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Chapter 6 Trigonometric Identities Exercise Ex. 6.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23(i)

Solution 23(i)

Question 23(ii)

Solution 23(ii)

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37. i

Solution 37. i

Question 37. ii

Prove the following trigonometric identities:

Solution 37. ii

L.H.S.,

Question 38(i)

Solution 38(i)

Question 38(ii)

Solution 38(ii)

Question 38(iii)

Solution 38(iii)

Question 38(iv)

Solution 38(iv)

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47 (i)

Solution 47 (i)

Question 47 (ii)

Solution 47 (ii)

Question 47 (iii)

Solution 47 (iii)

Question 47(iv)

Prove that:

(sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θSolution 47(iv)

Question 48

Solution 48

Question 49

Solution 49

Question 50

Prove that:

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55. i

Solution 55. i

Question 55. ii

Solution 55. ii

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Question 63

Solution 63

Question 64

Solution 64

Question 65. i

Solution 65. i

Question 65(ii)

Solution 65(ii)

Question 66

Solution 66

Question 67

Solution 67

Question 68

Solution 68

Question 69

Solution 69

Question 70

Solution 70

Question 71

Solution 71

Question 72

Solution 72

Question 73

Solution 73

Question 74

Solution 74

Question 75

Solution 75

Question 76

Solution 76

Question 77

Solution 77

Question 78

Solution 78

Question 79

Solution 79

Question 80

Solution 80

Question 81

Solution 81

Question 82

Solution 82

Question 83

Solution 83

Question 84

Solution 84

Question 85

Solution 85

Question 86

If sin θ + 2 cos θ = 1 prove that 2 sin θ – cos θ = 2.Solution 86

Question 23 (iii)

Solution 23 (iii)

To prove: 

Consider LHS

= RHS

Hence, proved.

Chapter 6 Trigonometric Identities Exercise Ex. 6.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

If 2sin² θ – cos² θ = 2, then find the value of θ.Solution 13

Question 14

If tan θ – 1 = 0, find the value of sin²θ – cos² θ.Solution 14

Question 4

If  evaluate  Solution 4

Given: 

Consider,

 … [Multiplying and dividing by tan θ]

Chapter 6 Trigonometric Identities Exercise 6.56

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

sec⁴ – sec²A  is equal to

(a)  tan²A –  tan⁴A     (b) tan⁴A –  tan²A 

(c) tan⁴A + tan²A      (d) tan²A –  tan⁴A  Solution 5

Question 6

cos⁴A-sin⁴A is equal  to

(a)  2cos²A+1    (b) 2cos²A-1   

(c)  2sin²A-1     (d) 2sin²A+1  Solution 6

Chapter 6 Trigonometric Identities Exercise 6.57

Question 7

Solution 7

Question 8

Solution 8

Question 9

The value of (1+cotθ – cosecθ )  (1+tanθ + secθ) is

 (a) 1     (b) 2      (c) 4      (d) 0Solution 9

Question 10

Solution 10

Question 11

(cosecθ – sinθ) (secθ – cosθ) (tanθ + cotθ) is equal

(a) 0          (b)   1       (c)-1        (d) None of theeSolution 11

Question 12

Solution 12

Question 13

If  x =a secθ  and  y=b tanθ  then  b²x²a²y²=

(a) ab      (b) a² – b²       (c) a² + b²    (d) a²b²Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

If acosθ + bsinθ = 4 and asinθ-bcosθ = 3, then a²+b²=

(a) 7    (b) 12      (c) 25      (d) None of theseSolution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Chapter 6 Trigonometric Identities Exercise 6.58

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

If sin θ – cos θ = 0, then the value of sin⁴ θ + cos⁴ θ is

1. 1
2. 
3. 
4. 

Solution 29

Question 30

The value of sin (45° + θ) – cos (45° – θ) is equal to

1. 2 cos θ
2. 0
3. 2 sin θ
4. 1

Solution 30

Question 31

If ΔABC is right angled at C, then the value of cos (A + B) is

1. 0
2. 1
3. 
4. 

Solution 31

Chapter 6 Trigonometric Identities Exercise 6.59

Question 32

If cos 8θ = sin θ and 8θ < 90°, then the value of tan 6θ is

1. 
2. 
3. 1
4. 0

Solution 32

Question 33

If cos (α + β) = 0, then sin (α  –  β) can be reduced to

1. cos β
2. cos 2β
3. sin α
4. sin 2α

Solution 33

Chapter 5 Trigonometric Ratios Exercise Ex. 5.1

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 1 (v)

Solution 1 (v)

Question 1 (vi)

Solution 1 (vi)

Question 1 (vii)

Solution 1 (vii)

Question 1 (viii)

Solution 1 (viii)

Question 1 (ix)

Solution 1 (ix)

Question 1 (x)

Solution 1 (x)

Question 1 (xi)

Solution 1 (xi)

Question 1 (xii)

Solution 1 (xii)

Question 2

In ABC right angled at B, AB = 24 cm, BC = 7 cm. Determine

(i)  sin A, cos A
(ii)  sin C, cos C

Solution 2

In ABC by applying Pythagoras theorem
AC² = AB² + BC²
= (24)² + (7)²
= 576 + 49
= 625
AC =  = 25 cm






Question 3

Solution 3

Question 4

Solution 4

Question 5

Given 15 cot A = 8. Find sin A and sec ASolution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 25

If  find the value of  Solution 25

Given: 

Chapter 5 Trigonometric Ratios Exercise Ex. 5.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26 (i)

Solution 26 (i)

Question 26 (ii)

Solution 26 (ii)

Question 26 (iii)

Solution 26 (iii)

Question 26 (iv)

Solution 26 (iv)

Question 27 (i)

If A = B = 60^o, verify that cos (A – B) = cos A cos B + sin A sin BSolution 27 (i)

Question 27 (ii)

Solution 27 (ii)

Question 27 (iii)

Solution 27 (iii)

Question 28 (i)

Solution 28 (i)

Question 28 (ii)

Solution 28 (ii)

Question 29

Solution 29

Question 30 (i)

Solution 30 (i)

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

If sin (A – B) = sin A cos B – cos A sin B and cos(A – B) = cos A cos B + sin A sin B, find the values of sin 15^o and cos 15^o.Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Prove that

Solution 40

Question 30 (ii)

If tan(A + B) = 1 and  0^o < A + B < 90^o, A > B, then find the values of A and B. Solution 30 (ii)

Given: tan(A + B) = 1 and 

Therefore,

A + B = 45^o … (i)

A – B = 30^o … (ii)

Adding the two equations, we get

Chapter 5 Trigonometric Ratios Exercise Ex. 5.3

Question 1

Solution 1

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 2 (v)

Solution 2 (v)

Question 2 (vi)

Solution 2 (vi)

Question 2 (vii)

Solution 2 (vii)

Question 2 (viii)

Solution 2 (viii)

Question 2 (ix)

Solution 2 (ix)

Question 2 (x)

Solution 2 (x)

Question 2 (xi)

Solution 2 (xi)

Question 3

Express each one of the following in terms of trigonometric ratios of angles lying between 0^o and 45^o

(i) sin 59^o + cos 56^o

(ii) tan 65^o + cot 49^o

(iii) sec 76^o + cosec 52^o

(iv) cos 78^o + sec 78^o

(v) cosec 54^o + sin 72^o

(vi) cot 85^o + cos 75^o

(vii) sin 67^o + cos 75^oSolution 3

Question 4

Solution 4

Question 5

If sin 3A = cos (A – 26^o), where 3A is an acute angle, find the value of A.Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 7 (i)

Solution 7 (i)

Question 7 (ii)

Solution 7 (ii)

Question 7 (iii)

Solution 7 (iii)

Question 7 (iv)

Solution 7 (iv)

Question 8 (i)

Solution 8 (i)

Question 8 (ii)

Solution 8 (ii)

Question 8 (iii)

Solution 8 (iii)

Question 8 (iv)

Solution 8 (iv)

Question 8 (v)

Solution 8 (v)

Question 9 (i)

Solution 9 (i)

Question 9 (ii)

Solution 9 (ii)

Question 9 (iii)

Solution 9 (iii)

Question 9 (iv)

Solution 9 (iv)

Question 9 (v)

Solution 9 (v)

Question 9 (vi)

Solution 9 (vi)

Question 9 (vii)

Solution 9 (vii)

Question 9 (viii)

Solution 9 (viii)

Question 9 (ix)

Solution 9 (ix)

Question 9 (x)

Solution 9 (x)

Question 10

Solution 10

Question 11 (ii)

If A, B,C are the interior angles of a ΔABC,show that

Solution 11 (ii)

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 6 (iii)

If A, B, C, are the interior angles of a triangle ABC, ∠A = 90^o, then find the value of  Solution 6 (iii)

Given: ∠A = 90^o

For a triangle ABC, ∠A + ∠B + ∠C = 90^o

Question 9 (xi)

Evaluate:  Solution 9 (xi)

Using the identities

Question 11 (i)

If A, B, C are the interior angles of a ∆ABC, show that:  Solution 11 (i)

For a triangle ABC, ∠A + ∠B + ∠C = 90^o

Question 18

If tan 2A = cot(A – 18^o), where 2A is an acute angle, find the value of A.Solution 18

Given: tan 2A = cot(A – 18^o) 

As tan x = cot(90^o – x), we have

cot(90^o – 2A) = cot(A – 18^o)

90^o – 2A = A – 18^o

3A = 108^o

Therefore, A = 36^o.

