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Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 12 Heron’s Formula MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 12 Heron’s Formula MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The sides of a triangle measure 8cm, 12cm and 6 cm. Then, the semiperimeter of the triangle is ……………………… cm.
A. 26
B. 52
C. 13
D. 6.5
Answer:
C. 13

Question 2.
Each side of an equilateral triangle measures 8 cm. Then, the semiperimeter of the triangle is ……………………….. cm.
A. 4
B. 24
C. 12
D.36
Answer:
C. 12

Question 3.
In a right angled triangle, the length of the hypotenuse is 15 cm and one of the sides forming right angle is 9 cm. Then, the semiperimeter of the triangle is ……………………….. cm.
A. 36
B. 18
C. 12
D. 15
Answer:
B. 18

Question 4.
The ratio of the measures of the sides of a triangle is 3:4:5. If the semiperimeter of the < triangle is 36 cm, the measure of the longest side of the triangle is ……………………. cm.
A. 12
B. 15
C. 20
D. 30
Answer:
D. 30

Question 5.
The area of a triangle is 48 cm2 and one of its sides measures 12 cm. Then, the length of the altitude corresponding to this side is …………………. cm.
A. 4
B. 8
C. 16
D. 6
Answer:
B. 8

Question 6.
The sides of a triangle measure 12 cm, 17 cm and 25 cm. Then, the area of the triangle is ……………………….. cm².
A. 54
B. 90
C. 180
D. 135
Answer:
B. 90

Question 7.
Two sides of a triangle measure 9 cm and 10 cm. If the perimeter of the triangle is 36cm, then its area is …………………. cm².
A. 17
B. 36
C. 72
D. 18
Answer:
B. 36

Question 8.
The area of an equilateral triangle with each side measuring 10 cm is ………………….. cm².
A. \(\frac{5 \sqrt{3}}{2}\)
B. 25√3
C. 5√3
D. 3√5
Answer:
B. 25√3

Question 9.
∆ ABC is an isosceles triangle in which BC = 8 cm and AB = AC = 5 cm. Then, area of ∆ ABC = ……………………….. cm².
A. 6
B. 12
C. 18
D. 24
Answer:
B. 12

Question 10.
ABCD is a parallelogram. If ar (ABC) = 18 cm², then ar(ABCD) = …………………. cm².
A. 18
B. 9
C. 36
D. 27
Answer:
C. 36

Question 11.
ABCD is a parallelogram. If ar (ABC) = 18 cm², then ar (ABCD) = …………………. cm².
A. 3.6
B. 7.2
C. 7.5
D. 6
Answer:
B. 7.2

Question 12.
In quadrilateral ABCD, AC = 10 cm. BM and DN are altitudes on AC from B and D respectively. If BM = 12cm and DN = 4 cm, then ar (ABCD) = …………………. cm².
A. 160
B. 80
C. 320
D. 480
Answer:
B. 80

Question 13.
The perimeter of rhombus ABCD is 40 cm and BD =16 cm. Then, ar (ABCD) = ……………………. cm².
A. 96
B. 48
C. 24
D. 72
Answer:
A. 96

Question 14.
The area of a rhombus is 72 cm² and one of its diagonals measures 16 cm. Then, the length of the other diagonal is ………………… cm.
A. 12
B. 9
C. 18
D. 15
Answer:
B. 9

Question 15.
PQRS is a square. If PQ = 10 cm, then PR = ……………………….. cm.
A. 10
B. 20
C. 10√2
D. 2√10
Answer:
C. 10√2

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.1

1. Measure the line segments using a ruler and a divider:

Solution:
(i) PQ = 4.4 cm
(ii) CD = 3.6 cm
(iii) XY = 2.5 cm
(iv) AB = 5.8 cm
(v) LM = 5 cm.

2. Compare the line segments in the figure and fill in the blanks:

Question (i)
AB _ AB
Solution:
AB = AB

Question (ii)
CD _ AC
Solution:
CD < AC Question (iii) AC _ AD Solution: AC > AD

Question (iv)
BC _ AC
Solution:
BC < AC Question (v) BD _ CD. Solution: BD > CD.

3. Draw any line segment AB. Take any point C between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
Solution:
If A, B, C are any three points on a line such that AC + CB = AB, then we are sure that C lies between A and B.

On measuring the lengths of AB, BC and AC, we get
AB = 6 cm, AC = 4 cm, CB = 2 cm
Now, AC + CB = 4 cm + 2 cm = 6 cm
Hence, AB = AC + CB.

