Class 6

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 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.

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.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 7 Algebra Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 7 Algebra Ex 7.2

1. Each side of equilateral triangle is denoted by ‘a’ then express the perimeter of the triangle using ‘a’.

Solution:
Each triangle of equilateral triangle = a
∴ Perimeter of equilateral triangle
= a + a + a = 3a

2. An isosceles triangle is shown. Express its perimeter in terms of ‘l’ and ‘b’.

Solution:
Perimeter of isosceles triangle = l + l + b
= 21 + b

3. Each side of regular hexagon is denoted by ‘S’ then express the perimeter of the regular hexagon using ‘S’.

Solution:
Each side of regular hexagon = S
Perimeter of regular hexagon
=S + S + S + S + S + S
= 6S

4. The cube has 6 faces and all of them are identify squares. If l is the length of an edge of a cube, find the total length of all edges of the cube in terms of ‘l’?

Solution:
Length of each edge of a cube = l
There are 12 edges of a cube
Total length of all edges of the cube
= 12 × l = 12l

5. Write commutative property of addition using variables x and y.
image
Solution:
According to commutative property of addition.
If the order of numbers, in addition, is changed it does not change their sum.
∴ x + y = y + x

6. Write associative property of multiplication using variables l, m and n.
Solution:
According to associative property of multiplication.
If three numbers can be multiplied in any order, it does not change their product.
∴ l × (m × n) = (l × m) × n

7. Write distributive property of multiplication over addition in terms of variables p, q and r respectively.
Solution:
According to Distributive property of multiplication over addition
p × (q + r) = p × q + p × r

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 7 Algebra Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 7 Algebra Ex 7.1

1. Find the rule which gives the number of matchsticks required to make the following ‘it’ matchstick patterns. Use a variables to write the rule:

Question (i)
A pattern of letter T as

Solution:
Number of matchsticks required in a pattern of letter T = 2
Number of matchsticks required in ‘n’ patterns = 2n

Question (ii)
A pattern of letter E as

Solution:
Number of matchsticks required in a pattern of letter E = 4

Number of matchsticks required in V patterns of letter E = 4n

Question (iii)
A pattern of letter F as

Solution
Number of matchsticks required in a pattern of letter F = 3

Number of matchsticks required in ‘n’ patterns of letter F = 3 n

Question (iv)
A pattern of letter C as

Solution:
Number of matchsticks required in a pattern of letter C = 3

Number of matchsticks required in ‘n’ patterns of letter C = 3n

Question (v)
A pattern of letter S as

Solution:
Number of matchsticks required in a pattern of letter S = 5

Number of matchsticks required in V patterns of letter S = 5 n

2. Students are sitting in rows. There are 12 students in row. What is the rule which gives the number of students in ‘n’ rows? (Represent by table)
Solution:
Let us make a table for the number of students in ‘n’ rows.

Number of Rows	1	2	3	4	…..	10	……	n
Number of Students	12	24	36	48	……	120	……	12 n

It is observed from the table that
Total number of students in ‘n’ number of rows
= (Number of Students) × (Number of rows)
= 12 × n = 12n

3. The teacher distributes 3 pencils to a student What is the rule which gives the number of pencils, if there are ‘a’ number of students?
Solution:
We know
Total number of pencils
= Number of pencils × Number of students
= 3 × a = 3a

4. There are 8 pens in a pen stand. What is the rule that gives the total cost of the pens if the cost of each pen is represented by a variable ‘c’?
Solution:
We know
Total cost of the pens in ₹
= Number of pens × cost of 1 pen
= 8 × c = 8c

5. Gurleen is drawing pictures by joining dots. To make one picture,’she has to join 5 dots. Find the rule that gives the number of dots, if the number of pictures is represented by the symbol ‘p’.
Solution:
We know
Total number of dots = Number of dots × Number of pictures
= 5p

6. The cost of a dozen bananas is ₹ 50. Find the rule of total cost of bananas if there are ‘d’ dozens bananas.
Solution:
We know
Total cost of bananas in ₹
= Cost of one dozen × Number of bananas
= 50 × d
= 50d

7. Look at the following matchsticks patterns of squares given below. The squares are not separate as there are two adjoined adjacent squares have a common match stick. Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares.

