PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.2

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’.
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.2 1
Solution:
Each triangle of equilateral triangle = a
∴ Perimeter of equilateral triangle
= a + a + a = 3a

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.2

2. An isosceles triangle is shown. Express its perimeter in terms of ‘l’ and ‘b’.
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.2 2
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’.
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.2 3
Solution:
Each side of regular hexagon = S
Perimeter of regular hexagon
=S + S + S + S + S + S
= 6S

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.2

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’?
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.2 4
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

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.2

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

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1

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
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1 1
Solution:
Number of matchsticks required in a pattern of letter T = 2
Number of matchsticks required in ‘n’ patterns = 2n
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1 2

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1

Question (ii)
A pattern of letter E as
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1 3
Solution:
Number of matchsticks required in a pattern of letter E = 4
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1 4
Number of matchsticks required in V patterns of letter E = 4n

Question (iii)
A pattern of letter F as
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1 4
Solution
Number of matchsticks required in a pattern of letter F = 3
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1 6
Number of matchsticks required in ‘n’ patterns of letter F = 3 n

Question (iv)
A pattern of letter C as
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1 7
Solution:
Number of matchsticks required in a pattern of letter C = 3
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1 8
Number of matchsticks required in ‘n’ patterns of letter C = 3n

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1

Question (v)
A pattern of letter S as
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1 9
Solution:
Number of matchsticks required in a pattern of letter S = 5
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1 10
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

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1

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

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1

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.
PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.1 11
(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.

PSEB 6th Class Maths MCQ Chapter 6 Decimals

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

PSEB 6th Class Maths MCQ Chapter 6 Decimals

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

PSEB 6th Class Maths MCQ Chapter 6 Decimals

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

PSEB 6th Class Maths MCQ Chapter 6 Decimals

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

PSEB 6th Class Maths MCQ Chapter 6 Decimals

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

PSEB 6th Class Maths MCQ Chapter 6 Decimals

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

PSEB 6th Class Maths MCQ Chapter 6 Decimals

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

PSEB 6th Class Maths MCQ Chapter 6 Decimals

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

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.4

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

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.4

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

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.4

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

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.4

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

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.4

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.

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.4

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.

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.4

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.

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.4

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.

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.4

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.

PSEB 6th Class Maths Solutions Chapter 7 Algebra Ex 7.4

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.

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4

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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 1
Hence 12.15 + 4.87 = 17.02

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4

Question (ii)
23.5 + 13.47
Solution:
We have 23.5 + 13.47
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 2
Hence 23.5 + 13.47 = 36.97

Question (iii)
12.56 + 6.234
Solution:
We have 12.56 + 6.234
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 3
Hence 12.56 + 6.234 = 18.794

Question (iv)
24.25 – 13.12
Solution:
We have 24.25 – 13.12
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 4
Hence 24.25 – 13.12 = 11.13

Question (v)
18.8 – 4.26
Solution:
We have 18.8 – 4.26
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 5.1
Hence 18.8 – 4.26 = 14.54

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4

Question (vi)
42.34 – 5.256
Solution:
We have 42.34 – 5.256
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 6
Hence 42.34 – 5.256 = 37.084

Question (vii)
45.4 + 13.25 + 28.68
Solution:
We have 45.4 + 13.25 + 28.68
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 7
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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 8
Hence 52.9 + 26.893 + 13.62 = 93.413

Question (ix)
42 – 27.563
Solution:
We have 42 – 27.563
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 9
Hence 42 – 27.563 = 14.437

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4

Question (x)
64.26 – 43.589 + 13.42
Solution:
We have 64.26 – 43.589 + 13.42
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 10
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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 11
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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 12

2.

Question (i)
Subtract 21.92 from 32.683
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 13

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4

Question (ii)
Subtract 14.812 from 23.
Solution:
Subtract 14.812 from 23.
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 14

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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 15
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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 16

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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 17

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4

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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 18

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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 19
Length of the third piece = 18.56 m – 11.875 m
= 6.685 m
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 20

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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 21
Total weight of vegetable = 20.000 kg
Weight of onions and potatoes = -11.775 kg
Weight of tomatoes = 8.225 kg
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 22

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4

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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 23

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
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.4 24

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.3

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 = ₹ \(\frac {1}{100}\))

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.3

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

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.3

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)

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.3

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)

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.3

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

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.3

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

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.3

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.

