaginary) it has are:

(a) 2 (b) 4 (c) 3 (d) None of these Q37. What is the difference between the values of the polynomial 7x-3x2+7 at x = 1 and x =2?

(a) –2 (b) +2 (c) 3 (d) None of these Q38. What is the remainder when the polynomial 3x4–4x3–3x –1 is divided by x–1?

(a) 5 (b) –5 (c) 8 (d) 6 Q39. The value of k, if x–1 is a factor of 4x3 + 3x2 – 4x + k is:

(a) –3 (b) 3 (c) 2 (d) 5 Q40. The product of (3–2x) (3+2x) is: (a) 8– 2x2 (b) 9 – 4x2 (c) 9–2x2 (d) 9–16x2

Q41. The number of zeroes which a polynomial of degree n can have is: (a) at most n (b) exactly n (c) n+1 (d) Can’t say

Q42. 1 1

7 79 9

x xy y

is equal to:

(a) 2

2

139

81x

y

(b) 2

2

149

81x

y

(c) 2

2

149

18x

y

(d) None of these

Q43. The term that should be added to 4x2 + 12xy to form a perfect square is: (a) 9y (b) 9xy (c) 9y2 (d) 4y2

Q44. The sum of the roots of the equation ax2+bx+c=0:

(a) c/a (b) –b/a (c) b/c (d) –c/a Q45. If the roots of the equation px2 + qx + 3 = 0 are reciprocal to each other, then:

Page | 35

(a) q = 3 (b) p = 3 (c) p – q = 0 (d) p + q = 0

Q46. If 4

4

1322x

x , then

1xx

is equal to:

(a) 4 (b) 6 (c) 5 (d) 2 Q47. If (x–2) is a factor of 2x3–6x2+5x+k, then the value of k is:

(a) –2 (b) 10 (c) 15 (d) None of these

Q48. If α, β, be the zero of the polynomials p(x) such that α+β+ =3, αβ+β+α= –10 and αβ= –24 then, p(x) is: (a) x3+3x2–10x+24 (b) x3+3x2+10x–24 (c) x3–3x2–10x+24 (d) None of these

Q49. If t2–4t+4 = 0, then the value of 3

3

1t

t

is:

(a) 8/56 (b) 8/65 (c) 56/8 (d) None of these Q50. The factors of x4+4 are: (a) (x2+2) (x2–2) (b) (x2+2x+2) (x2–2x+2) (c) (x+2) (x–2) (d) Not possible Q51. If (x+2) is a factor of p(x) =2x2+3x+k, then the value of k is:

(a) 2 (b) –2 (c) –14 (d) 14

Q52. The remaining zeroes of 3x4 –6x3–2x2–10x–5, if two of its zeroes are given as 5 / 3 , are: (a) –2, –1 (b) 2, –1 (c) –1, +1 (d) None of these Q53. If (x–4) is the HCF of (x2–x–12) and (x2–mx–8), then the value of m is:

(a) 0 (b) 1 (c) 2 (d) 6 Q54. The LCM of (x2+x –6) and 4(4 –x2) is: (a) 4(x+3) (x–2) (x+2) (b) –4(x+3) (x–2) (x+2) (c) –4(x–3) (x–2) (x–2) (d) 4(x–3) (x+2) (x+2)

Q55. If 1 1

2xx

, then the value of 2

2

14 x

x

is:

(a) 2 (b) 5 (c) 8 (d) 9 Q56. If x100+2x99+k, is divisible by (x+1), then the value of k is:

(a) 1 (b) 2 (c) –3 (d) –2

Q57. If 4

4

14 x

x then

1xx

is equal to:

(a) –1, 2 (b) –1, +1 (c) 8, –1 (d) None of these

Q58. The value of 2

2

1xx

, when 1

10 xx

:

(a) 96 (b) 97 (c) 98 (d) 102 Q59. A quadratic polynomial f(x), is such that:

( ) 0, for 3 2

0, otherwise

f x x

Which of the following can be the polynomial f(x)?

