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Chapter 20 Definite Integrals Exercise Ex. 20.1

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

Solution 23

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

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Evaluate the Integral in using substitution.

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Let cosx =u , Then

Hence

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Given :

Question 61

Solution 61

Question 62

Solution 62

Question 63

Solution 63

Question 64

Solution 64

We know , By reduction formula 

For n=2

For n=4

Hence

Note: Answer given at back is incorrect.Question 65

Solution 65

Using Integration By parts

Question 66

Solution 66

Question 67

Solution 67

Note: Answer given in the book is incorrect. Question 68

Solution 68

 =(1/4)log(2e)

Chapter 20 Definite Integrals Exercise Ex. 20.2

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

Using Integration By parts

Hence

Question 25

Solution 25

Question 26

Solution 26

Question 27

Evaluate 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

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

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

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Chapter 20 Definite Integrals Exercise Ex. 20.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

2x+3 is positive for x>-1.5 . Hence

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

Evaluate the integral Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

For

Using Integration By parts

For

Using Integration By parts

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 27

Solution 27

[x]=0 for 0 < x

and [x]=1 for 1< x < 2

Hence

Question 28

Solution 28

Question 26

Evaluate the following integrals:

Solution 26

NOTE: Answer not matching with back answer.

Chapter 20 Definite Integrals Exercise Ex. 20.4A

Question 1

Solution 1

We know

Hence

We know

If

Then also

Hence

Question 2

Solution 2

We know

Hence

If

Then

Question 3

Solution 3

We know

Hence

If

Then

So

Question 4

Solution 4

We know

Hence

If

Then

Hence

Question 5

Solution 5

We know

Hence

If

Then

So

We know

If f(x) is even

If f(x) is odd

Here

f(x) is even, hence

Note: Answer given in the book is incorrect.Question 6

Solution 6

We know

Hence

If

Then

So

Question 7

Solution 7

We know

Hence

If

Then

So

Question 8

Solution 8

We know

Hence

If

Then

So

Note: Answer given in the book is incorrect. Question 9

Solution 9

If f(x) is even

If f(x) is odd

Here

 is odd and

 is even. Hence

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

Chapter 20 Definite Integrals Exercise Ex. 20.4B

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

BQuestion 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

Hence

Question 20

Solution 20

Question 21

Solution 21

Now

Let cosx=t

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25 (i)

Solution 25 (i)

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 (i)

Solution 32 (i)

Question 33

Solution 33

Question 34

Evaluate the integral Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

We know

Also here

So

Hence

Question 39

Solution 39

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Chapter 20 Definite Integrals Exercise Ex. 20RE

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Evaluate the following integrals

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

Evaluate the integral 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

Evaluate the following integrals

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

Evaluate the following integrals

Solution 38

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

Solution 47

Question 48

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

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

Solution 65

Question 66

Solution 66

Question 67

Solution 67

Question 68

Solution 68

Question 69

Solution 69

Chapter 20 Definite Integrals Exercise Ex. 20.5

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

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Evaluate the following in tegrals as limit of sums

Solution 31

Question 32

Evaluate the following in tegrals as limit of sums

Solution 32

Question 33

Solution 33

Exercise Ex. 7A

Question 1

Solve each of the following systems of equations graphically:

2x + 3y = 2

x – 2y = 8Solution 1

Question 3

Solve each of the following systems of equations graphically:

2x + 3y = 8

x – 2y + 3 = 0Solution 3

Question 4

Solve each of the following systems of equations graphically:

2x – 5y + 4 = 0

2x + y – 8 = 0Solution 4

Question 5

Solve each of the following systems of equations graphically:

3x + 2y = 12, 5x – 2y = 4.Solution 5

Since the two graphs intersect at (2, 3),

x = 2 and y = 3.Question 6

Solve each of the following systems of equations graphically:

3x + y + 1 = 0

2x – 3y + 8 = 0Solution 6

Question 7

Solve each of the following systems of equations graphically:

2x + 3y + 5 = 0, 3x – 2y – 12 = 0.Solution 7

Since the two graphs intersect at (2, -3),

x = 2 and y = -3.Question 8

Solve each of the following systems of equations graphically:

2x – 3y + 13 = 0, 3x – 2y + 12 = 0.Solution 8

Since the two graphs intersect at (-2, 3),

x = -2 and y = 3.Question 9

Solve each of the following systems of equations graphically:

2x + 3y – 4 = 0, 3x – y + 5 = 0.Solution 9

Since the two graphs intersect at (-1, 2),

x = -1 and y = 2.Question 10

Solve each of the following systems of equations graphically:

x + 2y + 2 = 0

3x + 2y – 2 = 0Solution 10

Question 11

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – y + 3 = 0, 2x + 3y – 4 = 0.Solution 11

Question 12

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

2x – 3y + 4 = 0, x + 2y – 5 = 0.Solution 12

Question 13

Solve the following system of linear equations graphically:

4x – 3y + 4 = 0, 4x + 3y – 20 = 0

Find the area of the region bounded by these lines and the x-axis.Solution 13

Question 14

Solve the following system of linear equation graphically:

x – y + 1 = 0, 3x + 2y – 12 = 0

Calculate the area bounded by these lines and x-axis.Solution 14

Question 15

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – 2y + 2 = 0, 2x + y – 6 = 0.Solution 15

Question 16

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

2x – 3y + 6 = 0, 2x + 3y – 18 = 0.Solution 16

Question 17

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

4x – y – 4 = 0, 3x + 2y – 14 = 0.Solution 17

Question 18

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

x – y – 5 = 0, 3x + 5y – 15 = 0.Solution 18

Question 19

Solve the following system of linear equations graphically:

2x – 5y + 4 = 0, 2x + y – 8 = 0

Find the point, where these lines meet the y-axisSolution 19

Question 20

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

5x – y – 7 = 0, x – y + 1 = 0.Solution 20

Question 21

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

2x – 3y = 12, x + 3y = 6.Solution 21

Question 22

Show graphically that each of the following given systems of equations has infinitely many solutions:

2x + 3y = 6, 4x + 6y = 12.Solution 22

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions.Question 23

Show graphically that the system of equations 3x – y = 5, 6x – 2y = 10 has infinitely many solutions.Solution 23

Question 24

Show graphically that the system of equations 2x + y = 6, 6x + 3y = 18 has infinitely many solutions.Solution 24

Question 25

Show graphically that each of the following given systems of equations has infinitely many solutions:

x – 2y = 5, 3x – 6y = 15.Solution 25

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions. Question 26

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

x – 2y = 6, 3x – 6y = 0Solution 26

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 27

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

2x + 3y = 4, 4x + 6y = 12.Solution 27

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 28

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

2x + y = 6, 6x + 3y = 20.Solution 28

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 29

Draw the graphs of the following equations on the same graph paper:

2x + y = 2, 2x + y = 6.

Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.Solution 29

Exercise Ex. 7B

Question 1

Solve for x and y

x + y = 3

4x – 3y = 26Solution 1

Question 2

Solve for x and y:

Solution 2

Question 3

Solve for x and y

2x + 3y = 0

3x + 4y = 5Solution 3

Question 4

Solve for x and y

2x – 3y = 13

7x – 2y = 20Solution 4

Question 5

Solve for x and y

3x – 5y – 19 = 0

-7x + 3y + 1 = 0Solution 5

Question 6

Solve for x and y:

2x – y + 3 = 0, 3x – 7y + 10 = 0.Solution 6

Question 7

Solve for x and y:

Solution 7

Question 8

Solve for x and y

Solution 8

Question 9

Solve for x and y

Solution 9

Question 10

Solve for x and y

Solution 10

Question 11

Solve for x and y

Solution 11

Question 12

Solve for x and y

Solution 12

Question 13

Solve for x and y:

0.4x + 0.3y = 1.7, 0.7x – 0.2y = 0.8.Solution 13

Question 14

Solve for x and y:

0.3x + 0.5y = 0.5, 0.5x + 0.7y = 0.74.Solution 14

Question 15

Solve for x and y

7(y + 3) – 2(x + 2) = 14

4(y – 2) + 3(x – 3) = 2Solution 15

Question 16

Solve for x and y

6x + 5y = 7x + 2y + 1 = 2(x + 6y – 1)Solution 16

Question 17

Solve for x and y

Solution 17

Question 18

×Solve for x and y

Solution 18

Putting the given equations become

5u + 6y = 13—(1)

3u + 4y = 7 —-(2)

Multiplying (1) by 4 and (2) by 6, we get

20u + 24y = 52—(3)

18u + 24y = 42—(4)

Subtracting (4) from (3), we get

2u = 10 u = 5

Putting u = 5 in (1), we get

5 × 5 + 6y = 13

6y = 13 – 25

6y = -12

y = -2

Question 19

Solve for x and y

Solution 19

The given equations are and 

Putting 

x + 6v = 6 —-(1)

3x – 8v = 5—(2)

Multiplying (1) by 4 and (2) by 3

4x + 24v = 24—(3)

9x – 24v = 15 —(4)

Adding (3) and (4)

13x = 24 + 15 = 39

Puttingx = 3 in (1)

3 + 6v = 6

6v = 6 – 3 = 3

solution is x = 3, y = 2Question 20

Solve for x and y

Solution 20

Putting in the given equation

2x – 3v = 9 —(1)

3x + 7v = 2 —(2)

Multiplying (1) by7 and (2) by 3

14x – 21v = 63 —(3)

9x + 21v = 6 —(4)

Adding (3) and (4), we get

Putting x= 3 in (1), we get

2 × 3 – 3v = 9

         -3v = 9 – 6

     -3v= 3

       v = -1

the solution is x = 3, y = -1Question 21

Solve for x and y

Solution 21

Question 22

Solve for x and y

Solution 22

Putting in the equation

9u – 4v = 8 —(1)

13u + 7v = 101—(2)

Multiplying (1) by 7 and (2) by 4, we get

63u – 28v = 56—(3)

52u + 28v = 404—(4)

Adding (3) and (4), we get

Putting u = 4 in (1), we get

9 × 4 – 4v = 8

36 – 4v = 8

-4v = 8 – 36

-4v = -28

Question 23

Solve for x and y

Solution 23

Putting in the given equation, we get

5u – 3v = 1 —(1)

Multiplying (1) by 4 and (2) by 3, we get

20u – 12v = 4—-(3)

27u + 12v = 90—(4)

Adding (3) and (4), we get

Putting u = 2 in (1), we get

(5 × 2) – 3v = 1

10 – 3v = 1

-3v = 1 – 10 -3v = -9 

 v = 3

Question 24

Solve for x and y:

Solution 24

Question 25

Solve for x and y

4x + 6y = 3xy

8x + 9y = 5xy;Solution 25

4x + 6y = 3xy

Putting in (1) and (2), we get

4v + 6u = 3—(3)

8v + 9u = 5—(4)

Multiplying (3) by 9 and (4) by 6, we get

36v + 54u = 27 —(5)

48v + 54u = 30 —(6)

Subtracting (3) from (4), we get

12v = 3

Putting in (3), we get

the solution is x = 3, y = 4Question 26

Solve for x and y:

Solution 26

Question 27

Solve for x and y:

Solution 27

Question 28

Solve for x and y

Solution 28

Putting 

3u + 2v = 2—-(1)

9u – 4v = 1—-(2)

Multiplying (1) by 2 and (2) by 1. We get

6u + 4v = 4—-(3)

9u – 4v = 1—-(4)

Adding (3) and (4), we get

Adding (5) and (6), we get

Putting in (5). We get

the solution is Question 29

Solve for x and y

Solution 29

The given equations are

Putting 

Adding (1) and (2)

Putting value of u in (1)

Hence the required solution isx = 4, y = 5Question 30

Solve for x and y

Solution 30

Putting in the equation, we get

44u + 30v = 10—-(1)

55u + 40v = 13—-(2)

Multiplying (1) by 4 and (2) by 3, we get

176u + 120v = 40—(3)

165u + 120v = 39—(4)

Subtracting (4) from (3), we get

Putting in (1) we get

Adding (5) and (6), we get

Putting x = 8 in (5), we get

8 + y = 11 y = 11 – 8 = 3

the solution is x = 8, y = 3Question 31

Solve for x and y:

Solution 31

Question 32

Solve for x and y

71x + 37y = 253

37x + 71y = 287Solution 32

The given equations are

71x + 37y = 253—(1)

37x + 71y = 287—(2)

Adding (1) and (2)

108x + 108y = 540

108(x + y) = 540

—-(3)

Subtracting (2) from (1)

34x – 34y = 253 – 287 = -34

34(x – y) = -34

—(4)

Adding (3) and (4)

2x = 5 – 1= 4

Subtracting (4) from (3)

2y = 5 + 1 = 6

solution is x = 2, y = 3Question 33

Solve for x and y

217x + 131y = 913

131x + 217y = 827Solution 33

217x + 131y = 913—(1)

131x + 217y = 827—(2)

Adding (1) and (2), we get

348x + 348y = 1740

348(x + y) = 1740

x + y = 5—-(3)

Subtracting (2) from (1), we get

86x – 86y = 86

86(x – y) = 86

   x – y = 1—(4)

Adding (3) and (4), we get

2x = 6

x = 3

putting x = 3 in (3), we get

3 + y = 5

y = 5 – 3 = 2

solution is x = 3, y = 2Question 34

Solve for x and y:

23x – 29y = 98, 29x – 23y = 110.Solution 34

Question 35

Solve for x and y:

Solution 35

Question 36

Solve for x and y:

Solution 36

Question 37

Solve for x and y

where Solution 37

The given equations are

Multiplying (1) by 6 and (2) by 20, we get

Multiplying (3) by 6 and (4) by 5, we get

18u + 60v = -54—(5)