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.3

6. Express as litre using decimals:

(i) 2 ml
(ii) 80 ml
(iii) 725 ml
(iv) 3l 423 ml
(v) 8l 20 ml.
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.3 1
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.3 2

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 6 Decimals Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 6 Decimals Ex 6.2

1. Convert the following decimal numbers into fractions and reduce it to lowest form.

Question (i)
1.4
Solution:
1.4 = \(\frac{14}{10}=\frac{14 \div 2}{10 \div 2}\)
(H.C.F. of 14 and 10 is 2)
= \(\frac {7}{5}\)

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

Question (ii)
2.25
Solution:
2.25 = \(\frac{225}{100}=\frac{225 \div 25}{100 \div 25}\)
(H.C.F. of 225 and 100 is 25)
= \(\frac {9}{4}\)

Question (iii)
18.6
Solution:
18.6 = \(\frac{186}{10}=\frac{186 \div 2}{10 \div 2}\)
(H.C.F. of 186 and 10 is 2)
= \(\frac {93}{5}\)

Question (iv)
4.04
Solution:
4.04 = \(\frac{404}{100}=\frac{404 \div 4}{100 \div 4}\)
(H.C.F. of 404 and 100 is 4)
= \(\frac {101}{25}\)

Question (v)
21.6
Solution:
21.6 = \(\frac{216}{10}=\frac{216 \div 2}{10 \div 2}\)
(H.C.F. of 216 and 10 is 2)
= \(\frac {108}{5}\)

2. Convert the following fractions into decimal numbers:

Question (i)
\(\frac {7}{100}\)
Solution:
\(\frac {7}{100}\) = 0.07
(Here denominator is 100)

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

Question (ii)
\(\frac {12}{10}\)
Solution:
\(\frac {12}{10}\) = 1.2
(Here denominator is 10)

Question (iii)
\(\frac {215}{100}\)
Solution:
\(\frac {215}{100}\) = 2.15
(Here denominator is 100)

Question (iv)
\(\frac {18}{1000}\)
Solution:
\(\frac {18}{1000}\) = 0.018
(Here denominator is 1000)

Question (v)
\(\frac {245}{10}\)
Solution:
\(\frac {245}{10}\) = 24.5
(Here denominator is 10)

3. Convert the following fractions into decimal numbers by equivalent fraction method:

Question (i)
\(\frac {5}{2}\)
Solution:
Here denominator is 2.
Convert into equivalent fraction with denominator 10 by multiplying it by 5.
∴ \(\frac{5}{2}=\frac{5 \times 5}{2 \times 5}=\frac{25}{10}\) = 2.5

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

Question (ii)
\(\frac {3}{4}\)
Solution:
Here denominator is 4.
Convert into equivalent fraction with denominator 100 by multiplying it by 25.
∴ \(\frac{3}{4}=\frac{3 \times 25}{4 \times 25}=\frac{75}{100}\) = 0.75

Question (iii)
\(\frac {28}{5}\)
Solution:
Here denominator is 5.
Convert into equivalent fraction with denominator 10 by multiplying it by 2.
∴ \(\frac{28}{5}=\frac{28 \times 2}{5 \times 2}=\frac{56}{10}\) = 5.6

Question (iv)
\(\frac {135}{20}\)
Solution:
Here denominator is 20.
Convert into equivalent fraction with denominator 100 by multiplying it by 5.
∴ \(\frac{135}{20}=\frac{135 \times 5}{20 \times 5}=\frac{675}{100}\)
= 6.75

Question (v)
\(\frac {17}{4}\)
Here denominator is 4.
Convert into equivalent fraction with denominator 100 by multiplying it by 25.
∴ \(\frac{17}{4}=\frac{17 \times 25}{4 \times 25}=\frac{425}{100}\)
= 4.25

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

4. Convert the following fractions into decimals by long division method:

Question (i)
\(\frac {17}{2}\)
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 1
= 8.5

Question (ii)
\(\frac {33}{4}\)
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 2
= 8.25

Question (iii)
\(\frac {76}{5}\)
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 3
= 15.2

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

Question (iv)
\(\frac {24}{25}\)
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 4
= 0.96

Question (v)
\(\frac {5}{8}\)
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 5
= 0.625

5. Represent the following decimals on number line:

Question (i)
(i) 0.7
(ii) 1.6
(iii) 3.7
(iv) 6.3
(v) 5.4
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 6

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

6. Write three decimal numbers between:

Question (i)
1.2 and 1.6
Solution:
Three decimal numbers between 1.2 and 1.6 are:
1.3, 1.4, 1.5

Question (ii)
2.8 and 3.2
Solution:
Three decimal numbers between 2.8 and 3.2 are:
2.9, 3, 3.1

Question (iii)
5 and 5.5.
Solution:
Three decimal numbers between 5 and 5.5 are:
5.1, 5.2, 5.3, 5.4.