4. Draw a line segment AB = 5 cm and AC = 9 cm in such a way that points A, B, C are collinear. What is the length of BC?
Solution:

AB = 5 cm and AC = 9 cm
Since, A, B and C are collinear
∴ AB + BC = AC
⇒ 5 cm + BC = 9 cm
⇒ BC = 9 cm – 5 cm
= 4 cm.
Hence, Length of BC = 4 cm

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 14 Symmetry MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 14 Symmetry MCQ Questions

Multiple Choice Questions :

Question 1.
A polygon is said to be a regular polygon if its :
(a) All sides are equal
(b) All angles are equal
(c) Both (A) and (B)
(d) None of these.
Answer:
(c) Both (A) and (B)

Question 2.
The number of lines of symmetry for an equilater triangle is :
(a) One
(b) Two
(c) Three
(d) Four
Answer:
(c) Three

Question 3.
The number of lines of symmetry for a square will be :
(a) Two
(b) Four
(c) Three
(d) One
Answer:
(b) Four

Question 4.
What other name can you give to the line of symmetry of an isoscele triangle ?
(a) Perpendicular name
(b) Height
(c) Median
(d) Altitude
Answer:
(c) Median

Question 5.
Which letter has only one line of symmetry ?
(a) Z
(b) H
(c) E
(d) N
Answer:
(c) E

Fill in the blanks :

Question 1.
The objects or figures that do not have any line of symmetry are called ……………… figures.
Answer:
Asymmetrical

Question 2.
Mirror reflection leads to ………………
Answer:
Symmetry

Question 3.
The angle by which the object rotates is called the angle of ………………
Answer:
rotation

Question 4.
The number of lines of symmetry for regular pentagon is ………………
Answer:
five

Question 5.
The number of lines of symmetry scalar for scalar triangle is ………………
Answer:
none

Write True/False :

Question 1.
A square has four lines of symmetry. (True/False)
Answer:
True

Question 2.
An isosceles triangle has a line of symmetry but not rotational symmetry. (True/False)
Answer:
True

Question 3.
A square has both line symmetry as well as rotational symmetry. (True/False)
Answer:
True

Question 4.
Some figures have only line symmetry. (True/False)
Answer:
True

Question 5.
The number of lines of symmetry for a quadrilateral is four. (True/False)
Answer:
False

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 14 Symmetry Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3

1. In the following figures, find the number of lines of symmetry and angle of rotation for rotational symmetry.

Solution:
(a) Line of symmetry 3, angle of rotation 120°
(b) Line of symmetry 4, angle of rotation 90°.

2. Name any two figures that have both line of symmetry and rotational symmetry.
Solution:
Equilateral triangle and circle

3. If a figure has two or more lines of symmetry should it have a rotational symmetry of order more than 1 ?
Solution:
Yes, Square has four lines of symmetry and rotational symmetry of order 4.

4. Following shapes have both, line symmetry and rotational symmetry. Find the number of lines of symmetry, centre of rotation and order of rotational symmetry.



Solution:
(a) 3, centroid, 3
(b) 2, Intersection of diagonals, 2
(c) 6, centre of hexagon, 6

5. Some of the english alphabets have fascinating symmetrical structures. Which capital letters have just one line of symmetry (Like E) ? Which capital letters have a rotational symmetry of order 2 (Like I) ? By attempting to think on such lines, you will be able to fill in the following table.

Solution:

6. Multiple Choice Questions :

Question (i).
If 60° is the smallest angle of rotational for a given figure what will be the angle of rotation for same figure.
(a) 150°
(b) 180°
(c) 90°
(d) 330°
Answer:
(b) 180°

Question (ii).
Which of these can not be a measure of an angle of rotation for any figure.
(a) 120°
(b) 180°
(c) 17°
(d) 90°
Answer:
(c) 17°

Question (iii).
Which of the following have both line symmetry and rotational symmetry ?
(a) An isosceles triangle
(b) A scalene triangle
(c) A square
(d) A parallelogram
Answer:
(c) A square

Question (iv).
Which of the alphabet has both multiple line and rotational symmetries ?
(a) S
(b) O
(c) H
(d) L
Answer:
(b) O

Question (v).
In the word ‘MATHS’ which of the following pairs of letters shows rotational symmetry ?
(a) M and T
(b) H and S
(c) A and S
(d) T and S
Answer:
(b) H and S

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 14 Symmetry Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2

1. Write the order of rotation for the following figures.

Solution:
(a) 2
(b) 2
(c) 5
(d) 6

2. Specify the centre of rotation, direction of rotation, angle of rotation and order of rotation for the following.

(i)

Solution:
Centre of rotation is O, direction of rotation is clockwise, Angle of rotation is 120° and order of rotation is 3.

(ii)

Solution:
Centre of rotation is P, direction of rotation is clockwise, Angle of rotation is 90° and order of rotation is 4.

(iii)

Solution:
Centre of rotation is O, direction of rotation is clockwise, Angle of rotation is 90° and order of rotation is 4.

