R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

Chapter 10 - Quadratic Equations Ex. 4A

Solution 19

Hence, are the roots of

Solution 1(i)

(i)is a quadratic polynomial

= 0 is a quadratic equation

Solution 1(ii)

Clearly  is a quadratic polynomial

is a quadratic equation.

Solution 1(iii)

 is a quadratic polynomial

 = 0 is a quadratic equation

Solution 1(iv)

Clearly, is a quadratic equation

is a quadratic equation

Solution 1(v)

is not a quadratic polynomial since it contains in which power  of x is not an integer.

is not a quadratic equation

Solution 1(vi)

And Being a polynomial of degree 2, it is a quadratic polynomial.

Hence, is a quadratic equation.

Solution 1(vii)

And being a polynomial of degree 3, it is not a quadratic polynomial

Hence, is not a quadratic equation


Solution 1(viii)

is not a quadratic equation

Solution 1(ix)

Solution 1(x)

Solution 1(xi)

Solution 2

The given equation is


(i)On substituting x = -1 in the equation, we get


(ii)On substituting in the equation, we get

 

(iii)On substituting in the equation , we get

Solution 3(i)

Since x = 1 is a solution of it must satisfy the equation.

 

 

Hence the required value of k = -4

Solution 3(ii)

Since is a root of , we have

Again x = -2 being a root of , we have

Multiplying (2) by 4 adding the result from (1), we get

11a = 44 a = 4

Putting a = 4 in (1), we get

Solution 5

Solution 6

Solution 7

Hence, 9 and -9 are the roots of the equation

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Hence, are the roots of

Solution 14

Hence, are the roots of equation

Solution 15

Hence, and 1 are the roots of the equation .

Solution 16

are the roots of the equation

Solution 17

Hence, are the roots of the given equation

Solution 18

Hence, are the roots of given equation

Solution 20

Solution 21

Solution 22

Hence, are the roots of the given equation

Solution 23

Hence, are the roots of the given equation

Solution 24

Hence, are the roots of the given equation

Solution 25

Hence, are the roots of the given equation

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Hence, 1 and are the roots of the given equation

Solution 34

Solution 35

Solution 36

Solution 37

Hence, are the roots of the given equation

Solution 38

Hence, 2 and are the roots of given equation

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Hence, are the roots of the given equation

Solution 47

Solution 48

Hence, are the roots of given equation

Solution 49

Hence, are the roots of given equation

Solution 50

Hence, are the roots of given equation

Solution 51

Solution 52

Solution 53

Solution 54

Solution 55(i)

Solution 56

Solution 57

Solution 58

Solution 59(i)

Solution 60

Solution 61

Solution 62

Solution 63

Solution 64(i)

Solution 65

Solution 66

Solution 67

Solution 68

Putting the given equation become

Case I:

Case II:

Hence, are the roots of the given equation

Solution 69

The given equation

Hence, is the roots of the given equation

Solution 70

Solution 71

Hence, -2,0 are the roots of given equation

Solution 72

 

 

Hence, are the roots of given equation

Solution 73

Hence, 3 and 2 are roots of the given equation

Chapter 10 - Quadratic Equations Ex. 4B

Solution 4

Solution 5

Solution 1(i)

Solution 1(ii)

  

Solution 1(iii)

  

Solution 1(iv)

  

Solution 1(v)

 

Solution 1(vi)

  

Solution 2

  

Solution 3

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Chapter 10 - Quadratic Equations Ex. 10E

Solution 8

Let the required number be x and x - 3, then

Hence, the required numbers are (24,21) or (-21 and-24)

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Let the smaller part and larger part be x, 16 - x

Then,

-42 is not a positive part

Hence, the larger and smaller parts are 10, 6 respectively

Solution 21

Let the required number be x and y, hen

Solution 22

Let x, y be the two natural numbers and x > y

------(1)

Also, square of smaller number = 4 larger number

---------(2)

Putting value of from (1), we get

Thus, the two required numbers are 9 and 6

Solution 23

Let the three consecutive numbers be x, x + 1, x + 2

Sum of square of first and product of the other two

Required numbers are 4, 5 and 6

Solution 24

Let the tens digit be x and units digit be y

Hence, the tens digit is 3 and units digit is (2 3) is

Hence the required number is 36

Solution 25

Let the tens digit and units digits of the required number be x and y respectively