7. Which number is greater:

Question (i)
0.4 or 0.7
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 7
Since, 7 > 4
So, 0.7 > 0.4

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

Question (ii)
2.6 or 2.5
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 8
Since, 6 > 5
So, 2.6 > 2.5

Question (iii)
1.23 or 1.32
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 9
Since, 3 > 2
So, 1.32 > 1.23

Question (iv)
12.3 or 12.4
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 10
Since, 4 > 3
So, 12.4 > 12.3

Question (v)
18.35 or 18.3
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 11
Since, 5 > 0
So, 18.35 > 18.30

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

Question (vi)
12 or 1.2
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 12
Since, 12 > 1
So, 12 > 1.2

Question (vii)
5.06 or 5.061
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 13
Since, 1 > 0
So, 5.061 > 5.060

Question (viii)
2.34 or 23.3
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 14
Since, 23 > 2
So, 23.3 > 2.34

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

Question (ix)
13.08 or 13.078
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 15
Since, 8 > 7
So, 13.08 > 13.078

Question (x)
2.3 or 2.03.
Solution:
PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2 16
Since, 3 > 0
So, 2.3 > 2.03

8. Arrange the decimal numbers in ascending order:

Question (i)
2.5, 2, 1.8, 1.9
Solution:
Ascending order is :
1.8, 1.9, 2, 2.5

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

Question (ii)
3.4, 4.3, 3.1, 1.3
Solution:
Ascending order is :
1.3, 3.1, 3.4, 4.3

Question (iii)
1.24, 1.2, 1.42, 1.8.
Solution:
Ascending order is :
1.2, 1.24, 1.42, 1.8.

9. Arrange the decimal numbers in descending order:

Question (i)
4.1, 4.01, 4.12, 4.2
Solution:
Descending order is :
4.2, 4.12, 4.1, 4.01

Question (ii)
1.3, 1.03, 1.003, 13
Solution:
Descending order is :
13, 1.3, 1.03, 1.003

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.2

Question (iii)
8.02, 8.2, 8.1, 8.002.
Solution:
Descending order is :
8.2, 8.1, 8.02, 8.002.

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.1

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 6 Decimals Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 6 Decimals Ex 6.1

1. Write each of the following in figures:

Question (i)
Seventy-two point one four.
Solution:
Seventy-two point one four = 72.14

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.1

Question (ii)
Two hundred fifty-seven point zero eight
Solution:
Two hundred fifty-seven point zero eight = 257.08

Question (iii)
Eight point two five-six.
Solution:
Eight point two five six = 8.256

Question (iv)
Forty-five and twenty-three hundredths.
Solution:
Forty five and twenty three hundredths
= 45 + \(\frac {23}{100}\)
= 45.23

Question (v)
Six hundred twenty-one and two hundred fifty-three thousandths
Solution:
Six hundred twenty-one and two hundred fifty-three thousandths
= 621 + \(\frac {253}{1000}\)
= 621.253

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.1

Question (vi)
Twelve and eight thousandths.
Solution:
Twelve and eight thousandths
= 12 + \(\frac {8}{1000}\)
= 12.008

2. Write the following decimal numbers in words:

Question (i)
12.52
Solution:
12.52 = Twelve point five two or twelve and fifty-two hundredths.

Question (ii)
7.148
Solution:
7.148 = Seven point one four eight or seven and one hundred forty-eight thousandths.

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.1

Question (iii)
0.24
Solution:
0.24 = Zero Point two four or twenty-four hundredths.

Question (iv)
5.018
Solution:
5.018 = Five-point zero one eight or five and eighteen thousandths.

Question (v)
.009.
Solution:
.009 = Point zero zero nine or nine thousandths.