125u – 60v = —(6)

Adding (5) and (6), we get

Question 38

Solve for x and y:

Solution 38

Question 39

Solve for x and y:

Solution 39

Question 40

Solve for x and y:

x + y = a + b, ax – by = a² – b².Solution 40

Question 41

Solve for x and y

Solution 41

Question 42

Solve for x and y:

px + py = p – q, qx – py = p + q.Solution 42

Question 43

Solve for x and y:

Solution 43

Question 44

Solve for x and y

6(ax + by) = 3a + 2b

6(bx – ay) = 3b – 2aSolution 44

6(ax + by) = 3a + 2b

6ax + 6by = 3a + 2b —(1)

6(bx – ay) = 3b – 2a

6bx – 6ay = 3b- 2a —(2)

6ax + 6by = 3a + 2b —(1)

6bx – 6ay = 3b – 2a —(2)

Multiplying (1) by a and (2) by b

Adding (3) and (4), we get

Substituting  in(1), we get

Hence, the solution is  Question 45

Solve for x and y:

ax – by = a² + b², x + y = 2a.Solution 45

Question 46

Solve for x and y

bx – ay + 2ab = 0Solution 46

Question 47

Solve for x and y

x + y = 2abSolution 47

Taking L.C.M, we get

Multiplying (1) by 1 and (2) by 

Subtracting (4) from (3), we get

Substituting x = ab in (3), we get

solution is x = ab, y = abQuestion 48

Solve for x and y:

x + y = a + b, ax – by = a² – b².Solution 48

Question 49

Solve for x and y:

a²x + b²y = c², b²x + a²y = d².Solution 49

Question 50

Solve for x and y:

Solution 50

Exercise Ex. 7C

Question 1

Solve for x and y by method of cross multiplication:

x + 2y + 1 = 0

2x – 3y – 12 = 0Solution 1

x + 2y + 1 = 0 —(1)

2x – 3y – 12 = 0 —(2)

By cross multiplication, we have

Hence, x = 3 and y = -2 is the solutionQuestion 2

Solve for x and y by method of cross multiplication:

3x – 2y + 3 = 0

4x + 3y – 47 = 0Solution 2

3x – 2y + 3 = 0

4x + 3y – 47 = 0

By cross multiplication we have

the solution is x = 5, y = 9Question 3

Solve for x and y by method of cross multiplication:

6x – 5y – 16 = 0

7x – 13y + 10 = 0Solution 3

6x – 5y – 16 = 0

7x – 13y + 10 = 0

By cross multiplication we have

the solution is x = 6, y = 4Question 4

Solve for x and y by method of cross multiplication:

3x + 2y + 25 = 0

2x + y + 10 = 0Solution 4

3x + 2y + 25 = 0

2x + y + 10 = 0

By cross multiplication, we have

the solution is x = 5,y = -20Question 5

Solve for x and y by method of cross multiplication:

2x +5y = 1

2x + 3y = 3Solution 5

2x + 5y – 1 = 0 —(1)

2x + 3y – 3 = 0—(2)

By cross multiplication we have

the solution is x = 3, y = -1Question 6

Solve for x and y by method of cross multiplication:

2x + y – 35 = 0

3x + 4y – 65 = 0Solution 6

2x + y – 35 = 0

3x + 4y – 65 = 0

By cross multiplication, we have

Question 7

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

7x – 2y = 3, 22x – 3y = 16.Solution 7

Question 8

Solve for x and y by method of cross multiplication:

Solution 8

Question 9

Solve for x and y by method of cross multiplication:

Solution 9

Taking 

u + v – 7 = 0

2u + 3v – 17 = 0

By cross multiplication, we have

the solution is Question 10

Solve for x and y by method of cross multiplication:

Solution 10

Let in the equation

5u – 2v + 1 = 0

15u + 7v – 10 = 0

Question 11

Solve for x and y by method of cross multiplication:

Solution 11

Question 12

Solve for x and y by method of cross multiplication:

2ax + 3by – (a + 2b) = 0

3ax+ 2by – (2a + b) = 0Solution 12

2ax + 3by – (a + 2b) = 0

3ax+ 2by – (2a + b) = 0

By cross multiplication, we have

Question 13

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

Solution 13

Exercise Ex. 7D

Question 1

Show that the following system of equations has a unique solution:

3x + 5y = 12, 5x + 3y = 4

Also, find the solution of the given system of equations.Solution 1

Question 2

Show that each of the following systems of equations has a unique solution and solve it:

2x – 3y = 17, 4x + y = 13.Solution 2

Question 3

Show that the following system of equations has a unique solution:

Also, find the solution of the given system of equationsSolution 3

Question 4

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

2x + 3y – 5 = 0, kx – 6y – 8 = 0.Solution 4

Question 5

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

x – ky = 2, 3x + 2y + 5 = 0.Solution 5

Question 6

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

5x – 7y – 5 = 0, 2x + ky – 1 = 0.Solution 6

Question 7

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

4x + ky + 8 = 0, x + y + 1 = 0.Solution 7

Question 8

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

4x – 5y = k , 2x – 3y = 12Solution 8

4x – 5y – k = 0, 2x – 3y – 12 = 0

These equations are of the form

Thus, for all real value of k the given system of equations will have a unique solutionQuestion 9

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

kx + 3y = (k – 3),12x + ky = kSolution 9

kx + 3y – (k – 3) = 0

12x + ky – k = 0

These equations are of the form

Thus, for all real value of k other than , the given system of equations will have a unique solutionQuestion 10

Show that the system of equations

2x – 3y = 5, 6x – 9y = 15

has an infinite number of solutions.Solution 10

2x – 3y – 5 = 0, 6x – 9y – 15 = 0

These equations are of the form

Hence the given system of equations has infinitely many solutionsQuestion 11

Show that the system of equations  Solution 11

Question 12

For what value of k, the system of equations

kx + 2y = 5, 3x – 4y = 10

has (i) a unique solution (ii) no solution?Solution 12

kx + 2y – 5 = 0

3x – 4y – 10 = 0

These equations are of the form

This happens when 

Thus, for all real value of k other that , the given system equations will have a unique solution

(ii) For no solution we must have 

Hence, the given system of equations has no solution if Question 13

For what value of k, the system of equations

x + 2y = 5, 3x + ky + 15 = 0

has (i) a unique solution (ii) no solution?Solution 13

x + 2y – 5= 0

3x + ky + 15 = 0

These equations are of the form of

Thus for all real value of k other than 6, the given system ofequation will have unique solution

(ii) For no solution we must have

k = 6

Hence the given system will have no solution when k = 6.Question 14

For what value of k, the system of equations

x + 2y = 3, 5x + ky + 7 = 0

has (i) a unique solution (ii) no solution?

Is there any value of k for which the given system has an infinite number of solutions?Solution 14

x + 2y – 3 = 0, 5x + ky + 7 = 0

These equations are of the form

(i)For a unique solution we must have

Thus, for all real value of k other than 10

The given system of equation will have a unique solution.

(ii)For no solution we must have

Hence the given system of equations has no solution if 

For infinite number of solutions we must have

This is never possible since 

There is no value of k for which system of equations has infinitely many solutionsQuestion 15

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7

(k – 1)x + (k + 2)y = 3kSolution 15

2x + 3y – 7 = 0

(k – 1)x + (k + 2)y – 3k = 0

These are of the form

This hold only when

Now the following cases arises

Case : I

Case: II

Case III

For k = 7, there are infinitely many solutions of the given system of equationsQuestion 16

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

2x + (k – 2)y =k

6x + (2k – 1)y = (2k + 5)Solution 16

2x + (k – 2)y – k = 0

6x + (2k – 1)y – (2k + 5) = 0

These are of the form

For infinite number of solutions, we have

This hold only when

Case (1)

Case (2)

Case (3)

Thus, for k = 5 there are infinitely many solutionsQuestion 17

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

kx + 3y = (2k +1)

2(k + 1)x + 9y = (7k + 1)Solution 17

kx + 3y – (2k +1) = 0

2(k + 1)x + 9y – (7k + 1) = 0

These are of the form

For infinitely many solutions, we must have

This hold only when

Now, the following cases arise

Case – (1)

Case (2)

Case (3)

Thus, k = 2, is the common value for which there are infinitely many solutionsQuestion 18

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

5x + 2y = 2k

2(k + 1)x + ky = (3k + 4)Solution 18

5x + 2y – 2k = 0

2(k +1)x + ky – (3k + 4) = 0

These are of the form

For infinitely many solutions, we must have

These hold only when

Case I

Thus, k = 4 is a common value for which there are infinitely by many solutions.Question 19

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

(k – 1)x – y = 5

(k + 1)x + (1 – k)y = (3k + 1)Solution 19

(k – 1)x – y – 5 = 0

(k + 1)x + (1 – k)y – (3k + 1) = 0

These are of the form

For infinitely many solution, we must now

k = 3 is common value for which the number of solutions is infinitely manyQuestion 20

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

(k – 3)x + 3y = k, kx + ky = 12.Solution 20

Question 21

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

(a – 1)x + 3y = 2

6x + (1 – 2b)y = 6Solution 21

(a – 1)x + 3y – 2 = 0

6x + (1 – 2b)y – 6 = 0

These equations are of the form

For infinite many solutions, we must have

Hence a = 3 and b = -4Question 22

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

(2a – 1)x + 3y = 5

3x + (b – 1)y = 2Solution 22

(2a – 1)x + 3y – 5 = 0

3x + (b – 1)y – 2 = 0

These equations are of the form

These holds only when

Question 23

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x – 3y = 7

(a + b)x + (a + b – 3)y = (4a + b)Solution 23

2x – 3y – 7 = 0

(a + b)x + (a + b – 3)y – (4a + b) = 0

These equation are of the form

For infinite number of solution

Putting a = 5b in (2), we get

Putting b = -1 in (1), we get

Thus, a = -5, b = -1Question 24

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y=7, (a + b + 1)x +(a + 2b + 2)y = 4(a + b)+ 1.Solution 24

Question 25

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7, (a + b)x + (2a – b)y = 21.Solution 25

Question 26

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7, 2ax + (a + b)y = 28.Solution 26

Question 27

Find the value of k for which each of the following systems of equations no solution:

8x + 5y = 9, kx + 10y = 15.Solution 27

Question 28

Find the value of k for which each of the following systems of equations no solution:

kx + 3y = 3, 12x + ky = 6.Solution 28

Question 29

Find the value of k for which each of the following systems of equations no solution:

Solution 29

Question 30

Find the value of k for which each of the following systems of equations no solution:

kx + 3y = k – 3, 12x + ky = k.Solution 30

Question 31

Find the value of k for which the system of equations

5x – 3y = 0;2x + ky = 0

has a nonzero solution.Solution 31

We have 5x – 3y = 0 —(1)

2x + ky = 0—(2)

Comparing the equation with

These equations have a non – zero solution if 

Exercise Ex. 7E

Question 1

5 chairs and 4 tables together cost Rs.5600, while 4 chairs and 3 tables together cost Rs.4340. Find the cost of a chair and that of a table.Solution 1

Question 2

23 spoons and 17 forks together cost Rs.1770, while 17 spoons and 23 forks together cost Rs.1830. Find the cost of a spoon and that of a fork.Solution 2

Question 3

A lady has only 25-paisa and 50-paisa coins in her purse. If she has 50 coins in all Rs.19.50, how many coins of each kind does she have?Solution 3

Question 4

The sum of two numbers is 137 and their difference is 43. Find the numbers.Solution 4

Let the two numbers be x and y respectively.

Given:

x + y = 137 —(1)

x – y = 43 —(2)

Adding (1) and (2), we get

2x = 180

Putting x = 90 in (1), we get

90 + y = 137

y = 137 – 90

  = 47

Hence, the two numbers are 90 and 47.Question 5

Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.Solution 5

Let the first and second number be x and y respectively.

According to the question:

2x + 3y = 92 —(1)

4x – 7y = 2 —(2)

Multiplying (1) by 7 and (2) by 3, we get

14 x+ 21y = 644 —(3)

12x – 21y = 6 —(4)

Adding (3) and (4), we get

Putting x = 25 in (1), we get

2 × 25 + 3y = 92

50 + 3y = 92

 3y = 92 – 50

y = 14

Hence, the first number is 25 and second is 14Question 6

Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.Solution 6

Let the first and second numbers be x and y respectively.

According to the question:

3x + y = 142 —(1)

4x – y = 138 —(2)

Adding (1) and (2), we get

Putting x = 40 in (1), we get

3 × 40 + y = 142

y = 142 – 120

y = 22

Hence, the first and second numbers are 40 and 22.Question 7

If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the number.Solution 7

Let the greater number be x and smaller be y respectively.

According to the question:

2x – 45 = y

2x – y = 45—(1)

and

2y – x = 21

 -x + 2y = 21—(2)

Multiplying (1) by 2 and (2) by 1

4x – 2y = 90—(3)

-x + 2y = 21 —(4)

Adding (3) and (4), we get

3x = 111

Putting x = 37 in (1), we get

2 × 37 – y = 45

 74 – y = 45

y = 29

Hence, the greater and the smaller numbers are 37 and 29.Question 8

If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.Solution 8

Let the larger number be x and smaller be y respectively.

We know,

Dividend = Divisor × Quotient + Remainder

3x = y × 4 + 8

3x – 4y = 8 —(1)

And

5y = x × 3 + 5

-3x + 5y = 5 —(2)

Adding (1) and (2), we get

y = 13

putting y = 13 in (1)

Hence, the larger and smaller numbers are 20 and 13 respectively.Question 9

If 2 is added to each of two given numbers, their ratio becomes 1: 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5: 11. Find the numbers.Solution 9

Let the required numbers be x and y respectively.