Chapter 5 Trigonometric Ratios Exercise 5.56

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

So, the correct option is (a).Question 5

Solution 5

Question 6

Solution 6

Chapter 5 Trigonometric Ratios Exercise 5.57

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

So, the correct option is (d).Question 11

Solution 11

So, the correct option is (c).Question 12

Solution 12

Question 13

Solution 13

Question 14

If A and B are complementary angles, then

(a) sin A = sin B

(b) cos A and cos B

(c) tan A = tan B

(d) sec A = cosec BSolution 14

Question 15

Solution 15

So, the correct option is (b).Question 16

Solution 16

So, the correct option is (a).Question 17

Solution 17

So, the correct option is (b).Question 18

Solution 18

So, the correct option is (d).

Chapter 5 Trigonometric Ratios Exercise 5.58

Question 19

Solution 19

Question 20

Solution 20

So, the correct option is (b).Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

So, the correct option is (d).Question 28

Sin 2A = 2 sin A is true when A = 

(a) 0^o

(b) 30^o

(c) 45^o

(d) 60^oSolution 28

So, the correct option is (a).Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Chapter 5 Trigonometric Ratios Exercise 5.59

Question 33

Solution 33

So, the correct option is (c).Question 34

Solution 34

Question 35

Solution 35

Chapter 4 – Quadratic Equations Exercise Ex. 4.1

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 1 (v)

Solution 1 (v)

Question 1(vi)

Solution 1(vi)

Question 1 (vii)

Solution 1 (vii)

Question 1 (viii)

Solution 1 (viii)

Question 1 (ix)

Solution 1 (ix)

Question 1 (x)

Solution 1 (x)

Question 1 (xi)

Solution 1 (xi)

Question 1 (xii)

Solution 1 (xii)

Question 1 (xiii)

Solution 1 (xiii)

Question 1 (xiv)

Solution 1 (xiv)

Question 1 (xv)

Is X(x+1)+8=(x+2)(x-2) a quadratic equation?Solution 1 (xv)

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 2 (v)

Are x = 2 and x = 3, solutions of the equation 2x² – x + 9 = x² + 4x + 3 , Solution 2 (v)

= 2x² – x + 9 – x² + 4x + 3

= x² – 5x + 6 = 0

Here, LHS = x² – 5x + 6 and RHS = 0

Substituting x = 2 and x = 3

= x² – 5x + 6

= (2)² – 5(2) + 6

=10-10

=0

= RHS

= x² – 5x + 6

= (3)2 – 5(3) + 6

= 9 – 15 + 6

=15 – 15

=0

= RHS

x = 2 and x = 3 both are the solutions of the given quadratic equation.Question 2 (vi)

Solution 2 (vi)

Question 2 (vii)

Solution 2 (vii)

Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 3 (iii)

Solution 3 (iii)

Question 3 (iv)

Solution 3 (iv)

Question 4

Solution 4

Question 5

Solution 5

Chapter 4 – Quadratic Equations Exercise Ex. 4.2

Question 1

The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer.Solution 1

Question 2

John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they had to start with, if John had x marbles.Solution 2

Question 3

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation to find x.Solution 3

Question 4

The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle.Solution 4

Question 5

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.Solution 5

Question 6

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.Solution 6

Chapter 4 – Quadratic Equations Exercise Ex. 4.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solve the following quadratic equation by factorisation:

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solve the following quadratic equation by factorisation:

Solution 15

Question 16

Solution 16

Question 17

9x² – 6b²x – (a⁴ – b⁴) = 0Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solve the following quadratic equation by factorisation:

2x² + ax – a² = 0Solution 20

Question 21

Solve the following quadratic equation by factorisation:

Solution 21

Question 22

Solve the following quadratic equations by factorization:

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solve the following quadratic equation by factorisation:

Solution 29

Question 30

Solve the following quadratic equation by factorisation:

Solution 30

Question 31

Solve the following quadratic equation by factorisation:

Solution 31

Question 32

Solve the following quadratic equation by factorisation:

Solution 32

Question 33

Solve the following quadratic equation by factorisation:

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solve the following quadratic equation by factorisation:

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solve the following quadratic equations by factorization:

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solve the following quadratic equations by factorization:

Solution 47

Question 48

Solve the following quadratic equations by factorization:

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solve the following quadratic equation by factorisation:

Solution 58

Question 59

Solve the following quadratic equation by factorisation:

Solution 59

Question 60

Solve the following quadratic equation by factorisation:

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Chapter 4 – Quadratic Equations Exercise Ex. 4.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Find the roots of the following quadratic equations (if they exist) by the method of completing the square

x² – 8x + 18 = 0Solution 6

Given equation is x² – 8x + 18 = 0

x² – 2 × x × 4 + + 4² – 4² + 18 = 0

(x – 4)² – 16 + 18 = 0

(x – 4)² = 16 – 18

(x – 4)² = -2

Taking square root on both the sides, we get 

Therefore, real roots does not exist.Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 4 – Quadratic Equations Exercise Ex. 4.5

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1(iii)

Solution 1(iii)

Question 1 (iv)

Solution 1 (iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1 (vii)

Write the discriminant of the following quadratic equations:

(x + 5)² = 2(5x – 3)Solution 1 (vii)

x² + 2 × x × 5 + 5² = 10x – 6

x² + 10x + 25 = 10x – 6

x² + 31 = 0

Here, a = 1, b = 0 and c = 31

Therefore, the discriminant is

D = b² – 4ac

 = 0 – 4 × 1 × 31

 = -124Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 2 (v)

Solution 2 (v)

Question 2 (vi)

Solution 2 (vi)

Question 2 (vii)

Solution 2 (vii)

Question 2 (viii)

Solution 2 (viii)

Question 2 (ix)

Solution 2 (ix)

Question 2(x)

Solution 2(x)

Question 2(xi)

Solution 2(xi)

Question 2(xii)

3x² – 5x + 2 = 0Solution 2(xii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solve for x:

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solve for x:

Solution 3(iv)

Question 3(v)

Solve for x:

Solution 3(v)

Chapter 4 – Quadratic Equations Exercise Ex. 4.6

Question 1(i)

Determine the nature of the roots of the following quadratic equations:

2x² – 3x + 5 = 0Solution 1(i)

Question 1(ii)

Determine the nature of the roots of the following quadratic equations:

2x² – 6x + 3 = 0Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1 (vi)

Determine the nature of the roots of the following quadratic equations:

Solution 1 (vi)

Given quadratic equation is 

Here, 

Therefore, we have

As D = 0, roots of the given equation are real and equal.Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 2(xi)

Solution 2(xi)

Question 2(xii)

Solution 2(xii)

Question 2(xiii)

Solution 2(xiii)

Question 2(xiv)

Solution 2(xiv)

Question 2(xv)

Solution 2(xv)

Question 2(xvi)

Solution 2(xvi)

Question 2(xvii)

Solution 2(xvii)

Question 2(xviii)

Find the values of k for which the roots are real and equal in each of the following equations:

4x² – 2 (k + 1)x + (k + 1) = 0 Solution 2(xviii)

4x² – 2 (k + 1)x + (k + 1) = 0 Comparing with ax² + bx + c = 0, we get a = 4, b = -2(k + 1), c = k + 1 According to the question, roots are real and equal. Hence, b² – 4ac = 0Question 3(i)

Solution 3(i)

Question 3(ii)

In the following, determine the set of values of k for which the given quadratic equation has real roots:

2x² + x + k = 0Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Find the values of k for which the following equations have real and equal roots:

x² + k(2x + k – 1) + 2 = 0 Solution 4(iv)

x² + k(2x + k – 1) + 2 = 0 x² + 2kx + k(k – 1) + 2 = 0 Comparing with ax² + bx + c = 0, we get a = 1, b = 2k, c = k(k – 1) + 2 According to the question, roots are real and equal. Hence, b² – 4ac = 0Question 5(i)

Find the values of k for which the roots are real and equal in each of the following equations:

2x² + kx + 3 = 0Solution 5(i)

Question 5(ii)

Find the values of k for which the roots are real and equal in each of the following equations:

kx (x – 2) + 6 = 0Solution 5(ii)

Question 5(iii)

Find the values of k for which the roots are real and equal in each of the following equations:

x² – 4kx + k = 0Solution 5(iii)

Question 5(iv)

Find the value of k for which the roots are real and equal in the following equation:

Solution 5(iv)

Question 5(v)

Find the value of p for which the roots are real and equal in the following equation:

px(x – 3) + 9 = 0Solution 5(v)

Question 5(vi)

Find the values of k for which the following equations have real roots.

4x² + kx + 3 = 0 Solution 5(vi)

4x² + kx + 3 = 0 Comparing with ax² + bx + c = 0, we get a = 4, b = k, c = 3 According to the question, roots are real and equal. Hence, b² – 4ac = 0Question 6 (i)

Solution 6 (i)

Question 6 (ii)

Solution 6 (ii)

Question 7

Solution 7

Question 8

Solution 8

Question 9(i)

Find the values of k for which the quadratic equation (3k + 1)x² + 2(k + 1)x + 1 = 0 has equal roots. Also, find the roots.Solution 9(i)

Question 9 (ii)

Write all the values of k for which the quadratic equation x² + kx + 16 = 0 has equal roots. Find the roots of the equation so obtained.Solution 9 (ii)

Given quadratic equation is x² + kx + 16 = 0

As it has equal roots, the discriminant will be 0.