3. Which of the following figures have rotational symmetry about the marked point (×) give the angle of rotation and order of the rotation of the figures.
(a)

Solution:
It has rotational symmetry, angle of rotation 180° and order of rotation 2.

(b)

Solution:
It has rotational symmetry, angle of rotation 90° and order of rotation 4.

(c)

Solution:
It has rotational symmetry, angle of rotation 72° and order of rotation 5.

(d)

Solution:
It has rotational symmetry, angle of rotation 60° and order of rotation 6.

(e)

Solution:
It has rotational symmetry, angle of rotation 90° and order of rotation 4.

4. Multiple choice questions :

Question (i).
The angle of rotation in an equilateral triangle is :
(a) 60°
(b) 70°
(c) 90°
(d) 120°
Answer:
(d) 120°

Question (ii).
A square has a rotational symmetry of order 4 about its centre what is the angle of rotation ?
(a) 45°
(b) 90°
(c) 180°
(d) 270°
Answer:
(b) 90°

Question (iii).
What is the order of rotational symmetry of the english alphabet Z ?
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2

Question (iv).
Which of these letters has only rotational symmetry ?
(a) S
(b) E
(c) B
(d) P
Answer:
(a) S

Question (v).
If the smallest angle of rotation is 90° then order of symmetry is ?
(a) 1
(b) 3
(c) 4
(d) 2
Answer:
(c) 4

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 8 Basic Geometrical Concepts MCQ Questions

Multiple Choice Questions

Question 1.
How many lines can pass through a point?
(a) 1
(b) 2
(c) 4
(d) Infinite.
Answer:
(d) Infinite.

Question 2.
The number of points lie on a line are
(a) 2
(b) 4
(c) 1
(d) Infinite.
Answer:
(d) Infinite.

Question 3.
The number of lines passes through two points are ……………… .
(a) 1
(b) 2
(c) 3
(d) Infinite.
Answer:
(a) 1

Question 4.
In how many parts, a closed curve divides the plane?
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(c) 3

Question 5.
A quadrilateral has……………. diagonals.
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(b) 2

Question 6.
Which of the following is not a polygon?
(a) Triangle
(b) Pentagon
(c) Circle
(d) Quadrilateral.
Answer:
(c) Circle

Question 7.
A triangle has…………… parts.
(a) 3
(b) 6
(c) 9
(d) 2.
Answer:
(b) 6

Question 8.
Which of file following is not a quadrilateral?

Answer:

Question 9.
A line segment joining the opposite vertices of a quadrilateral is called its …………. .
(a) Diagonal
(b) Side
(c) Angle
(d) Region.
Answer:
(a) Diagonal

Question 10.
The radius of a circle is 4 cm then the diameter is …………….. .
(a) 8 cm
(b) 2 cm
(c) 6 cm
(d) 12 cm.
Answer:
(a) 8 cm

Question 11.
The diameter of a circle is 12 cm then the radius is …………….. .
(a) 24 cm
(b) 6 cm
(c) 18 cm
(d) 4 cm.
Answer:
(b) 6 cm

Question 12.
The longest chord of a circle is…………….
(a) Arc
(b) Perimeter
(c) Diameter
(d) Radius.
Answer:
(c) Diameter

Question 13.
Which of the following figure is not a polygon?

Answer:

Question 14.
Which of the following figure is a polygon?

Answer:

Question 15.
Which of the following is a closed curve?

Answer:

Question 16.
Which of the following is an open curve?

Answer:

Question 17.
Two different line when intersect each other at some point, they are called:
(a) Intersecting lines
(b) Parallel lines
(c) Concurrent lines
(d) None of these.
Answer:
(a) Intersecting lines

Fill in the blanks:

Question (i)
In the environment, a railway track is an example of ……………. .
Answer:
parallel lines

Question (ii)
In the environment, a nail fixed in the wall is an example of ……………….. .
Answer:
a point

Question (iii)
The intersection of the three walls of a room, is an example of …………… .
Answer:
concurrent lines

Question (iv)
……………….. is the longest chord of a circle.
Answer:
Diameter

Question (v)
A triangle has ……………… part.
Answer:
six

Write True/False:

Question (i)
Three lines can pass through a point. (True/False)
Answer:
False

Question (ii)
The number of lines passing through two points is one. (True/False)
Answer:
True

Question (iii)
Every circle has a centre. (True/False)
Answer:
True

Question (iv)
The diameter is twice the radius. (True/False)
Answer:
True

Question (v)
Every chord of a circle is also a diameter. (True/False)
Answer:
False

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 14 Symmetry Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.1

1. Which of the following figures are asymmetrical ?


Solution:
Figure (a) and (c) are asymmetrical.