Then,

Therefore,

2x – y =-2—(1)

11x – 5y = 24 —(2)

Multiplying (1) by 5 and (2) by 1

10x – 5y = -10—(3)

11x – 5y = 24—(4)

Subtracting (3) and (4) we get

x = 34

putting x = 34 in (1), we get

 2 × 34 – y = -2

 68 – y = -2

-y = -2 – 68

y = 70

Hence, the required numbers are 34 and 70.Question 10

The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.Solution 10

Let the numbers be x and y respectively.

According to the question:

x – y = 14 —(1)

From (1), we get

x = 14 + y —(3)

putting x = 14 + y in (2), we get

Putting y = 9 in (1), we get

x – 9 = 14

  x = 14 + 9 = 23

Hence the required numbers are 23 and 9Question 11

The sum of the digits of a two digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.Solution 11

Let the ten’s digit be x and units digit be y respectively.

Then,

x + y = 12—(1)

Required number = 10x + y

Number obtained on reversing digits = 10y + x

According to the question:

10y + x – (10x + y) = 18

10y + x – 10x – y = 18

9y – 9x = 18

y – x = 2 —-(2)

Adding (1) and (2), we get

Putting y = 7 in (1), we get

x + 7 = 12

x = 5

Number= 10x + y

                  = 10 × 5 + 7

                  = 50 + 7

                  = 57

Hence, the number is 57.
Question 12

A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.Solution 12

Let the ten’s digit of required number be x and its unit’s digit be y respectively

Required number = 10x + y

 10x + y = 7(x + y)

     10x + y = 7x + 7y

       3x – 6y = 0—(1)

Number found on reversing the digits = 10y + x

(10x + y) – 27 = 10y + x

      10x – x + y – 10y = 27

        9x – 9y = 27

         (x – y) = 27

           x – y = 3—(2)

Multiplying (1) by 1 and (2) by 6

3x – 6y = 0—(3)

6x – 6y = 18 —(4)

Subtracting (3) from (4), we get

Putting x = 6 in(1), we get

3 × 6 – 6y = 0

18 – 6y = 0

Number = 10x + y

            = 10 × 6 + 3

            = 60 + 3

            = 63

Hence the number is 63.
Question 13

The sum of the digits of a two-digit number is 15. The number of obtained by interchanging the digits exceeds the given number by 9. Find the number.Solution 13

Let the ten’s digit and unit’s digits of required number be x and y respectively.

Then,

x + y = 15—(1)

Required number = 10x + y

Number obtained by interchanging the digits = 10y + x

10y + x – (10x + y) = 9

10y + x – 10x – y = 9

       9y – 9x = 9

Add (1) and (2), we get

Putting y = 8 in (1), we get

x + 8 = 15

x = 15 – 8 = 7

Required number = 10x + y

                         = 10 × 7 + 8

                         = 70 + 8

                         = 78

Hence the required number is 78.
Question 14

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.Solution 14

Let the ten’s and unit’s of required number be x and y respectively.

Then,required number =10x + y

According to the given question:

 10x + y = 4(x + y) + 3

 10x + y = 4x + 4y + 3

6x – 3y = 3

  2x – y = 1 —(1)

And

10x + y + 18 = 10y + x

9x – 9y = -18

 x – y = -2—(2)

Subtracting (2) from (1), we get

x = 3

Putting x = 3 in (1), we get

2 × 3 – y = 1

y = 6 – 1 = 5

x = 3, y = 5

Required number = 10x + y

                         = 10 × 3 + 5

                         = 30 + 5

                         = 35

Hence, required number is 35.
Question 15

A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.Solution 15

Let the ten’s digit and unit’s digit of required number be x and y respectively.

We know,

Dividend = (divisor × quotient) + remainder

According to the given questiion:

10x + y = 6 × (x + y) + 0

10x – 6x + y – 6y = 0

4x – 5y = 0 —(1)

Number obtained by reversing the digits is 10y + x

 10x + y – 9 = 10y + x

9x – 9y = 9

9(x – y) =9

(x – y) = 1—(2)

Multiplying (1) by 1 and (2) by 5, we get

4x – 5y = 0 —(3)

5x – 5y = 5 —(4)

Subtracting (3) from (4), we get

x = 5

Putting x = 5 in (1), we get

x =5 and y= 4

Hence, required number is 54.Question 16

A two – digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchanged their places. Find the number.Solution 16

Let the ten’s and unit’s digits of the required number be x and y respectively.

Then, xy = 35

Required number = 10x + y

Also,

(10x + y) + 18 = 10y + x

9x – 9y = -18

9(y – x) = 18—(1)

y – x = 2

Now, 

Adding (1) and (2),

2y = 12 + 2 = 14

y = 7

Putting y = 7 in (1),

7 – x = 2

x = 5

Hence, the required number = 5 × 10 + 7

                                        = 57Question 17

A two-digit number is such that the product of its digit is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.Solution 17

Let the ten’s and unit’s digits of the required number be x and y respectively.

Then, xy = 18

Required number = 10x + y

Number obtained on reversing its digits = 10y + x

(10x + y) – 63 = (10y + x)

9x – 9y = 63

x – y = 7—(1)

Now,

Adding (1) and (2), we get

Putting x = 9 in (1), we get

9 – y = 7

y = 9 – 7

y ₌2

x = 9, y = 2

Hence, the required number = 9 × 10 + 2

                                        = 92.Question 18

The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.Solution 18

Question 19

The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes Find the fraction.Solution 19

Let the numerator and denominator of fraction be x and y respectively.

According to the question:

x + y = 8—(1)

And

Multiplying (1) be 3 and (2) by 1

3x + 3y = 24—(3)

4x – 3y = -3 —(4)

Add (3) and (4), we get

Putting x = 3 in (1), we get

3 + y= 8

y = 8 – 3

y = 5

x = 3, y = 5

Hence, the fraction is Question 20

If 2 is added to the numerator of a fraction, it reduces to and if 1 is subtracted from the denominator, it reduces to . Find the fractionSolution 20

Let the numerator and denominator be x and y respectively.

Then the fraction is .

Subtracting (1) from (2), we get

x = 3

Putting x = 3 in (1), we get

2 × 3 – 4

-y = -4 -6

y = 10

x = 3 and y = 10

Hence the fraction is Question 21

The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes Find the fraction.Solution 21

Let the numerator and denominator be x and y respectively.

Then the fraction is _.

According to the given question:

y = x + 11

y- x = 11—(1)

and

-3y + 4x = -8 —(2)

Multiplying (1) by 4 and (2) by 1

4y – 4x = 44—(3)

-3y + 4x = -8—(4)

Adding (3) and (4), we get

y = 36

Putting y = 36 in (1), we get

y – x = 11

36 – x = 11 

  x = 25

x = 25, y = 36

Hence the fraction is Question 22

Find a fraction which becomes when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes when 7 is subtracted from the numerator and 2 is subtracted from the denominator.Solution 22

Let the numerator and denominator be x and y respectively.

Then the fraction is 

Subtracting (1) from (2), we get

x = 15

Putting x = 15 in (1), we get

2 × 15 – y = 4

 30 – y = 4

 y = 26

x = 15 and y = 26

Hence the given fraction is Question 23

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.Solution 23

Question 24

The sum of two numbers is 16 and the sum of their reciprocals is Find the numbers.Solution 24

Let the two numbers be x and y respectively.

According to the given question:

x + y = 16—(1)

And

—(2)

From (2),

xy = 48

We know,

Adding (1) and (3), we get

2x = 24

x = 12

Putting x = 12 in (1),

y = 16 – x

   = 16 – 12

   = 4

The required numbers are 12 and 4Question 25

There are two classrooms A and B. If 10 students are sent from A to B, the number of students in each room becomes the same. If 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in each room.Solution 25

Let the number of student in class room A and B be x and y respectively.

When 10 students are transferred from A to B:

x – 10 = y + 10

 x – y = 20—(1)

When 20 students are transferred from B to A:

 2(y – 20) = x + 20

 2y – 40 = x + 20

-x + 2y = 60—(2)

Adding (1) and (2), we get

y = 80

Putting y = 80 in (1), we get

x – 80 = 20

 x = 100

Hence, number of students of A and B are 100 and 80 respectively.Question 26

Taxi charges in a city consist of fixed charges and the remaining depending upon the distance travelled in kilometres. If a man travels 80 km, he pays Rs.1330, and travelling 90 km, he pays Rs.1490. Find the fixed charges and rate per km.Solution 26

Question 27

A part of monthly hostel charges in a college hostel 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 25 days, he has to pay Rs.4500, whereas a student B who takes food for 30 days, has to pay Rs.5200. Find the fixed charges per month and the cost of the food per day.Solution 27

Question 28

A man invested an amount at 10% per annum and another amount at 8% per annum simple interest. Thus, he received Rs. 1350 as an annual interest. Had he interchanged the amounts invested, he would have received Rs.45 less as interest. What amounts did he invest at different rates?Solution 28

Question 29

The monthly incomes of A and B are in the ratio 5 : 4 and their monthly expenditures are in the ratio 7 : 5. If each saves Rs.9000 per month, find the monthly income of eachSolution 29

Question 30

A man sold a chair and a table together for Rs.1520, thereby making a profit of 25% on chair and 10% on table. By selling them together for Rs.1535, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.Solution 30

Question 31

Points A and B are 70km apart on a highway. A car starts from A and another starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car.Solution 31

Let P and Q be the cars starting from A and B respectively and let their speeds be x km/hr and y km/hr respectively.

Case- I

When the cars P and Q move in the same direction.

Distance covered by the car P in 7 hours = 7x km

Distance covered by the car Q in 7 hours = 7y km

Let the cars meet at point M.

AM = 7x km and BM = 7y km

AM – BM = AB

 7x – 7y = 70

7(x – y) = 70

x – y = 10 —-(1)

Case II

When the cars P and Q move in opposite directions.

Distance covered by P in 1 hour = x km

Distance covered by Q in 1 hour = y km

In this case let the cars meet at a point N.

AN = x km and BN = y km

AN + BN = AB

 x + y = 70—(2)

Adding (1) and (2), we get

2x = 80

 x = 40

Putting x = 40 in (1), we get

40 – y = 10

 y = (40 – 10) = 30

x = 40, y = 30

Hence, the speeds of these cars are 40 km/ hr and 30 km/ hr respectively.Question 32

A train covered a certain distance at a uniform speed. If the train had been 5kmph faster, it would have taken 3 hours less than the scheduled time. And, if the train were slower by 4 kmph, it would have taken 3hours less than the scheduled time. Find the length of the journey.Solution 32

Let the original speed be x km/h and time taken be y hours

Then, length of journey = xy km

Case I:

Speed = (x + 5)km/h and time taken = (y – 3)hour

Distance covered = (x + 5)(y – 3)km

(x + 5) (y – 3) = xy

xy + 5y -3x -15 = xy

5y – 3x = 15 —(1)

Case II:

Speed (x – 4)km/hr and time taken = (y + 3)hours

Distance covered = (x – 4)(y + 3) km

(x – 4)(y + 3) = xy

xy -4y + 3x -12 = xy

 3x – 4y = 12 —(2)

Multiplying (1) by 4 and (2) by 5, we get

20y – 12x = 60 —(3)

-20y + 15x = 60 —(4)

Adding (3) and (4), we get

3x = 120

or x = 40

Putting x = 40 in (1), we get

5y – 3 × 40 = 15 

  5y = 135  

  y = 27

Hence, length of the journey is (40 × 27) km = 1080 kmQuestion 33

Abdul travelled 300 km by train and 200 km by taxi taking 5 hours 30 minutes. But, if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi.Solution 33

Question 34

Places A and B are 160 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car.Solution 34

Question 35

A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current.Solution 35

Question 36

A boat goes 12km upstream and 40km downstream in 8hours. It can go 16km upstream and 32 downstream in the same times. Find the speed of the boat in still water and the speed of the stream.Solution 36

Let the speed of the boat in still water be x km/hr and speed of the stream be y km/hr.

Then,

Speed upstream = (x – y)km/hr

Speed downstream = (x + y) km/hr

Time taken to cover 12 km upstream = 

Time taken to cover 40 km downstream = 

Total time taken = 8hrs

Again, time taken to cover 16 km upstream = 

Time taken to taken to cover 32 km downstream = 

Total time taken = 8hrs

Putting 

12u + 40v = 8 

   3u + 10v = 2 —(1)

and

16u + 32v = 8 

   2u + 4v = 1—(2)

Multiplying (1) by 4 and (2) by 10, we get

12u + 40v = 8—(3)

20u + 40v = 10 —(4)

Subtracting (3) from (4), we get

Putting in (3), we get

On adding (5) and (6), we get

2x = 12 

  x = 6

Putting x = 6 in (6) we get

6 + y = 8

y = 8 – 6 = 2

x = 6, y = 2

Hence, the speed of the boat in still water = 6 km/hr and speedof the stream = 2km/hrQuestion 37

2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.Solution 37

Let man’s 1 day’s work be and 1 boy’s day’s work be 

Also let  and 

Multiplying (1) by 6 and (2) by 5 we get

Subtracting (3) from (4), we get

Putting in (1), we get

x = 18, y = 36

The man will finish the work in 18 days and the boy will finish the work in 36 days when they work alone.Question 38

The length of a room exceeds its breadth by 3 meters. If the length is increased by 3 meters and the breadth is decreased by 2 meters, the area remains the same. Find the length and breadth of the room.Solution 38

Let the length = x meters and breadth = y meters

Then,

x = y + 3

 x – y = 3 —-(1)

Also,

(x + 3)(y – 2) = xy

 3y – 2x = 6 —-(2)

Multiplying (1) by 2 and (2) by 1

-2y + 2x = 6 —(3)

3y – 2x = 6 —(4)

Adding (3) and (4), we get

y = 12

Putting y = 12 in (1), we get

x – 12 = 3

 x= 15

x = 15, y = 12

Hence length = 15 metres and breadth = 12 metresQuestion 39

The area of a rectangle gets reduced by 8 , when its length is reduced by 5m and its breadth is increased by 3m. If we increase the length by 3m and breadth by 2m, the area is increased by 74 . Find the length and breadth of the rectangle.Solution 39

Let the length of a rectangle be x meters and breadth be y meters.