(Hint: If you remove the vertical stick at the end you will get a patterns of C)
Solution:

Fig. No.	No. of Squares	Number of matchsticks	Pattern
(i)	1	4	3 x 1+ 1
(ii)	2	7	3 × 2 + 1
(iii)	3	10	3 × 3 + 1

Thus, we get the rule the number of matchsticks = 3x + 1 or 1 + 3x where x is the number of squares.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 6 Decimals MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 6 Decimals MCQ Questions

Multiple Choice Questions

Question 1.
3 + \(\frac {2}{10}\) = ………….
(a) 302
(b) 3.2
(c) 3.02
(d) 30.2.
Answer:
(b) 3.2

Question 2.
200 + 4 + \(\frac {5}{10}\) = …………
(a) 24.5
(b) 204.05
(c) 204.5
(d) 24.05.
Answer:
(c) 204.5

Question 3.
\(\frac {7}{100}\) = …………..
(a) .07
(b) 700
(c) .007
(d) 7.
Answer:
(a) .07

Question 4.
50 + \(\frac {3}{1000}\) = ………….
(a) 50.3
(b) 503000
(c) 50.0003
(d) 50.003.
Answer:
(d) 50.003.

Question 5.
Seventy and four thousandths = …………….
(a) 74000
(b) 70.004
(c) .00074
(d) .074.
Answer:
(b) 70.004

Question 6.
2.03 in expanded form = ……….
(a) 2 + \(\frac {3}{10}\)
(b) 20 + \(\frac {3}{10}\)
(c) 2 + \(\frac {3}{100}\)
(d) 20 + \(\frac {3}{100}\)
Answer:
(c) 2 + \(\frac {3}{100}\)

Question 7.
2.5 = ……….. .
(a) \(\frac {5}{2}\)
(b) \(\frac {25}{2}\)
(c) \(\frac {5}{10}\)
(d) \(\frac {1}{4}\)
Answer:
(a) \(\frac {5}{2}\)

Question 8.
\(\frac {13}{2}\) = …………….
(a) 6
(b) 6.1
(c) 1.3
(d) 6.5.
Answer:
(d) 6.5.

Question 9.
Which of the following decimals is largest?
(a) 0.5
(b) 0.05
(c) 0.51
(d) 0.005.
Answer:
(c) 0.51

Question 10.
Which of the following decimals is smallest?
(a) 2.13
(b) .213
(c) 21.3
(d) 213.
Answer:
(b) .213

Question 11.
75 g = ……. kg.
(a) .075 kg
(b) .75 kg
(c) 7.5 kg
(d) 75 kg.
Answer:
(a) .075 kg

Question 12.
27 mm = ………….. cm.
(a) .27 cm
(b) 27 cm
(c) 2.7 cm
(d) .027 cm.
Answer:
(c) 2.7 cm

Question 13.
2.5 + 4.23 = ……………
(a) 4.48
(b) 6.73
(c) 4.73
(d) 6.48.
Answer:
(b) 6.73

Question 14.
15 + 3.84 = ………… .
(a) 3.99
(b) 18.99
(c) 3.84
(d) 18.84
Answer:
(d) 18.84

Question 15.
13.5 – 4.23 = …………….
(a) 2.87
(b) 7.29
(c) 9.27
(d) 9.37.
Answer:
(c) 9.27

Question 16.
20 – 12.56 = …………..
(a) 7.44
(b) 8.44
(c) 9.44
(d) 6.44.
Answer:
(a) 7.44

Question 17.
14.8 + 2.62 – 8.4 = …………….. .
(a) 8.02
(b) 9.12
(c) 9.02
(d) 6.44.
Answer:
(c) 9.02

Question 18.
517 ml = …………… l.
(a) 5.07 l
(b) 5.7 l
(c) 5.70 l
(d) 5.007 l.
Answer:
(d) 5.007 l

Question 19.
12 kg 85 g = ……………. kg.
(a) 12.085 kg
(b) 12.85 kg
(c) 128.5 kg
(d) 12.0085 kg.
Answer:
(a) 12.085 kg