2. Draw the lines of symmetry in the following figures.


Solution:

3. Draw all lines of symmetry if any in each of the following figures.


Solution:

4. Copy the figures with punched holes and find the axes of symmetry for the following.

Solution:

5. In the following figures mark the missing holes in order to make them symmetrical about the dotted line.

Solution:
The missing holes are marked by dark punches (small circles) in each of following figures

6. In each of the following figures, the mirror line (i.e. the line of symmetry) is given as dotted line complete each figure performing reflection in the dotted (mirror) line. (You might perhaps place a mirror along the dotted line and look into the mirror for the image). Are you able to recall the name of figure you complete.

Solution:

7. Draw the reflection of the following letter in the given mirror line.

Solution:

8. Copy each diagram on a squared paper and complete each shape to be symmetrical about the mirror lines shown dotted.

Solution:

9. State the number of lines of symmetry for the following figures :

(a) A scalene triangle
(b) A rectangle
(c) A rhombus
(d) A parallelogram
(e) A regular hexagon
(f) A circle
Answer:
(a) 0
(b) 2
(c) 2
(d) 0
(e) 6
(f) Infinite.

10. Multiple Choice Questions :

Question (i).
Which of the following triangles have no line of symmetry ?
(a) An equilateral triangle
(b) An Isosceles triangle
(c) A scalene triangle
(d) All of above
Answer:
(a) An equilateral triangle

Question (ii).
What is the other name for a line of symmetry of a circle ?
(a) An arc
(b) A sector
(c) A diameter
(d) A radius
Answer:
(c) A diameter

Question (iii).
How many lines of symmetry does a regular polygon have ?
(a) Infinitely many
(b) As many as its sides
(c) One
(d) Zero
Answer:
(b) As many as its sides

Question (iv).
In the given figure, the dotted line is the line of symmetry which figure is formed if the given figure is reflected in the dotted line fig.

(a) Square
(b) Rhombus
(c) Triangle
(d) Pentagon
Answer:
(b) Rhombus

Question (v).
What is other name for a line of symmetry of an Isosceles triangle ?
(a) Side
(b) Median
(c) Radius
(d) Angle
Answer:
(b) Median

Question (vi).
Which of the following alphabets has a vertical line of symmetry ?
(a) M
(b) Q
(c) E
(d) B
Answer:
(a) M

Question (vii).
Which of the following alphabets has a horizontal line of symmetry ?
(a) C
(b) D
(c) K
(d) All the above
Answer:
(d) All the above

Question (viii).
Which of the following alphabets has no line of symmetry ?
(a) A
(b) B
(c) P
(d) O
Answer:
(c) P

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 13 Exponents and Powers MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 13 Exponents and Powers MCQ Questions

Multiple Choice Questions :

Question 1.
The value of (-1)¹⁰¹
(a) 1
(b) -1
(c) 101
(d) -101
Answer:
(b) -1

Question 2.
The value of (-1)¹⁰⁰ is:
(a) 100
(b) -100
(c) 1
(d) -1
Answer:
(c) 1

Question 3.
Find the value of 2⁶ will be :
(a) 32
(b) 64
(c) 16
(d) 8
Answer:
(b) 64

Question 4.
The exponential form of 6 × 6 × 6 × 6 is :
(a) 6²
(b) 6°
(c) 6⁴
(d) 6⁵
Answer:
(c) 6⁴

Question 5.
The exponential notation of 512 is :
(a) 2⁶
(b) 2⁷
(c) 2⁸
(d) 2⁹
Answer:
(d) 2⁹

Question 6.
The exponential form of a × a × a × c × c × c × c × d is :
(a) a³c⁴d
(b) a⁸cd
(c) a³c⁵d
(d) ab⁵d³
Answer:
(a) a³c⁴d

Question 7.
Simplify: (-3)² × (-5)².
(a) 45
(b) 75
(c) 15
(d) 225
Answer:
(d) 225

Question 8.
Choose correct expanded form of 47051 out of the following :
(a) 4 × 10⁶ + 7 × 10⁵ + 5 × 10³ + 1 × 10²
(b) 4 × 10⁵ + 7 × 10⁴ + 5 × 10 + 1
(c) 4 × 10⁴ + 7 × 10³ + 5 × 10 + 1
(d) 4 × 10⁴ + 7 × 10³ + 5 × 10² + 1
Answer:
(c) 4 × 10⁴ + 7 × 10³ + 5 × 10 + 1

Question 9.
Find the number for the following form :
3 × 10⁴ + 7 × 10² + 5 × 10°
(a) 3075
(b) 30705
(c) 375
(d) 3750
Answer:
(b) 30705

Question 10.
The value of (2° + 3° + 4°) will be :
(a) 9
(b) 3
(c) 5
(d) 24
Answer:
(b) 3