Then, area = xy sq.m

Now,

xy – (x – 5)(y + 3) = 8

xy – [xy – 5y + 3x -15] = 8

xy – xy + 5y – 3x + 15 = 8 

 3x – 5y = 7 —(1)

And

(x + 3)(y + 2) – xy = 74

 xy + 3y +2x + 6 – xy = 74

2x + 3y = 68—(2)

Multiplying (1) by 3 and (2) by 5, we get

9x – 15y = 21—(3)

10x + 15y = 340—(4)

Adding (3) and (4), we get

Putting x = 19 in (3) we get

x = 19 meters, y = 10 meters

Hence, length = 19m and breadth = 10mQuestion 40

The area of a rectangle gets reduced by 67 square metres, when its length is increased by 3 m and breadth is decreased by 4 m. If the length. is reduced by 1 m and breadth is increased by 4 m, the area is increased by 89 square metres. Find the dimensions of the rectangle.Solution 40

Question 41

A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Delhi costs Rs.4150 while one full and one half reserved first class tickets cost Rs.6255. What is the basic first class full fare and what is the reservation charge?Solution 41

Question 42

Five years hence, a man’s age will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present ages.Solution 42

Question 43

Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of man and his son.Solution 43

Let the present ages of the man and his son be x years and y years respectively.

Then,

Two years ago:

(x – 2) = 5(y – 2)

x – 2 = 5y – 10

  x – 5y = -8 —(1)

Two years later:

(x + 2) = 3(y + 2) + 8

x + 2 = 3y + 6 + 8

 x – 3y = 12 —(2)

Subtracting (2) from (1), we get

-2y = -20

  y = 10

Putting y = 10 in (1), we get

x – 5 10 = -8

x – 50 = -8

 x = 42

Hence the present ages of the man and the son are 42 years and 10 respectively.Question 44

If twice the son’s age in years is added to the mother’s age, the sum is 70 years. But, if twice the mother’s age is added to the son’s age, the sum is 95years. Find the age of the mother and that of the son.Solution 44

Let the present ages of  the mother and her son be x and y respectively.

According to the given question:

x + 2y = 70—(1)

and

2x + y = 95—(2)

Multiplying (1) by 1 and (2) by 2, we get

x + 2y = 70 —(3)

4x + 2y = 190—(4)

Subtracting (3) from (4), we get

Putting x = 40 in (1), we get

40 + 2y = 70

2y = 30

 y = 15

x = 40, y = 15

Hence, the ages of the mother and the son are 40 years and 15 years respectively.Question 45

The present age of a woman is 3 years more than three times the age of her daughter; three years hence, the woman’s age will be 10 years more than twice the age of her daughter. Find their present ages.Solution 45

Let the present ages of woman and daughter be x and y respectively.

Then,

Their present ages:

x = 3y + 3

x – 3y = 3—(1)

Three years later:

 (x + 3) = 2(y + 3) + 10

x + 3 = 2y + 6 + 10

 x – 2y = 13—(2)

Subtracting (2) from (1), we get

 y = 10

Putting y = 10 in (1), we get

x – 3 × 10 = 3

 x = 33

x = 33, y = 10

Hence, present ages of woman and daughter are 33 and 10 years.Question 46

On selling a tea set at 5% loss and a lemon set at 15% gain, a crockery seller gains Rs.7. If he sells the tea set at 5% gain and the lemon set at 10% gain, he gains Rs.13. Find the actual price of each of the tea set and the lemon set.Solution 46

Question 47

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Mona paid Rs.27 for a book kept for 7 days, while Tanvy paid Rs. 21 for the book she kept for 5 days. Find the fixed charge and the charge for each extra day.Solution 47

Question 48

A chemist has one solution containing 50% acid and a second one containing 25% arid. How much of each should be used to make 10 litres of a 40% arid solution?Solution 48

Question 49

A jeweller has bars of 18-carat gold and 12-carat gold. How much of each must be melted together to obtain a bar of 16-carat gold, weighing 120 g? (Given: Pure gold is 24-carat)Solution 49

Question 50

90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution. Find the quantity of each type of adds to be mixed to form the mixture.Solution 50

Question 51

The larger of the two supplementary angles exceeds the smaller by 18^o.  Find them.Solution 51

Question 52

Solution 52

Question 58

Solution 58

Exercise Ex. 7F

Question 1

Write the number of solutions of the following pair of linear equations:

x + 2y – 8 = 0, 2x + 4y = 16.Solution 1

Question 2

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

2x + 3y = 7, (k – 1) x + (k + 2) y = 3k.Solution 2

Question 3

For what value of k does the following pair of linear equations have infinitely many solutions?

10x + 5y – (k – 5) = 0 and 20x + 10y – k = 0.Solution 3

Question 4

For what value of k will the following pair of linear equations have no ‘solution?

2x + 3y = 9, 6x + (k – 2)y = (3k – 2).Solution 4

Question 5

Write the number of solutions of the following pair of linear equations:

x + 3y – 4 = 0 and 2x + 6y – 7 = 0.Solution 5

Question 6

Write the value of k for which the system of equations 3x + ky = 0, 2x – y = 0 has a unique solution.Solution 6

Question 7

The difference between two numbers is 5 and the difference between their squares is 65. Find the numbers.Solution 7

Question 8

The cost of 5 pens and 8 pencils is Rs.120, while the cost of 8 pens and 5 pencils is Rs.153. Find the cost of 1 pen and that of I pencil.Solution 8

Question 9

The sum of two numbers is 80. The larger number exceeds four times the smaller one by 5. Find the numbers.Solution 9

Question 10

A number consists of two digits whose sum is 10. If 18 is subtracted from the number, its digits are reversed. Find the number.Solution 10

Question 11

A man purchased 47 stamps of 20 p and 25 p for Rs.10. Find the number of each type of stamps.Solution 11

Question 12

A man has some hens and cows. If the number of heads be 48 and number of feet be 140, how many cows are there?Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

If 12x + 17y = 53 and 17x + 12y = 63 then find the value of (x +y).Solution 15

Question 16

Find the value of k for which the system 3x + 5y = 0, kx + 10y = 0 has a nonzero solution.Solution 16

Question 17

Find k for which the system kx – y = 2 and 6x – 2y = 3 has a unique solution.Solution 17

Question 18

Find k for which the system 2x + 3y – 5 = 0, 4x + ky – 10 = 0 has an infinite number of solutions.Solution 18

Question 19

Show that the system 2x + 3y – 1 = 0, 4x + 6y – 4 = 0 has no solution.Solution 19

Question 20

Find k for which the system x + 2y = 3 and 5x + ky + 7 = 0 is inconsistent.Solution 20

Question 21

Solution 21

Exercise MCQ

Question 1

If 2x + 3y = 12 and 3x – 2y = 5 then

(a) x = 2, y = 3

(b) x = 2, y = -3

(c) x = 3, y = 2

(d) x = 3, y = -2Solution 1

Question 2

(a) x = 4, y = 2

(b) x = 5, y = 3

(c) x = 6, y = 4

(d) x = 7, y = 5Solution 2

Question 3

(a) x = 2, y = 3

(b) x = -2, y = 3

(c) x = 2, y = -3

(d) x = -2, y = -3Solution 3

Question 4

Solution 4

Question 5

(a) x = 1, y = 1

(b) x = -1, y = -1

(c) x = 1, y = 2

(d) x = 2, y = 1Solution 5

Question 6

Solution 6

Question 7

If 4x+6y=3xy and 8x+9y=5xy then

(a) x=2, y=3

(b) x=1, y=2

(c) x=3, y=4

(d) x=1, y=-1Solution 7

Question 8

If 29x+37y=103 and 37x+29y=95 then

(a) x=1, y=2

(b) x=2, y=1

(c) x=3, y=2

(d) x=2, y=3Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The system kx – y = 2 and 6x – 2y = 3  has a unique solution only when

(a) k = 0

(b) k ≠ 0

(c) k = 3

(d) k ≠ 3Solution 11

Question 12

The system x – 2y = 3 and 3x + ky = 1 has a unique solution only when

(a) k = -6

(b) k ≠ -6

(c) k = 0

(d) k ≠ 0Solution 12

Question 13

The system x+2y=3 and 5x+ky+7=0 has no solution, when

(a) k=10

(b) k≠10

(c) 

(d) K=-21Solution 13

Question 14

If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel then the value of k is

Solution 14

Question 15

For what value of k do the equations kx – 2y = 3 and

3x + y = 5 represent two lines intersecting at a unique point?

(a) k=3

(b) k=-3

(c) k=6

(d) all real values except -6Solution 15

Question 16

The pair of equations x + 2y + 5 = 0 and -3x – 6y + 1 = 0 has

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solutionSolution 16

Question 17

The pair of equations 2x + 3y = 5 and 4x + 6y = 15 has

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solutionSolution 17

Question 18

If a pair of linear equations is consistent then their graph lines will be

(a) parallel

(b) always coincident

(c) always intersecting

(d) intersecting or coincidentSolution 18

Question 19

If a pair of linear equations is inconsistent then their graph lines will be

(a) parallel

(b) always coincident

(c) always intersecting

(d) intersecting or coincidentSolution 19

Question 20

In a ΔABC, ∠C = 3 ∠B = 2 (∠A + ∠B), then ∠B = ?

(a) 20° 

(b) 40° 

(c) 60° 

(d) 80° Solution 20

Question 21

In a cyclic quadrilateral ABCD, it is being given that

∠A = (x + y + 10) °, ∠B = (y + 20) °,

∠C = (x + y – 30)° and ∠D = (x + y)°. Then, ∠B = ?

(a) 70° 

(b) 80° 

(c) 100° 

(d) 110° Solution 21

Question 22

The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. The number is

(a) 96

(b) 69

(c) 87

(d) 78Solution 22

Question 23

Solution 23

Question 24

5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man is

(a) 45 years

(b) 50 years

(c) 47 years

(d) 40 yearsSolution 24

Question 25

The graphs of the equations 6x – 2y + 9 = 0 and

3x – y + 12 = 0 are two lines which are

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 25

Question 26

The graphs of the equations 2x+3y-2=0 and x-2y-8=0 are two lines which are

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 26

Question 27

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 27

Exercise FA

Question 1

The graphic representation of the equations x+2y=3 and 2x+4y+7=0 gives a pair of

(a) parallel lines

(b) intersecting lines

(c) coincident lines

(d) none of theseSolution 1

Question 2

If 2x – 3y = 7 and (a + b) x – (a + b – 3) y = 4a+b have an infinite number of solutions then

(a) a= 5, b = 1

(b) a = -5, b = 1

(c) a = 5, b = -1

(d) a = -5, b = -1Solution 2

Question 3

The pair of equations 2x+y=5, 3x+2y=8 has

(a) a unique solution

(b) two solutions

(c) no solution

(d) infinitely many solutionsSolution 3

Question 4

If x = -y and y > 0, which of the following is wrong?

(a) x²y > 0

(b) x + y = 0

(c) xy < 0

(d) Solution 4

Question 5

Solution 5

Question 6

For what values of k is the system of equations kx + 3y = k – 2, 12x + ky = k inconsistent?Solution 6

Question 7

Solution 7

Question 8

Solve the system of equations x – 2y = 0, 3x + 4y = 20.Solution 8

Question 9

Show that the paths represented by the equations x – 3y = 2 and -2x + 6y = 5 are parallel.Solution 9

Question 10

The difference between two numbers is 26 and one number is three times the other. Find the numbers.Solution 10

Question 11

Solve : 23x+29y=98, 29x+23y=110.Solution 11

Question 12

Solve : 6x+3y=7xy and 3x+9y = 11xy.Solution 12

Question 13

Find the value of k for which the system of equations 3x+y=1 and kx+2y=5 has (i) a unique solution, (ii) no solution.Solution 13

Question 14

In a ΔABC, ∠C =3∠B =2(∠A+∠B). Find the measure of each one of the ∠A, ∠B and ∠C. Solution 14

Question 15

5 pencils and 7 pens together cost Rs. 195 while 7 pencils and 5 pens together cost Rs. 153. Find the cost of each one of the pencil and the pen.Solution 15

Question 16

Solve the following system of equations graphically :

2x-3y=1, 4x-3y+1=0.Solution 16

Since the intersection of the lines is the point with coordinates (-1, -1), x = -1 and y = -1.Question 17

Find the angles of a cyclic quadrilateral ABCD in which ∠A =(4x+20)°, ∠B=(3x-5)°, ∠C=(4y)° and ∠D=(7y+5)° Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Exercise Ex. 2.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Exercise Ex. 2.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Exercise Ex. 2.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Exercise Ex. 2.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Exercise Ex. 2.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Exercise Ex. 2.3

Solution 1

Solution 2

Solution 3

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Solution 5

Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Exercise Ex. 4.1

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 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Exercise Ex. 5.1

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 14

Exercise Ex. 5.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Exercise Ex. 5.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Exercise Misc. Ex.

