Class 6

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 8 Basic Geometrical Concepts Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.2

1.

Question (i)
(a) Which of the following are simple curves?
(b) Classify the following as open or closed curve.

Solution:
(a) Simple curves :
(i), (iii), (iv), (vi), (vii), (iii)

(b) Open curves :
(iii) , (vi), (viii)
Closed curves :
(i), (ii), (iv), (v), (vii)

2. Identify the polygons:

Question (i)

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

3. Draw any polygon and shade its interior.
Solution:

4. Name the points which are:

Question (i)
In the interior of the closed figure.
Solution:
Points in the interior of closed figure are :
A B, Q

Question (ii)
In the exterior of the closed figure,
Solution:
Points in the exterior of closed figure are :
R, N

Question (iii)
On the boundary of the closed figure.
Solution:
Points in the boundary of closed figure are :
P, M

5. In the given figure, name :

Question (i)
The vertices
Solution:
Vertices are :
D, E, A, B, C

Question (ii)
The sides
Solution:
Sides are :
AB, BC, CD, DE, EA

Question (iii)
The diagonals
Solution:
Diagonals are :
AC, AD, BE, BD, CE

Question (iv)
Adjacent sides of AB
Solution:
Adjacent sides of AB are :
AE and BC

Question (v)
Adjacent vertices of E.
Solution:
Adjacent vertices of E are :
A and D.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.1

1. Give the examples of:

Question (i)
A point
Solution:
A point. Point A •
Examples:
(i) A small dot marked by a sharp pencil on a sheet of paper.
(ii) A tiny prick made by a fine needle or pin on a paper.
(iii) Bindi.
(iv) A star in the sky.

Question (ii)
A line segment
Solution:
A line segment
Examples:
(i) An edge of a box.
(ii) A tube light.
(iii) The edge of a postcard.
(iii) Parallel lines.

Question (iii)
Parallel lines
Solution:
Examples:
(i) The opposite edges of ruler (scale).
(ii) The crossbars of window.
(iii) The opposite edges of blackboard.
(iv) Rail lines.
(iv) Interescting lines.

Question (iv)
Intersecting lines
Solution:
Examples:
(i) Two adjacent edges of your notebook.
(ii) The letter X of the English alphabet.
(iii) Crossing roads.

Question (v)
Concurrent lines.
Solution:
Concurrent lines.
(i) Three angle bisectors of a triangle.
(ii) Three medians of a triangle.
(iii) Three perpendiculars of a triangle.
(iv) The intersection of the three walls of a room.

2. Name the lines segments in given lines.

Solution:
AB, AC, AD, BC, BD, CD are line segments.

3. How many lines can pass through a point?
Solution:
Infinite lines can pass through a point.

4. How many points lie on line?
Solution:
Infinite points lie on a line.

5. How many lines pass through two points?
Solution:
One and only one line passes through two points.

6. Use the figure to name:

Question (i)
Five Points
Solution:
Five Points are :
O, A, B, C, D, or E

Question (ii)
A line
Solution:
BE is the line.

Question (iii)
Four rays
Solution:
Four rays are :
\(\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}}, \overrightarrow{\mathrm{OC}}, \overrightarrow{\mathrm{OD}} \text { or } \overrightarrow{\mathrm{OE}}\)

Question (iv)
Five line segments.
Solution:
Five line segments are :
OA, OB, OC, OD, OE, DE.

7. Name the given ray in all possible ways.

Solution:
The possible rays are:
\(\overrightarrow{\mathrm{PQ}}, \overrightarrow{\mathrm{PR}}, \overrightarrow{\mathrm{QR}}\)

8. Use the figure to name :

Question (i)
Pair of parallel lines.
Solution:
Pair of parallel lines are :
l and m.

Question (ii)
All pairs of intersecting lines.
Solution:
All pairs of intersecting lines are:
p and n, n and l, n and m, p and l, p and m.

Question (iii)
Lines whose point of intersection is S.
Solution:
Lines whose point of intersection is S :
m and n.

Question (iv)
Collinear points.
Solution:
Collinear points are :
P, Q, S and P, R, T.

9. Use the figure to name:

Question (i)
All pairs of parallel lines.
Solution:
All pairs of parallel lines are : n and p, q and p, n and q.

Question (ii)
All pairs of intersecting lines.
Solution:
All pairs of intersecting lines are :
m and l, m and n, m and p, m and q, l and n, l and p, l and q.

Question (iii)
Lines whose point of intersection is D.
Solution:
Lines whose point of intersection is D are :
p and l.

Question (iv)
Point of intersection of lines m and p.
Solution:
Point of intersection of lines m and p is E.

Question (v)
All sets of collinear points.
Solution:
All sets of collinear points are :
G, E, C, A and F, D, C, B.

10. Use the figure to name:

Question (i)
Line containing point P.
Solution:
Lines containing point are : l, n.

Question (ii)
Lines whose point of intersection is B.
Solution:
Lines whose point of intersection is B are : l and m.

Question (iii)
Point of intersection of lines m and l.
Solution:
Point of intersection of lines m and l is : B.

Question (iv)
All pairs of intersecting lines.
Solution:
All pairs of intersecting lines are : m and l, n and l.

11. State which of the following statements are True (T) or False (F):

Question (i)
Two lines in a plane, always intersect at a point
Solution:
False

Question (ii)
If four lines intersect at a point, those are called concurrent lines.
Solution:
True

Question (iii)
Point has a size because we can see it as a thick dot on the paper.
Solution:
False

Question (iv)
Through a given point, only one line can be drawn.
Solution:
False

Question (v)
Rectangle is a part of the plane.
Solution:
True.

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 7 Algebra MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 7 Algebra MCQ Questions

Multiple Choice Questions

Question 1.
Each side of square is represented by ‘s’ then perimeter of square is:
(a) 4 + s
(b) s – 4
(c) 4s
(d) s.
Answer:
(c) 4s

Question 2.
Write commutative property of multiplication using variables x and y:
(a) xy = yx
(b) x + y = y + x
(c) x + y
(d) xy.
Answer:
(a) xy = yx

Question 3.
How many terms in expression 7l – 3l?
(a) 1
(b) 3
(c) 2
(d) 4.
Answer:
(c) 2

Question 4.
5 is subtracted from m = ……………….. .
(a) 5 – m
(b) m + 5
(c) 5 + m
(d) m – 5.
Answer:
(d) m – 5.

Question 5.
Multiply p by 3 then 2 is added = ……………. .
(a) 2p + 3
(b) 3p – 2
(c) 3p + 2
(d) 2p – 3.
Answer:
(c) 3p + 2

Question 6.
If Armaan’s present age is x years then what will be his age after 4 years?
(a) x – 4
(b) x + 4
(c) 4x
(d) 4 – x.
Answer:
(b) x + 4

Question 7.
Write as algebraic equation: 7 more than 4 times ofy gives 23 :
(a) 4 + 7y = 23
(b) 7 + y = 23
(c) 4y – 7 = 23
(d) 4y + 7 = 23.
Answer:
(d) 4y + 7 = 23.

Question 8.
Find x if x – 3 = 2 :
(a) 3
(b) 6
(c) 5
(d) 2.
Answer:
(c) 5

Question 9.
Solve:
4l – 3 = 5
(a) 3
(b) 4
(c) 1
(d) 2.
Answer:
(d) 2.

Question 10.
If \(\frac {a}{4}\) = 5 then a = ……………. .
(a) 5
(b) 20
(c) 4
(d) 18.
Answer:
(b) 20

Question 21.
What is algebraic expression for subtracting 7 from – m?
(a) m – 1
(b) m + 7
(c) 7 – m
(d) – m – 7.
Answer:
(d) – m – 7.

Question 22.
What is algebraic expression for subtracting 7 from p?
(a) p – 7
(b) p + 7
(c) 7 – p
(d) 7 × p.
Answer:
(a) p – 7

Question 23.
What is algebraic expression for multiplying p by 16?
(a) 16 p
(b) p + 6
(c) p – 16
(d) \(\frac {p}{16}\)
Answer:
(a) 16 p

Question (iv)
What is algebraic expression for first multiplying x by 3 and then adding 2 to the product?
(a) x + 6
(b) 3x + 2
(c) 3x – 2
(d) 6x
Answer:
(b) 3x + 2

Question (v)
What is algebraic expression for first multiplying y by 2 and then subtracting 5 from the product?
(a) 2y + 5
(b) y + 10
(c) 2y – 5
(d) 10y.
Answer:
(c) 2y – 5

Fill in the blanks:

Question (i)
The algebraic expression for first multiplying y by 10 and then adding 7 to the product is ………… .
Answer:
10y + 7

Question (ii)
The algebraic expression for first multiplying n by 2 and then subtracting l from the product ………… .
Answer:
2n – l

Question (iii)
7 × 20 – 82 is expression of only ………………. .
Answer:
Numbers

Question (iv)
Each side of a square is l, then perimeter of square is ……………. .
Answer:
4l

Question (v)
5 is added to x = …………….. .
Answer:
x + 5

Write True/False:

Question (i)
If \(\frac {a}{5}\) = 4, then a = 20. (True/False)
Answer:
True

Question (ii)
If x – 3 = 2, then x = l. (True/False)
Answer:
False

Question (iii)
If 4l – 3 = 5, then l = 2. (True/False)
Answer:
True

Question (iv)
Number of terms in expression 3p + 2 is two. (True/False)
Answer:
True