3. Write the following decimals in the place value table:

Question (i)
(i) 21.569
(ii) 0.64
(iii) 3.51
(iv) 14.087
(v) 3.002.
Solution:

Number Thousands Hundreds Tens Ones Tenths Hundredths Thousandths
1. 21.569 2 1 5 6 9
2. 0.64 0 6 4
3. 3.51 3 5 1
4. 14.087 1 4 0 8 7
5. 3.002 3 0 0 2

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.1

4. Write the following as decimals:

Question (i)
40 + \(\frac {2}{10}\)
Solution:
40 + \(\frac {2}{10}\) = 40.2

Question (ii)
700 + 5 + \(\frac {3}{10}\) + \(\frac {4}{100}\)
Solution:
700 + 5 + \(\frac {3}{10}\) + \(\frac {4}{100}\) = 705.34

Question (iii)
100 + \(\frac {5}{100}\) + \(\frac {3}{1000}\)
Solution:
100 + \(\frac {5}{100}\) + \(\frac {3}{1000}\) = 10.053

Question (iv)
100 + \(\frac {7}{10}\) + \(\frac {4}{1000}\)
Solution:
100 + \(\frac {7}{10}\) + \(\frac {4}{1000}\) = 0.704

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.1

Question (v)
\(\frac {5}{1000}\)
Solution:
\(\frac {5}{1000}\) = 0.005

5. Write the decimals shown in the following place value table:

Question (i)

Thousands Hundreds Tens Ones Tenth Hundredths Thousandths
5 2 4 1 2
2 0 3 4 2 1
6 1 0 2 3
4 0 0 1
1 0 0 0 3

Solution:
(i) 524.12
(ii) 2034.21
(iii) 61.023
(iv) 4.001
(v) 100.03

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.1

6. Expand the following decimals.

Question (i)
2.5
Solution:
2.5 = 2 + 0.5
= 2 + \(\frac {5}{10}\)

Question (ii)
18.43
Solution:
18.43 = 10 + 8 + 0.4 + 0.03
= 10 + 8 + \(\frac {4}{10}\) + \(\frac {3}{100}\)

Question (iii)
4.05
Solution:
4.05 = 4 + 0.05
= 4 + \(\frac {5}{100}\)

Question (iv)
13.123
Solution:
13.123 = 10 + 3 + 0.1 + 0.02 + 0.003
= 10 + 3 + \(\frac {1}{10}\) + \(\frac {2}{100}\) + \(\frac {3}{1000}\)

PSEB 6th Class Maths Solutions Chapter 6 Decimals Ex 6.1

Question (v)
245.456
Solution:
245.456 = 200 + 40 + 5 + 0.4 + 0.05 + 0.006
= 200 + 40 + 5 + \(\frac {4}{10}\) + \(\frac {5}{100}\) + \(\frac {6}{1000}\)

Question (vi)
20.057
Solution:
20.057 = 20 + 0.05 + 0.007
= 20 + \(\frac {5}{100}\) + \(\frac {7}{1000}\)

PSEB 6th Class Maths MCQ Chapter 5 Fractions

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 5 Fractions MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 5 Fractions MCQ Questions

Multiple Choice Questions

Question 1.
Which of the following does not represent any fraction?
PSEB 6th Class Maths MCQ Chapter 5 Fractions 1.1
Answer:
PSEB 6th Class Maths MCQ Chapter 5 Fractions 1

PSEB 6th Class Maths MCQ Chapter 5 Fractions

Question 2.
Which of the following is a proper fraction?
(a) \(\frac {5}{5}\)
(b) \(\frac {12}{11}\)
(c) \(\frac {7}{9}\)
(d) 7
Answer:
(c) \(\frac {7}{9}\)

Question 3.
Which of the following is an improper fraction?
(a) \(\frac {5}{8}\)
(b) \(2\frac {3}{4}\)
(c) \(\frac {7}{11}\)
(d) \(\frac {15}{16 }\)
Answer:
(b) \(2\frac {3}{4}\)

Question 4.
The fractions having las numerator are called …………. fractions.
(a) Like
(b) Unlike
(c) Unit
(d) Proper.
Answer:
(c) Unit

Question 5.
The fractions having same denominators are called …………. fractions.
(a) Proper
(b) Unit
(c) Improper
(d) Like.
Answer:
(d) Like.