(a) 2 6x x (b) 2 6x x (c) 2 6x x (d) 2 6x x

Page | 36

03. LINEAR EQUATIONS IN TWO VARIABLES One cannot escape the feeling that these mathematical formulas have an independent existence and an

intelligence of their own, that they are wiser than we are, wiser even than their discoverers...

Q01. The solutions of the equation 2x–y–5=0 are:

(a) 2, 1 x y (b) 2, 1x y (c) 1, 1 x y (d) 2, 1 x y

Q02. The sum of digit of a two digit number is 9. Also, 9 times this number is twice the number obtained by reversing the order of the digit. The number is:

(a) 20 (b) 16 (c) 18 (d) None of these

Q03. The system of equations kx –y =2 and 6x–2y =3 has a unique solution when:

(a) k = 0 (b) k ≠ 0 (c) k = 3 (d) k ≠ 3

Q04. A boat can row 1km with stream in 10minutes and 1km against the stream in 20minutes. The speed of the boat in still water is:

(a) 1.5km/hr (b) 3km/hr (c) 3.4km/hr (d) 4.5km/hr

Q05. A boat goes 24km upstream and 28km downstream in 6hours. It goes 30km upstream and 21km downstream in 6hours and 30minutes. The speed of the boat in still water is:

(a) 4km/hr (b) 6km/hr (c) 10km/hr (d) 14km/hr

Q06. Point (4, 3) lies on the line:

(a) 3x + 7y =27 (b) 7x + 2y = 47 (c) 3x +4y = 24 (d) 5x – 4y =1

Q07. The speed of train 150m long is 50km/hr. The time it will take to cross a platform 600m long is:

(a) 50sec (b) 54sec (c) 60sec (d) None of these

Q08. The graph of an equation y = –3 is a line which will be:

(a) parallel to x-axis (b) parallel to y-axis (c) passing through origin (d) on x-axis

Q09. The value of k for which kx + 2y = 5 and 3x + y =1 have unique solution, is:

(a) k = –1 (b) k ≠ 6 (c) k = 6 (d) k =2

Q10. The graph of the equation x–y =0 is:

(a) parallel to x-axis (b) parallel to y-axis

(c) passing through origin (d) None of these

Q11. Five years hence, father’s age will be three times the age of his daughter. Five years ago, father was seven times as old as his daughter. Their present ages are:

(a) 20 years, 10 years (b) 40 years, 20 years (c) 40 years,10 years (d) 30 years, 10 years

Q12. In a two digit number, the unit’s digit is twice the ten’s digit. If 27 is added to the number, the digits interchange their places. The number is:

(a) 22 (b) 46 (c) 36 (d) 63

Q13. The pair of equations 16

3 4 18, 4 243

x y x y has:

A. no solution B. unique solution

C. infinitely many solution D. can’t say

Q14. The pair of equations 3 2 5, 2 3 7 x y x y has:

A. no solution B. one solution C. many solutions D. two solutions

Q15. If the pair of equation 9

2 3 7, 122

x y kx y have no solution, then value of k is:

A. 2/3 B. 3/2 C. 3 D. –3

Page | 37

Q16. The equations 0.9 x y and 11

2x y

have the solution:

A. 5, 1x y B. 2.3 & 3.2x y C. 3.2 & 2.3x y D. 3, 2x y

Q17. If 2 2 bx ay a b and 0 ax by then, the value of x y is:

A. b a B. a b C. 2 2a b D. 2 2b a

Q18. If 2 3 0, 4 3 0 x y x y then, x y equals:

A. 0 B. –1 C. 1 D. 2

Q19. If ax by b a and 0 bx a y then, value of x y is:

A. a b B. a b C. a b D. b a

Q20. If 2 3

13 x y

and 5 4

2 x y

then, x y equals:

A. 1/6 B. –1/6 C. 5/6 D. –5/6

Q21. If 31 43 117 x y and 43 31 105 x y then, the value of x y is:

A. –3 B. 1/3 C. –1/3 D. 3

Q22. If 19 17 55 x y and 17 19 53 x y then, the value of x y is:

A. –3 B. 1/3 C. 3 D. 5

Q23. If 0.82 x

y and 7

10

2

y

x then, the value of x y is:

A. 1 B. 0.6 C. –0.8 D. 0.5

Q24. If 6,k is a solution of the equation 3 22 x y then, the value of k is:

A. –4 B. 4 C. 3 D. –3

Q25. If 2

3 5 1, 4

xx y

x y then, the value of x y is:

A. 3 B. –3 C. 1/3 D. –1/3

Q26. If the pair of equation 2 3 5 x y and 10 15 2 x y k represent two coincident lines then, the

value of k is:

A. –25/2 B. –5 C. 25/2 D. –5/2

Q27. Rs.4900 was divided among a group of 150 children. If each girl gets Rs.50 and each boy gets Rs.25 then, the number of boys in the group is:

A. 100 B. 102 C. 104 D. 105

Q28. Every linear equation in two variables has ___________ solution(s).

A. no B. one C. two D. infinitely many

Q29. 1 1 1

2 2 2

a b c

a b c is the condition for:

A. intersecting lines B. parallel lines C. coincident lines D. none of these

Q30. For a pair of equation to be consistent and dependent, the pair must have:

A. no solution B. unique solution

C. infinitely many solution D. none of these

Q31. Graph of every linear equation in two variables represents a _________.

A. point B. straight line C. curve D. triangle

Q32. Each point on the graph of pair of two lines is a common solution of the lines in case of:

A. infinitely many solution B. only one solution

C. no solution D. none of these

Page | 38

Q33. One of the common solution of ax by c and y-axis is:

A. 0, /c b B. 0, /b c C. / ,0c b D. 0, /c b

Q34. If the value of x in the equation 2 8 12 x y is 2 then, the corresponding value of y will be:

A. –1 B. 1 C. 0 D. 2

Q35. The pair of linear equations is said to be inconsistent if they have:

A. only one solution B. no solution

C. infinitely many solution D. both a and c

Q36. On representing x=a and y=b graphically, we get:

A. parallel lines B. coincident lines

C. intersecting lines at ,a b D. intersecting lines at ,b a

Q37. How many real solutions of 2 3 5 x y are possible:

A. no B. one C. two D. infinitely many

Q38. The value of k for which the system of equation 3 2 5, 2 x y x ky has a unique solution is:

A. 2/3k B. 2/3k C. 2/3 k D. 2/3 k

Q39. If the lines represented by the pair of linear equations 2 5 3, 2 2 1 2 x y k y k x k are

coincident then, the value of k is:

A. –3 B. 3 C. 1 D. –2

Q40. The coordinates of the point where x-axis and the line 12 3 x y

intersect, are:

A. (0,3) B. (3,2) C. (2,0) D. (0,2)

Q41. Graphically 2 0x represents a line:

A. parallel to x-axis at a distance 2 units from x-axis

B. parallel to y-axis at a distance 2 units from y-axis

C. parallel to x-axis at a distance 2 units from y-axis

D. parallel to y-axis at a distance 2 units from x-axis

Q42. If ax by c and lx my n has unique solution then the relation between the coefficients will

be of the form:

A. am lb B. am lb C. ab lm D. ab lm

Q43. The value of a for which 3,a lies on 2 3 5 x y :

A. 1/3 B. 3 C. –1/3 D. None of these

Q44. If 2x–y=8 and 2x+y=64, then value of x and y will be:

(a) 9/2, 3/2 (b) –9/2, 3/2 (c) 9/2, –3/2 (d) 3, 2

Q45. On solving x–y=3, x+y=5, we have value of y as:

(a) 1 (b) 2 (c) 3 (d) 4

Q46. The solution of the equations 7x–2y =3 & 11x–1.5y =8 is:

(a) x =2, y =1 (b) x=1, y =2 (c) x = –1, y =2 (d) None of these

Q47. If 3x–y=9 and 3x+y=81, then value of y is:

(a) 1 (b) 2 (c) 3 (d) None of these

Q48. If 1 is added in numerator and denominator then a fraction changes to 4. If 1 is subtracted from the numerator and denominator, fraction changes to 7. Numerator of the fraction is:

(a) 2 (b) 3 (c) 7 (d) 15

Q49. If system of equations a1x+b1y+c1=0 and a2x+b2y+c2=0 has infinitely many solutions, then:

Page | 39

(a) 1 1

2 2

a b

a b (b) 1 1 1

2 2 2

a b c

a b c (c) 1 1 1

2 2 2

a b c

a b c (d) None of these

Q50. The value of y obtained on solving the equations 2 2 8 x y x y is:

(a) 0 (b) 1/4 (b) 1/2 (d) 3/4

Q51. The value of k for which the system of equation 2x+3y=5 & 4x+ky=10 has an infinite number of solutions, is:

(a) 1 (b) 3 (c) 6 (d) 0

Q52. Half the perimeter of a rectangular garden, whose length is 4m more than its width is 36m. The dimensions of the garden are:

(a) l=20m; b=16m (b) l=16m; b=20m (c) l=24m; b =20m (d) l=30m; b=16m

Q53. A system of two simultaneous linear equations in two variables is inconsistent, if their graphs:

(a) are parallel (b) are coincident (c) intersect at one point (d) None of these

Q54. Ritu can row downstream 20km in 2hours, and upstream 4km in 2hours. Her speed of rowing in still water and the speed of the current respectively are:

(a) 4km/h, 4km/h (b) 6km/h, 4km/h (c) 6km/h, 6km/h (d) 4km/h, 6 km/h

Q55. A boat is rowed downstream at 15.5km/h and upstream at 8.5km/h. The speed of the stream is:

(a) 3.5km/h (b) 5.75km/h (c) 6.5km/h (d) 7km/h

Q56. On solving 3x+y=81 and 81x–y =3, we observe that:

(a) No solution (b) 1 1

2 , 12 2

x y

(c) x =2, y = 2 (d) 1 7

2 , 18 8

x y

Q57. The sum of two digits of a two digits number is 12. If the digits are reversed, then the number so formed exceeds the original number by 18. The original number is:

(a) 64 (b) 56 (c) 79 (d) 57

Q58. If 6 12

7 x y

and 2 3

2 x y

then, the solution is:

(a) 6, 12 (b) 2, 4 (c) 2, 3 (d) None of these

Page | 40

04. TRIGONOMETRIC RATIOS & IDENTITIES

Black holes result from God dividing the universe by zero.

Q01. If sinθx r and cosθy r then, the value of 2 2x y is:

(a) r (b) r 2 (c) 1/ r (d) 1

Q02. The value of o ocosec70 sec 20 is:

(a) 0 (b) 1 (c) 90 (d) 50

Q03. If 3secθ 5 0 then, cot θ is equal to:

(a) 5/3 (b) 4/5 (c) 3/4 (d) 3/5

Q04. If oθ 45 then, secθcot θ cosecθ tan θ is:

(a) 0 (b) 1 (c) 2 2 (d) 2

Q05. If osin(90 θ)cosθ 1 and, θ is an acute angle then θ is:

(a) 90o (b) 60o (c) 30o (d) None of these

Q06. Triangle TRY is a right angled isosceles triangle then, cosT+ cosR+ cosY is:

(a) 2 (b) 2 2 (c) 1+ 2 2 (d) 1 [1/ 2]

Q07. If triangles ABC and PRT are similar such that o90 C R and 3

5

AC

AB then, sinT is:

(a) 3/5 (b) 5/3 (c) 4/5 (d) 5/4

Q08. If 2 o 2 o o7sec 62 7cot 28 7sec0 k then, the value of k is:

(a) 1 (b) 0 (c) 7 (d) 1/7

Q09. The value of o ocot θ sin 90 θ cos 90 θ is:

(a) cot θ (b) 2cos θ (c) 2cot θ (d) 2cot θcos θ

Q10. 2

sin θ

1 sin θ can also be written as:

(a) cot θ (b) sin θ (c) sin θ

cos θ (d) tanθ

Q11. If

2 o 2 o o

2 o 2 o

sin 20 sin 70 sec60

2 cos 69 cos 21

k then, the value of k is:

(a) 1 (b) 2 (c) 3 (d) 4

Q12. 1+tan2 equals:

(a) sec (b) sec2 (c) sec2 (d) cot2

Q13. If cosec =13/12, then

(a) tan =12/5 (b) tan = –5/12 (c) tan =12/25 (d) tan = ±12/25

Q14. cot + tan equals:

(a) cosec sec (b) sin sec (c) cos tan (d) sin2

Q15. cos 1º. cos2º.cos3º….cos180º =_____?