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 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Exercise Ex. 6.1

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 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

Exercise Ex. 6.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Exercise Ex. 6.3

Solution 1

Solution 2

Solution 3

Solution 4

x + y ≥ 4   … (1)

2x – y < 0  … (2)

The graph of the lines, x + y = 4 and 2x – y = 0 are drawn in the figure below.

Inequality (1) represents the region above the line x + y = 4. (including the line x + y = 4)

It is observed that (–1, 0) satisfies the inequality, 2x – y < 0.

[2(-1) – 0 = -2< 0]

Therefore, inequality (2) represents the half plane corresponding to the line, 2x – y = 0 containing the point (-1, 0). [excluding the line 2x – y < 0]

Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the line x + y = 4 and excluding the points on line 2x – y = 0 as follows:

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Exercise Misc. Ex.

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 14

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Exercise Ex. 1.1

Solution 1

 Hence: this collection is a set. (vii) The collection of all even integers is a well-defined collection because one can definitely identify an even integer that belongs to this collection.  Hence: this collection is a set.

(viii) The collection of questions in this chapter is a well-defined collection because one can definitely identify a question that belongs to this chapter.

 Hence: this collection is a set.

(ix) The collection ofmost dangerous animals of the world is not a well-defined collection because the criteria for determining the dangerousness of an animal can vary from person to person.

 Hence: this collection is not a set.Solution 2

Solution 3

(i) A = { x : x is an integer and -3 ≤ x < 7}.

The elements of this set are -3, -2, -1, 0, 1, 2, 3, 4, 5 and 6 only.

Therefore, the given set can written in roster form as

A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = { x : x is a natural number less than 6}

The elements of this set are 1, 2, 3, 4 and 5 only.

Therefore, the given set can written in roster form as

B = {1, 2, 3, 4, 5}

(iii) C = { x : x is a two-digit number such that the sum of its digits is 8}

The elements of this set are 17, 26, 35, 44, 53, 62, 71 and 80 only.

Therefore, the given set can written in roster form as

C = {17, 26, 35, 44, 53, 62, 71, 80}

(iv) D = { x : x is a prime number which is divisor of 60}.

2	60
2	30
3	15
	5

∴ 60 = 2 × 2 × 3 × 5

The elements of this set are 2, 3 and 5 only.

Therefore, this set can written in roster form as D = {2, 3, 5}.

(v) E = The set of all letters in the word TRIGONOMETRY.

There are 12 letters in the word TRIGONOMETRY, out of which letters T, R, and O are repeated.

Therefore, this set can written in roster form as

E = {T, R, I, G, O, N, M, E, Y}

(vi) F = The set of all letters in the word BETTER.

There are 6 letters in the word BETTER, out of which letters E and T are repeated.

Therefore, this set can written in roster form as

F = {B, E, T, R}Solution 4

Solution 5

Solution 6

Exercise Ex. 1.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Exercise Ex. 1.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

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Exercise Ex. 1.4

Solution 1

Solution 2

Solution 3

Solution 4

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Solution 6

Solution 7

Solution 8

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Solution 10

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Solution 12

Exercise Ex. 1.5

Solution 1

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Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Exercise Ex. 1.6

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Exercise Misc. Ex.

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 14

Solution 15

Solution 16

Exercise MCQ

Question 1

The length, breadth and height of a cuboid are 15 cm, 12 cm and 4.5 cm respectively. Its volume is

1. 243 cm³
2. 405 cm³
3. 810 cm³
4. 603 cm³

Solution 1

Question 2

A cuboid is 12 cm long, 9 cm broad and 8 cm high. Its total surface area is

1. 864 cm²
2. 552 cm²
3. 432 cm²
4. 276 cm²

Solution 2

Question 3

The length breadth and height of a cuboid are 15m, 6m, and 5 dm respectively. The lateral surface area of the cuboid is

1. 45 m²
2. 21 m²
3. 201 m²
4. 90 m²

Solution 3

Question 4

A beam 9 m long, 40 cm wide and 20 cm high is made up of iron which weighs 50 kg per cubic metre. The weight of the beam is

1. 27 kg
2. 48 kg
3. 36 kg
4. 56 kg

Solution 4

Question 5

The length of the longest rod that can be placed in a room of dimensions (10 m × 10 m × 5 m) is

1. 15 m
2. 16 m
3.  
4. 12 m

Solution 5

Question 6

What is the maximum length of a pencil that can be placed in a rectangular box of dimensions (8 cm × 6 cm × 5 cm)? 

1. 8 cm
2. 9.5 cm
3. 19 cm
4. 11.2 cm

Solution 6

Question 7

The number of planks of dimensions (4 m × 5 m × 2 m) that can be stored in a pit which is 40 m long, 12 m wide and 16 m deep is

1. 190
2. 192
3. 184
4. 180

Solution 7

Question 8

How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 50 cm deep?

1. 480
2. 450
3. 320
4. 360

Solution 8

Question 9

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?

1. 4800
2. 5600
3. 6400
4. 5200

Solution 9

Question 10

How many persons can be accommodated in a dining hall of dimensions (20 m × 15 m × 4.5 m), assuming that each person requires 5 m³ of air?

1. 250
2. 270
3. 320
4. 300

Solution 10

Question 11

A river 1.5 m deep and 30 m wide is flowing at the rate of 3 km per hour. The volume of water that runs into the sea per minute is

1. 2000 m³
2. 2250 m³
3. 2500 m³
4. 2750 m³

Solution 11

Question 12

The lateral surface area of a cube is 256 m². The volume of the cube is

1. 64 m³
2. 216 m³
3. 256 m³
4. 512 m³

Solution 12

Question 13

The total surface area of a cube is 96m². The volume of the cube is

1. 8 cm³
2. 27cm³
3. 64cm³
4. 512 cm³

Solution 13

Question 14

The volume of a cube is 512 cm³. Its surface area is

1. 256 cm²
2. 384 cm²
3. 512 cm²
4. 64 cm²

Solution 14

Question 15

The length of the longest rod that can fit in a cubical vessel of side 10 cm is

1. 10 cm
2. 20 cm
3.  
4. 

Solution 15

Question 16

If the length of diagonal of a cube is  cm, then its surface area is

1. 192 cm²
2. 384 cm²
3. 512 cm²
4. 768 cm²

Solution 16

Question 17

If each edge of a cube is increased by 50%, then the percentage increase in its surface area is

1. 50%
2. 75%
3. 100%
4. 125%

Solution 17

Question 18

Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. The lateral surface area of the new cube formed is

1. 72 cm²
2. 144 cm²
3. 128 cm²
4. 256 cm²

Solution 18

Question 19

In a shower, 5 cm of rain falls, what is the volume of water that falls on 2 hectors of ground?

1. 500 m³
2. 750 m³
3. 800 m³
4. 1000 m³

Solution 19

Question 20

Two cubes have their volumes in the ratio 1:27. The ratio of their surface area is

1. 1:3
2. 1:8
3. 1:9
4. 1:18

Solution 20

Question 21

If each side of a cube is doubled, then its volume

1. is doubled
2. becomes 4 times
3. becomes 6 times
4. becomes 8 times

Solution 21

Question 22

The diameter of a base of a cylinder is 6 cm and its height is 14 cm. The volume of the cylinder is

a. 198 cm³

b. 396 cm³

c. 495 cm³

d. 297 cm³Solution 22

Question 23

The diameter of a cylinder is 28 cm and its height is 20 cm, then its curved surface area is

1. 880 cm²
2. 1760 cm²
3. 3520 cm²
4. 2640 cm²

Solution 23

Question 24

If the curved surface area of a cylinder is 1760 cm² and its base radius is 14 cm, then its height is

1. 10 cm
2. 15 cm
3. 20 cm
4. 40 cm

Solution 24

Question 25

The height of a cylinder is 14 cm and its curved surface area is 264 cm². The volume of the cylinder is

1. 308 cm²
2. 396 cm²
3. 1232 cm²
4. 1848 cm²

Solution 25

Question 26

The curved surface area of the cylindrical pillar is 264 m² and its volume is 924m³. The height of the pillar is

1. 4 m
2. 5 m
3. 6 m
4. 7 m

Solution 26

Question 27

The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3. The ratio of their surface area is

1. 2:5
2. 8:7
3. 10:9
4. 16:9

Solution 27

Question 28

The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3. The ratio of their volumes is

1. 27:20
2. 20:27
3. 4:9
4. 9:4

Solution 28

Question 29

The ratio between the radius of the base and height of a cylinder is 2:3. If its volume is 1617 cm³, then its total surface area is

1. 308 cm²
2. 462 cm²
3. 540 cm²
4. 770 cm²

Solution 29

Question 30

Two circular cylinders of equal volume have their heights in the ratio 1:2. The ratio of their radii is

Solution 30

Question 31

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1:2. If the total surface area is 616 cm², then the volume of the cylinder is

1. 1078 cm³
2. 1232 cm³
3. 1848 cm³
4. 924 cm³

Solution 31

Question 32

In a cylinder, if the radius is halved and the height is doubled, then the volume will be

1. The same
2. Doubled
3. Halved
4. Four times

Solution 32

Question 33

The number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm is

1. 540
2. 450
3. 380
4. 472

Solution 33

Question 34

The radius of a wire is decreased to one-third. If volume remains the same, the length will become

1. 2 times
2. 3 times
3. 6 times
4. 9 times

Solution 34

Question 35

The diameter of a roller, 1m long is 84 cm. If it takes 500 complete revolutions to level a playground, the area of the playground is

1. 1440 m²
2. 1320 m²
3. 1260 m²
4. 1550 m²

Solution 35

Question 36

2.2 dm³ of lead is to be drawn into a cylindrical wire 0.50 cm in diameter. The length of the wire is

1. 110 m
2. 112 m
3. 98 m
4. 124 m

Solution 36

Question 37

The lateral surface area of a cylindrical is

Solution 37

Question 38

The height of a cone is 24 cm and the diameter of its base is 14 cm. The curved surface area of the cone is

1. 528 cm²
2. 550 cm²
3. 616 cm²
4. 704 cm²

Solution 38

Question 39

The volume of a right circular cone of height is 12 cm and base radius 6 cm, is

1. (12π) cm³
2. (36π) cm³
3. (72π) cm³
4. (144π) cm³

Solution 39

Question 40

How much cloth 2.5 m wide will be required to make a conical tent having base radius 7 m and height 24 m?

1. 120 m
2. 180 m
3. 220 m
4. 550 m

Solution 40

Question 41

The volume of a cone is 1570 cm³ and its height is 15 cm. What is the radius of the cone? (Use π = 3.14)

1. 10 cm
2. 9 cm
3. 12 cm
4. 8.5 cm

Solution 41

Question 42

The height of cone is 21 cm and its slant height is 28 cm. The volume of the cone is

1. 7356 cm³
2. 7546 cm³
3. 7506 cm³
4. 7564 cm³

Solution 42

Correct option: (b)

Question 43

The volume of a right circular cone of height 24 cm is 1232 cm³. Its curved surface area is

1. 1254 cm²
2. 704 cm²
3. 550 cm²
4. 462 cm²

Solution 43

Question 44

If the volumes of two cones be in the ratio 1:4 and the radii of their bases be in the ratio 4:5, then the ratio of their heights is

1. 1:5
2. 5:4
3. 25:16
4. 25:64

Solution 44

Question 45

If the height of a cone is doubled, then its volume is increased by

1. 100%
2. 200 %
3. 300 %
4. 400 %

Solution 45

Question 46

The curved surface area of the cone is twice that of the other while the slant height of the latter is twice that of the former. The ratio of their radii is

1. 2:1
2. 4:1
3. 8:1
4. 1:1

Solution 46

Question 47

The ratio of the volumes of a right circular cylinder and a right circular cone of the same base and same height will be

1. 1:3
2. 3:1
3. 4:3
4. 3:4

Solution 47

Question 48

A right circular cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is

1. 3:5
2. 2:5
3. 3:1
4. 1:3

Solution 48

Question 49

The radii of the bases of a cylinder and a cone are in the ratio 3:4 and their heights are in the ratio 2:3. Then their volumes are in the ratio

1. 9:8
2. 8:9
3. 3:4
4. 4:3

Solution 49

Question 50

If the height and the radius of cone are doubled, the volume of the cone becomes

1. 3 times
2. 4 times
3. 6 times
4. 8 times

Solution 50

Question 51

A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to make n solid cones of height 1 cm and base radius 1 mm. The value of n is

1. 450
2. 1350
3. 4500
4. 13500

Solution 51

Question 52

A conical tent is to accommodate 11 persons such that each person occupies 4 m² of space on the ground. They have 220m³ of air to breathe. The height of the cone is

1. 14m
2. 15 m
3. 16 m
4. 20 m

Solution 52

Question 53

The volume of a sphere of radius 2r is

Solution 53

Question 54

The volume of a sphere of a radius 10.5 cm is

1. 9702 cm³
2. 4851 cm³
3. 19404 cm³
4. 14553 cm³

Solution 54

Question 55

The surface area of a sphere of radius 21 cm is

1. 2772 cm²
2. 1386 cm²
3. 4158 cm²
4. 5544 cm²

Solution 55

Question 56

The surface area of a sphere is 1386 cm². Its volume is

1. 1617 cm³
2. 3234 cm³
3. 4851 cm³
4. 9702 cm³

Solution 56

Question 57

If the surface area of a sphere is (144 π) m², then its volume is

1. (288 π) m³
2. (188 π) m³
3. (300 π) m³
4. (316 π) m³

Solution 57

Question 58

The volume of a sphere is 38808 cm³. Its curved surface area is

1. 5544 cm²
2. 8316 cm²
3. 4158 cm²
4. 1386 cm²

Solution 58

Question 59

If the ratio of the volumes of two spheres is 1:8, then the ratio of their surface area is

1. 1:2
2. 1:4
3. 1:8
4. 1:16

Solution 59

Question 60

A solid metal ball of radius 8 cm is melted and cast into smaller balls, each of radius 2 cm, The number of such balls is

1. 8
2. 16
3. 32
4. 64

Solution 60

Question 61

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is

1. 4.2 cm
2. 2.1 cm
3. 2.4 cm
4. 1.6 cm

Solution 61

Question 62

A solid lead ball of radius 6 cm is melted and then drawn into a wire of diameter 0.2 cm. The length of wire is

1. 272 m
2. 288 m
3. 292 m
4. 296 m

Solution 62

Question 63

A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. The number of such cones will be

1. 21
2. 63
3. 126
4. 130

Solution 63

Question 64

How many lead shots, each 0.3 cm in diameter, can be made from a cuboid of dimensions 9 cm × 11 cm × 12 cm?