Question (v)
The letters which can take any numerical value are called variables. (True/False)
Answer:
True

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 7 Algebra Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 7 Algebra Ex 7.4

1. Write the following statements as algebraic equations:

Question (i)
The sum of x and 3 gives 10.
Solution:
The sum of x and 3 = x + 3
It gives 10.
∴ The algebraic equation is x + 3 = 10

Question (ii)
5 less than a number ‘a’ is 12.
Solution:
5 less than a number ‘a’ = a – 5
It is 12
∴ The algebraic equation is a – 5 = 12

Question (iii)
2 more than 5 times of p gives 32.
Solution:
2 more than 5 times of p = 5p + 2
It gives 32
∴ The algebraic equation is 5p + 2 = 32

Question (iv)
Half of a number is 10.
Solution:
Let Half of a number x = \(\frac {x}{2}\)
It is 10
∴ The algebraic equations is
\(\frac {x}{2}\) = 10

Question (v)
Twice of a number added to 3 gives 17.
Solution:
Let the number be x
Twice of a number added to 3 = 2x + 3
It gives 17
∴ The algebraic equation is 2x + 3 = 17

2. Write the L.H.S. and R.H.S. for the following equations:

Question (i)
l + 5 = 8
Solution:
L.H.S. = l + 5, R.H.S. = 8

Question (ii)
13 = 2m + 3
Solution:
L.H.S. = 13, R.H.S. = 2m + 3

Question (iii)
\(\frac {t}{4}\) = 6
Solution:
L.H.S. = \(\frac {t}{4}\), R.H.S. = 6

Question (iv)
2h – 5 = 13
Solution:
L.H.S. = 2h – 5, R.H.S. = 13

Question (v)
\(\frac {5x}{7}\) = 15.
Solution:
L.H.S. = \(\frac {5x}{7}\), R.H.S. = 15.

3. Solve the following equations by trial and error method:

Question (i)
x + 2 = 7
Solution:
x + 2= l
We try different values of x to make L.H.S. = R.H.S.

Value of JC	L.H.S. = x + 2	R.H.S. = 7	L.H.S. = R.H.S.
1	1+ 2 = 3	7	No
2	2 + 2 = 4	7	No
3	3 + 2 = 5	7	No
4	4 + 2 = 6	7	No
5	5 + 2 = 7	7	Yes

From the above table we find that L.H.S. = R.H.S. When x = 5

Question (ii)
5p = 20
Solution:
5p = 20
We try different values of p to make L.H.S. = R.H.S.

Value of p	L.H.S. = 5p	R.H.S. = 20	L.H.S. = R.H.S.
1	5 × 1 = 5	20	No
2	5 × 2 = 10	20	No
3	5 × 3 = 15	20	No
4	5 × 4 = 20	20	Yes

From the above table we find that L.H.S. = R.H.S. When p = 4

Question (iii)
\(\frac {a}{5}\) = 2
Solution:
We try different values of a to make L.H.S. = R.H.S.
image
From the above table we find that L.H.S. = R.H.S. When a = 10

Question (iv)
2l – 4 = 8
Solution:
21-4 = 8
We try different values of l to make L.H.S. = R.H.S.

Value of a	L.H.S.	R.H.S. = 8	L.H.S. ff R.H.S.
1	2 × 1 – 4 = 2 – 4 = – 2	8	No
2	2 × 2 – 4 = 4 – 4 = 0	8	No
3	2 × 3 – 4 = 6 – 4 = 2	8	No
4	2 × 4 – 4 = 8 – 4 = 4	8	No
5	2 × 5 – 4 = 10 – 4 = 6	8	No
6	2 × 6 – 4  = 12 – 4 = 8	8	Yes

From the above table we find that L.H.S. = R.H.S. When l = 6

Question (v)
3x + 2 = 11.
Solution:
3x + 2 = 11
We try different values of x to make L.H.S. = R.H.S.

Value of p	L.H.S. = 3x + 2	R.H.S. = 11	L.H.S. = R.H.S.
1	3 × 1 + 2 = 3 + 2 = 5	11	No
2	3 × 2 + 2 = 6 + 2 = 8	11	No
3	3 × 3 + 2 = 9 + 2 = 11	11	Yes

From the above table we find thatL.H.S. = R.H.S. When x = 3

4. Solve the following equations by systematic method.

Question (i)
z – 4 = 10
Solution:
Given Equation is z – 4 = 10 Adding 4 on both sides, we get
2 – 4 + 4 = 10 + 4
⇒ z = 14 is the required solution.

Question (ii)
a + 3 = 15
Solution:
Given equation is a + 3 = 15
Subtracting 3 from both sides, we get
a + 3- 3 = 15 – 3
⇒ a = 12 is the required solution.