The ten digit is 2 and unit digit is 7

Hence, the required number is 27

Solution 26

Let the numerator and denominator be x, x + 3

Then,

Hence, numerator and denominator are 2 and 5 respectively and fraction is

Solution 27

Solution 28

Solution 29

Let there be x rows and number of student in each row be x

Then, total number of students =

Hence total number of student

Total number of students is 600

Solution 30

Let the number of students be x, then

Hence the number of students is 50

Solution 31

Let the marks obtained by Kamal in Mathematics and English be x and y

The marks obtained by Kamal in Mathematics and English respectively are (21,19) or (12,28)

Solution 38

Let the age of son be x and age of man = y

1 year ago

Solution 41

Let the present age of Meena be x

Then,

Hence the present age of Meena is 7 years

Solution 43

 

  

Solution 44

Solution 45

Solution 46

  

Solution 47

Solution 48

Let the original speed of the train be x km.hour

Then speed increases by 15 km/ph = (x + 15)km/hours

Then time taken at original speed =

Then, time taken at in increased speed =

Difference between the two lines taken

Then, original speed of the train = 45km / h

Solution 49

Solution 50

Let the speed of the Deccan Queen = x kmph

The, speed of other train = (x - 20)kmph

Then, time taken by Deccan Queen =

Time taken by other train =

Difference of time taken by two trains is 

Hence, speed of Deccan Queen = 80km/h

Solution 51

Let the speed of stream be x km/h

Speed of boat in still stream = 18 km/h

Speed of boat up the stream = 18 - x km/h

Time taken by boat to go up the stream 24 km =

Time taken by boat to go down the stream =

Time taken by the boat to go up the stream is 1 hour more that the time taken down the stream

Speed of the stream = 6 km/h

Solution 52

Let the speed of the stream be x kmph

Then the speed of boat down stream = (8 + x) kmph

And the speed of boat upstream = (8 - x)kmph

Time taken to cover 15 km upstream =

Time taken to cover 22 km downstream =

Total time taken = 5 hours

Hence, the speed of stream is 3 kmph

Solution 53

Let the speed of the stream be = x km/h

Speed of boat in still waters = 9 km/h

Speed of boat down stream = 9 + x

time taken by boat to go 15 km downstream =

Speed of boat upstream = 9 - x

time taken by boat to go 15 km of stream =

Solution 54

 

Solution 55

Let the faster pipe takes x minutes to fill the cistern

Then, the other pipe takes (x + 3) minute

The faster pipe takes 5 minutes to fill the cistern

Then, the other pipe takes (5 + 3) minutes = 8 minutes

Solution 56

Solution 57

Solution 58

Let the breadth of a rectangle = x cm

Then, length of the rectangle = 2x cm

Thus, breadth of rectangle = 12 cm

And length of rectangle = (2 12) = 24 cm

Solution 59

Let the breadth of a rectangle = x meter

Then, length of rectangle = 3x meter

Thus, breadth of rectangle = 7 m

And length of rectangle = (3 7)m = 21 m

Solution 60

Let the breadth of hall = x meters

Then, length of the hall = (x + 3) meters

Area = length breadth =

Thus, the breadth of hall is 14 cm

And length of the hall is (14 + 3) = 17 cm

Solution 61

Solution 62

Let the width of the path be x meters,

Then,

Area of path = 16 10 - (16 - 2x) (10 - 2x) = 120

Hence the required width is 3 meter as x cannot be 10m

Solution 63

Let x and y  be the lengths of the two square fields.

4x - 4y = 64

x - y = 16------(2)

From (2),

x = y + 16,

Putting value of x in (1)

Sides of two squares are 24m and 8m respectively

Solution 64

Let the side of square be x cm

Then, length of the rectangle = 3x cm

Breadth of the rectangle = (x - 4) cm

Area of rectangle = Area of square x

Thus, side of the square = 6 cm

And length of the rectangle = (3 6) = 18 cm

Then, breadth of the rectangle = (6 - 4) cm = 2 cm

Solution 65

Let the length = x meter

Area = length breadth =

If ength of the rectangle = 15 m

Also, if length of rectangle = 24 m

Solution 66

Let the altitude of triangle be x cm

Then, base of triangle is (x + 10) cm

Hence, altitude of triangle is 30 cm and base of triangle 40 cm

Solution 67

Let the altitude of triangle be x meter

Hence, base = 3x meter

Hence, altitude of triangle is 8 cm

And base of triangle = 3x = (3 8) cm = 24 cm

Solution 68

Let the base of triangle be x meter

Then, altitude of triangle = (x + 7) meter

Thus, the base of the triangle = 15 m

And the altitude of triangle = (15 + 7) = 22 m

Solution 69

Let the other sides of triangle be x and (x -4) meters

By Pythagoras theorem, we have

Thus, height of triangle be = 16 cm

And the base of the triangle = (16 - 4) = 12 cm

Solution 70

Let the base of the triangle be x

Then, hypotenuse = (x + 2) cm

Thus, base of triangle = 15 cm

Then, hypotenuse of triangle = (15 +2 )= 17 cm

And altitude of triangle =

Solution 71

Let the shorter side of triangle be x meter

Then, its hypotenuse = (2x - 1)meter

And let the altitude = (x + 1) meter

Chapter 10 - Quadratic Equations Ex. 10F

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

 

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Solution 54

Solution 55

Solution 14

Chapter 4 - Quadratic Equations MCQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 14

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 30

Solution 31

 

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Solution 54

Solution 55

Solution 29

Chapter 4 - Quadratic Equations Ex. 4C

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1(vi)

Solution 2

Solution 3

The given equation is

This is the form of

Now .

So, the roots of the given equation are real for all real value of p and q.

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8(i)

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

The given equation is


For real and equal roots, we must have D = 0

Solution 15

The given equation is


For real and equal roots , we must have D =0

Solution 16

Solution 17

Solution 18

Solution 19(i)

Solution 19(ii)

Solution 19(iii)

Solution 19(iv)

Solution 20

Solution 21

Solution 23

Chapter 4 - Quadratic Equations Ex. 4D

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Let the required number be x and x - 3, then

Hence, the required numbers are (24,21) or (-21 and-24)

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Let the smaller part and larger part be x, 16 - x

Then,

-42 is not a positive part

Hence, the larger and smaller parts are 10, 6 respectively

Solution 21

Let the required number be x and y, hen

Solution 22

Let x, y be the two natural numbers and x > y

------(1)

Also, square of smaller number = 4 larger number

---------(2)

Putting value of from (1), we get

Thus, the two required numbers are 9 and 6

Solution 23

Let the three consecutive numbers be x, x + 1, x + 2

Sum of square of first and product of the other two

Required numbers are 4, 5 and 6

Solution 24

Let the tens digit be x and units digit be y

Hence, the tens digit is 3 and units digit is (2 3) is

Hence the required number is 36

Solution 25

Let the tens digit and units digits of the required number be x and y respectively

The ten digit is 2 and unit digit is 7

Hence, the required number is 27

Solution 26

Let the numerator and denominator be x, x + 3

Then,

Hence, numerator and denominator are 2 and 5 respectively and fraction is

Solution 27

Solution 28

Solution 29

Let there be x rows and number of student in each row be x

Then, total number of students =

Hence total number of student

Total number of students is 600

Solution 30

Let the number of students be x, then

Hence the number of students is 50

Solution 31

Let the marks obtained by Kamal in Mathematics and English be x and y

The marks obtained by Kamal in Mathematics and English respectively are (21,19) or (12,28)

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38(i)

Let the age of son be x and age of man = y

1 year ago

Solution 39

Solution 40

Solution 41

Let the present age of Meena be x

Then,

Hence the present age of Meena is 7 years

Solution 42

Solution 43

 

  

Solution 44

Solution 45

Solution 46

  

Solution 47

Solution 48

Let the original speed of the train be x km.hour

Then speed increases by 15 km/ph = (x + 15)km/hours

Then time taken at original speed =

Then, time taken at in increased speed =

Difference between the two lines taken

Then, original speed of the train = 45km / h

Solution 49

Solution 50

Let the speed of the Deccan Queen = x kmph

The, speed of other train = (x - 20)kmph

Then, time taken by Deccan Queen =

Time taken by other train =

Difference of time taken by two trains is 

Hence, speed of Deccan Queen = 80km/h

Solution 51

Let the speed of stream be x km/h

Speed of boat in still stream = 18 km/h

Speed of boat up the stream = 18 - x km/h

Time taken by boat to go up the stream 24 km =

Time taken by boat to go down the stream =

Time taken by the boat to go up the stream is 1 hour more that the time taken down the stream

Speed of the stream = 6 km/h

Solution 52

Let the speed of the stream be x kmph

Then the speed of boat down stream = (8 + x) kmph

And the speed of boat upstream = (8 - x)kmph

Time taken to cover 15 km upstream =

Time taken to cover 22 km downstream =

Total time taken = 5 hours

Hence, the speed of stream is 3 kmph

Solution 53

Let the speed of the stream be = x km/h

Speed of boat in still waters = 9 km/h

Speed of boat down stream = 9 + x

time taken by boat to go 15 km downstream =

Speed of boat upstream = 9 - x

time taken by boat to go 15 km of stream =

Solution 54

 

Solution 55

Let the faster pipe takes x minutes to fill the cistern

Then, the other pipe takes (x + 3) minute

The faster pipe takes 5 minutes to fill the cistern

Then, the other pipe takes (5 + 3) minutes = 8 minutes

Solution 56

Solution 57

Solution 58

Let the breadth of a rectangle = x cm

Then, length of the rectangle = 2x cm

Thus, breadth of rectangle = 12 cm

And length of rectangle = (2 12) = 24 cm

Solution 59

Let the breadth of a rectangle = x meter

Then, length of rectangle = 3x meter

Thus, breadth of rectangle = 7 m

And length of rectangle = (3 7)m = 21 m

Solution 60

Let the breadth of hall = x meters

Then, length of the hall = (x + 3) meters

Area = length breadth =

Thus, the breadth of hall is 14 cm

And length of the hall is (14 + 3) = 17 cm

Solution 61

Solution 62

Let the width of the path be x meters,

Then,

Area of path = 16 10 - (16 - 2x) (10 - 2x) = 120

Hence the required width is 3 meter as x cannot be 10m

Solution 63

Let x and y  be the lengths of the two square fields.

4x - 4y = 64

x - y = 16------(2)

From (2),

x = y + 16,

Putting value of x in (1)

Sides of two squares are 24m and 8m respectively

Solution 64

Let the side of square be x cm

Then, length of the rectangle = 3x cm

Breadth of the rectangle = (x - 4) cm

Area of rectangle = Area of square x

Thus, side of the square = 6 cm

And length of the rectangle = (3 6) = 18 cm

Then, breadth of the rectangle = (6 - 4) cm = 2 cm

Solution 65

Let the length = x meter

Area = length breadth =

If ength of the rectangle = 15 m

Also, if length of rectangle = 24 m

Solution 66

Let the altitude of triangle be x cm

Then, base of triangle is (x + 10) cm

Hence, altitude of triangle is 30 cm and base of triangle 40 cm

Solution 67

Let the altitude of triangle be x meter

Hence, base = 3x meter

Hence, altitude of triangle is 8 cm

And base of triangle = 3x = (3 8) cm = 24 cm

Solution 68

Let the base of triangle be x meter

Then, altitude of triangle = (x + 7) meter

Thus, the base of the triangle = 15 m

And the altitude of triangle = (15 + 7) = 22 m

Solution 69

Let the other sides of triangle be x and (x -4) meters

By Pythagoras theorem, we have

Thus, height of triangle be = 16 cm

And the base of the triangle = (16 - 4) = 12 cm

Solution 70

Let the base of the triangle be x

Then, hypotenuse = (x + 2) cm

Thus, base of triangle = 15 cm

Then, hypotenuse of triangle = (15 +2 )= 17 cm

And altitude of triangle =

Solution 71

Let the shorter side of triangle be x meter

Then, its hypotenuse = (2x - 1)meter

And let the altitude = (x + 1) meter

Chapter 4 - Quadratic Equations 29

Solution 29