Question 20.
235 paise = …………..
(a) ₹ 235
(b) ₹ 23.5
(c) ₹ 2.35
(d) ₹ .235.
Answer:
(c) ₹ 2.35

Question 21.
Express 88 m as km using decimals:
(a) 0.88 km
(b) 8.8 km
(c) 0.088 km
(d) 0.0088 km.
Answer:
(c) 0.088 km

Question 22.
In the following lists which numbers are in the descending order?
(a) 0.355, 0.4, 0.43, 0.355
(b) 0.4, 0.43, 0.444, 0.355
(c) 0.43, 0.355, 0.444, 0.4
(d) 0.444, 0.43, 0.4, 0.355.
Answer:
0.444, 0.43, 0.4, 0.355.

Question 23.
In the following lists which numbers are in the descending order?
(a) 19.4, 0.3, 10.6, 205.9
(b) 205.9, 10.6, 0.3, ,19.4
(c) 205.9, 19.4, 10.6, 0.3
(d) 0.3, 10.6, 19.4, 205.9.
Answer:
205.9, 19.4, 10.6, 0.3

Question (iv)
In the following lists which numbers are in the ascending order?
(a) 0.7, 20.9, 14.6, 600.8
(b) 0.7, 14.6, 20.9, 600.8
(c) 600.8, 14.6, 20.9, 0.7
(d) 14.6,20.9,0.7,600.8.
Answer:
0.7, 14.6, 20.9, 600.8

Question (v)
Express 30 mm as cm using decimals:
(a) 3,0 cm
(b) 0.30 cm
(c) 0.03 cm
(d) 0.003 cm.
Answer:
3,0 cm

Fill in the blanks:

Question (i)
15 cm as m using decimals is …………… m.
Answer:
0.15

Question (ii)
75 paise as ₹ using decimals is ₹ ………….. .
Answer:
₹ 0.75

Question (iii)
9 cm 8 mm as cm using decimals is …………… m.
Answer:
0.98

Question (iv)
27 m = ………….. cm.
Answer:
2.7

Question (v)
15 + 3.84 = ……………… .
Answer:
18.84

Write True/False:

Question (i)
The word decimal comes from Latin word “Decem.” (True/False)
Answer:
True

Question (ii)
\(\frac {1}{10}\) is read as one tenth. (True/False)
Answer:
True

Question (iii)
10 + 3 + \(\frac{2}{10}=\frac{15}{10}\) (True/False)
Answer:
False

Question (iv)
Seven and three-tenths is written as 7.3. (True/False)
Answer:
True

Question (v)
Twenty-four point five is written as 24.5. (True/False)
Answer:
True

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 6 Decimals Ex 6.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 6 Decimals Ex 6.4

1. Solve the following:

Question (i)
12.15 + 4.87
Solution:
We have 12.15 + 4.87

Hence 12.15 + 4.87 = 17.02

Question (ii)
23.5 + 13.47
Solution:
We have 23.5 + 13.47

Hence 23.5 + 13.47 = 36.97

Question (iii)
12.56 + 6.234
Solution:
We have 12.56 + 6.234

Hence 12.56 + 6.234 = 18.794

Question (iv)
24.25 – 13.12
Solution:
We have 24.25 – 13.12

Hence 24.25 – 13.12 = 11.13

Question (v)
18.8 – 4.26
Solution:
We have 18.8 – 4.26

Hence 18.8 – 4.26 = 14.54

Question (vi)
42.34 – 5.256
Solution:
We have 42.34 – 5.256

Hence 42.34 – 5.256 = 37.084

Question (vii)
45.4 + 13.25 + 28.68
Solution:
We have 45.4 + 13.25 + 28.68

Hence 45.4 + 13.25 + 28.68 = 87.33

Question (viii)
52.9 + 26.893 + 13.62
Solution:
We have 52.9 + 26.893 + 13.62

Hence 52.9 + 26.893 + 13.62 = 93.413

Question (ix)
42 – 27.563
Solution:
We have 42 – 27.563

Hence 42 – 27.563 = 14.437

Question (x)
64.26 – 43.589 + 13.42
Solution:
We have 64.26 – 43.589 + 13.42

Hence 64.26 – 43.589 + 13.42
= 34.091

Question (xi)
18.3 + 2.56 – 11.643
Solution:
We have 18.3 + 2.56 – 11.643

Hence 18.3 + 2.56 – 11.643
= 9.217

Question (xii)
66.5 – 13.49 – 29.712.
Solution:
We have 66.5 – 13.49 – 29.712
= 66.5 – (13.49 + 29.712)
= 66.5 – 43.202 = 23.298

2.