Here, a = 1, b = k, c = 16

Therefore, D = k² – 4(1)(16) = 0

i.e. k² – 64 = 0

i.e. k = ± 8

When k = 8, the equation becomes x² + 8x + 16 = 0

 or x² – 8x + 16 = 0

As D = 0, roots of the given equation are real and equal.Question 10

Find the values of p for which the quadratic equation (2p + 1)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also, find these roots.Solution 10

Question 11

 If -5 is a root of the quadratic equation, 2x² + px – 15 = 0, and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.Solution 11

Question 12

 If 2 is a root of the quadratic equation 3x² + px – 8 = 0 and the quadratic equation 4x² – 2px + k = 0 has equal roots, find the value of k.Solution 12

Question 13

If 1 is root of the quadratic equation 3x² + ax – 2 = 0 and the quadratic equation a(x² + 6x) – b = 0 has equal roots, find the value of b.Solution 13

Question 14

Find the value of p for which the quadratic equation (p + 1)x² – 6(p + 1)x + 3(p + 9) = 0, p ≠ -1 has equal roots. Hence, find the roots of equation.Solution 14

Question 15(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 15(iii)

Solution 15(iii)

Question 15(iv)

Solution 15(iv)

Question 16(i)

Solution 16(i)

Question 16(ii)

Solution 16(ii)

Question 16(iii)

Solution 16(iii)

Question 16(iv)

Solution 16(iv)

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Chapter 4 – Quadratic Equations Exercise Ex. 4.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The sum of a number and its square is 63/4. Find the numbers.Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

A natural number when increased by 12 equals 160 times its reciprocal. Find the number.Solution 24

Let x be the natural number.

As per the question, we have

Therefore, x = 8 as x is a natural number.

Hence, the required natural number is 8.Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger, find the numbers.Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

The sum of two numbers is 9. The sum of their reciprocals is 1/2. Find the numbers.Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Find two consecutives odd positive integers, sum of whose squares is 970.Solution 34

Question 35

The difference of two natural numbers is 3 and the difference of their reciprocal is . Find the numbers.Solution 35

Question 36

The sum of the squares of two consecutive odd numbers is 394. Find the numbers.Solution 36

Question 37

The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.Solution 37

Question 38

The sum of the squares of two consecutive even numbers is 340. Find the numbers.Solution 38

Question 39

The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is. Find the original fraction.Solution 39

Question 40

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.Solution 40

Chapter 4 – Quadratic Equations Exercise Ex. 4.8

Question 1

Solution 1

Question 2

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed, If it takes 3 hours to complete total journey, what is its original average speed?Solution 8

Question 9

Solution 9

Question 10

Solution 10

Concept Insight: Use the relation s =d/t to crack this question and remember here distance is constant so speed and time will vary inversely.Question 11

Solution 11

Question 12

An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.Solution 12

Question 13

While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hr. Find the original speed/hour of the plane.Solution 13

Question 14

A motor boat whose speed in still water is 18 km/hr takes 1 hour more to go 24 km up stream than to return downstream to the same spot. Find the speed of the stream.Solution 14

Question 15

A car moves a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.Solution 15

Let the speed of a car be x km/hr. According to the question, time is hr. Distance = Speed × Time 2592 =  x = 72 km/hr Hence, the time taken by a car to cover a distance of 2592 km is 36 hrs.Question 16

A motor boat whose speed instill water is 9 km/hr, goes 15 km downstream and comes back to the same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream.Solution 16

Let x km/hr be the speed of the stream.

Therefore, we have

Downstream speed = (9 + x) km/hr

Upstream speed = (9 – x) km/hr

Distance covered downstream = distance covered upstream

Total time taken = 3 hours 45 minutes = hours

Therefore, x = 3 as the speed can’t be negative.

Hence, the speed of the motor boat is 3 km/hr.

Chapter 4 – Quadratic Equations Exercise Ex. 4.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than 5 times her actual age. What is her age now?Solution 8

Question 9

At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.Solution 9

Chapter 4 – Quadratic Equations Exercise Ex. 4.10

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?Solution 4

Chapter 4 – Quadratic Equations Exercise Ex. 4.11

Question 1

Solution 1

Question 2

Solution 2

Question 3

Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm². Find the sides of the squares.Solution 3

Question 4

Solution 4

Question 5

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.Solution 5

Question 6

Solution 6

Question 7

Sum of the areas of two squares is 640 m². If the difference of their perimeter is 64 m, find the sides of the two squares.Solution 7

Question 8

Sum of the areas of two squares is 400 cm². If the difference of their perimeters is 16 cm, find the sides of two squares.Solution 8

Question 9

The area of a rectangular plot is 528 m². The length of the plot (in meters) is one meter more than twice its breadth. Find the length and the breadth of the plot.Solution 9

Question 10

In the centre of a rectangular lawn of dimension 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m². Find the length and breadth of the pond.Solution 10

Chapter 4 – Quadratic Equations Exercise Ex. 4.12

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?Solution 5

Let us assume that the larger pipe takes ‘x’ hours to fill the pool.

So, as per the question, the smaller pipe takes ‘x + 10’ hours to fill the same pool.

Question 6

Two water taps together can fill a tank in  hours. The tap with longer diameter takes 2 hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately. Solution 6

Let the tap with smaller diameter takes x hours to completely fill the tank.

So, the other tap takes (x – 2) hours to fill the tank completely.

Total time taken to fill the tank  hours

As per the question, we have

When x = 5, then (x – 2) = 3

When  which can’t be possible as the time becomes negative.

Hence, the smaller diameter tap fills in 5 hours and the larger diameter tap fills in 3 hours.

Chapter 4 – Quadratic Equations Exercise Ex. 4.13

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 marks more in Mathematics and 4 marks less in Science. The product of his marks, would have been 180. Find his marks in the two subjects.Solution 9

Question 10

Solution 10

Question 11

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.Solution 11

Question 12

At t minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than  minutes. Find t.Solution 12

Chapter 4 – Quadratic Equations Exercise 4.82

Question 1

If the equation x² + 4x + k = 0 has real and distinct root, then

(a) k < 4

(b) k > 4

(c) k ≥ 4

(d) k ≤ 4Solution 1

We know for the quadratic equation

ax² + bx + c = 0

condition for roots to be real and distinct is

D = b² – 4ac > 0       ……….(1)

for the given question

x² + 4x + k = 0

a = 1, b = 4, c = k

from (1)

16 – 4k > 0

k < 4

So, the correct option is (a).

Chapter 4 – Quadratic Equations Exercise 4.83

Question 2

If the equation x² – ax + 1 = 0 has two distinct roots, then

(a) |a| = 2

(b) |a| < 2

(c) |a| > 2

(d) None of theseSolution 2

For the equation x² – ax + 1 = 0 has two distinct roots, condition is

(-a)² – 4 (1) (1) > 0

a² – 4 > 0

a² > 4

|a| > 2

So, the correct option is (c).Question 3

Solution 3

Question 4

Solution 4

Question 5

If the equation ax² + 2x + a = 0 has two distinct roots, if

(a) a = ± 1

(b) a = 0

(c) a = 0, 1

(d) a = -1, 0Solution 5

For any quadratic equation

ax²  + bx + c = 0

having two distinct roots, condition is

b² – 4ac > 0

For the equation ax² + 2x + a = 0 to have two distinct roots,

(2)² – 4 (a) (a) > 0

4 – 4a² > 0

4(1 – a²) > 0

1 – a² > 0 since 4 > 0

that is, a² – 1 < 0

Hence -1 < a < 1, only integral solution possible is a = 0

So, the correct option is (b).Question 6

The positive value of k for which the equation x² + kx + 64 = 0 and x² – 8x + k = 0 will both have real roots, is

(a) 4

(b) 8

(c) 12

(d) 16Solution 6

For any quadratic equation

ax² + bx + c = 0

having real roots, condition is

b² – 4ac ≥ 0     …….(1)

According to question 

x² + kx + 64 = 0 have real root if

k² – 4 × 64 ≥ 0

k² ≥ 256

|k|≥ 16             ……..(2)

Also, x² – 8x + k = 0 has real roots if

64 – 4k ≥ 0

k ≤ 16    ………(3)

from (2), (3) the only positive solution for k is

k = 16

So, the correct option is (d).Question 7

Solution 7

Question 8

If 2 is a root of the equation x² + bx + 12 = 0 and the equation x² + bx + q = 0 has equal roots, then q =

(a) 8

(b) -8

(c) 16

(d) -16Solution 8

It is given that 2 is a root of equation x²  + bx + 12 = 0

Hence

(2)² + b(2) + 12 = 0

4 + 2b + 12 = 0

2b + 16 = 0

b = -8        ………(1)

It is also given that x² + bx + q = 0 has equal root 

so, b² – 4(q) = 0         ……..(2)

from (1) & (2)