Question 6.
The fraction having different denominators are called ……………. fractions.
(a) Unlike
(b) Like
(c) Improper
(d) Unit.
Answer:
(a) Unlike

PSEB 6th Class Maths MCQ Chapter 5 Fractions

Question 7.
Express 8 hours as a fraction of 1 day.
(a) \(\frac {2}{3}\)
(b) \(\frac {1}{3}\)
(c) \(\frac {8}{1}\)
(d) \(\frac {1}{8}\)
Answer:
(b) \(\frac {1}{3}\)

Question 8.
Find : \(\frac {2}{5}\) of ₹ 20.
(a) ₹ 8
(b) ₹ 10
(c) ₹ 12
(d) ₹ 40.
Answer:
(a) ₹ 8

Question 9.
Write \(\frac {19}{4}\) as mixed fraction.
(a) \(3\frac {4}{5}\)
(b) \(4\frac {4}{3}\)
(c) \(4\frac {3}{4}\)
(d) \(5\frac {1}{4}\)
Answer:
(c) \(4\frac {3}{4}\)

Question 10.
\(7\frac {2}{3}\) = ……………….
(a) \(\frac {17}{3}\)
(b) \(\frac {23}{3}\)
(c) \(\frac {13}{3}\)
(d) \(\frac {42}{3}\)
Answer:
(b) \(\frac {23}{3}\)

PSEB 6th Class Maths MCQ Chapter 5 Fractions

Question 11.
Which of the following represents an improper fraction?
PSEB 6th Class Maths MCQ Chapter 5 Fractions 2.1
Answer:
PSEB 6th Class Maths MCQ Chapter 5 Fractions 2

Question 12.
Which of the following is an equivalent of \(\frac {5}{7}\)?
(a) \(\frac {25}{49}\)
(b) \(\frac {20}{35}\)
(c) \(\frac {35}{49}\)
(d) \(\frac {35}{28}\)
Answer:
(c) \(\frac {35}{49}\)

Question 13.
PSEB 6th Class Maths MCQ Chapter 5 Fractions 3.1
(a) 32
(b) 24
(c) 40
(d) 16
Answer:
(a) 32

Question 14.
Which of the following are in ascending order?
(a) \(\frac{2}{3}, \frac{2}{7}, \frac{2}{5}\)
(b) \(\frac{2}{3}, \frac{2}{5}, \frac{2}{7}\)
(c) \(\frac{2}{7}, \frac{2}{3}, \frac{2}{5}\)
(d) \(\frac{2}{7}, \frac{2}{5}, \frac{2}{3}\)
Answer:
(d) \(\frac{2}{7}, \frac{2}{5}, \frac{2}{3}\)

PSEB 6th Class Maths MCQ Chapter 5 Fractions

Question 15.
Which of the following are in descending order?
(a) \(\frac{1}{8}, \frac{1}{3}, \frac{1}{9}\)
(b) \(\frac{1}{3}, \frac{1}{8}, \frac{1}{9}\)
(c) \(\frac{1}{8}, \frac{1}{9}, \frac{1}{3}\)
(d) \(\frac{1}{3}, \frac{1}{9}, \frac{1}{8}\)
Answer:
(b) \(\frac{1}{3}, \frac{1}{8}, \frac{1}{9}\)

Question 16.
\(\frac{4}{6}+\frac{3}{6}\) = …………. .
(a) \(\frac {7}{12}\)
(b) \(\frac {7}{8}\)
(c) \(1\frac {1}{6}\)
(d) \(1\frac {1}{12}\)
Answer:
(c) \(1\frac {1}{6}\)

Question 17.
\(\frac{4}{9}+\frac{5}{9}-\frac{2}{9}\) = ……………. .
(a) \(\frac {7}{9}\)
(b) \(\frac {7}{18}\)
(c) \(\frac {11}{9}\)
(d) \(\frac {5}{9}\)
Answer:
(a) \(\frac {7}{9}\)

Question 18.
\(\frac{2}{3}+\frac{1}{6}\) = ………………
(a) \(\frac {3}{9}\)
(b) \(\frac {5}{6}\)
(c) \(\frac {7}{6}\)
(d) \(\frac {5}{9}\)
Answer:
(b) \(\frac {5}{6}\)

PSEB 6th Class Maths MCQ Chapter 5 Fractions

Question 19.
4 – \(\frac {1}{3}\) = …………….
(a) \(4\frac {1}{3}\)
(b) \(3\frac {1}{3}\)
(c) \(4\frac {2}{3}\)
(d) \(3\frac {2}{3}\)
Answer:
(d) \(3\frac {2}{3}\)

Question 20.
Divide \(\frac {1}{6}\) by 2
(a) \(\frac {1}{3}\)
(b) \(\frac {1}{12}\)
(c) \(\frac {1}{18}\)
(d) 12
Answer:
(b) \(\frac {1}{12}\)

