Class 6

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.3

1. Identify the shape having perpendicular lines:

Solution:
(ii), (iii), (v).

2. Identify the examples having perpendicular lines:

Question (i)
(i) Lines of railway track.
(ii) Adjacent edges of a table.
(iii) Line segment forming letter ‘L’.
Solution:
(ii) Adjacent edges of a table.
(iii) Line segment forming letter ‘L’.

3. Let \(\overrightarrow{\mathbf{AB}}\) be perpendicular to \(\overrightarrow{\mathbf{PQ}}\) and they intersect at O. What is the measure of \(\angle \mathbf{AOP}\)?
Solution:

\(\angle \mathbf{AOP}\) = 90°, because AB ⊥ PQ.

4. Line m is perpendicular to line l in the given figure. Each point on the line l is marked at equal intervals. Study the diagram and state true or false.

Question (i)
Line m is ⊥ bisector of line segment AI.
Solution:
True

Question (ii)
CE = EG
Solution:
True

Question (iii)
DF = 2DE.
Solution:
True.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.2

1. Classify the angles as acute, obtuse, right, straight or reflex angles:

Solution:
(i) Acute angle
(ii) Obtuse angle
(iii) Reflex angle
(iv) Straight angle
(v) Acute angle
(vi) Right angle
(vii) Obtuse angle
(viii) Right angle
(ix) Reflex angle
(x) Acute angle.

2. Classify the angles:

Question (i)
80°
Solution:
80° is between 0° and 90°.
∴ It is an acute angle.

Question (ii)
172°
Solution:
172° is between 90° and 180°
∴ It is an obtuse angle.

Question (iii)
90°
Solution:
90° is a right angle.

Question (iv)
0°
Solution:
0° is a zero angle.

Question (v)
179°
Solution:
179° is between 90° and 180°.
∴ It is an obtuse angle.

Question (vi)
215°
Solution:
215° is between 180° and 360°.
∴ It is an reflex angle.

Question (vii)
360°
Solution:
360° is a complete angle.

Question (viii)
350°
Solution:
350° is between 180° and 360°.
∴ It is a reflex angle.

Question (ix)
15°
Solution:
15° is between 0° and 90°.
∴ It is an acute angle.

Question (x)
180°
Solution:
180° is a straight angle.

3. Measure the following angles with protractor and write their measurement:

Solution:
(i) 60°
(ii) 125°
(iii) 110°
(iv) 80°
(v) 120°
(vi) 105°
(vii) 80°
(viii) 135°
(ix) 88°
(x) 90°.

4. How many degrees are there in

Question (i)
Two right angles
Solution:
1 right angle = 90°
∴ Two right angles = 2 × 90°
= 180°

Question (ii)
\(\frac {2}{3}\) right angles
Solution:
1 right angle = 90°
∴ \(\frac {2}{3}\) right angles = \(\frac {2}{3}\) × 90°
= 2 × 30°
= 60°

Question (iii)
Four right angles?
Solution:
1 right angle = 90°
∴ Four right angles = 4 × 90°
= 360°

5. What fraction of a clockwise revolution does the hour hand of a clock turn through when it goes from:

Question (i)
3 to 9
Solution:
3 to 9 : Half or \(\frac {1}{2}\)

Question (ii)
5 to 8
Solution:
5 to 8 : Quarter or \(\frac {1}{4}\)

Question (iii)
10 to 4
Solution:
10 to 4 : Half or \(\frac {1}{2}\)

Question (iv)
2 to 11
Solution:
2 to 11 : 3 Quarters or \(\frac {3}{4}\)

Question (v)
6 to 3
Solution:
6 to 3 : 3 Quarters or \(\frac {3}{4}\)

Question (vi)
2 to 7.
Solution:
2 to 7 : \(\frac {5}{12}\)

6. Find the number of right angles turned through by the hour hand of a dock when it goes from

Question (i)
5 to 8
Solution:
5 to 8 : 1 right angle

Question (ii)
1 to 7
Solution:
1 to 7 : 2 right angles

Question (iii)
4 to 10
Solution:
4 to 10 : 2 right angles

Question (iv)
9 to 12
Solution:
9 to 12 : 1 right angles

Question (v)
11 to 2
Solution:
11 to 2 : 1 right angles

Question (vi)
9 to 6
Solution:
9 to 6 : 3 right angles

Question (vii)
2 to 11
Solution:
2 to 11 : 3 right angles

Question (viii)
10 to 1
Solution:
10 to 1 : 1 right angles

Question (ix)
12 to 6
Solution:
12 to 6 : 2 right angles

Question (x)
5 to 2.
Solution:
5 to 2 : 3 right angles.

7. Where will be the hand of a clock stop if it starts at:

Question (i)
12 and make \(\frac {1}{4}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours.
For \(\frac {1}{4}\) revolution, the hour hand takes \(\frac {1}{4}\) × 12 hours = 3 hours.
If hour hand starts at 12 and make \(\frac {1}{4}\) revolution clockwise it will stop at 3.

Question (ii)
2 and make \(\frac {1}{2}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours.
For \(\frac {1}{2}\) revolution, the hour hand takes \(\frac {1}{2}\) × 12 hours = 6 hours.
If hour hand starts at 2 and make \(\frac {1}{2}\) revolution clockwise it will stop at 8.

Question (iii)
5 and make \(\frac {1}{4}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours.
For \(\frac {1}{4}\) revolution, the hour hand takes \(\frac {1}{4}\) × 12 hours = 3 hours
If hour hand starts at 5 and make \(\frac {1}{4}\) revolution clockwise it will stop at 8.

Question (iv)
5 and make \(\frac {3}{4}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours .
For \(\frac {3}{4}\) revolution, the hour hand takes \(\frac {3}{4}\) × 12 hours = 9 hours.
If hour hand starts at 5 and make \(\frac {3}{4}\) revolution clockwise it will stop at 2.

8. What part of revolution have you turned through if you stand facing:

Question (i)
East and turn clockwise to North
Solution:
I turned through \(\frac {3}{4}\) part of a revolution.

Question (ii)
South and turn clockwise to North
Solution:
I turned through \(\frac {1}{2}\) part of a revolution.

Question (iii)
South and turn clockwise to East
Solution:
I turned through \(\frac {3}{4}\) part of a revolution.

Question (iv)
West and turn clockwise to East
Solution:
I turned through \(\frac {1}{2}\) part of at revolution.

9. Find the angle measure between the hands of the clock in each figure:

Solution:
(i) Angle measure between the hands of the clock at 3.00 a.m.
= \(\frac {3}{12}\) × 360° = 90°
(ii) Angle measure between the hands of the clock at 6.00 a.m.
= \(\frac {6}{12}\) × 360° = 180°
(iii) Angle measure between the hands of the clock at 2.00 a.m.
= \(\frac {2}{12}\) × 360° = 60°

10. Draw the following angles by protractor:

Question (i)
(i) 40°
(ii) 75°
(iii) 105°
(iv) 90°
(v) 130°
Solution:

11. State true or false:

Question (i)
The sum of two right angles is always a straight angle.
Solution:
True

Question (ii)
The sum of two acute angles is always a reflex angle.
Solution:
False

Question (iiii)
The obtuse angle has measurement between 90° to 180°.
Solution:
True

Question (iv)
A complete revolution has four right angles.
Solution:
True

12. Fill in the blanks:

Question (i)
The angle which is greater than 0° and less than 90° is called ………….. .
Solution:
acute angle

Question (ii)
The angle whose measurement equal to two right angle is …………….. .
Solution:
straight angle

Question (iii)
The angle between 90° and 180° is ……………. .
Solution:
obtuse angle.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.6

1. In the given figure, write the name of:

Question (i)
Centre
Solution:
Centre: O

Question (ii)
Radii
Solution:
Radii: OX, OY, OP

Question (iii)
Diameter
Solution:
Diameter: XY

Question (iv)
Chord.
Solution:
Chord: QR.

2. In the given figure, write the name

Question (i)
minor arc
Solution:
Minor arc : PAQ

Question (ii)
major arc
Solution:
Major arc : PBQ

Question (iii)
minor sector
Solution:
Minor sector: OPAQ

Question (iv)
major sector.
Solution:
Major sector : OPBQ.

3. In the given figure, write the name

Question (i)
Minor segment
Solution:
Minor segment: ACBA

Question (ii)
Major segment.
Solution:
Major segment: ADBA

4. In the given figure, name the points:

Question (i)
In its interior
Solution:
Points in its interior : O, A, D, F

Question (ii)
On its boundary (circumference)
Solution:
Points on its boundary (circumference) C

Question (iii)
In its exterior.
Solution:
Points in its exterior : B, E

5. Find the diameter of the circle whose radius is:

Question (i)
5 cm
Solution:
Given Radius of the circle = 5 cm
∴ Diameter of the Circle = 2 × radius = 2 × 5 cm = 10 cm

Question (ii)
4 cm
Solution:
Given radius of the circle = 4 cm
∴ Diameter of the circle = 2 × radius = 2 × 4m = 8m

Question (iii)
10 cm.
Solution:
Given radius of the circle = 10 cm
∴ Diameter of circle = 2 × Radius = 2 × 10 cm = 20 cm

6. If the diameter of the circle is 12 cm. Find the radius:
Solution:
Given diameter of a circle = 12 cm
∴ Radius of circle = Diameter + 2
= 12 cm + 2
= 6 cm

7. Fill in the blanks:

Question (i)
The distance around a circle is called ……………… .
Solution:
Circumference

Question (ii)
The diameter of a circle is ……………… times its radius.
Solution:
two

Question (iii)
The longest chord of circle is …………….. .
Solution:
diameter

Question (iv)
All the radii of a circle are of ……………… length.
Solution:
equal

Question (v)
The diameter of a circle passes through ……………. .
Solution:
centre

Question (vi)
A circle divides all the points in a plane into ……………… parts.
Solution:
three.

8. State true or false:

Question (i)
The diameter of a circle is equal to its radius.
Solution:
False

Question (ii)
The diameter is a chord of circle.
Solution:
True

Question (iii)
A radius is a chord of the circle.
Solution:
False

Question (iv)
Every circle has a centre.
Solution:
True

Question (v)
The region enclosed by a chord and arc is called a segment
Solution:
True

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.5

1. Out of the following, Identify the quadrilateral:

Solution:

2. Name the given quadrilaterals:

Solution:
(i) ABCD
(ii) PQRS
(iii) XYZW.

3. Write the name of all vertices, angles, sides, diagonals of the following quadrilaterals:

Solution:
(i) Vertices = O, N, M, L;
Angles = \(\angle \mathrm{O}, \angle \mathrm{N}, \angle \mathrm{M}, \angle \mathrm{L}\)
Sides = ON, NM, ML, LO;
Diagonals = OM, NL

(ii) Vertices = H, G, F, E;
Angles = \(\angle \mathrm{H}, \angle \mathrm{G}, \angle \mathrm{F}, \angle \mathrm{E}\)
Sides = HG, GF, FE, EH;
Diagonals = EG, FH.

4. For the given quadrilateral ABCD, name:

Question (i)
(i) Side opposite to AB
(ii) Angles adjacent to B
(iii) Diagonal joining B and D
(iv) Angle opposite to A
(v) Sides adjacent to CD.

Solution:
(i) CD
(ii) \(\angle \mathrm{A} \text { and } \angle \mathrm{C}\)
(iii) BD
(iv) \(\angle \mathrm{C}\)
(v) AD and BC.

5. In the given quadrilateral JUMP, name the points.

Question (i)
In its interior
Solution:
Points in the interior of quad. JUMP are :
L, N, C

Question (ii)
In its exterior
Solution:
Points in its exterior of quad. JUMP are :
B, O, X

Question (iii)
On its boundary.
Solution:
Points on the boundary of quad. JUMP are :
P, M, U, Y, J, A.

6. Fill in the blanks:

Question (i)
A quadrilateral has …………….. vertices.
Solution:
4

Question (ii)
A quadrilateral has …………… sides.
Solution:
4

Question (iii)
A quadrilateral has …………… angles.
Solution:
4

Question (iv)
A quadrilateral has ………….. diagonals.
Solution:
2

Question (v)
A diagonal divides the quadrilateral into……………… triangles.
Solution:
2

Question (vi)
A line segment joining the opposite vertices of a quadrilateral is called its ………. .
Solution:
Diagonal

Question (vii)
The interior and the boundary of a quadrilateral together constitute the ……………. region.
Solution:
Quadrilateral.

7. State True or False:

Question (i)
A diagonal divides quadrilateral into four triangles.
Solution:
False

Question (ii)
The angle that have a common vertex are called adjacent angles.
Solution:
True

Question (iii)
The sides that have a common vertex are called adjacent sides.
Solution:
True

Question (iv)
A quadrilateral has four diagonals.
Solution:
False

Question (v)
The quadrilateral region consists of the exterior and the boundary of the quadrilateral.
Solution:
False

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.4

1. Write all the names of the following triangles in all orders:

Solution:
(i) ∆ABC, ∆ACB, ∆BAC, ∆BCA, ∆CAB, ∆CBA.
(ii) ∆XYZ, ∆XZY, ∆YZX, ∆YXZ, ∆ZXY, ∆ZYX
(iii) ∆LMN, ∆LNM, ∆MNL, ∆MLN, ∆NML, ∆NLM.

2. Write the name of vertices, sides and angles of the following triangles:

Solution:

	(i)	(ii)	(iii)
Vertices	P, R, Q	D, E, F	T, P, S
Sides	PR, QR, PQ	DE, EF, DF	TP, PS, TS
Angles	\(\angle \mathrm{P}, \angle \mathrm{R}, \angle \mathrm{Q}\)	\(\angle \mathrm{D}, \angle \mathrm{E}, \angle \mathrm{F}\)	\(\angle \mathrm{T}, \angle \mathrm{P}, \angle \mathrm{S}\)

3. In the given figure, name the points that lie:

Question (i)
On the boundary of ∆GEM
Solution:
Points on the boundary of AGEM are: G, A, E, C, M

Question (ii)
In the interior of ∆GEM
Solution:
Points in the interior of AGEM are : P, X, D

Question (iii)
In the exterior of ∆GEM.
Solution:
Points in the exterior of AGEM are : Y, B.

4. In the given figure, write the name of:

Question (i)
All different triangles
Solution:
All different triangles are :
∆AOD, ∆DOC, ∆BOC, ∆AOB, ∆ABD, ∆BCD, ∆ACD, ∆ABC

Question (ii)
Triangles having O as the vertex
Solution:
Triangles having O as the vertex are :
∆AOB, ∆BOC, ∆COD, ∆AOD

Question (iii)
Triangles having A as the vertex.
Solution:
Triangles having A as the vertex are:
∆AOB, ∆AOD, ∆ABD, ∆ABC, ∆ACD.

5. Fill in the blanks of the following:

Question (i)
A triangle has …………… vertices.
Solution:
3

Question (ii)
A triangle has …………… angles.
Solution:
3

Question (iii)
A triangle has ……………. sides.
Solution:
3

Question (iv)
A triangle divide the plane into ……………. parts.
Solution:
3

Question (v)
A triangle has …………… parts.
Solution:
6

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.3

1. Name the given angles in all ways:

Solution:
(i) \(\angle \mathrm{DEF}, \angle \mathrm{FED}, \angle \mathrm{E}, \angle a\)
(ii) \(\angle \mathrm{XOY}, \angle \mathrm{YOX}, \angle \mathrm{O}, \angle 1\)
(iii) \(\angle N O M, \angle M O N, \angle O, \angle x\)

2. Name the vertex and the arms of given angles:

Solution:

	(i)	(ii)	(iii)
Vertex	B	Q	o
Arm	\(\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{BA}}\)	\(\overrightarrow{\mathrm{QP}}, \overrightarrow{\mathrm{QR}}\)	\(\overrightarrow{\mathrm{OS}}, \overrightarrow{\mathrm{OP}}\)

3. Name all the angles of the given figure:

Solution:
(i) \(\angle \mathrm{X}, \angle \mathrm{Y}, \angle \mathrm{Z}\)
(ii) \(\angle \mathrm{P}, \angle \mathrm{Q}, \angle \mathrm{R}, \angle \mathrm{S}\)
(iii) \(\angle \mathrm{AOB}, \angle \mathrm{BOC}, \angle \mathrm{AOC}\)

4. In the given figure, name the points that lie:

Question (i)
In the interior of \(\angle \mathrm{DOE}\)
Solution:
Points in the interior of \(\angle \mathrm{DOE}\) are :
A, X, M

Question (ii)
In the exterior of \(\angle \mathrm{DOE}\)
Solution:
Points in the exterior of \(\angle \mathrm{DOE}\) are :
H, L

Question (iii)
On the \(\angle \mathrm{DOE}\)
Solution:
Points on the \(\angle \mathrm{DOE}\) are :
D, B, O, E.

5. In the given figure, write another name for the following angles :

Question (i)
\(\angle \mathrm{1}\)
Solution:
\(\angle S \text { or } \angle PSR \text { or } \angle RSP\)

Question (ii)
\(\angle \mathrm{2}\)
Solution:
\(\angle \mathrm{RPQ} \text { or } \angle \mathrm{QPR}\)

Question (iii)
\(\angle \mathrm{3}\)
Solution:
\(\angle \mathrm{SRP} \text { or } \angle \mathrm{PRS}\)

Question (iv)
\(\angle \mathrm{a}\)
Solution:
\(\angle \mathrm{Q} \text { or } \angle \mathrm{RQP} \text { or } \angle \mathrm{PQR}\)

Question (v)
\(\angle \mathrm{b}\)
Solution:
\(\angle \mathrm{PRQ} \text { or } \angle \mathrm{QRP}\)

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 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 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

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 7 Algebra Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 7 Algebra Ex 7.3

1. Pick the algebraic expressions and the arithmetic expressions from the following:

Question (i)
(i) 2l – 3
(ii) 5 × 3 + 8
(iii) 6 – 3x
(iv) 51
(v) 2 × (21 – 18) + 9
(vi) \(\frac {6a}{5}\) + 2
(vii) 7 × 20 + 5 + 3
(viii) 8.
Solution:
Algebraic Expressions :
2l – 3, 6 – 3x, 51, \(\frac {6a}{5}\) + 2
Arithmetic Expressions :
5 × 3 + 8, 2 × (21 – 18) + 9, 7 × 20 + 5 + 3, 8.

2. Write the terms for the following expressions:

Question (i)
2y + 5z
Solution:
Terms of 2y + 5z = 2y, 5z

Question (ii)
6x – 3y + 8
Solution:
Terms of 6x – 3y + 8 = 6x, -3y, 8

Question (iii)
7a
Solution:
Terms of 7a = 7a

Question (iv)
3l – 5m + 2n
Solution:
Terms of 31 – 5m + 2n = 31, -5m, 2n

Question (v)
\(\frac {2l}{3}\) + x.
Solution:
Terms of = \(\frac {2l}{3}\) + x = \(\frac {2l}{3}\), x

3. Tell how the following expressions are formed.

Question (i)
a + 11
Solution:
a + 11 = a is increased by 11

Question (ii)
12 – x
Solution:
12 – x = x is subtracted from 12

Question (iii)
3z + 8
Solution:
3z + 8 = Three time of z is increased by 8

Question (iv)
6 – 5l
Solution:
6 – 5l = 5 times of l is subtracted from 6

Question (v)
\(\frac {5a}{4}\)
Solution:
\(\frac {5a}{4}\) = 5 times a is divided by 4.

4. Give expressions for the following:

Question (i)
10 is added to p
Solution:
10 is added to p = p + 10

Question (ii)
5 is subtracted from y
Solution:
S is subtracted from y = y – 5

Question (iii)
d is divided by 3
Solution:
d is divided by 3 = \(\frac {d}{3}\)

Question (iv)
l is multiplied by – 6
Solution:
l is multiplied by – 6 = – 6l

Question (v)
m is subtracted from l
Solution:
m is subtracted from 1 = 1 – m

Question (vi)
11 is added to 3x
Solution:
11 is added to 6x = 6x + 11

Question (vii)
y is divided by -2 and then 2 is added to the result
Solution:
y is multiplied by – 2 and then 2 is added to the result = – 2y + 2

Question (viii)
c is divided by 5 and then 7 is multiplied to the result
Solution:
c is divided by 5 and thus 7 is multiplied to the result = \(\frac {7c}{5}\)

Question (ix)
x is multiplied by 3 then subtracted this result from y
Solution:
x is multiplied by 3 then subtracted this result from y = y – 3x

Question (x)
a is added to b then c is multiplied with this result.
Solution:
a is added to b then c is multiplied by this result = (a + b) c

5. Write the number which is 15 less than y.
Solution:
The number which is 15 less than y = y – 15

6. Write the number which is 3 more than a.
Solution:
The number which is 3 more than a = a + 3

7. Find the number which is 1 more than twice of x.
Solution:
The number which is 1 more than twice of x = 2x + 1

8. Find the number which is 7 less than 5 times of y.
Solution:
The number which is 7 less than 5 times of y = 5y – 7

9. Somi’s present age is ‘a’ years. Express the following in algebraic form:

Question (i)
Her age after 15 years.
Solution:
Somi’s present age = ‘a’ years
Her age after 15 years = (a + 15) years

Question (ii)
Her age 2 years ago.
Solution:
Her age 2 years ago = (a – 2) years

Question (iii)
If Somi’s father’s age is 5 more than twice of her present age, express her father’ age.
Solution:
Somi’s father’s age is 5 more than twice of her present age
∴ Her father’s age = (2a + 5) years.

Question (iv)
If Somi’s sister is 4 years younger to her. Express her sister’s age.
Solution:
Somi’s sister is 4 years younger to her
∴ Her sister’s age = (a – 4) years

Question (v)
If Somi’s mother is 3 less than 3 times her present age. Express her mother’s age.
Solution:
Somi’s mother is 3 less than 3 times her present age
∴ Her mother’s age = (3a – 3) years.

10. The length of a floor is 10 more than two times of breadth what is the length if breadth is l meters?
Solution:
The breadth of floor = l metres
The length of the floor is 10 more than two times of its breadth
∴ Length of floor = (2l + 10) metres.