1. 7200
2. 8400
3. 72000
4. 84000

Solution 64

Question 65

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. The length of the wire is

1. 12 m
2. 18 m
3. 36 m
4. 66 m

Solution 65

Question 66

A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm. The radius of the base of the cone is

1. 6.3 cm
2. 2.1 cm
3. 6 cm
4. 4 cm

Solution 66

Question 67

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. The radius of the third ball is

1. 1 cm
2. 1.5 cm
3. 2.5 cm
4. 0.5 cm

Solution 67

Question 68

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloons in two cases is

1. 1:4
2. 1:3
3. 2:3
4. 1:2

Solution 68

Question 69

The volumes of the two spheres are in the ratio 64:27 and the sum of their radii is 7 cm. The difference of their total surface areas is

1. 38 cm²
2. 58 cm²
3. 78 cm²
4. 88 cm²

Solution 69

Question 70

A hemispherical bowl of radius 9 cm contains a liquid. This liquid is to be filled into cylindrical small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty the bowl?

1. 27
2. 35
3. 54
4. 63

Solution 70

Question 71

A cone and a hemisphere have equal bases and equal volumes. The ratio of their heights is

1. 1:2
2. 2:1
3. 4:1
4. 

Solution 71

Question 72

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is

1. 1:2:3
2. 2:1:3
3. 2:3:1
4. 3:2:1

Solution 72

Question 73

If the volumes and the surface area of sphere are numerically the same, then its radius is

1. 1 units
2. 2 units
3. 3 units
4. 4 units

Solution 73

Exercise Ex. 15C

Question 1

Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.Solution 1

Radius of a cone, r = 5.25 cm

Slant height of a cone, l = 10 cm

Question 2

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.Solution 2

Radius of a cone, r = 12 m

Slant height of a cone, l = 21 cm

Question 3

A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.Solution 3

Radius of a conical cap, r = 7 cm

Height of a conical cap, h = 24 cm

Thus, 5500 cm² sheet will be required to make 10 caps.Question 4

The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.Solution 4

Let r be the radius of a cone.

Slant height of a cone, l = 14 cm

Curved surface area of a cone = 308 cm²

Question 5

The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of Rs.12 per m².Solution 5

Radius of a cone, r = 7 m

Slant height of a cone, l = 25 m

Cost of whitewashing = Rs. 12 per m²

⇒ Cost of whitewashing 550 m² area = Rs. (12 × 550) = Rs. 6600 Question 6

A conical tent is 10 m high and radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m² canvas is Rs.70, find the cost of canvas required to make the tent.Solution 6

Radius of a conical tent, r = 24 m

Height of a conical tent, h = 10 m

Question 7

A bus stop is barricaded from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each one has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs.25 per m², what will be the cost of painting all these cones? (Use π = 3.14 and 1.02.)Solution 7

Question 8

Find the volume, curved surface area and the total surface area of a cone having base radius 35 cm and height 12 cm.Solution 8

Question 9

_{Find the volume, curved surface area and the total surface area of a cone whose height and slant heights are 6 cm and 10 cm respectively. (Take =3.14)}Solution 9

Question 10

A conical pit of diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

HINT 1 m³ = 1 kilolitre.Solution 10

Question 11

A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (Use π = 3.14.)Solution 11

Radius of a conical heap, r = 4.5 m

Height of a conical tent, h = 3.5 m

Question 12

A man uses a piece of canvas having an area of 551 m², to make a conical tent of base radius 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amount to approximately 1 m², find the volume of the tent that can be made with it.Solution 12

Radius of a conical tent, r = 7 m

Area of canvas used in making conical tent = (551 – 1) m² = 550 m²

⇒ Curved surface area of a conical tent = 550 m²

Question 13

_{How many meters of cloth , 2.5 m wide , will be required to make conical tent whose base radius is 7 m and height 24 metres?}Solution 13

Question 14

_{Two cones have their height in the ratio 1:3 and the radii of their bases in the ratio3: 1. Show that their volumes are in the ratio 3:1.}Solution 14

Question 15

_{A cylinder and a cone have equal radii of their bases and equal height s. If their curved surface areas are in the ratio 8:5, show that the radius and height of each has the ratio 3:4.}Solution 15

Question 16

_{A right circular cone is 3.6 cm height and the radius of its base is 1.6 cm. It is melted and recast into a right circular cone having base radius 1.2 cm. Find its height.}Solution 16

Question 17

_{A circus tent is cylindrical to a height of 3 meters and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.}Solution 17

Question 18

_{An iron pillarconsistsof a cylindricalportion2.8 m highand 20cm indiameterand a cone42 cm high is surmounting it . Find the weight of the pillar, given that 1 cm³ of iron weights 7.5 g.}Solution 18

Question 19

_{From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and the base is removed .find the volume of the remaining solid. (Take =3.14)}Solution 19

Question 20

_{Water flows at the rate of 10 meters per minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the surface 40 cm and depth 24 cm?}Solution 20

Question 21

A cloth having an area of 165 m² is shaped into the form of a conical tent of radius 5 m. (i) How many students can sit in the tent if a student, on an average, occupies m² on the ground? (ii) Find the volume of the cone.Solution 21

Curved surface area of the tent = Area of the cloth = 165 m²

Exercise Ex. 15A

Question 1(iv)

_{Find the volume, the lateral surface area and the total surface area cuboid whose dimensions are:}

Length =24 m, breadth =25 cmand height =6mSolution 1(iv)

_{Length = 24 m, breadth = 25 cm =0.25 m, height = 6m.}

_{Volume of cuboid= l x b x h}

_{= (24 x 0.25 x 6) m³.}

_{= 36 m³.}

_{Lateral surface area= 2(l + b) x h}

_{= [2(24 +0.25) x 6] m²}

_{= (2 x 24.25 x 6) m²}

_{= 291 m².}

_{Total surface area =2(lb+ bh + lh)}

_{=2(24 x 0.25+0.25x 6 +24 x 6) m²}

_{= 2(6+1.5+144) m²}

_{= (2 x151.5) m²=303 m².}Question 1(iii)

_{Find the volume, the lateral surface area and the total surface area cuboid whose dimensions are:}

_{Length =15m, breadth =6 m and height =5 dm}Solution 1(iii)

_{Length = 15 m, breadth = 6m and height = 5 dm = 0.5 m}

_{Volume of a cuboid = l x b x h}

_{= (15 x 6 x 0.5) m³=45 m³.}

_{Lateral surface area = 2(l + b) x h}

_{= [2(15 + 6) x 0.5] m²}

_{= (2 x 21×0.5) m²=21 m²}

_{Total surface area =2(lb+ bh + lh)}

_{= 2(15 x 6 +6 x 0.5+ 15 x 0.5) m²}

_{= 2(90+3+7.5) m²}

_{= (2 x 100.5) m²}

_{=201 m²}Question 1(ii)

_{Find the volume, the lateral surface area and the total surface area cuboid whose dimensions are:}

_{Length =26 m, breadth =14m and height =6.5m}Solution 1(ii)

Length 26 m, breadth =14 m and height =6.5 m

Volume of a cuboid= l x b x h

= (26 x 14 x 6.5) m³

= 2366 m³

Lateral surface area of a cuboid =2 (l + b) x h

= [2(26+14) x 6.5] m²

= (2 x 40 x 6.5) m²

= 520 m²

_{Total surface area= 2(lb+ bh + lh)}

_{= 2(26 x 14+14 x6.5 +26 x6.5)}

_{= 2 (364+91+169) m²}

_{= (2 x 624) m2= 1248 m².}Question 1(i)

_{Find the volume, the lateral surface area and the total surface area cuboid whose dimensions are:}

_{Length=12 cm,breadth=8 cm and height =4.5 cm}Solution 1(i)

length =12cm, breadth = 8 cm and height = 4.5 cm

Volume of cuboid = l x b x h

= (12 x 8 x 4.5) cm³= 432 cm³

Lateral surface area of a cuboid = 2(l + b) x h

= [2(12 + 8) x 4.5] cm²

= (2 x 20 x 4.5) cm² = 180 cm²

Total surface area cuboid = 2(lb +b h+ l h)

= 2(12 x 8 + 8 x 4.5 + 12 x 4.5) cm²

= 2(96 +36 +54) cm²

= (2 x186) cm²

= 372 cm²Question 2

A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 12 such matchboxes?Solution 2

For a matchbox,

Length = 4 cm

Breadth = 2.5 cm

Height = 1.5 cm

Volume of one matchbox = Volume of cuboid

= Length × Breadth × Height

= (4 × 2.5 × 1.5) cm³

= 15 cm³

Hence, volume of 12 such matchboxes = 12 × 15 = 180 cm³Question 3

A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (Given, 1 m³ = 1000 litres.)Solution 3

For a cuboidal water tank,

Length = 6 m

Breadth = 5 m

Height = 4.5 m

Now,

Volume of a cuboidal water tank = Length × Breadth × Height

= (6 × 5 × 4.5) m³

= 135 m³

= 135 × 1000 litres

= 135000 litres

Thus, a tank can hold 135000 litres of water. Question 4

The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively 10 m and 2.5 m. (Given, 1000 litres = 1 m³.)Solution 4

For a cuboidal water tank,

Length = 10 m

Breadth = 2.5 m

Volume = 50000 litres = 50 m³

Now,

Volume of a cuboidal tank = Length × Breadth × Height

⇒ 50 = 10 × 2.5 × Height

⇒ Height = 2 m = Depth

Thus, the depth of a tank is 2 m. Question 5

A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates, each measuring 1.5 m × 1.25 m × 0.5 m, that can be stored in godown.Solution 5

For a godown,

Length = 40 m

Breadth = 25 m

Height = 15 m

Volume of a godown = Length × Breadth × Height

= (40 × 25 × 15) m³

For each wooden crate,

Length = 1.5 m

Breadth = 1.25 m

Height = 0.5 m

Volume of each wooden crate = Length × Breadth × Height

= (1.5 × 1.25 × 0.5) m³

Question 6

_{How many planks of dimensions (5mx25cmX10cm) can be stored in a pit which is 20 m long , 6 m wide and 80 cm deep ?}Solution 6

Question 7

_{How many bricks will be required to construct a wall 8 m long , 6 m high and 22.5 cm thick if each brick measures (25cm x11.25cm x 6cm)?}Solution 7

Question 8

_{Find the capacity of a closed rectangular cistern whose length is 8 m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet required to make the cistern.}Solution 8

_{Length of Cistern = 8 m}

_{Breadth of Cistern = 6 m}

_{And Height (depth) of Cistern =2.5 m}

 _{Capacity of the Cistern = Volume of cistern}

 _{Volume of Cistern = (l x b x h)}

_{= (8 x 6 x2.5) m³}

_{=120 m³}

_{Area of the iron sheet required = Total surface area of the cistem.}

 _{Total surface area = 2(lb +bh +lh)}

_{= 2(8 x 6 + 6×2.5+ 2.5×8) m²}

_{= 2(48 + 15 + 20) m²}

_{= (2 x 83) m²=166 m²}Question 9

The dimensions of a room are (9 m × 8 m × 6.5 m). It has one door of dimensions (2 m × 1.5 m) and two windows, each of dimensions (1.5 m × 1 m). Find the cost of whitewashing the walls at Rs.25 per square metre.Solution 9

Area of four walls of the room = 2(length + breadth) × Height

= [2(9 + 8) × 6.5] m²

= (34 × 6.5) m²

= 221 m²

Area of one door = Length × Breadth = (2 × 1.5) m² = 3 m²

Area of two windows = 2 × (Length × Breadth)

= [2 × (1.5 × 1)] m²

= (2 × 1.5) m²

= 3 m²

Area to be whitewashed

= Area of four walls of the room – Area of one door – Area of two windows

= (221 – 3 – 3) m²

= 215 m²

Cost of whitewashing = Rs. 25 per square metre

⇒ Cost of whitewashing 215 m² = Rs. (25 × 215) = Rs. 5375Question 10

_{A wall 15 m long , 30 cm wide and 4 m high is made of bricks, each measuring (22cm x12.5cm x7.5cm) if  of the total volume of the wall consists of mortar , how many bricks are there in the wall ?}Solution 10

_LQuestion 11

How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm, 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cm³ of iron weighs 15 g, find the weight of the empty box in kilograms.Solution 11

External length of the box = 36 cm

External breadth of the box = 25 cm

External height of the box = 16.5 cm

∴ External volume of the box = (36 × 25 × 16.5) cm³ = 14850 cm³

Internal length of the box = [36 – (1.5 × 2)] cm = 33 cm

Internal breadth of the box = [25 – (1.5 × 2)] cm = 22 cm

Internal height of the box = (16.5 – 1.5) cm = 15 cm

∴ Internal volume of the box = (33 × 22 × 15) cm³ = 10890 cm³

Thus, volume of iron used in the box

= External volume of the box – Internal volume of the box

= (14850 – 10890) cm³

= 3960 cm³

Question 12

A box made of sheet metal costs Rs.6480 at Rs.120 per square metre. If the box is 5 m long and 3 m wide, find its height.Solution 12

Question 13

_{The volume of a cuboid is 1536m³. Its length is 16m, and its breadth and height are in the ratio 3:2. Find the breadth and height of the cuboid.}Solution 13

Question 14

_{How many persons can be accommodated in a dining hall of dimensions (20m x16mx4.5m), assuring that each person’s requires 5 cubic metres of air?}Solution 14

Question 15

_{A classroom is 10m long, 6.4 m wide and 5m high. If each student be given 1.6 m² of the floor area, how many students can be accommodated in the room? How many cubic metres of air would each student get?}Solution 15

Question 16

_{The surface of the area of a cuboid is 758 cm². Its length and breadth are 14 cm and 11cm respectively. Find its height.}Solution 16

Question 17In  shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground.Solution 17

Question 18

_{Find the volume, the lateral surface area, the total surface area and the diagonal of cube, each of whose edges measures 9m. [Take ]}Solution 18

Question 19

_{The total surface area of a cube is 1176 cm². Find its volume.}Solution 19

Question 20

_{The lateral surface area of a cube is 900 cm². Find its volume.}Solution 20

Question 21

_{The volume of a cube is 512 cm³. Find its surface area.}Solution 21

Question 22

_{Three cubes of metal with edges 3cm, 4 cm and 5 cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.}
Solution 22

Question 23

_{Find the length of the longest pole that can be put in a room of dimensions (10mx 10m x5m).}Solution 23

Question 24

The sum of length, breadth and depth of a cuboid is 19 cm and length of its diagonal is 11 cm. Find the surface area of the cuboid.Solution 24

Question 25

Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.Solution 25

Let the edge of the cube = ‘a’ cm

Then, surface area of cube = 6a² cm²

Question 26

If V is the volume of a cuboid of dimensions a, b, c and S is its surface area then prove that Solution 26

Question 27

Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocity of 20 km per hour. How much area will it irrigate, if 9 cm of standing water is desired?Solution 27

Question 28

A solid metallic cuboid of dimensions (9 m × 8 m × 2 m) is melted and recast into solid cubes of edge 2 m. Find the number of cubes so formed.Solution 28

Volume of a cuboid = (9 × 8 × 2) m³ = 144 m³

Volume of each cube of edge 2 m = (2 m)³ = 8 m³

Exercise Ex. 15B

Question 1

_{The diameter of a cylinder is 28 cm and its height is 40 cm. find the curved surface area, total surface area and the volume of the cylinder.}Solution 1

Question 2

A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?Solution 2

Radius (r) of cylindrical bowl =

Height (h) up to which the bowl is filled with soup = 4 cm

Volume of soup in 1 bowl = pr²h  = 154 cm³

Hence, volume of soup in 250 bowls = (250 × 154) cm³ = 38500 cm³ = 38.5 litres

Thus, the hospital will have to prepare 38.5 litres of soup daily to serve 250 patients.  Question 3

The pillars of a temple are cylindrically shaped. Each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?Solution 3

Radius (r) of pillar = 20 cm =  m

Height (h) of pillar = 10 m

Question 4

A soft drink is available in two packs: (i) a tin can with a rectangular base of length 5 cm, breadth 4 cm and height 15 cm, and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?Solution 4

For a tin can of rectangular base,

Length = 5 cm

Breadth = 4 cm

Height = 15 cm

∴ Volume of a tin can = Length × Breadth × Height

= (5 × 4 × 15) cm³

= 300 cm³

For a cylinder with circular base,

Diameter = 7 ⇒ Radius = r = cm

Height = h = 10 cm

⇒ Volume of plastic cylinder is greater than volume of a tin can.

Difference in volume = (385 – 300) = 85 cm³

Thus, a plastic cylinder has more capacity that a tin can by 85 cm³.Question 5

There are 20 cylindrical pillars in a building, each having a diameter of 50 cm and height 4 m. Find the cost of cleaning them at Rs.14 per m².Solution 5

Radius (r) of 1 pillar = 

Height (h) of 1 pillar = 4 m

Question 6

The curved surface area of a right circular cylinder is 4.4 m². If the radius of its base is 0.7 m, find its (i) height and (ii) volume.Solution 6

Curved surface area of a cylinder = 4.4 m²

Radius (r) of a cylinder = 0.7 m

Question 7

The lateral surface area of a cylinder is 94.2 cm² and its height is 5 cm. Find (i) the radius of its base and (ii) its volume. (Take π = 3.14.)Solution 7

Lateral surface area of a cylinder = 94.2 cm²

Height (h) of a cylinder = 5 cm

Question 8

The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. Find the area of the metal sheet needed to make it.Solution 8

Volume of a cylinder = 15.4 litres = 15400 cm³

Height (h) of a cylinder = 1 m = 100 cm

Question 9

The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.Solution 9

Internal diameter of a cylinder = 24 cm

⇒ Internal radius of a cylinder, r = 12 cm

External diameter of a cylinder = 28 cm

⇒ External radius of a cylinder, R = 14 cm

Length of the pipe, i.e height, h = 35 cm

Question 10

In a water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.Solution 10

Diameter of a cylindrical pipe = 5 cm

⇒ Radius (r) of a cylindrical pipe = 2.5 cm

Height (h) of a cylindrical pipe = 28 m = 2800 cm

Question 11

_{Find the weight of a solid cylinder of radius10.5 cm and height 60 cm if the material of the cylinder weights 5 g per cm²}Solution 11

Question 12

_{The curved surface area of a cylinder is 1210 cm² and its diameter is 20 cm. find its height and volume.}Solution 12

Question 13

_{The curved surface area of a cylinder is 4400 cm}² _{and the circumferences of its base are 110 cm. Find the height and the volume of the cylinder.}Solution 13

Question 14

_{The radius of the base and the height of a cylinder are in the ratio 2:3. If its volume is 1617 cm³, find the total surface area of the cylinder}Solution 14

Question 15

_{The total surface area of the cylinder is 462 cm}²_{. And its curved surface area is one third of its total surface area. Find the volume of the cylinder.}Solution 15

Question 16

_{The total surface area of the solid cylinder is 231 cm² and its curved surface area is  of the total surface area. Find the volume of the cylinder.}Solution 16

Question 17

_{The ratio between the curved surface area and the total surface area of a right circular cylinder is 1:2. Find the volume of the cylinder if its total surface area is 616 cm².}Solution 17

Question 18

_{A cylindrical bucket , 28 cm in diameter and 72 cm high , is full of water .The water is emptied into a rectangular tank, 66 cm long and 28 cm wide. Find the height of the water level in the tank}Solution 18

Question 19

_{The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words would use up a bottle of ink containing one fifth of a liter?}Solution 19

Question 20

_{1 cm³ of gold is drawn into a wire 0.1 mm is diameter. Find the length of a wire.}Solution 20

Question 21

_{Ifs 1 cm³ of cast iron weighs 21 g, find the weight of a cast iron pipe of length 1 m with a bore of 3 cm in which the thickness of the metal is 1 cm.}Solution 21

Question 22

_{A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.}Solution 22

Question 23

It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?Solution 23

Diameter of a cylinder = 140 cm

⇒ Radius, r = 70 cm

Height (h) of a cylinder = 1 m = 100 cm

Question 24

A juiceseller has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses of radius 3 cm up to a height of 8 cm, and sold for Rs.15 each, How much money does he receive by selling the juice completely?Solution 24

Radius (r) of cylindrical vessel = 15 cm

Height (h) of cylindrical vessel = 32 m

Radius of small cylindrical glass = 3 cm

Height of a small cylindrical glass = 8 cm

Question 25

A well with inside diameter 10 m is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.Solution 25

Radius of the well = 5 m

Depth of the well = 8.4 m

Width of the embankment = 7.5 m

External radius of the embankment, R = (5 + 7.5) m = 12.5 m

Internal radius of the embankment, r = 5 m

Area of the embankment = π (R² – r²)

Volume of the embankment = Volume of the earth dug out = 660 m²

Question 26

How many litres of water flows out of a pipe having an area of cross section of 5 cm² in 1 minute, if the speed of water in the pipe is 30 cm/sec?Solution 26

Speed of water = 30 cm/sec

∴ Volume of water that flows out of the pipe in one second

= Area of cross-section × Length of water flown in one second

= (5 × 30) cm³

= 150 cm³

Hence, volume of water that flows out of the pipe in 1 minute

= (150 × 60) cm³

= 9000 cm³

= 9 litresQuestion 27

A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 m per second. In how much time will the tank be filled?Solution 27

Suppose the tank is filled in x minutes. Then,

Volume of the water that flows out through the pipe in x minutes

= Volume of the tank

Hence, the tank will be filled in 28 minutes.Question 28

A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions (32 cm × 22 cm × 14 cm). Find the rise in the level of water when the solid is completely submerged.Solution 28

Let the rise in the level of water = h cm

Then,

Volume of the cylinder of height h and base radius 28 cm

= Volume of rectangular iron solid

Thus, the rise in the level of water is 4 cm.Question 29

Find the cost of sinking a tube-well 280 m deep, having a diameter 3 m at the rate of Rs.15 per cubic metre. Find also the cost of cementing its inner curved surface at Rs.10 per square metre.Solution 29

Radius, r = 1.5 m

Height, h = 280 m

Question 30

Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic centimetre of copper weights 8.4 g.Solution 30

Let the length of the wire = ‘h’ metres

Then,

Volume of the wire × 8.4 g = (13.2 × 1000) g

Thus, the length of the wire is 125 m.Question 31

It costs Rs.3300 to paint the inner curved surface of a cylindrical vessel 10 m deep at the rate of Rs.30 per m². Find the

(i) inner curved surface area of the vessel,

(ii) inner radius of the base, and

(iii) capacity of the vessel.Solution 31

Question 32

The difference between inside and outside surfaces of a cylindrical tube 14 cm long, is 88 cm². If the volume of the tube is 176 cm³, find the inner and outer radii of the tube.Solution 32

Let R cm and r cm be the outer and inner radii of the cylindrical tube.

We have, length of tube = h = 14 cm

Now,

Outside surface area – Inner surface area = 88 cm²

⇒ 2πRh – 2πrh = 88

⇒ 2π(R – r)h = 88

It is given that the volume of the tube = 176 cm³

⇒ External volume – Internal volume = 176 cm³

⇒ πR²h – πr²h = 176

⇒ π (R² – r²)h = 176

Adding (i) and (ii), we get

2R = 5

⇒ R = 2.5 cm

⇒ 2.5 – r = 1

⇒ r = 1.5 cm

Thus, the inner and outer radii of the tube are 1.5 cm and 2.5 cm respectively.Question 33

A rectangular sheet of paper 30 cm × 18 cm can be transformed into the curved surface of a right circular cylinder in two ways namely, either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders, thus formed.Solution 33

When the sheet is folded along its length, it forms a cylinder of height, h₁ = 18 cm and perimeter of base equal to 30 cm.

Let r₁ be the radius of the base and V₁ be is volume.

Then,

Again, when the sheet is folded along its breadth, it forms a cylinder of height, h₂ = 30 cm and perimeter of base equal to 18 cm.

Let r₂ be the radius of the base and V₂ be is volume.

Then,

Exercise Ex. 15D

Question 1(iii)

_{Find the volume and the surface area of a sphere whose radius is:}

_{5 m}Solution 1(iii)

Question 1(ii)

_{Find the volume and the surface area of a sphere whose radius is}

_{4.2 cm}Solution 1(ii)

Question 1(i)

_{Find the volume and the surface area of a sphere whose radius is}

_{3.5 cm}Solution 1(i)

Question 2

_{The volume of a sphere is 38808 cm³. Find the radius and hence its surface area.}Solution 2

Question 3

_{Find the surface area of a sphere whose volume is 606.375 m³}Solution 3

Question 4

Find the volume of a sphere whose surface area is 154 cm².Solution 4

Surface area of sphere = 154 cm²

⇒ 4πr² = 154

Question 5

_{The surface area of a sphere is (576) cm². Find its volume.}Solution 5

Question 6

_{How many leads shots, each 3 mm in diameter, can be made from cuboid with dimensions  (12cm x 11cm x 9cm)?}Solution 6

Question 7

_{How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm?}Solution 7

Question 8

_{A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameters 0.6 cm. find the number of small balls thus obtained.}Solution 8

Question 9

_{A metallic sphere of radius 10.5 cm is melted an then recast into smaller cones , each of radius 3.5 cm and height 3 cm. How many cones are obtained?}Solution 9

Question 10

_{How many spheres 12 cm in diameter can be made from a metallic cylinder of diameter 8 cm and}

_{height 90 cm ?}Solution 10

Question 11

_{The diameter of sphere is 6 cm. It is melted and drawn into wire of diameter 2 mm. Find the length of the wire.}Solution 11

Question 12

_{The diameter of the copper sphere is 18cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter.}Solution 12

Question 13

_{A sphere of a diameter 15.6 cm is melted and cast into a right circular cone of height 31.2 cm. find the diameter of the base of the cone.}Solution 13

Question 14

_{A spherical cannonball 28 cm in diameter is melted and recast into a right circular cone mould,  whose base is 35 cm in diameter. Find the height of the cone.}Solution 14

Question 15

_{A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2cm. Find the radius of the third ball.}Solution 15

Question 16

The radii of two spheres are in the ratio 1:2. Find the ratio of their surface areas.Solution 16

Question 17

The surface areas of two spheres are in the ratio 1:4. Find the ratio of their volumes.Solution 17

Question 18

A cylindrical tub of a radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75cm.what is the radius of the ball?Solution 18

Question 19

A cylindrical bucket with base radius 15 cm is filled with water to up height of 20 cm. a heavy iron spherical ball of radius 9 cm is dropped into the bucket to submerge completely in the water . Find the increase in the level of waterSolution 19

Question 20

_{The outer diameter of a spherical shell is 12 cm and its inner diameter is 8 cm. Find the volume of metal contained in the shell. Also, find its outer surface area.}Solution 20

Question 21

A hollow spherical shell is made of a metal of density 4.5 g per cm³. If it’s internal and external radii are 8 cm and 9cm respectively, find the weight of the shell.Solution 21

Question 22

A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm . Find the radius of the base of the cone.Solution 22

Question 23

A hemisphere bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped bottles of diameter 3 cm and height 4 cm. How many bottles are required to empty the bowl?Solution 23

Question 24

A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. Find the volume of steel used in making the bowl.Solution 24

Question 25

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.Solution 25

Inner radius = 5 cm

⇒ Outer radius = 5 + 0.25 = 5.25 cm

Question 26

A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs.32 per 100 cm².Solution 26

Inner diameter of the hemispherical bowl = 10.5 cm

Question 27

The diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?Solution 27

Let the diameter of earth = d

⇒ Radius of the earth = 

Then, diameter of moon = .

⇒ Radius of moon = 

Volume of moon 

Volume of earth 

Thus, the volume of moon is of volume of earth.Question 28

Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere?Solution 28

Volume of a solid hemisphere = Surface area of a solid hemisphere

Exercise MCQ

Question 1

The range of the data

12,25,15,18,17,20,22,6,16,11,8,19,10,30,20,32 is

1. 10
2. 15
3. 18
4. 26

Solution 1

Correct option: (d)

Range = maximum value – minimum value

 = 32 – 6

 = 26Question 2

The class mark of the class 100-120 is

1. 100
2. 110
3. 115
4. 120

Solution 2

Question 3

In the class intervals 10-20, 20-30, the number 20 is included in

1. 10-20
2. 20-30
3. In each of 10-20 and 20-30
4. In none of 10-20 and 20-30

Solution 3

Question 4

The class marks of a frequency distribution are 15, 20, 25, 30………. The class corresponding to the mark 20 is

1. 12.5-17.5
2. 17.5-22.5
3. 18.5-21.5
4. 19.5-20.5

Solution 4

Question 5

In a frequency distribution, the mid-value of a class is 10 and width of each class is 6. The lower limit of the class is

1. 6
2. 7
3. 8
4. 12

Solution 5

Question 6

The mid – value of a class interval is 42 and the class size is 10. The lower and upper limits are

1. 37-47
2. 37.5-47.5
3. 36.5-47.5
4. 36.5-46.5

Solution 6

Question 7

Let m be in the midpoint and u be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is

1. 2m – u
2. 2m + u
3. m – u
4. m + u

Solution 7

Question 8

The width of each of the five continuous classes in a frequency distribution is 5 and the lower class limit of the class is

1. 45
2. 25
3. 35
4. 40

Solution 8

Question 9

Let L be the lower class boundary of a class in a frequency distribution and m be the midpoint of the class. Which one of the following is the upper class boundary of the class?

Solution 9

Exercise Ex. 16

Question 1

Define statistics as a subject.Solution 1

Statistics is a branch of science which deals with the collection, presentation, analysis and interpretation of numerical data.Question 2

Define some fundamental characteristics of statistics.Solution 2

Fundamental characteristics of statistics :

(i) It deals only with the numerical data.

(ii) Qualitative characteristic such as illiteracy, intelligence, poverty etc cannot be measured numerically

(iii) Statistical inferences are not exact.Question 3

What are the primary data and secondary data? Which of the two is more reliable and why?Solution 3

Primary data: Primary data is the data collected by the investigator himself with a definite plan in his mind. These data are very accurate and reliable as these being collected by the investigator himself.

Secondary Data: Secondary data is the data collected by a person other than the investigator.

Secondary Data is not very reliable as these are collected by others with purpose other than the investigator and may not be fully relevant to the investigation.
Question 4

Explain the meaning of each of the following terms.

(i)Variate(ii) Class interval(iii)Class size

(iv)Class mark (v)Class limit(vi)True class limits

(vii)Frequency of a class(viii) Cumulative frequency of a classSolution 4

(i)Variate : Any character which can assume many different values is called a variate.

(ii)Class Interval :Each group or class in which data is condensed is calleda class interval.

(iii)Class-Size :The difference between the true upper limitand the true lower limit of a class is called class size.

(iv)Classmark : The average of upper and lower limit of a class interval is called its class mark.

i.e Class mark=

(v) Class limit: Class limits are the two figures by which a class is bounded . The figure on the left side of a class is called lower lower limit and on the right side is called itsupper limit.

(vi)True class limits: In the case of exclusive form of frequency distribution, the upper class limits and lower classlimits are the true upper limits and thetrue lower limits. But in the case of inclusive form of frequency distribution , the true lower limit of a class is obtained by subtracting 0.5 from the lower limit of the class. And the true upper limit of the class is obtained by adding 0.5 to the upper limit.

(vii)Frequency of a class : The number of observations falling in aclass determines its frequency.

(viii)Cumulative frequency of a class: The sum of all frequenciesup to and including that class is called , the cumulative frequency of that class.Question 5

The blood groups of 30 students of a class are recorded as under:

A, B, O, O, AB, O, A, O, A, B, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.

(i) Represent this data in the form of a frequency distribution table.

(ii) Find out which is the most common and which is the rarest blood group among these students.Solution 5

(i) Frequency Distribution Table:

(ii) The most common blood group is ‘O’ and the rarest blood group is ‘AB’.Question 6

Three coins are tossed 30 times. Each time the number of heads occurring was noted down as follows:

0, 1, 2, 2, 1, 2, 3, 1, 3, 0, 1, 3, 1, 1, 2, 2, 0, 1, 2, 1, 0, 3, 0, 2, 1, 1, 3, 2, 0, 2.

Prepare a frequency distribution table.Solution 6

Frequency Distribution Table:

Question 7

Following data gives the number of children in 40 families :

1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 4, 4, 3, 2, 2, 0, 0, 1,2, 2, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 6, 2, 2.

Represent in the form of a frequency distribution, taking classes 0-2, 2-4, etc.Solution 7nimum observation is 0 and maximum observation is 6. The classes of equal size covering the given data are : (0-2), (2-4), (4-6) and (6-8).

Thus , the frequency distribution may be given as under:

Question 8

Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as under:

8, 4, 8, 5, 1, 6, 2, 5, 3, 12, 3, 10, 4, 12, 2, 8, 15, 1, 6, 17, 5, 8, 2, 3, 9, 6, 7, 8, 14, 12.

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class interval as 5 – 10.

(ii) How many children watched television for 15 or more hours a week?Solution 8

(i) Grouped Frequency Distribution Table:

(ii) 2 children watch television for 15 or more hours a week. Question 9

The marks obtained by 40 students of a class in an examination are given below .

3, 20, 13, 1, 21, 13, 3, 23, 16, 13, 18, 12, 5, 12, 5, 24, 9, 2, 7, 18, 20, 3, 10, 12, 7, 18, 2, 5, 7, 10, 16, 8, 16, 17, 8, 23, 24, 6, 23, 15.

Present the data in the form of a frequency distribution using equal class size, one such class being 10-15(15 not included).Solution 9

Minimum observation is 1 and minimum observation is 24. The classes of equal size converging the given data are : (0-5), (5-10), (10-15), (15-20), (20-25)

Thus, the frequency distribution may be given as under :Question 10

Construct a frequency table for the following ages (in years) of 30 students using equal class intervals, one of them being 9-12, where 12 is not included.

18, 12, 7, 6, 11, 15, 21, 9, 8, 13, 15, 17, 22, 19, 14, 21, 23, 8, 12, 17, 15, 6, 18, 23, 22, 16, 9, 21, 11, 16.Solution 10

Grouped Frequency Distribution Table:

Question 11

Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of 28 labourers working in a factory, taking one of the class intervals as 210-230 (230 not included).

220, 268, 258, 242, 210, 268, 272, 242, 311, 290, 300, 320, 319, 304, 302, 318, 306, 292, 254, 278, 210, 240, 280, 316, 306, 215, 256, 236.Solution 11

Minimum observation is 210 and maximum observation =320

So the range is (320-210)=110

The classes of equal size covering the given data are :

(210-230), (230-250), (250-270) , (270-290), (290-310), (310-330)

Thus the frequency distribution may be given as under :

Question 12

The weights (in grams ) of 40 oranges picked at random from a basket are as follow :

40, 50, 60, 65, 45, 55, 30, 90, 75, 85, 70,85, 75, 80, 100, 110, 70, 55, 30, 35, 45, 70, 80, 85, 95, 70, 60, 70, 75, 40, 100, 65, 60, 40, 100, 75, 110, 30, 45, 84.

Construct a frequency table as well as a cumulative frequency table.Solution 12

Minimum observation is 30 and maximum observation is 110

So, range is 100-30=80

The classes of equal size covering the given data are :

(30-40) ,(40-50) , (50-60) ,(60-70) , (70-80), (80-90),(90-100),(100-110), (110-120)

Thus , the frequency and cumulative frequency table may be given as under :

Question 13

The heights (in cm) of 30 students of a class are given below:

161, 155, 159, 153, 150, 158, 154, 158, 160, 148, 149, 162, 163, 159, 148, 153, 157, 151, 154, 157, 153, 156, 152, 156, 160, 152, 147, 155, 155, 157.

Prepare a frequency table as well as a cumulative frequency table with 160-165 (165 not included) as one of the class intervals.Solution 13

Grouped Frequency Distribution Table and Cumulative Frequency Table:

Question 14

Following are the ages (in years ) of 360 patients , getting medical treatment in a hospital:

Age (in years)	10-20	20-30	30-40	40-50	50-60	60-70
Number of patients	90	50	60	80	50	30

Construct the cumulative frequency table for the above data.Solution 14

Age (in years)(age)	No of patients (Frequency)	Cumulative Frequency
10-2020-3030-4040-5050-6060-70	905060805030	90140200280330360
Total	360

Question 15

Present the following as an ordinary grouped frequency table :

Marks(below)	10	20	30	40	50	60
Number of students	5	12	32	40	45	48

Solution 15

Marks (below)	No of students(Cumulative Frequency.)	Class Intervals	Frequency
102030405060	51232404548	0-1010-2020-3030-4040-5050-60	512 – 5 = 732 – 12 = 2040 – 32 = 845 – 40 = 548 – 45 = 3
		Total	48

Question 16

Given below is a cumulative frequency table ;

Marks	Number of students
Below 10	17
Below 20	22
Below 30	29
Below 40	37
Below 50	50
Below 60	60

Extract a frequency table from the above .Solution 16

Marks (below)	No of students(Cumulative Frequency)	Class Intervals	Frequency
102030405060	172229375060	0-1010-2020-3030-4040-5050-60	1722 – 17 = 529 – 22 = 737 – 29 = 850 – 37 = 1360 – 50 = 10
		Total	60

Question 17

Make a frequency table from the following ;

Marks obtained	Number of students
More than 60	0
More than50	16
More than40	40
More than30	75
More than20	87
More than10	92
More than0	100

Solution 17

Marks (below)	No of student s(C.F.)	Class Intervals	Frequency
More than 60More than 50More than 40More than 30More than 20More than 10More than 0	01640758792100	More than 6050-6040-5030-4020-3010-200-10	016-0=1640-16=2475-40=3587-75=1292-87=5100-92=8
		Total	100

Question 18

The marks obtained by 17 students in a mathematics test (out of 100) are given below:

90, 79, 76, 82, 65, 96, 100, 91, 82, 100, 49, 46, 64, 48, 72, 66, 68.

Find the range of the above data.Solution 18

Arranging data in ascending order, we have

46, 48, 49, 64, 65, 66, 68, 72, 76, 79, 82, 82, 90, 91, 96, 100, 100

Minimum marks = 46

Maximum Marks = 100

∴ Range of the above data = Maximum Marks – Minimum Marks

 = 100 – 46

= 54Question 19

(i) Find the class mark of the class 90 – 120.

(ii) In a frequency distribution, the mid-value of the class is 10 and width of the class is 6. Find the lower limit of the class.

(iii) The width of each of five continuous classes in a frequency distribution is 5 and lower class limit of the lowest class is 10. What is the upper class limit of the highest class?

(iv) The class marks of a frequency distribution are 15, 20, 25, … Find the class corresponding to the class mark 20.

(v) In the class intervals 10-20, 20-30, find the class in which 20 is included.Solution 19

Question 20

Find the values of a, b, c, d, e, f, g from the following frequency distribution of the heights of 50 students in a class:

Height (in cm)	Frequency	Cumulativefrequency
160-165	15	a
165-170	b	35
170-175	12	c
175-180	d	50
180-185	e	55
185-190	5	f
	g

Solution 20

Height(in cm)	Frequency	Cumulativefrequency
160-165	15	a = 15
165-170	b = 35 – 15 = 20	35
170-175	12	c = 35 + 12 = 47
175-180	d = 50 – 47 = 3	50
180-185	e = 55 – 50 = 5	55
185-190	5	f = 55 + 5 = 60
	g = 60