(a) 1 (b) –1 (c) 0 (d) None of these

Q16. If 1

sin A B2

and 1

cos A B2

then, A and B will be:

Page | 41

(a) 15º, 45º (b) 45º, 15º (c) 45º, 45º (d) 30º, 60º

Q17. If sin + sin2=1, then the value of cos2 +cos4 will be:

(a) 1 (b) 2 sin2 (c) 1+2 sin2 (d) Can’t be determined

Q18. If osin A cosA 2 cos(90 A) , then cot A is equal to:

(a) 1 2 (b) 2 1 (c) 2 1 (d) 2 2

Q19. If sin A B 0.5 , cos A B 0.5 ; 0º A+B ≤ 90º, A > B then, values of A and B are:

(a) A=45º and B = 15º (b) A=55º and B=25º (c) A=35º and B =25º (d) None

Q20. An isosceles triangle ABC in which AB=AC and Angle B =70 then angle A is:

(a) 30 (b) 40 (c) 70 (d) 140

Q21. If sin3A= cos(A–26º), where 3A is an acute angle, then the value of A is:

(a) A=29º (b) A=15º (c) A=30º (d) None of these

Q22.

2

2

1 tan A__________?

1 cot A

(a) sec2A (b) –1 (c) cot2A (d) tan2A

Q23. o o

o o

cos60 sin60................

cos60 sin60

:

(a) 3 2 (b) 2 3 (c) 3 2 (d) None of these

Q24. The value of tan 5º tan 10º tan 15º tan 20º tan 70º tan 75º tan 80º tan 85º is:

(a) 0 (b) 1 (c) 2 (d) None of these

Q25. If sin A =12/13, then the value of 13sin A 5sec A

5tan A 12cosec A

will be:

(a) 9 (b) 8 (c) 4 (d) None of these

Q26. The value of tan 30º sin 30º cot 60º cosec 30º will be:

(a) 1 (b) 1

3 (c)

1

3 (d) 3

Q27. If = 45º, then the value of cos2 –sin2 will be:

(a) 0 (b) –1/2 (c)1/2 (d) None of these

Q28. If sin = cos, then the value of will be:

(a) 60º (b) 30º (c) 45º (d) None of these

Q29. If is an acute angle and 7+4 sin = 9, then the value of is:

(a) 90º (b) 30º (c) 45º (d) 60º

Q30. If increases from o0 to o90 , sin changes according to:

(a) from –∞ to 0 (b) from 0 to 1 (c) from –∞ to 1 (d) None of these

Q31. If sin2A = cos3A, then correct statement is:

(a) A=110º (b) A=30º (c) A=20º (d) A=18º

Q32. If α+β =90º and α =2β, then cos2α+sin2β is equal to:

(a) 1 (b) 1/2 (c) 0 (d) 2

Q33. If A + B =45o, the value of (cosA cosB – sinA sinB) is:

(a) 3 / 2 (b) 0 (c) 1/ 2 (d) None of these

Q34. The value of , for o osin 2θ 1, 0 θ 90 is:

A. o60 B. o55 C. o45 D. o135

Q35. Value of 2 o 2 osec 26 cot 64 is:

A. 0 B. 1 C. –1 D. 2

Page | 42

Q36. The value of o o o otan1 tan 2 tan 3 . ....tan89 is:

A. 0 B. 1 C. –1 D. 90

Q37. 21 tan θ is equal to:

A. cot θ B. cosθ C. cosecθ D. secθ

Q38. If oα β 90 , cotβ 3/4 then, tan α is equal to:

A. 3/4 B. 4/3 C. 1/4 D. 1/3

Q39. Maximum value of 1

cosecθ, o o0 θ 90 is:

A. –1 B. 2 C. 1 D. Can’t be determined

Q40. If cosθ =1/2, sinβ= 1/2 then value of θ β :

A. o30 B. o60 C. o90 D. o120

Q41. If sin A B 1 cos A B then:

A. oA B 90 B. oA B 0 C. oA B 45 D. A 2B

Q42. The maximum value of (sin + cos ) is:

(a) 1 (b) 2 (c) 2 (d) 2 2

Q43. If sec4A= cosec (A –20º), the value of A:

(a) A = 25º (b) A = 15º (c) A = 22º (d) A = 35º

Q44. 9sec2A – 9 tan2 A = ______?

(a) 1 (b) 9 (c) 8 (d) 0

Q45. 2 o

2 o

1 tan 45___?

1 tan 45

(a) tan 90º (b) 1 (c) sin 45º (d) 0

Q46. ocos0 ____?

(a) 0 (b) 1 (c) not defined (d) None of these

Q47. tan x + sin x = m and tan x –sin x = n, then (m2 – n2) is equal to:

(a) 4 mn (b) mn (c) 2 mn (d) None of these

Q48. If x = r sinA cosC and y = r sinA sinC and z = r cosA, then the value of x2 + y2 + z2 is:

(a) 2

1

r (b) r2 (c)

2

2

r (d)

2

r

Page | 43

05.DATA ANALYSIS AND STATISTICS

Mathematics may not teach us how to deal with the complexity of life. But it surely gives us a hope that every problem has solution!

Q01. Mean of first 10 natural numbers is:

(a) 5 (b) 6 (c) 5.5 (d) 6.5

Q02. If mean of 4, 6, 8, 10, x, 14, 16 is 10 then, the value of x is:

(a) 11 (b) 12 (c) 13 (d) 9

Q03. The mean of x, x+1, x+2, x+3, x+4, x+5 and x+6 is:

(a) x (b) x+4 (c) 3 (d) x+3

Q04. The median of 2, 3, 2, 5, 6, 9, 10, 12, 16, 18 and 20 is:

(a) 9 (b) 10 (c) 20 (d) 9.5

Q05. The median of 2, 3, 6, 0, 1, 4, 8, 2, 5 is:

(a) 4 (b) 1 (c) 3 (d) 2

Q06. Mode of 1, 0, 2, 2, 3, 1, 4, 5, 1, 0 is:

(a) 5 (b) 0 (c) 2 (d) 1

Q07. If the mode of 2, 3, 5, 4, 2, 6, 3, 5, 5, 2 and x is 2 then, the value of x is:

(a) 4 (b) 3 (c) 2 (d) 5

Q08. The modal class of the following distribution is:

Class interval 10–15 15–20 20–25 25–30 30–35

Frequency 4 7 12 8 2

(a) 30–35 (b) 20–25 (c) 25–30 (d) 15–20

Q09. A teacher asks one of his student to find the average marks obtained by all the class students in Mathematics, the student will have to find:

(a) Mean (b) Mode (c) Median (d) Sum

Q10. The mean of 11 observations is 50. If the mean of first six observations is 49 and that of the last six observations is 52, then the sixth observation is:

(a) 56 (b) 55 (c) 54 (d) None of these

Q11. Mean of x, 3, 4 and 5 is 3 then, the value of x will be:

(a) 4 (b) 6 (c) 10 (d) None of these

Q12. Mean of 6, 4, p, 7, 10 is 8, then the value of p is:

(a) 13 (b) 14 (c) 19 (d) 20

Q13. The correct empirical relationship is:

(a) Median + A.M. = 2Mode (b) Median – A.M. = Mode

(c) 2(A.M.) – 2(Median) = Mode (d) 3(Median) – 2(A.M.) = Mode

Q14. Consider the following distribution of data:

Class interval Frequency

35-45 8

45-55 12

55-65 20

65-70 10

The median of this distribution is:

(a) 56.5 (b) 57.5 (c) 58.7 (d) None of these

Q15. The median of 15, 17, 19, 14, 12 will be:

Page | 44

(a) 15 (b) 17 (c) 14 (d) 13

Q16. What is the class size of 40 – 60?