Fill in the blanks :

Question 1.
The value of (1000)° is ………………
Answer:
1

Question 2.
The value of (1)¹⁰⁰⁰ is ………………
Answer:
1

Question 3.
The value of 2⁵ is ………………
Answer:
32

Question 4.
The exponential notation of 512 is ………………
Answer:
29

Question 5.
The exponential form of 5 × 5 × 5 × 5 × 5 × 5 is ………………
Answer:
56

Write True or False :

Question 1.
The value of a⁰ is 1. (True/False)
Answer:
True

Question 2.
The value of 2⁰ × 3⁰ × 4⁰ will be 24. (True/False)
Answer:
False

Question 3.
The value of (3⁰ + 5⁰ × 2⁰ will be 2. (True/False)
Answer:
True

Question 4.
a^m ÷ aⁿ = a^mn (True/False)
Answer:
False

Question 5.
(a^m)ⁿ = a^mn (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 9 Rational Numbers Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.1

1. Write two equivalent rational numbers of the following :

Question (i).
\(\frac {4}{5}\)
Solution:
\(\frac {4}{5}\) = \(\frac {4}{5}\) × \(\frac {2}{2}\)
= \(\frac {8}{10}\)
\(\frac {4}{5}\) = \(\frac {4}{5}\) × \(\frac {3}{3}\)
= \(\frac {12}{15}\)
∴ Equivalent rational numbers of \(\frac {4}{5}\) are \(\frac {8}{10}\) and \(\frac {12}{15}\)

Question (ii).
\(\frac {-5}{9}\)
Solution:
\(\frac {-5}{9}\) = \(\frac {-5}{9}\) × \(\frac {2}{2}\)
= \(\frac {-10}{18}\)
\(\frac {-5}{9}\) = \(\frac {-5}{9}\) × \(\frac {3}{3}\)
= \(\frac {-15}{27}\)
∴ Equivalent rational numbers of \(\frac {-5}{9}\) are \(\frac {-10}{18}\) and \(\frac {-15}{27}\)

Question (iii).
\(\frac {3}{-11}\)
Solution:
\(\frac {3}{-11}\) = \(\frac {3}{-11}\) × \(\frac {2}{2}\)
= \(\frac {6}{-22}\)
\(\frac {3}{-11}\) = \(\frac {3}{-11}\) × \(\frac {3}{3}\)
= \(\frac {9}{-33}\)
∴ Equivalent rational numbers of \(\frac {3}{-11}\) are \(\frac {6}{-22}\) and \(\frac {9}{-33}\)

2. Find the standard form of the following rational numbers :

Question (i).
\(\frac {35}{49}\)
Solution:
\(\frac {35}{49}\)
∵ H.C.F. of 35 and 49 is 7

So dividing both the numerator and denominator by 7 we get.
\(\frac {35}{49}\) = \(\frac{35 \div 7}{49 \div 7}\) = \(\frac {5}{7}\)
∴ Standard form of \(\frac {35}{49}\) is \(\frac {5}{7}\)

Question (ii).
\(\frac {-42}{56}\)
Solution:
\(\frac {-42}{56}\)
∵ H.C.F. of -42 and 56 is 14

So dividing both the numerator and denominator by 14 we get.
\(\frac {-42}{56}\) = \(\frac{-42 \div 14}{56 \div 14}\) = \(\frac{-3}{4}\)
∴ Standard form of \(\frac {-42}{56}\) is \(\frac{-3}{4}\)

Question (iii).
\(\frac {19}{-57}\)
Solution:
\(\frac {19}{-57}\)
∵ H.C.F. of 59 and 57 is 19

So dividing both the numerator and denominator by 19 we get.
\(\frac {19}{-57}\) = \(\frac{-19 \div 19}{-57 \div 19}\) = \(\frac{1}{-3}\)
∴ Standard form of \(\frac {19}{-57}\) is \(\frac{1}{-3}\)

Question (iv).
\(\frac{-12}{-36}\)
Solution:
\(\frac{-12}{-36}\)
∵ H.C.F. of 12 and 36 is 12.

So dividing both the numerator and denominator by 12 we get.
\(\frac{-12}{-36}\) = \(\frac{-12 \div 12}{-36 \div 12}\) = \(\frac{1}{3}\)
Standard form of \(\frac{-12}{-36}\) is \(\frac{1}{3}\)

3. Which of the following pairs represent same rational number ?

Question (i).
\(\frac{-15}{25}\) and \(\frac{18}{-30}\)
Solution:
\(\frac{-15}{25}\) = \(\frac{-15 \div 5}{25 \div 5}\)
= \(\frac{-3}{5}\)
\(\frac{18}{-30}\) = \(\frac{18 \div-6}{-30 \div-6}\)
= \(\frac{-3}{5}\)
∴ \(\frac{-15}{25}\) and \(\frac{18}{-30}\) represents the same number.