Question (iii)
4m = 20
Solution:
Given equation is 4m = 20
Dividing both sides by 4, we get
\(\frac{4 m}{4}=\frac{20}{4}\)
⇒ m = 5 is the required solution.

Question (iv)
3x – 3 = 15
Solution:
Given equation is 3x – 3 = 15
Adding 3 on both sides, we get
3x – 3 + 3 = 15 + 3
⇒ 3x = 18
Dividing both sides by 3, we get
\(\frac{3x}{3}=\frac{18}{3}\)
⇒ x = 6 is the required solution.

Question (v)
4x + 5 = 13.
Solution:
Given equation is 4x + 5 = 13
Subtracting 5 from both sides, we get
4x + 5 – 5 = 13 – 5
⇒ 4x = 8
Dividing both sides by 4, we get
\(\frac{4 x}{4}=\frac{8}{4}\)
⇒ x = 2 is the required solution.

5. Solve the following equation by transposition:

Question (i)
x – 5 = 6
Solution:
Given equation : x – 5 = 6
∴ x = 6 + 5
(Transposing – 5 to other side, it becomes + 5)
⇒ x = 11 is the required solution.

Question (ii)
y + 2 = 3
Solution:
Given equation : y + 2 = 3
⇒ y = 3 – 2
(Transposing + 2 to other side, it becomes – 2)
∴ y = 1 is the required solution.

Question (iii)
5x = 10
Solution:
Given equation : 5x = 10
⇒ x = \(\frac {10}{5}\)
(Transposing ‘multiplication’, it becomes ‘division’)
∴ x = 2 is the required solution.

Question (iv)
\(\frac {a}{6}\) = 4
Solution:
Given equation : \(\frac {a}{6}\) = 4
⇒ a = 4 × 6
(Transposing ‘division’, it becomes ‘multiplication’)
∴ a = 24 is the required solution.

Question (v)
4y – 2 = 30.
Solution:
Given equation : 4y – 2 = 30
⇒ 4y = 30 + 2
(Transposing – 2, it becomes + 2)
⇒ 4y = 32
⇒ y = \(\frac {32}{4}\)
(Transposing ‘multiplication’, it becomes division)
∴ y = 8 is the required solution.

6. Solve the following equations:

Question (i)
x + 7 = 11
Solution:
Given equation :
x + 7 = 11
⇒ x = 11 – 7
(Transposing 7 to R.H.S.)
⇒ x = 4 is the required solution.

Question (ii)
x – 3 = 15
Solution:
Given equation : x – 3 = 15
⇒ x = 15 + 3
(Transposing – 3 to L.H.S. it becomes + 3)
∴ x = 18 is the required solution

Question (iii)
x – 2 = 13
Solution:
Given equation : x – 2 = 13
⇒ x = 13 + 2
(Transposing – 2 to L.H.S.)
∴ x = 15 is the required solution.

Question (iv)
6x = 18
Solution:
Given equation is 6x = 18
Dividing both sides by 6 we get
\(\frac{6x}{6}=\frac{18}{6}\)
∴ x = 3 is the required solution.

Question (v)
3x = 24
Solution:
Given equation 3x = 24
Dividing both sides by 3, we get
\(\frac{3x}{3}=\frac{24}{3}\)
∴ x = 8 is the required solution.

Question (vi)
\(\frac {x}{4}\) = 7
Solution:
Given equation :
\(\frac {x}{4}\) = 7
Multiplying both sides by 4, we get
4 × \(\frac {x}{4}\) = 4 × 7
∴ x = 28 is the required solution.

Question (vii)
\(\frac {x}{8}\) = 5
Solution:
Given equation : \(\frac {x}{8}\) = 5
Multiplying both sides by 8, we get
8 × \(\frac {x}{8}\) = 8 × 5
∴ x = 40 is the required solution.

Question (viii)
2x – 5 = 17
Solution:
Given equation : 2x – 5 = 17
⇒ 2x = 17 – 5
(Transposing – 5 to R.H.S.)
⇒ 2x = 22
⇒ x = \(\frac {22}{2}\)
(Dividing both sides by 2)
∴ x = 11 is the required solution.

Question (ix)
4x + 5 = 21
Solution:
Given equation : 4x + 5 = 21
⇒ 4x = 21 + 5
(Transposing 5 to R.H.S.)
⇒ 4x = 16
⇒ x = \(\frac {16}{4}\)
(Dividing both sides by 4)
∴ x = 4 is the required solution.

Question (x)
5x – 2 = 13.
Solution:
Given equation : 5x – 2 = 13
⇒ 5x = 13 + 2
(Transposing – 2 to R.H.S.)
⇒ 5x = 15
⇒ x = \(\frac {15}{5}\)
(Dividing both sides by 5)
∴ x = 3 is the required solution.