(a) 40 (b) 50 (c) 60 (d) 100

Q17. There are 45 students in a class out of which 15 are girls. The average weight of 15 girls is 45kg and that of 30 boys is 52kg. The mean weight of entire class is:

(a) 46.67kg (b) 47.67kg (c) 48.67kg (d) 49.67kg

Q18. Measure of central tendency is represented by the abscissa of the point where the ‘less than ogive’ and ‘more than ogive’ intersects is:

(a) Mean (b) Mode (c) Median (d) None of these

Q19. The mean of 20 numbers is 17. If 3 is added to each number, then the new mean is:

(a) 20 (b) 21 (c) 24 (d) None of these

Q20. The mean of 5 numbers is 18. If a particular number is excluded then their mean becomes 16. The number excluded is:

(a) 23 (b) 24 (c) 25 (d) 26

Q21. The mean of first 5 prime numbers is:

(a) 5.5 (b) 5.6 (c) 5.7 (d) 5

Q22. The sum of deviations of the values 3, 4, 6, 8, 14 from their mean is:

(a) 0 (b) 1 (c) 2 (d) 3

Q23. If median= 15 and mean= 16, then the mode is:

(a) 10 (b) 11 (c) 12 (d) 13

Page | 45

06.EUCLIDIAN GEOMETRY & SIMILAR TRIANGLES

In most sciences one generation tears down what another has built and what one has established another undoes. In mathematics alone each generations adds a new story to the old structure.

Q01. Given that ΔABC~ΔDEF . If DE= 2AB and BC= 3cm then, EF is equal to ___________. (a) 12cm (b) 2cm (c) 1.5cm (d) 6cm Q02. The straight line distance between A and B is (see the Fig.1):

Fig.1

(a) 5 3 (b) 5 (c) 3 5 (d) 5 2

Q03. In a triangle ABC, oA 25 , oB 35 and AB=16units. In triangle PQR, oP 35 , oQ 120

and PR=4units. Which of the following is true? (a) ar( ABC)= 2ar( PQR) (b) ar( ABC)= 4ar( PQR) (c) ar( ABC)= 8ar( PQR) (d) ar( ABC)= 16ar( PQR) Q04. The altitude of an equilateral triangle, having the length of its side as 12cm, is:

(a) 6 2 cm (b) 6cm (c) 8.5cm (d) 6 3 cm

Q05. The areas of two similar triangles are 49cm2 and 64cm2 respectively. The ratio of their corresponding sides is (a) 49:64 (b) 7:8 (c) 64:49 (d) None of these Q06. If ∆ABC is similar to ∆DEF such that BC =3cm, EF=4cm and area of ∆ABC=54cm2. The area of ∆DEF is: (a) 106cm2 (b) 96cm2 (c) 120cm2 (d) 132cm2

Q07. All the equilateral triangles are ________________. (a) Similar (b) Congruent (c) Both (a) and (b) (d) None Q08. A triangle PQR is similar to another triangle ABC such that ar(PQR)=4ar(ABC). The ratio of their perimeters is given as: (a) 2:1 (b) 1:2 (c) 4:1 (d) None of these Q09. In a right triangle ABC right angled at C, AC=BC. Then AB2= ___ AC2. (a) 1 (b) 2 (c) 4 (d) None of these

Q10. If the three sides of a triangle are a , 3a , 2a then the measure of the angle opposite to the

longest side is:

(a) o60 (b) o90 (c) o45 (d) o30 Q11. QA and PB are perpendicular on AB, if AO = 10cm, BO=6cm and PB=9cm, then measure of AQ (see the Fig.2):

Fig.2

(a) 15cm (b) 25cm (c) 10cm (d) None of these

Q12. In right triangles ABD and BDC (see the figure given in Fig.3), AB = x and CD=y, then PQ is:

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Fig.3 Fig.4

(a) xy/(x+y) (b) (x–y)/xy (c) (x+y)/xy (d) None of these

Q13. In the figure (see the Fig.4) FGH and PQR are two triangles. If the measurements are as shown in the figure, then PR is equal to: (a) 16cm (b) 12cm (c) 8cm (d) 4cm Q14. If ABC ~ PQR , ar(PQR)=100cm2 and AB/PQ= 1/2 then, ar(ABC) is:

(a) 50cm2 (b) 25cm2 (c) 4cm2 (d) None of these Q15. The areas of two similar triangles are 144cm2 and 81cm2. If one median of the first triangle is 16cm, length of corresponding median of the second triangle is: (a) 9cm (b) 27cm (c) 12cm (d) 16cm Q16. The ratio of the areas of two similar triangles is equal to the: (a) ratio of their corresponding sides (b) ratio of their corresponding altitudes (c) ratio of the squares of their perimeters (d) ratio of the squares of their corresponding sides

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Answers Of Objective Mathematicia

Chapter 01

Q01. c Q02. d Q03. b Q04. d Q05. c Q06. c Q07. b Q08. c Q09. a Q10. b Q11. c Q12. d Q13. d Q14. c Q15. c Q16. b Q17. b Q18. b Q19. a Q20. b Q21. b Q22. c Q23. b Q24. b Q25. c Q26. d Q27. c Q28. a Q29. b Q30. A Q31. b Q32. d Q33. d Q34. b Q35. b Q36. b Q37. A Q38. c Q39. d

Chapter 02

Q01. d Q02. b Q03. b Q04. a Q05. a Q06. a Q07. c Q08. c Q09. a Q10. a Q11. c Q12. c Q13. c Q14. b Q15. d Q16. d Q17. c Q18. d Q19. c Q20. c Q21. a Q22. d Q23. d Q24. d Q25. b Q26. a Q27. b Q28. a Q29. c Q30. c Q31. b Q32. a Q33. b Q34. d Q35. b Q36. a Q37. a Q38. b Q39. a Q40. b Q41. b Q42. b Q43. c Q44. b Q45. b Q46. a Q47. a Q48. c Q49. d Q50. d Q51. b Q52. d Q53. c Q54. a Q55. d Q56. a Q57. d Q58. c Q59. d

Chapter 03

Q01. a Q02. c Q03. d Q04. d Q05. c Q06. c Q07. b Q08. a Q09. b Q10. c Q11. c Q12. c Q13. c Q14. b Q15. c Q16. c Q17. a Q18. a Q19. d Q20. c Q21. d Q22. c Q23. a Q24. b Q25. a Q26. c Q27. c Q28. d Q29. c Q30. c Q31. b Q32. a Q33. a Q34. a Q35. b Q36. a Q37. d Q38. d Q39. b Q40. c Q41. c Q42. a Q43. a Q44. a Q45. a Q46. b Q47. a Q48. d Q49. b Q50. a Q51. c Q52. a Q53. a Q54. b Q55. a Q56. d Q57. d Q58. c

Chapter 04

Q01. b Q02. a Q03. c Q04. a Q05. d Q06. a Q07. a Q08. b Q09. d Q10. d Q11. d Q12. b Q13. a Q14. a Q15. c Q16. b Q17. a Q18. c Q19. a Q20. b Q21. a Q22. d Q23. b Q24. b Q25. d Q26. b Q27. a Q28. c Q29. b Q30. b Q31. d Q32. b Q33. c Q34. c Q35. b Q36. b Q37. d Q38. a Q39. c Q40. c Q41. c Q42. b Q43. c Q44. b Q45. b Q46. b Q47. a Q48. b

Chapter 05 Q01. c Q02. b Q03. d Q04. a Q05. c Q06. d Q07. c Q08. b Q09. a Q10. a Q11. d Q12. a Q13. d Q14. d Q15. a Q16. b Q17. d Q18. c Q19. a Q20. d Q21. b Q22. a Q23. d

Chapter 06 Q01. d Q02. c Q03. b Q04. d Q05. b Q06. b Q07. a Q08. a Q09. b Q10. b Q11. a Q12. a Q13. c Q14. b Q15. c Q16. d.