Question 21.
Which fraction is represented by point P on the adjoining number line?
PSEB 6th Class Maths MCQ Chapter 5 Fractions 3
(a) \(\frac {4}{5}\)
(b) \(\frac {5}{6}\)
(c) \(\frac {3}{4}\)
(d) \(\frac {3}{5}\)
Answer:
(a) \(\frac {4}{5}\)

Question 22.
Which fraction is represented by point P on the adjoining number line?
PSEB 6th Class Maths MCQ Chapter 5 Fractions 4
(a) \(\frac {1}{2}\)
(b) \(\frac {3}{5}\)
(c) \(\frac {4}{5}\)
(d) \(\frac {2}{5}\)
Answer:
(d) \(\frac {2}{5}\)

PSEB 6th Class Maths MCQ Chapter 5 Fractions

Question 23.
Which of the following fractions are in ascending order?
(a) \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\)
(b) \(\frac{5}{10}, \frac{5}{11}, \frac{5}{12}\)
(c) \(\frac{1}{10}, \frac{1}{100}, \frac{1}{1000}\)
(d) \(\frac{3}{10}, \frac{4}{10}, \frac{7}{10}\)
Answer:
(d) \(\frac{3}{10}, \frac{4}{10}, \frac{7}{10}\)

Question 24.
Which of the following fractions are in descending order?
(a) \(\frac{1}{3}, \frac{1}{4}, \frac{1}{5}\)
(b) \(\frac{4}{11}, \frac{4}{12}, \frac{4}{13}\)
(c) \(\frac{1}{10}, \frac{1}{100}, \frac{1}{1000}\)
(d) \(\frac{7}{10}, \frac{9}{10}, \frac{11}{10}\)
Answer:
(a) \(\frac{1}{3}, \frac{1}{4}, \frac{1}{5}\)

Question 25.
Which of the following fraction is shown in the shaded portion of the figure?
PSEB 6th Class Maths MCQ Chapter 5 Fractions 5
(a) \(\frac {1}{3}\)
(b) \(\frac {2}{4}\)
(c) \(\frac {3}{4}\)
(d) \(\frac {2}{5}\)
Answer:
(b) \(\frac {2}{4}\)

Fill in the blanks:

Question (i)
The fraction shown in the shaded portion of the figure is ……………..,
PSEB 6th Class Maths MCQ Chapter 5 Fractions 6
Answer:
\(\frac {1}{4}\)

PSEB 6th Class Maths MCQ Chapter 5 Fractions

Question (ii)
The fraction of the shaded portion shown by the following figure is …………. .
PSEB 6th Class Maths MCQ Chapter 5 Fractions 7
Answer:
\(\frac {1}{8}\)

Question (iii)
The fraction of the shaded portion shown by the given figure is …………… .
PSEB 6th Class Maths MCQ Chapter 5 Fractions 8
Answer:
\(\frac {3}{7}\)

Question (iv)
The upper part of the fraction is called ……………. .
Answer:
numerator

Question (v)
The lower part of the fraction is called ……………. .
Answer:
denominator

Write True/False:

Question (i)
The fraction \(\frac {3}{5}\) is read as three fifth. (True/False)
Answer:
True

PSEB 6th Class Maths MCQ Chapter 5 Fractions

Question (ii)
The fraction whose numerator is less than the denominator is called the proper fraction. (True/False)
Answer:
True

Question (iii)
\(\frac {7}{8}\) is an improper fraction. (True/False)
Answer:
False

Question (iv)
Mixed fraction is a combination of a whole number and a proper fraction. (True/False)
Answer:
True

Question (v)
Fraction with the same denominator are called unlike fractions. (True/False)
Answer:
False

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.6

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 5 Fractions Ex 5.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 5 Fractions Ex 5.6

1. Multiply:

Question (i)
(i) \(\frac {1}{5}\) × 4
(ii) \(\frac {2}{7}\) × 3
(iii) \(\frac {5}{8}\) × 2
(iv) \(\frac {7}{12}\) × 4
(iv) 10 × \(\frac {4}{5}\)
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.5 1

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.6

2. Divide:

Question (ii)
(i) \(\frac {1}{4}\) ÷ 5
(ii) \(\frac {3}{5}\) ÷ 3
(iii) \(\frac {5}{8}\) ÷ 3
(iv) \(\frac {6}{7}\) ÷ 2
(iv) \(\frac {12}{15}\) ÷ 6.
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.5 2
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.5 3