Question (ii).
\(\frac{2}{3}\) and \(\frac{-4}{6}\)
Solution:
\(\frac{2}{3}\) = \(\frac{2}{3}\) × \(\frac{1}{1}\)
= \(\frac{2}{3}\)
\(\frac{-4}{6}\) = \(\frac{-4 \div 2}{6 \div 2}\)
= \(\frac{-2}{3}\)
∴ \(\frac{-2}{3}\) and \(\frac{-4}{6}\) doesnot represents the same rational numbers.

Question (iii).
\(\frac{-3}{4}\) and \(\frac{-12}{16}\)
Solution:
\(\frac{-3}{4}\) = \(\frac{-3}{4}\) × \(\frac{4}{4}\)
= \(\frac{-12}{16}\)
\(\frac{-12}{16}\) = \(\frac{-12}{16}\)
∴ \(\frac{-3}{4}\) and \(\frac{-12}{16}\) represents the same rational number.

Question (iv).
\(\frac{-3}{-7}\) and \(\frac{3}{7}\)
Solution:
\(\frac{-3}{4}\) = \(\frac{-3 \div-1}{-7 \div-1}\)
= \(\frac{-3}{4}\)
∴ \(\frac{-3}{-7}\) and \(\frac{3}{7}\) represents the same rational number.

4. Which is greater in each of the following ?

Question (i).
\(\frac{3}{7}\), \(\frac{4}{5}\)
Solution:
Given rational nrnnbere are \(\frac{3}{7}\) and \(\frac{4}{5}\)
L.C.M. of 7 and 5 is 35
∴ \(\frac{3}{7}\) = \(\frac{3 \times 5}{7 \times 5}\)
= \(\frac{15}{35}\)
and \(\frac{4}{5}\) = \(\frac{4 \times 7}{5 \times 7}\)
= \(\frac{28}{35}\)
∵ Numerator of second is greater than first i.e. 28 > 15
So \(\frac{4}{5}\) > \(\frac{3}{7}\)

Question (ii).
\(\frac{-4}{12}\), \(\frac{-8}{12}\)
Solution:
Given rational numbere are \(\frac{-4}{12}\) and \(\frac{-8}{12}\)
∵ Numerator of first is greater than second i.e. -4 > – 8
∴ \(\frac{-4}{12}\) > \(\frac{-8}{12}\)

Question (iii).
\(\frac{-3}{9}\), \(\frac{4}{-18}\)
Solution:
Given rational numbers are \(\frac{-3}{9}\), \(\frac{4}{-18}\)
\(\frac{-3}{9}\) = \(\frac{-3 \times 2}{9 \times 2}\)
= \(\frac{-6}{18}\)
\(\frac{4}{-18}\) = \(\frac{4 \times-1}{-18 \times-1}\)
\(\frac{-4}{18}\)
Since -4 > – 6.
\(\frac{4}{-18}\) > \(\frac{-3}{9}\)

Question (iv).
-2\(\frac{3}{5}\), -3\(\frac{5}{8}\)
Solution:
-2\(\frac{3}{5}\) = \(\frac{-13}{5} \times \frac{8}{8}\)
= \(\frac{-104}{40}\)
-3\(\frac{5}{8}\) = \(\frac{-29}{8} \times \frac{5}{5}\)
= \(\frac{-135}{40}\)
∵ -104 > -135
∴ \(\frac{-13}{5}\) > \(\frac{-29}{8}\)
Thus, -2\(\frac{3}{5}\) > -3\(\frac{5}{8}\)

5. Write the following rational numbers in ascending order.

Question (i).
\(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)
Solution:
\(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)
Here -5 < -3 < -1
i.e. \(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)
Therefore, the ascending order is:
\(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)

Question (ii).
\(\frac{-1}{5}, \frac{-2}{15}, \frac{-4}{5}\)
Solution:
\(\frac{-1}{5}, \frac{-2}{15}, \frac{-4}{5}\)
L.C.M of 5, 15, 5 is 15

Question (iii).
\(\frac{-3}{8}, \frac{-2}{4}, \frac{-3}{2}\)
Solution:
\(\frac{-3}{8}, \frac{-2}{4}, \frac{-3}{2}\)
L.C.M of 8, 4, 2 is 8
∴ \(\frac{-3}{8}=\frac{-3}{8} \times \frac{1}{1}=\frac{-3}{8}\)
\(\frac{-2}{4}=\frac{-2 \times 2}{4 \times 2}=\frac{-4}{8}\)
\(\frac{-3}{2}=\frac{-3 \times 4}{2 \times 4}=\frac{-12}{8}\)
∴ -12 < -4 < -3
or \(\frac {-12}{8}\) < \(\frac {-4}{8}\) < \(\frac {-3}{8}\)
Hence assending order is \(\frac{-3}{2}, \frac{-2}{4}, \frac{-3}{8}\)