Question (i)
Subtract 21.92 from 32.683
Solution:

Question (ii)
Subtract 14.812 from 23.
Solution:
Subtract 14.812 from 23.

3. What should be added to 3.412 to get 7?
Solution:
Let x should be added to 3.412 to get 7
3.412 + x = 7
x = 7 – 3.412
= 3.588

Hence, 3.588 should be added to 3.412 to get 7

4. Khan spent ₹ 63.25 for Maths book and ₹ 48.99 for English book. Find the total amount spent by Khan.
Solution:
Amount spent for Maths book = ₹ 63.25
Amount spent for English book = 48.99
Total amount spent by khan = ₹ 112.24

5. Samar walked 3 km 450 m in morning and 2 km 585 m in evening. How much distance did he walk in all ?
Solution:
Distance walked in morning = 3 km 450 m
Distance walked in evening = 2 km 585 m
Distance Samar Walked in all = 6 km 035 m

6. Sheetal has ₹ 190.50 in her pocket. She buys a school bag for ₹ 123.99. How much money is left with her now?
Solution:
Total amount Sheetal has = ₹ 190.50
Amount spent on school bag = – ₹ 123.99
Money left with her = ₹ 66.51

7. A piece of 18.56 m long ribbon is cut into three pieces. If the length of two pieces are 8.75 m and 3.125 m respectively. Find the length of the third piece.
Solution:
Total length of ribbon = 18.56 m
Length of two pieces = 8.75 m + 3.125 m

Length of the third piece = 18.56 m – 11.875 m
= 6.685 m

8. Veerpal bought vegetables weighing 20 kg. Out of this 6 kg 750 g are onions, 5 kg 25 g are potatoes and rest are tomatoes. What is the weight of the tomatoes?
Solution:
Total weight of vegetables
Veerpal bought = 20 kg
Weight of onions = 6 kg 750 g = 6.750 kg
Weight of potatoes = 5 kg 25 g = 5.025 kg
Weight of onions and potatoes = 11.775 kg

Total weight of vegetable = 20.000 kg
Weight of onions and potatoes = -11.775 kg
Weight of tomatoes = 8.225 kg

9. Ashish’s school is 28 km far from his house. He covers 14 km 250 m by bus, 12 km 650 m by car and the remaining distance by foot. How much distance does he cover on foot?
Solution:
Distance covered by bus 14 km 250 m = 14.250 km
Distance covered by car 12 km 650 m = 12.650 km
Distance covered by bus and car = 26.900 km

Total distance of school form Ashish house = 28 km
Distance covered by bus and car = 26.900 km
Distance covered on foot = 28 km – 26.900 km
= 1 km 100 m

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 6 Decimals Ex 6.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 6 Decimals Ex 6.3

1. Express as rupee using decimals:

Question (i)
35 paise
Solution:
35 paise = ₹ \(\frac {35}{100}\)
= ₹ 0.35
(∵ 1 paise = ₹ \(\frac {1}{100}\))

Question (ii)
4 paise
Solution:
4 paise = ₹ \(\frac {4}{100}\)
= ₹ 0.04
(∵ 1 paise = ₹ \(\frac {1}{100}\))

Question (iii)
240 paise
Solution:
240 paise = ₹ \(\frac {240}{100}\)
= ₹ 2.40
(∵ 1 paise = ₹ \(\frac {1}{100}\))

Question (iv)
12 rupees 25 paise
Solution:
= (12 rupees) + 25 paise
= ₹ 12 + ₹ \(\frac {25}{100}\)
(∵ 1 paise = ₹ \(\frac {1}{100}\))
= ₹ 12 + ₹ 0.25
= ₹ 12.25