(-8)² – 4q =0

q = 16

So, the correct option is (c).Question 9

Solution 9

For any quadratic equation

ax² + bx + c = 0

Having equal roots, the condition is

b²  – 4ac = 0

For the equation

(a² + b²) x² – 2 (ac + bd)x + c² + d² = 0

to have equal roots, we have

(-2(ac + bd))² – 4 (c² + d²) (a² + b²) = 0

4 (ac + bd)² – 4 (a²c² + b²c² + d²a² + b²d²) = 0

(a²c² + b²d² + 2abcd) – (a²c² + b²c² + a²d²+ b²d²) = 0

2abcd – b²c² – a²d² = 0

b²c² – a²d² – 2abcd = 0

(bc – ad)² = 0

bc = ad

So, the correct option is (b).Question 10

Solution 10

For any quadratic equation

ax² + bx + c = 0

having equal roots, condition is

b² – 4ac = 0

According to question, quadratic equation is

(a² + b²)x² – 2b(a + c)x + b² + c² = 0

having equal roots, so

(2b(a + c))² – 4(a² + b²) (b² + c²) = 0

4b² (a + c)² – 4(a²b² + a²c² + b⁴ + b²c²) = 0

b²(a² + c² + 2ac) – (a²b² + a²c² + b²c² + b⁴) = 0

a²b² + b²c² + 2acb² – a²b² – a²c² – b²c² – b⁴ = 0

2acb² – a²c² – b⁴ = 0

a²c² + b⁴ – 2acb² = 0

(ac – b²)² = 0

ac = b²

So, the correct option is (b).Question 11

If the equation x² – bx + 1 = 0 does not possess real roots, then

(a) -3 < b < 3

(b) -2 < b < 2

(c) b < 2

(d) b < -2Solution 11

For any quadratic equation ax² + bx + c = 0 having no real roots, condition is

b² – 4ac < 0

For the equation, x² – bx + 1 = 0 having no real roots

b² – 4 < 0

b² < 4

-2 < b < 2

So, the correct option is (b).Question 12

If x = 1 is a common root of the equations ax² + ax + 3 = 0 and x² + x + b = 0, then ab =

(a) 3

(b) 3.5

(c) 6

(d) -3Solution 12

Question 13

If p and q are the roots of the equation x² -px + q = 0, then

(a) p = 1, q = -2

(b) b = 0, q = 1

(c) p = -2, q = 0

(d) p = -2, q = 1Solution 13

If p, q are the roots of equation x² – px + q = 0, then p and q satisfies the equation

Hence

(p)² – p(p) + q = 0

p² – p² + q = 0

q = 0

and (q)² – p(q) + q = 0

q² – p(q) + q = 0

0 = 0

p can take any value

p = -2 and q = 0

So, the correct option is (c).Question 14

If a and b can take values 1, 2, 3, 4. Then the number of the equations of the form ax² + bx + 1 = 0 having real roots is

(a) 10

(b) 7

(c) 6

(d) 12Solution 14

For the ax² + bx + 1 = 0 having real roots condition is

b² – 4(a) (1) ≥ 0

b² ≥ 4a

For a = 1

b² ≥ 4

b ≥ 2

b can take value 2, 3, 4

Here, 3 possible solutions are possible            …….(1)

For a = 2

b² ≥ 8

Here, b can take value 3, 4            ……(2)

Here, 2 solutions are possible

For a = 3

b² ≥ 12

possible value of b is 4

Hence, only 1 possible solution      ……(3)

For a = 4

b² ≥ 16

possible value of b is 4

Hence, only 1 possible solution    …….(4)

from (1), (2), (3), (4)

Total possible solutions are 7

So, the correct option is (b).Question 15

The number of quadratic equations having real roots and which do not change by squaring their roots is

(a) 4

(b) 3

(c) 2

(d) 1Solution 15

Any quadratic equation having roots 0 or 1 are only possible quadratic equation because on squaring 0 or 1, it remains same.

Hence, 2 solutions are possible, one having roots 1 and 1, while the other having roots 0 and 1.

So, the correct option is (c).

Chapter 4 – Quadratic Equations Exercise 4.84

Question 16

If (a² + b²) x² + 2(ab + bd) x + c² + d² = 0 has no real roots, then

(a) ad = bc

(b) ab = cd

(c) ac = bd

(d) ad ≠ bcSolution 16

If any quadratic equation ax² + bx + c has no real roots then b² – 4ac < 0    ……(1)

According to the question, the equation is

(a² + b²) x² + 2(ac + bd) x + c² + d² = 0

from (1)

4(ac + bd)² – 4(a² + b²) (c² + d²) < 0

a²c² + b²d² + 2abcd – (a²c² + a²d² + b²c² + b²d²) < 0

a²c² + b²d² + 2abcd – a²c² – a²d² – b²c² – b²d² < 0

2abcd – a²d² – b²c² < 0

-(ad – bc)² < 0

(ad – bc)² > 0

For this condition to be true ad ≠ bc

So, the correct option is (d).Question 17

Solution 17

Question 18

If x = 1 is a common root of ax² + ax + 2 = 0 and x² + x + b = 0 then, ab =

(a) 1

(b) 2

(c) 4

(d) 3Solution 18

It is given that x = 1 is root of equation ax² + ax + 2 = 0

Hence, a(1)² + a(1) + 2 = 0

           2a + 2 = 0

           a = -1         ……(1)

It is given that x = 1 is also root of x² + x + b = 0

Hence, (1)² + (1) + b = 0

           b = -2     ……(2)

from (1) & (2)

ab = (-1) (-2)

ab = 2

So, the correct option is (b).Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

If sin α and cos α are the roots of the equation ax² + bx + c = 0, then b² = 

(a) a² – 2ac

(b) a² + 2ac

(c) a² – ac

(d) a² + acSolution 22

Question 23

If 2 is a root of the equation x² + ax + 12 = 0 and the quadratic equation x²  + ax + q = 0 has equal roots, then q =

(a) 12

(b) 8

(c) 20

(d) 16Solution 23

Given, 2 is a root of equation x² + ax + 12 = 0

so (2)² + a(2) + 12 = 0

4 + 2a + 12 = 0

a = -8     …..(1)

Given x² + ax + q = 0 has equal roots so

a² – 4q = 0   ……(2)

from (1) & (2)

(-8)² – 4q = 0

4q = 64

q = 16

So, the correct option is (d).Question 24

If the sum of the roots of the equation x² – (k + 6)x + 2(2k – 1) = 0 is equal to half of their product, then k =

(a) 6

(b) 7

(c) 1

(d) 5Solution 24

Question 25

If a and b are roots of the equation x² + ax + b = 0, then a + b =

(a) 1

(b) 2

(c) -2

(d) -1Solution 25

If a and b are roots of the equation x² + ax + b = 0

Then, sum of roots = -a

a + b = -a

2a + b = 0          …….(1)

product of roots = b

ab = b

ab – b = 0

b(a – 1) = 0         ……..(2)

from (1) and (2)

-2a(a – 1) = 0…(From (1), we have b = -2a)

a(a – 1) = 0

a = 0 or a = 1

if a = 0  b = 0

if a = 1  b = -2

Now a and b can’t be zero at same time, so correct solution is

a = 1 and b = -2

a + b = -1

So, the correct option is (d).Question 26

A quadratic equation whose one root is 2 and the sum of whose roots is zero, is

(a) x² + 4 = 0

(b) x² – 4 = 0

(c) 4x² – 1 = 0

(d) x² – 2 = 0Solution 26

Given sum of roots is zero and one root is 2.

So the other root must be -2

so any quadratic equation having root 2 and -2 is

(x – 2) (x – (-2)) = 0

(x – 2) (x + 2) = 0

x² – 4 = 0

So, the correct option is (b).Question 27

If one root of the equation ax² + bx + c = 0 is three times the other, then b² : ac =

(a) 3 : 1

(b) 3 : 16

(c) 16 : 3

(d) 16 : 1Solution 27

Question 28

If one root of the equation 2x² + kx + 4 = 0 is 2, then the other root is

(a) 6

(b) -6

(c) -1

(d) 1Solution 28

Question 29

If one root of the equation x² + ax + 3 = 0 is 1, then its other root is

(a) 3

(b) -3

(c) 2

(d) -2Solution 29

x² + ax + 3 = 0

product of roots = 3

One root is 1. Hence other root is 3.

So, the correct option is (a).

Chapter 4 – Quadratic Equations Exercise 4.85

Question 30

If one root of the equation 4x² – 2x + (λ – 4) = 0 is a reciprocal of the other, then λ =

(a) 8

(b) -8

(c) 4

(d) -4Solution 30

Question 31

Solution 31

Question 32

Solution 32

Any quadratic equation, ax² + bx + c = 0 has real and equal roots if b² – 4ac = 0

For the question, equation is 16x² + 4kx + 9 = 0,

(4k)² – 4 × 16 × 9 = 0

16k² – 36 × 16 = 0

k² – 36 = 0

k² = 36

k = ± 6

So, the correct option is (c).

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.1

Question 1Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and graphically.Solution 1

Question 2Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’ this interesting?) Represent this situation algebraically and graphically.Solution 2Let the present age of Aftab and his daughter be x and y respectively.

Seven years ago,
Age of Aftab = x – 7
Age of his daughter = y – 7

According to the given condition,

Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3

According to the given condition,

Thus, the given conditions can be algebraically represented as:
x – 7y = -42
x – 3y = 6

Three solutions of this equation can be written in a table as follows:

x	-7	0	7
y	5	6	7

Three solutions of this equation can be written in a table as follows:

x	6	3	0
y	0	-1	-2

The graphical representation is as follows:

Concept insight: In order to represent a given situation mathematically, first see what we need to find out in the problem. Here, Aftab and his daughter’s present age needs to be found so, so the ages will be represented by variables x and y. The problem talks about their ages seven years ago and three years from now. Here, the words ‘seven years ago’ means we have to subtract 7 from their present ages, and ‘three years from now’ or ‘three years hence’ means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.Question 3

Solution 3

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 7The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.Solution 7Let the cost of 1 kg of apples and 1 kg grapes be Rsx and Rsy.
The given conditions can be algebraically represented as:

Three solutions of this equation can be written in a table as follows:

x	50	60	70
y	60	40	20

Three solutions of this equation can be written in a table as follows:

x	70	80	75
y	10	-10	0

The graphical representation is as follows:

Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1 kg grapes will be taken as the variables. From the given conditions of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then, in order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are large so take the suitable scale.

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Since, the graph of the two lines coincide, the given system of equations have infinitely many solutions.Question 13

Solution 13

Question 14

Solution 14

Question 15Show graphically that each one of the following systems of equations is in-coinsistent (i.e. has no solution):

3x – 5y = 20

6x – 10y = – 40Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19(i)

Solution 19(i)

Question 19(ii)

Solution 19(ii)

Question 20

Solution 20

Question 21(i)

Solution 21(i)

Question 21(ii)

Solution 21(ii)

Question 22(i)

Solution 22(i)

Question 22(ii)

Solution 22(ii)

Question 22(iii)Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.

2x + y – 11 = 0

x – y – 1 = 0Solution 22(iii)

Question 22(iv)Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.

x + 2y – 7 = 0

2x – y – 4 = 0Solution 22(iv)

Question 22(v)

Solution 22(v)

Question 22(vi)

Solution 22(vi)

Question 23(i)

Solution 23(i)

Question 23(ii)

Solution 23(ii)

Question 23(iii)

Solution 23(iii)

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28(i)

Solution 28(i)

Question 28(ii)

Solution 28(ii)

Question 28(iii)

Solution 28(iii)

Question 28(iv)

Solution 28(iv)

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis. Calculate the area of the triangle so formed.Solution 32

Three solutions of this equation can be written in a table as follows:

x	0	1	2
y	-5	0	5

x	0	1	2
y	-3	0	3

The graphical representation of the two lines will be as follows:

It can be observed that the required triangle is ABC.
The coordinates of its vertices are A (1, 0), B (0, -3), C (0, -5).

Concept insight: In order to find the coordinates of the vertices of the triangle so formed, find the points where the two lines intersects the y-axis and also where the two lines intersect each other. Here, note that the coordinates of the intersection of lines with y-axis is taken and not with x-axis, this is because the question says to find the triangle formed by the two lines and the y-axis.

Question 33

Form the pair of linear equations in the following problems, and find their solutions graphically.

(i)  10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii)  5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

(iii) Champa went to a ‘Sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased”. Help her friends to find how many pants and skirts Champa a bought.Solution 33

(i) Let the number of girls and boys in the class be x and y respectively.
According to the given conditions, we have:
x + y = 10
x – y = 4
x + y = 10  x = 10 – y
Three solutions of this equation can be written in a table as follows:

x	4	5	6
y	6	5	4

x – y = 4  x = 4 + y
Three solutions of this equation can be written in a table as follows:

x	5	4	3
y	1	0	-1

The graphical representation is as follows:

From the graph, it can be observed that the two lines intersect each other at the point (7, 3).
So, x = 7 and y = 3.

Thus, the number of girls and boys in the class are 7 and 3 respectively.


(ii)    Let the cost of one pencil and one pen be Rs x and Rs y respectively.

According to the given conditions, we have:
5x + 7y = 50
7x + 5y = 46

Three solutions of this equation can be written in a table as follows:

x	3	10	-4
y	5	0	10

Three solutions of this equation can be written in a table as follows:

x	8	3	-2
y	-2	5	12

The graphical representation is as follows:


From the graph, it can be observed that the two lines intersect each other at the point (3, 5).
So, x = 3 and y = 5.

Therefore, the cost of one pencil and one pen are Rs 3 and Rs 5 respectively.



(iii)

Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are:

y = 2x – 2 …(1)

and y = 4x – 4 …(2)

Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations.

They are given in Table

x	2	0
y = 2x – 2	2	-2

x	0	1
y = 4x – 4	-4	0

Plot the points and draw the lines passing through them to represent the equations, as shown in fig.,

The two lines intersect at the point (1,0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.

Concept insight: Read the question carefully and examine what are the unknowns. Represent the given conditions with the help of equations by taking the unknowns quantities as variables. Also carefully state the variables as whole solution is based on it. On the graph paper, mark the points accurately and neatly using a sharp pencil. Also, take at least three points satisfying the two equations in order to obtain the correct straight line of the equation. Since joining any two points gives a straight line and if one of the points is computed incorrect will give a wrong line and taking third point will give a correct line. The point where the two straight lines will intersect will give the values of the two variables, i.e., the solution of the two linear equations. State the solution point.Question 34(i)

Solution 34(i)

Question 34(ii)

Solution 34(ii)

Question 35

Solution 35

Question 36

Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) Intersecting lines

(ii) parallel lines

(iii) Coincident linesSolution 36

(i) For the two lines a₁x + b₁x + c₁ = 0 and a₂x + b₂x + c₂ = 0, to be intersecting, we must have

So, the other linear equation can be 5x + 6y – 16 = 0
 
(ii)  For the two lines a₁x + b₁x + c₁ = 0 and a₂x + b₂x + c₂ = 0, to be parallel, we must have

So, the other linear equation can be 6x + 9y + 24 = 0,

(iii)  For the two lines a₁x + b₁x + c₁ = 0 and a₂x + b₂x + c₂ = 0 to be coincident, we must have

So, the other linear equation can be 8x + 12y – 32 = 0,


Concept insight: In order to answer such type of problems, just remember the conditions for two lines to be intersecting, parallel, and coincident. This problem will have multiple answers as their can be many equations satisfying the required conditions.Question 37(i)

Solution 37(i)

Question 37(ii)

Solution 37(ii)

Question 38

Graphically, solve the following pair of equations:

2x + y = 6

2x – y + 2 = 0

Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.Solution 38

The lines AB and CD intersect at point R(1, 4). Hence, the solution of the given pair of linear equations is x = 1, y = 4.

From R, draw RM ⊥ X-axis and RN ⊥ Y-axis.

Then, from graph, we have

RM = 4 units, RN = 1 unit, AP = 4 units, BQ = 4 units

Question 39

Determine, graphically, the vertices of the triangle formed by the lines y = x, 3y = x, x + y = 8.Solution 39

From the graph, the vertices of the triangle AOP formed by the given lines are A(4, 4), O(0, 0) and P(6, 2).Question 40

Draw the graph of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also, find the area of the quadrilateral formed by the lines and the x-axis.Solution 40

The graph of x = 3 is a straight line parallel to Y-axis at a distance of 3 units to the right of Y-axis.

The graph of x = 5 is a straight line parallel to Y-axis at a distance of 5 units to the right of Y-axis.

Question 41

Draw the graphs of the lines x = -2, and y = 3. Write the vertices of the figure formed by these lines, the x-axis and the y-axis. Also, find the area of the figure.Solution 41

The graph of x = -2 is a straight line parallel to Y-axis at a distance of 2 units to the left of Y-axis.

The graph of y = 3 is a straight line parallel to X-axis at a distance of 3 units above X-axis.

Question 42

Draw the graphs of the pair of linear equations x – y + 2 = 0 and 4x – y – 4 = 0. Calculate the area of the triangle formed by the lines so drawn and the x-axis.Solution 42

Question 5

Solve the following equations graphically:

x – y + 1 = 0

3x + 2y – 12 = 0Solution 5

Given equations are:

x – y + 1 = 0 … (i)

3x + 2y – 12 = 0 … (ii)

From (i) we get, x = y – 1

When x = 0, y = 1

When x = -1, y = 0

When x = 1, y = 2

We have the following table:

x	0	-1	1
y	1	0	2

From (ii) we get,  

When x = 0, y = 6

When x = 4, y = 0

When x = 2, y = 3

We have the following table:

x	0	4	2
y	6	0	3

Graph of the given equations is:

As the two lines intersect at (2, 3). 

Hence, x = 2, y = 3 is the solution of the given equations.

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solve the following systems of equation:

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solve the pair of equations:

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solve the following systems of equation:

21x + 47y = 110

47x + 21y = 162Solution 48

Question 49

If x + 1 is a factor of 2x³ + ax² + 2bx + 1, the find the values of a and b given that 2a – 3b = 4.Solution 49

Question 50

Find the solution of the pair of equations  and . Hence, find λ, if y = λx + 5.Solution 50

Question 51

Find the values of x and y in the following rectangle.

Solution 51

Question 52

Write an equation of a line passing through the point representing solution of the pair of linear equations x + y = 2 and 2x – y = 1. How many such lines can we find?Solution 52

Question 53

Write a pair of linear equations which has the unique solution x = -1, y = 3. How many such pairs can you write?Solution 53

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.5

Question 29

Find c if the system of equations cx + 3y + 3 – c = 0, 12x + cy – c = 0 has infinitely many solutions.Solution 29

The given system of equations will have infinite number of solutions if

Question 1

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

x – 3y = 3

3x – 9y = 2Solution 1

Question 2

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

2x + y = 5

4x + 2y = 10Solution 2

Question 3

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

3x – 5y = 20

6x – 10y = 40Solution 3

Question 4

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

x – 2y = 8

5x – 10y = 10Solution 4

Question 5

Find the value of k for which the following system of equations has a unique solution:

kx + 2y = 5

3x + y = 1Solution 5

Question 6Find the value of k for which the following system of equations has a unique solution:

4x + ky + 8 = 0

2x + 2y + 2 = 0Solution 6

Question 7

Find the value of k for which the following system of equations has a unique solution:

4x – 5y = k

2x – 3y = 12Solution 7

Question 8

Find the value of k for which the following system of equations has a unique solution:

x + 2y = 3

5x + ky + 7 = 0Solution 8

Question 9

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y – 5 = 0

6x + ky – 15 = 0Solution 9

Question 10

Find the value of k for which the following systems of equations have infinitely many solutions:

4x + 5y = 3

kx + 15y = 9Solution 10

Question 11

Find the value of k for which the following systems of equations have infinitely many solutions:

kx – 2y + 6 = 0

4x – 3y + 9 = 0Solution 11

Question 12

Find the value of k for which the following systems of equations have infinitely many solutions:

8x + 5y = 9

kx + 10y = 18Solution 12

Question 13

Find the value of k for which the following systems of equations have infinitely many solutions:

2x – 3y = 7

(k + 2)x – (2k + 1)y = 3(2k – 1)Solution 13

Question 14

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y = 2

(k + 2)x + (2k + 1)y = 2(k – 1)Solution 14

Question 15

Find the value of k for which the following systems of equations have infinitely many solutions:

x + (k + 1)y = 4

(k + 1)x + 9y = (5k + 2)Solution 15

Question 16

Find the value of k for which the following systems of equations have infinitely many solutions:

kx + 3y = 2k + 1

2(k + 1)x + 9y = 7k + 1Solution 16

Question 17

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + (k – 2)y = k

6x + (2k – 1)y = 2k + 5Solution 17

Question 18

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y = 7

(k + 1)x + (2k – 1)y = 4k + 1Solution 18

Question 19

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y = k

(k – 1)x + (k + 2)y = 3kSolution 19

Question 20Find the value of k for which the following system of equations has no solution:

kx – 5y = 2

6x + 2y = 7Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Find he value of k for which of the following system of equation has no solution:

kx + 3y = k – 3

12x + ky = 6Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 30

Solution 30

Question 31For what value of k, the following system of equations will represent the coincident lines?

x + 2y + 7 = 0

2x + ky + 14 = 0Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36 (i)

Solution 36 (i)

Question 36 (ii)

Solution 36 (ii)

Question 36 (iii)

Solution 36 (iii)

Question 36 (iv)

Solution 36 (iv)

Question 36 (v)

Find the values of a and b for which the following system of equations has infinitely many solutions:

2x + 3y = 7

(a – b) x + (a + b)y = 3a + b – 2Solution 36 (v)

Question 36 (vi)

Solution 36 (vi)

Question 36 (vii)

Solution 36 (vii)

Question 36(viii)

Find the values of a and b for which the following system of equations has infinitely many solutions:

x + 2y = 1

(a – b)x + (a + b)y = a + b – 2Solution 36(viii)

Question 36(ix)

Find the values of a and b for which the following system of equations has infinitely many solutions:

2x + 3y = 7

2ax + ay = 28 – bySolution 36(ix)

Question 37(i)

For which value(s) of λ, do the pair of linear equations λx + y = λ² and x + λy = 1 have no solution?Solution 37(i)

Question 37(ii)

For which value(s) of λ, do the pair of linear equations λx + y = λ² and x + λy = 1 have infinitely many solutions?Solution 37(ii)

Question 37(iii)

For which value(s) of λ, do the pair of linear equations λx + y = λ² and x + λy = 1 have a unique solution?Solution 37(iii)

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.8

Question 9

A fraction becomes 1/3 when 2 is subtracted from the numerator and it becomes 1/2 when 1 is subtracted from the denominator. Find the fraction.Solution 9

Let the fraction be 

According to the given conditions, we have

Subtracting (ii) from (i), we get x = 7

Substituting the value of x in (ii), we get

y = 15Question 1

Solution 1

Question 2A fraction becomes 9/11 if 2 is added to both numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.Solution 2

Question 3

Solution 3

Question 4If we add 1 to the numerator and subtract 1 from the denominator, a fraction becomes 1. It also becomes 1/2 if we only add 1 to the denominator. What is the fraction?Solution 4

Question 5

The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9Old

Solution 9Old

Question 10

Solution 10

Question 11

Solution 11

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solve each of the following systems of equations by the method of cross-multiplication:

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Jamila sold a table and a chair for Rs.1050, thereby making a profit of 10% on a table and 25% on the chair. If she had taken profit of 25% on the table and 10% on the chair she would have got Rs.1065. Find the cost price of each.Solution 7

Question 8

Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received Rs.1860 as annual interest. However, had she interchanged the amount of investment in the two schemes, she would have received Rs.20 more as annual interest. How much money did she invest in each scheme?Solution 8

Question 9

The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, he buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.Solution 9

Question 10

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.Solution 10

Question 11

The cost of 4 pens and 4 pencils boxes is Rs.100. Three times the cost of a pen is Rs.15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.Solution 11

Question 12

One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital?Solution 12

Question 13

A and B each have a certain number of mangoes. A says to B, “if you give 30 of your mangoes, I will have twice as many as left with you. “B replies, “if you give me 10, I will have thrice as many as left with you. “How many mangoes does each have?Solution 13

Question 14

Vijay had some bananas, and he divided them into two lots A and B. He sold first lot at the rate of Rs.2 for 3 bananas and the second lot at the rate of Rs.1 per banana and got a total of Rs.400. If he had sold the first lot at the rate of Rs.1 per banana and the second lot at the rate of Rs.4 per five bananas, his total collection would have been Rs.460. Find the total number of bananas he had.Solution 14

Question 15

Solution 15

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5The sum of two-digit number and the number formed by reversing the order of digits is 66. If the two digits differ by 2, find the number. How many such numbers are there?Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14The sum of 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.Solution 14

Question 15

Solution 15

Question 16

Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.Solution 16

Question 17

A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.Solution 17

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6The present age of a father is three more than three times the age of the son. Three years hence father’s age will be 10 years more than twice the age of the son. Determine their present ages.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differs by 30 years. Find the ages of Ani and Biju.Solution 11

The difference between the ages of Ani and Biju is given as 3 years. So, either Biju is 3 years older than Ani or Ani is 3 years older than Biju.

Let the age of Ani and Biju be x years and y years respectively.
Age of Dharam = 2 × x = 2x years
Case I: Ani is older than Biju by 3 years
x – y = 3        … (1)

4x – y = 60     ….(2)
Subtracting (1) from (2), we obtain: 3x = 60 – 3 = 57

Age of Ani = 19 years
Age of Biju = 19 – 3 = 16 years

Case II: Biju is older than Ani by 3 years
y – x = 3        … (3)

4x – y = 60        … (4)

Adding (3) and (4), we obtain:
3x = 63
x = 21

Age of Ani = 21 years
Age of Biju = 21 + 3 = 24 years

Concept Insight: In this problem, ages of Ani and Biju are the unknown quantities. So, we represent them by variables x and y. Now, note that here it is given that the ages of Ani and Biju differ by 3 years. So, it is not mentioned that which one is older. So, the most important point in this question is to consider both cases  Ani is older than Biju and  Biju is older than Ani. For second condition the relation  on the ages of Dharam and Cathy can be implemented . Pair of linear equations can be solved using a suitable algebraic method.

Question 12

Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now?Solution 12

Question 13

The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.Solution 13

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.10

Question 1

Solution 1

Question 2

Solution 2

Question 3The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boat in still water.Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

A person rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of stream.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.Solution 12

Question 13

A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.Solution 13

Question 14

Solution 14

Question 15A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time.  Find the distance covered by the train.Solution 15Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.
Or, d = xt        … (1)

According to the question,

By using equation (1), we obtain:
3x – 10t = 30        … (3)

Adding equations (2) and (3), we obtain:
x = 50
Substituting the value of x in equation (2), we obtain:
(-2) x (50) + 10t = 20
-100 + 10t = 20
10t = 120
t = 12
From equation (1), we obtain:
d = xt = 50 x 12 = 600

Thus, the distance covered by the train is 600 km.

Concept insight: To solve this problem, it is very important to remember the relation . Now, all these three quantities are unknown. So, we will represent these

by three different
variables. By using the given conditions, a pair of equations will be obtained. Mind one thing that the equations obtained will not be linear. But they can be reduced to linear form by using the fact that . Then two linear equations can be formed which can

be solved easily by elimination method.

Question 16Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hours. What are the speeds of two cars?Solution 16

Question 17

Solution 17

Question 18

Solution 18

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.11

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

ABCD is a cyclic quadrilateral such that A = (4y + 20)^o, B = (3y – 5)^o, C = (-4x)^o and D = (7x + 5)^o. Find the four angles.Solution 6

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.

A + C = 180
4y + 20 – 4x = 180
-4x + 4y = 160
x – y = -40            … (1)
Also, B + D = 180
3y – 5 – 7x + 5 = 180
-7x + 3y = 180        … (2)
Multiplying equation (1) by 3, we obtain:

3x – 3y = -120        … (3)
Adding equations (2) and (3), we obtain:
-4x = 60
x = -15
Substituting the value of x in equation (1), we obtain:
-15 – y = -40
y = -15 + 40 = 25

A = 4y + 20 = 4(25) + 20 = 120^o
B = 3y – 5 = 3(25) – 5 = 70^o
C = -4x = -4(-15) = 60^o
D = -7x + 5 = -7(-15) + 5 = 110^o

Concept insight: The most important idea to solve this problem is by using the fact that the sum of the measures of opposite angles in a cyclic quadrilateral is 180^o. By using this relation, two linear equations can be obtained which can be solved easily by eliminating a suitable variable.Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12The car hire charges in a city comprise of fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is Rs 89 and for a journey of 20 km, the charge paid is Rs. 145. What will a person have to pay for travelling a distance of 30 km?Solution 12

Question 13A part of monthly hostel charges in a college are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, he has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charge and the cost of food per day.Solution 13

Question 14

Solution 14

Question 15The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.Solution 15

Question 162 Women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroideery, and that taken by 1 man alone.Solution 16

Question 17

Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes Rs 50 and Rs 100 she received.Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row there would be 2 rows more. Find number of students in the class.Solution 21

Question 22

One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital?Solution 22

Let the money with the first person and second person be Rs x and Rs y respectively.

According to the question,
x + 100 = 2(y – 100)
x + 100 = 2y – 200
x – 2y = -300        … (1)

6(x – 10) = (y + 10)
6x – 60 = y + 10
6x – y = 70            … (2)

Multiplying equation (2) by 2, we obtain:
12x – 2y = 140        … (3)
Subtracting equation (1) from equation (3), we obtain:
11x = 140 + 300
11x = 440
x = 40
Putting the value of x in equation (1), we obtain:
40 – 2y = -300
40 + 300 = 2y
2y = 340
y = 170

Thus, the two friends had Rs 40 and Rs 170 with them.

Concept insight: This problem talks about the amount of capital with two friends. So, we will represent them by variables x and y respectively. Now, using the given conditions, a pair of linear equations can be formed which can then be solved easily using elimination method.

Question 23

A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby getting a sum of Rs.1008. If she had sold the saree at 10% profit and sweater at 8% discount, she would have got Rs.1028. Find the cost price of the saree and the list price (price before discount) of the sweater.Solution 23

Question 24

In a competitive examination, one mark is awarded for each correct answer while ½ mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?Solution 24

Question 25

A shopkeeper gives book on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid Rs.22 for a book kept for 6 days, while Rs.16 for the book kept for four days. Find the fixed charges and charge for each extraday.Solution 25

Chapter 3 Pairs of Linear Equations in Two Variables Exercise 3.114

Question 1

Solution 1

So, the correct option is (b).Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 3 Pairs of Linear Equations in Two Variables Exercise 3.115

Question 6

Solution 6

Question 7

If   am ≠ bl, then the system of equations

ax + by = c

lx + my = n

(a)  has a unique solution        

(b) has no solution

(c) has infinite many solution    

(d) may or may not have a solutionSolution 7

So, the correct option is (a).Question 8

Solution 8

So, the correct option is (b).Question 9

Solution 9

So, the correct option is (a).Question 10

If 2x-3y=7 and (a+b)x – (a+b-3)y = 4a+b represent coincident lines than a and b satisfy the equation

(a) a+5b=0    (b) 51+b=0    (c) a-5b=0     (d) 5a-b=0Solution 10

So, the correct option is (c).Question 11

If a pair of linear equations in two variables is consistent, then the lines represented by two equations are

(a) Intersecting             (b) parallel

(c) always coincident       (d) intersecting or coincidentSolution 11

Consistent solution means either linear equations have unique solutions or infinite solutions.

⇒ In case of unique solution; lines are intersecting

⇒ If solutions are infinite, lines are coincident.

So, lines are either intersecting or coincident

So, the correct option is (d).Question 12

Solution 12

So, the correct option is (c).Question 13

The area of the triangle formed by the lines

y=x,  x=6, and  y=0  is

(a) 36 sq. units      

(b) 18 sq. units

(c) 9 sq. units         

(d) 72 sq. unitsSolution 13

So, the correct option is (b).Question 14

If the system of equations 2x + 3y=5,  4x + ky =10 has infinitely many solutions, then k=

(a) 1     

(b)     

(c) 3    

(d) 6Solution 14

So, the correct option is (d).Question 15

If the system of equations  kx – 5y = 2,  6x +2y=7 has no solution, then k=

(a) -10     

(b) -5      

(c) -6     

(d)-15Solution 15

So, the correct option is (d).Question 16

The area of the triangle formed by the lines

x = 3, y = 4 and x = y is

(a)  sq. unit    

(b) 1 sq. unit

(c) 2 sq. unit     

(d) None of theseSolution 16

So, the correct option is (a).

Chapter 3 – Pairs of Linear Equations in Two Variables Exercise 3.116

Question 17

The area of the triangle formed by the lines

2x + 3y = 12,     x – y – 1= 0  and x = 0

(a) 7 sq. units          

(b) 7.5 sq. units

(c) 6.5 sq. units        

(d) 6 sq. units   Solution 17

So, the correct option is (b).Question 18

The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the number get reversed. The number is

1. 25
2. 72
3. 63
4. 36

Solution 18

Question 19

If x = a, y = b is the solution of the system of equations x – y = 2 and x + y = 4, then the values of a and b are, respectively

1. 3 and 1
2. 3 and 5
3. 5 and 3
4. -1 and -3

Solution 19

Since x = a and y = b is the solution of given system of equations x – y = 2 and x + y = 4, we have

a – b = 2 ….(i)

a + b = 4 ….(ii)

Adding (i) and (ii), we have

2a = 6 ⇒ a = 3

⇒ b = 4 – 3 = 1

Hence, correct option is (a).Question 20

For what value k, do the equations 3x – y + 8 = 0 and 6x – ky + 16 = 0 represent coincident lines?

1. 
2. 
3. 2
4. -2

Solution 20

Question 21

Aruna has only Rs.1 and Rs.2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs.75, then the number of Rs.1 and Rs.2 coins are, respectively

1. 35 and 15
2. 35 and 20
3. 15 and 35
4. 25 and 25

Solution 21

Chapter 2 Polynomials Exercise Ex. 2.1

Question 1(i)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

f(x) = x² – 2x – 8 Solution 1(i)

x² – 2x – 8 = x² – 4x + 2x – 8 = x(x – 4) + 2(x – 4) = (x – 4)(x + 2) The zeroes of the quadratic equation are 4 and -2. Let ∝ = 4 and β = -2 Consider f(x) = x² – 2x – 8 Sum of the zeroes = …(i) Also, ∝ + β = 4 – 2 = 2 …(ii) Product of the zeroes = …(iii) Also, ∝ β = -8 …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients isverified.Question 1(ii)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

g(s) = 4s² – 4s + 1 Solution 1(ii)

4s² – 4s + 1 = 4s² – 2s – 2s + 1 = 2s(2s – 1) – (2s – 1) = (2s – 1)(2s – 1) The zeroes of the quadratic equation are and . Let ∝ =  and β =  Consider4s² – 4s + 1 Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(iii)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

h (t) = t² – 15 Solution 1(iii)

h (t) = t² – 15 = (t + √15)(t – √15)  The zeroes of the quadratic equation areand . Let ∝ =  and β =  Considert² – 15 = t² – 0t – 15 Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(iv)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

f(s) = 6x² – 3 – 7x Solution 1(iv)

f(s) = 0 6x² – 3 – 7x =0 6x² – 9x + 2x – 3 = 0 3x (2x – 3) + (2x – 3) = 0 (3x + 1) (2x – 3) = 0 The zeroes of a quadratic equation are  and . Let ∝ =  and β =  Consider6x² – 7x – 3 = 0 Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(v)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(v)

 The zeroes of a quadratic equation are and . Let ∝ =  and β =  Consider Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(vi)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(vi)

 The zeroes of a quadratic equation are and . Let ∝ =  and β =  Consider Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(vii)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(vii)

 The zeroes of a quadratic equation are  and 1. Let ∝ =  and β = 1 Consider Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(viii)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(viii)

 The zeroes of a quadratic equation are a and. Let ∝ = a and β = Consider Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(ix)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(ix)

 The zeroes of a quadratic equation are and . Let ∝ =  and β =  Consider Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(x)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(x)

 The zeroes of a quadratic equation are and . Let ∝ =  and β =  Consider=0 Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(xi)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(xi)

 The zeroes of a quadratic equation are  and. Let ∝ =  and β = Consider=0 Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(xii)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(xii)

 The zeroes of a quadratic equation are  and. Let ∝ =  and β = Consider=0 Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 2(i)

For the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorization.

Solution 2(i)

Question 2(ii)

For the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorization.

Solution 2(ii)

Question 2(iii)

For the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorization.

Solution 2(iii)

Question 2(iv)

For the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorization.

Solution 2(iv)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

(i)



(ii)



(iii)



(iv)



(v)



(vi)



(vii)



(viii)

 

Chapter 2 Polynomials Exercise Ex. 2.2

Question 1Verify that the numbers given along side of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in each case:

Solution 1

On comparing the given polynomial with the polynomial ax³ + bx² + cx + d, we obtain a = 2, b = 1, c = -5, d = 2

Thus, the relationship between the zeroes and the coefficients is verified.

On comparing the given polynomial with the polynomial ax^₃ + bx^₂ + cx + d, we obtain a = 1, b = -4, c = 5, d = -2.

Thus, the relationship between the zeroes and the coefficients is verified.

Concept insight: The zero of a polynomial is that value of the variable which makes the polynomial 0. Remember that there are three  relationships between the zeroes of a cubic polynomial and its coefficients which involve the sum of zeroes, product of all zeroes and the product of zeroes taken two at a time.

Question 2

Solution 2

Question 3

Find all zeroes of the polynomial 3x³ + 10x² – 9x – 4, if one of its zeroes is 1.Solution 3

Let f(x) = 3x³ + 10x² – 9x – 4 

As 1 is one of the zeroes of the polynomial, so (x – 1) becomes the factor of f(x).

Dividing f(x) by (x – 1), we have

Hence, the zeroes are  Question 4

If 4 is a zero of the cubic polynomial x³ – 3x² – 10x + 24, find its other two zeroes.Solution 4

Let f(x) = x³ – 3x² – 10x + 24 

As 4 is one of the zeroes of the polynomial, so (x – 4) becomes the factor of f(x).

Dividing f(x) by (x – 4), we have

Hence, the zeroes are 4, -3 and 2.Question 5

Solution 5

Question 6

Solution 6

Chapter 2 Polynomials Exercise Ex. 2.3

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Find all zeros of the polynomial 2x⁴ – 9x³ + 5x² + 3x – 1, if two of its zeros are  Solution 11

Let f(x) = 2x⁴ – 9x³ + 5x² + 3x – 1 

As  are two of the zeroes of the polynomial, so  becomes the factor of f(x).

Dividing f(x) by  we have

Hence, the other zeros  Question 12

For what value of k, is the polynomial f(x) = 3x⁴ – 9x³ + x² + 15x + k completely divisible by 3x² – 5?Solution 12

Let f(x) = 3x⁴ – 9x³ + x² + 15x + k 

As f(x) is completely divisible by 3x² – 5, it becomes one of the factors of f(x).

Dividing f(x) by 3x² – 5, we have

As (3x² – 5) is one of the factors, the remainder will be 0.

Therefore, k + 10 = 0

Thus, k = -10.Question 13

Solution 13

Question 14

Solution 14

Chapter 2 Polynomials Exercise 2.61

Question 1

(a) 1

(b) -1

(c) 0

(d) None of theseSolution 1

Question 2

Solution 2

So, the correct option is (d).Question 3

If one zero of the polynomial f(x) = (k² + 4)x² + 13x + 4k is reciprocal of the other, then k =

(a) 2

(b) -2

(c) 1

(d) -1Solution 3

Chapter 2 Polynomials Exercise 2.62

Question 4

If the sum of the zeros of the polynomial f(x) = 2x³ – 3kx² + 4x – 5 is 6, then value of k is

(a) 2

(b) 4

(c) -2

(d) -4Solution 4

So, the correct option is (b).Question 5

(a) x² + qx + p

(b) x² – px + q

(c) qx² + px + 1

(d) px² + qx + 1Solution 5

Question 6

If α, β are the zeros of polynomial f(x) = x² – p(x + 1) – c, then (α + 1) (β + 1) =

(a) c – 1

(b) 1 – c

(c) c

(d) 1 + cSolution 6

Question 7

If α, β are the zeros of the polynomial f(x) = x² – p(x + 1) – c such that (α + 1) (β + 1) = 0 then c =

(a) 1

(b) 0

(c) -1

(d) 2Solution 7

Given that (α + 1) (β + 1) = 0

So, the correct option is (a).Question 8

If f(x) = ax² + bx + c has no real zeros and a + b + c < 0, then

(a) c = 0

(b) c > 0

(c) c < 0

(d) None of theseSolution 8

We know that, if the quadratic equation ax² + bx + c = 0 has no real zeros

then

Case 1:

a > 0, the graph of quadratic equation should not intersect x – axis

It must be of the type

Case 2 :

a < 0, the graph will not intersect x – axis and it must be of type

According to the question,

a + b + c < 0

This means,

f(1) = a + b + c

f(1) < 0

Hence, f(0) < 0 [as Case 2 will be applicable]

 So, the correct option is (c).Question 9

If the diagram in figure show the graph of the polynomial f(x) = ax² + bx + c then

(a) a < 0, b < 0 and c > 0

(b) a < 0, b < 0 and c < 0

(c) a < 0, b > 0 and c > 0

(d) a < 0, b > 0 and c < 0Solution 9

Question 10

Figure shows the graph of the polynomial f(x) = ax² + bx + c for which

(a) a < 0, b > 0 and c > 0

(b) a > 0, b < 0 and c > 0

(c) a < 0, b < 0 and c < 0

(d) a > 0, b > 0 and c < 0Solution 10

Question 11

Solution 11

Question 12

If zeros of the polynomial f(x) = x³ – 3px² + qx – r are in A.P, then

(a) 2p³ = pq – r

(b) 2p³ = pq + r

(c) p³ = pq – r

(d) None of theseSolution 12

Chapter 2 Polynomials Exercise 2.63

Question 13

Solution 13

So, the correct option is (b).Question 14

Solution 14

Question 15

In Q. No. 14, c =

(a) b

(b) 2b

(c) 2b²

(d) -2bSolution 15

So, the correct option is (c).Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

If two of the zeros of the cubic polynomial ax³  + bx² + cx + d are each equal to zero then the third zero is

Solution 21

Question 22

Solution 22

Question 23

The product of the zeros of x³ + 4x² + x – 6 is 

(a) -4

(b) 4

(c) 6

(d) -6Solution 23

Chapter 2 Polynomials Exercise 2.64

Question 24

What should be added to the polynomial x² – 5x + 4, so that 3 is the zero of the resulting polynomial ?

(a) 1

(b) 2

(c) 4

(d) 5Solution 24

We know that, if α and β are roots of ax² + bx + c = 0 then they must satisfy the equation.

According to the question, the equation is

x² – 5x + 4 = 0

If 3 is the root of equation it must satisfy equation.

x² – 5x + 4 = 0

but f(3) = 3² – 5(3) + 4 = -2

so, 2 has to be added in the equation.

So, the correct option is (b).Question 25

What should be subtracted to the polynomial x² – 16x + 30, so that 15 is the zero of resulting polynomial?

(a) 30

(b) 14

(c) 15

(d) 16Solution 25

We know that, if α and β are roots of ax²  + bx + c = 0, then α and β must satisfy the equation.

According to the question, the equation is

x² – 16x + 30 = 0

If 15 is a root, then it must satisfy the equation x² – 16x + 30 = 0, 

But f(15) = 15² – 16(15) + 30 = 225 – 240 + 30 = 15

and so 15 should be subtracted from the equation.

So, the correct option is (c).Question 26

A quadratic polynomial, the sum of whose zeros is 0 and one zero is 3, is 

(a) x² – 9

(b) x² + 9

(c) x²  + 3

(d) x² – 3Solution 26

Question 27

If two zeros of the polynomial x³  + x² – 9x – 9 are 3 and -3, then its third zero is

(a) -1

(b) 1

(c) -9

(d) 9Solution 27

Question 28

Solution 28

Question 29

If x + 2 is a factor of x²  + ax + 2b and a + b = 4, then

(a) a = 1, b = 3

(b) a = 3, b = 1

(c) a = -1, b = 5

(d) a = 5, b = -1Solution 29

Question 30

The polynomial which when divided by -x² + x – 1 gives a quotient x – 2 and remainder is 3, is

(a) x³ – 3x² + 3x – 5

(b) -x³ – 3x² – 3x – 5

(c) -x³ + 3x²  – 3x + 5

(d) x³ – 3x² – 3x + 5Solution 30

We know that 

Dividend = Divisor × quotient  + remainder

Then according to question,

Required polynomial

= (-x² + x – 1) (x – 2) + 3

= -x³ + 2x² + x² -2x – x + 2 + 3

= -x³ + 3x² – 3x + 5

So, the correct option is (c).Question 31

The number of polynomials having zeroes -2 and 5 is

a. 1

b. 2

c. 3

d. more than 3Solution 31

Correct option: (d)

The polynomials having -2 and 5 as the zeroes can be written in the form 

k(x + 2)(x – 5), where k is a constant. 

Thus, number of polynomials with roots -2 and 5 are infinitely many, since k can take infinitely many values.Question 32

If one of the zeroes of the quadratic polynomial (k – 1)x² + kx + 1 is – 3, then the value of k is

a. 

b. 

c. 

d. Solution 32

Question 33

The zeroes of the quadratic polynomial x² + 99x + 127 are

a. Both positive

b. Both negative

c. both equal

d. One positive and one negativeSolution 33

The zeroes of the quadratic polynomial x² + 99x + 127 are both negative since all terms are positive.

Hence, correct option is (b).Question 34

If the zeroes of the quadratic polynomial x² + (a + 1)x + b are 2 and -3, then

a. a = -7, b = -1

b. a = 5, b = -1

c. a = 2, b = -6

d. a = 0, b = -6Solution 34

Question 35

Given that one of the zeroes of the cubic polynomial ax³ + bx² + cx + d is zero, the product of the other two zeroes is

a. 

b. 

c. 0

d. Solution 35

Question 36

The zeroes of the quadratic polynomial x² + ax + a, a ≠ 0,

a. cannot both be positive

b. cannot both be negative

c. are always unequal

d. are always equalSolution 36

Question 37

If one of the zeros of the cubic polynomial x³ + ax² + bx + x is -1, then the product of other two zeroes is

a. b – a + 1

b. b – a – 1

c. a – b + 1

d. a – b – 1Solution 37

Chapter 2 Polynomials Exercise 2.65

Question 38

Given that two of the zeros of the cubic polynomial ax³ + bx² + cx + d are 0, the third zero is

a. 

b. 

c. 

d. Solution 38

Question 39

If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is

a. 10

b. -10

c. 5

d. -5Solution 39

Question 40

If the zeros of the quadratic polynomial ax² + bx + c, c ≠ 0 are equal , then

a. c and a have opposite signs

b. c and b have opposite signs

c. c and a have the same sign

d. c and b have the same signSolution 40

It is given that the zeros of the quadratic polynomial ax² + bx + c, c ≠ 0 are equal.

⇒ Discriminant = 0

⇒ b² – 4ac = 0

⇒ b² = 4ac

Now, b² can never be negative,

Hence, 4ac also can never be negative.

⇒ a and c should have same sign.

Hence, correct option is (c).Question 41

If one of the zeros of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it

a. has no linear term and constant term is negative.

b. has no linear term and the constant term is positive

c. can have a linear term but the constant term is negative

d. can have a linear term but the constant term is positiveSolution 41

Question 42

Which of the following is not the graph of a quadratic polynomial?

Solution 42

The graph of a quadratic polynomial crosses X-axis at atmost two points.

Hence, correct option is (d).