6. Write five rational numbers between following rational numbers.

Question (i).
-2 and -1
Solution:
Given rational numbers are -2 and -1
Let us write -2 and -1 as rational numbers with 5 + 1 = 6 as denominator.
We have -2 = -2 × \(\frac {6}{6}\)
= \(\frac {-6}{6}\)
\(\frac {-12}{6}\) < \(\frac {-11}{6}\) < \(\frac {-10}{6}\) < \(\frac {-9}{6}\) < \(\frac {-8}{6}\) < \(\frac {-7}{6}\) < \(\frac {-6}{6}\)
Hence five rational numbers between -2 and -1 are :
\(\frac {-11}{6}\),\(\frac {-10}{6}\),\(\frac {-9}{6}\),\(\frac {-8}{6}\),\(\frac {-7}{6}\)
i.e. \(\frac {-11}{6}\),\(\frac {-5}{3}\),\(\frac {-3}{2}\),\(\frac {-4}{3}\),\(\frac {-7}{6}\)

Question (ii).
\(\frac {-4}{5}\) and \(\frac {-2}{3}\)
Solution:
Given rational numbers are \(\frac {-4}{5}\) and \(\frac {-2}{3}\)
First we find equivalent rational numbers having same denominator
Thus \(\frac {-4}{5}\) = \(\frac{-4 \times 9}{5 \times 9}\)
= \(\frac {-36}{45}\)
and \(\frac {-2}{3}\) = \(\frac{-2 \times 15}{3 \times 15}\)
= \(\frac {-30}{45}\)
Now, we choose any five integers -35, -34, -33, -32, -31 between the numerators -36 and -30
Then the five rational numbers between \(\frac {-36}{45}\) and \(\frac {-30}{45}\) are:
\(\frac{-35}{45}, \frac{-34}{45}, \frac{-33}{45}, \frac{-32}{45}, \frac{-31}{45}\)
Hence, five rational numbers between \(\frac {-4}{5}\) and \(\frac {-2}{3}\) are
\(\frac{-35}{45}, \frac{-34}{45}, \frac{-33}{45}, \frac{-32}{45}, \frac{-31}{45}\)
i.e. \(\frac{-7}{9}, \frac{-34}{45}, \frac{-11}{15}, \frac{-32}{45}, \frac{-31}{45}\)

Question (iii).
\(\frac {1}{3}\) and \(\frac {5}{7}\)
Solution:
Given rational numbers are \(\frac {1}{3}\) and \(\frac {5}{7}\)
First we find equivalent rational numbers having same denominator

\(<\frac{4}{7}<\frac{13}{21}<\frac{2}{3}<\frac{5}{7}\)
Hence, five rational numbers between \(\frac {1}{3}\) and \(\frac {5}{7}\) are
\(\frac{8}{21}, \frac{3}{7}, \frac{10}{21}, \frac{4}{7}, \frac{13}{21}\).

7. Write four more rational numbers in each of the following.

Question (i).
\(\frac{-1}{5}, \frac{-2}{10}, \frac{-3}{15}, \frac{-4}{20}, \ldots\)
Solution:
The given rational numbers are :
\(\frac{-1}{5}, \frac{-2}{10}, \frac{-3}{15}, \frac{-4}{20}, \ldots\)
\(\frac {-1}{5}\) is the rational number in its lowest form
Now, we can write
\(\frac{-2}{10}=\frac{-1}{-5} \times \frac{2}{2}\),
\(\frac{-3}{15}=\frac{-1}{5} \times \frac{3}{3}\) and \(\frac{-1}{5}=\frac{-1}{5} \times \frac{4}{4}\)
Thus, we observe a pattern in these numbers.
The next four rational numbers would be
\(\frac{-1}{5} \times \frac{5}{5}=\frac{-5}{25}\),
\(\frac{-1}{5} \times \frac{6}{6}=\frac{-6}{30}\),
\(\frac{-1}{5} \times \frac{7}{7}=\frac{-7}{35}\)
\(\frac{-1}{5} \times \frac{8}{8}=\frac{-8}{40}\)
Hence required four more rational numbers are :
\(\frac{-5}{25}, \frac{-6}{30}, \frac{-7}{35}, \frac{-8}{40}\)

Question (ii).
\(\frac{-1}{7}, \frac{2}{-14}, \frac{3}{-21}, \frac{4}{-28}, \ldots\)
Solution:
The given rational numbers are
\(\frac{-1}{7}, \frac{2}{-14}, \frac{3}{-21}, \frac{4}{-28}, \ldots\)
\(\frac {-1}{7}\) is the rational number in its lowest form
Now, we can write
\(\frac{2}{-14}=\frac{-1}{7} \times \frac{-2}{-2}=\frac{2}{-14}, \frac{3}{-21}\)
= \(\frac{-1}{7} \times \frac{-3}{-3}\) and \(\frac{4}{-28}=\frac{-1}{7} \times \frac{-4}{-4}\)
Thus, we observe a pattern in these numbers.
The next four rational numbers would be :
\(-\frac{1}{7} \times \frac{-5}{-5}=\frac{5}{-35}\), \(\frac{-1}{7} \times \frac{-6}{-6}=\frac{6}{-42}\),
\(\frac{-1}{7} \times \frac{-7}{-7}=\frac{7}{-49}\), \(\frac{-1}{7} \times \frac{-8}{-8}=\frac{8}{-56}\)
Hence required four more rational numbers are :
\(\frac{5}{-35}, \frac{6}{-42}, \frac{7}{-49}, \frac{8}{-56}\)

8. Draw a number line and represent the following rational number on it.

Question (i).
\(\frac {2}{4}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0). We choose a point A to the right of 0 to represent 1.
Divide the segment OA into four equal parts. Second part from O to the right represents the rational number \(\frac {2}{4}\) as shown in the figure.

Question (ii).
\(\frac {-3}{4}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0). We choose a point A to the right of 0 to represent -1.
Divide the segment OA into four equal parts. Third part from O to the left represents the rational number \(\frac {-3}{4}\) as shown in the figure.

Question (iii).
\(\frac {5}{8}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0).
We choose a point A to the right of 0 to represent 1.
Divide the segment OA into eight equal parts. Fifth part from O to the right represents the rational number as shown in the figure.

Question (iv).
\(\frac {-6}{4}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0).
We choose a point A to the left of O to represent -2.
Divide the segment OA into eight equal parts. Sixth part from O to the left represents the rational number \(\frac {-6}{4}\) as shown in the figure.

9. Multiple Choice Questions :

Question (i).
\(\frac{3}{4}=\frac{?}{12}\) then ? =
(a) 3
(b) 6
(c) 9
(d) 12.
Answer:
(c) 9

Question (ii).
\(\frac{-4}{7}=\frac{?}{14}\) then ? =
(a) -4
(b) -8
(c) 4
(d) 8
Answer:
(b) -8

Question (iii).
The standard form of rational number \(\frac {-21}{28}\) is
(a) \(\frac {-3}{4}\)
(b) \(\frac {3}{4}\)
(c) \(\frac {3}{7}\)
(d) \(\frac {-3}{7}\)
Answer:
(a) \(\frac {-3}{4}\)

Question (iv).
Which of the following rational number is not equal to \(\frac {7}{-4}\) ?
(a) \(\frac {14}{-8}\)
(b) \(\frac {21}{-12}\)
(c) \(\frac {28}{-16}\)
(d) \(\frac {7}{-8}\)
Answer:
(d) \(\frac {7}{-8}\)

Question (v).
Which of the following is correct ?
(a) 0 > \(\frac {-4}{9}\)
(b) 0 < \(\frac {-4}{9}\)
(c) 0 = \(\frac {4}{9}\)
(d) None
Answer:
(a) 0 > \(\frac {-4}{9}\)

Question (vi).
Which of the following is correct ?
(a) \(\frac{-4}{5}<\frac{-3}{10}\)
(b) \(\frac{-4}{5}>\frac{3}{-10}\)
(c) \(\frac{-4}{5}=\frac{3}{-10}\)
(d) None
Answer:
(a) \(\frac{-4}{5}<\frac{-3}{10}\)

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.3

1. Find prime factors of the following numbers by factor tree method:

Question (i)
96
Solution:

∴ Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3

Question (ii)
120
Solution:

∴ Prime factorisation of 120 = 2 × 2 × 2 × 3 × 5

Question (iii)
180.
Solution:

∴ Prime factorisation of 180 = 2 × 2 × 3 × 3 × 5

2. Complete each factor tree:

Question (i)

Solution:

Question (ii)

Solution:

Question (iii)

Solution:

3. Find the prime factors of the following numbers by division method:

Question (i)
420
Solution:

∴ 420 = 2 × 2 × 3 × 5 × 7

Question (ii)
980
Solution:

∴ 980 = 2 × 2 × 5 × 7 × 7

Question (iii)
225
Solution:

∴ 225 = 3 × 3 × 5 × 5

Question (iv)
150
Solution:

∴ 150 = 2 × 3 × 5 × 5

Question (v)
324
Solution:

∴ 324 = 2 × 2 × 3 × 3 × 3 × 3