Question (v)
24 rupees 5 paise.
Solution:
(24 rupees) + (5 paise)
= ₹ 24 + ₹ \(\frac {5}{100}\)
(∵ 1 paise = ₹ \(\frac {1}{100}\))
= ₹ 24 + ₹ 0.05
= ₹ 24.05

2. Express as metre using decimals

Question (i)
5 cm
Solution:
5 cm = \(\frac {5}{100}\) m
= 0.05 m
(∵ 1cm =\(\frac {1}{100}\)m)

Question (ii)
62 cm
Solution:
62 cm = \(\frac {62}{100}\) m
= 0.62 m
(∵ 1cm =\(\frac {1}{100}\)m)

Question (iii)
135 cm
Solution:
135 cm = \(\frac {135}{100}\) m
= 1.35 m
(∵ 1cm =\(\frac {1}{100}\)m)

Question (iv)
5 m 20 cm
Solution:
= 5 m + 20 cm
(∵ 1cm = \(\frac {1}{100}\)m)
= 5m + 0.20m
= 5.20m

Question (v)
12 m 8 cm
Solution:
= 12 m + 8 cm
= 12 m + \(\frac {8}{100}\)m
(∵ 1cm = \(\frac {1}{100}\)m)
12 cm + 0.08 m
= 12.08 m

3. Express as centimetre using decimals:

Question (i)
2 mm
Solution:
2 mm = \(\frac {2}{10}\) cm
(∵ 1mm = \(\frac {1}{10}\)m)
= 0.2 cm

Question (ii)
28 mm
Solution:
28mm = \(\frac {28}{10}\)cm
(∵ 1mm = \(\frac {1}{10}\)m)

Question (iii)
8 cm 4 mm.
Solution:
8 cm 4 mm = 8 cm + 4 mm
= 8cm + \(\frac {4}{10}\)
= 8 cm + 0.4 cm
= 8.4 cm

4. Express as kilometre using decimals:

Question (i)
7 m
Solution:
= \(\frac {7}{1000}\)
= (∵ 1m = \(\frac {1}{1000}\) km)
= 0.007 km

Question (ii)
50 m
Solution:
= \(\frac {50}{1000}\)
= (∵ 1m = \(\frac {1}{1000}\) km)
= 0.050 km

Question (iii)
425 m
Solution:
= \(\frac {425}{1000}\)
= (∵ 1m = \(\frac {1}{1000}\) km)
= 0.425 km

Question (iv)
2475 m
Solution:
= \(\frac {2475}{1000}\) km
= (∵ 1m = \(\frac {1}{1000}\) km)
= 2.475 km

Question (v)
3 km 225 m.
Solution:
= 3 km + 225 m
= 3 km + \(\frac {225}{1000}\)
= (∵ 1m = \(\frac {1}{1000}\) km)
= 3.225 km

5. Express as kilogram using decimals:

Question (i)
5g
Solution:
5g = \(\frac {5}{1000}\)
= (∵ 1g = \(\frac {1}{1000}\) km)
= 0.005 kg

Question (ii)
75g
Solution:
75g = \(\frac {75}{1000}\)
= (∵ 1g = \(\frac {1}{1000}\) km)
= 0.075 kg

Question (iii)
423 g
Solution:
423 g = \(\frac {423}{1000}\)
= (∵ 1g = \(\frac {1}{1000}\) km)
= 0.423 kg

Question (iv)
1265 g
Solution:
1265 g = \(\frac {1265}{1000}\)
(∵ 1g = \(\frac {1}{1000}\) km)
= 1.265 kg

Question (v)
5 kg 418 g.
Solution:
= 5 kg + 418 g.
= 5 kg + \(\frac {418}{1000}\)
(∵ 1g = \(\frac {1}{1000}\) km)
= 5 kg + 0.418 kg
= 5.418 kg.

6. Express as litre using decimals:

(i) 2 ml
(ii) 80 ml
(iii) 725 ml
(iv) 3l 423 ml
(v) 8l 20 ml.
Solution: