Class 6

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:

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

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

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.

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

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

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

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 + \(\frac {1}{10}\) + \(\frac {2}{100}\) + \(\frac {3}{1000}\)

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 + \(\frac {5}{100}\) + \(\frac {7}{1000}\)

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)
= \(\frac {7}{5}\)

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)

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

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

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

Question (i)
\(\frac {17}{2}\)
Solution:

= 8.5

Question (ii)
\(\frac {33}{4}\)
Solution:

= 8.25

Question (iii)
\(\frac {76}{5}\)
Solution:

= 15.2

Question (iv)
\(\frac {24}{25}\)
Solution:

= 0.96

Question (v)
\(\frac {5}{8}\)
Solution:

= 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:

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:

Since, 7 > 4
So, 0.7 > 0.4

Question (ii)
2.6 or 2.5
Solution:

Since, 6 > 5
So, 2.6 > 2.5

Question (iii)
1.23 or 1.32
Solution:

Since, 3 > 2
So, 1.32 > 1.23

Question (iv)
12.3 or 12.4
Solution:

Since, 4 > 3
So, 12.4 > 12.3

Question (v)
18.35 or 18.3
Solution:

Since, 5 > 0
So, 18.35 > 18.30

Question (vi)
12 or 1.2
Solution:

Since, 12 > 1
So, 12 > 1.2

Question (vii)
5.06 or 5.061
Solution:

Since, 1 > 0
So, 5.061 > 5.060

Question (viii)
2.34 or 23.3
Solution:

Since, 23 > 2
So, 23.3 > 2.34

Question (ix)
13.08 or 13.078
Solution:

Since, 8 > 7
So, 13.08 > 13.078

Question (x)
2.3 or 2.03.
Solution:

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

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

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

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.5

1. Which of the following are polygons and there is no polygon. Give the reason:

Solution:
(i) It is not a closed figure. Therefore it is not a polygon.
(ii) It is made up of lines segment. Therefore it is polygon.
(iii) It is not a polygon, because it is not made of line segments.
(iv) It is not closed by line segment. Therefore, it is not a polygon.
(v) It is not polygon because line segments are intersecting each other.
(vi) It is made up of line segments, therefore it is a polygon.

2. Classify the following as concave or convex polygons:

Solution:
(i) Concave Polygon
(ii) Convex Polygon
(iii) Concave Polygon
(iv) Concave Polygon
(v) Convex Polygon
(vi) Convex Polygon.

3. Tick in the boxes, if the property holds true for a particular quadrilateral otherwise eroes out ‘x’:

Quadrilateral Properties	Rectangle	Parallelogram	Rhombus	Trapezium	Square
All sides are equal
Only opposite sides are equal
Diagonals are equal
Diagonals bisect each other
Diagonals are perpendicular to each other
Each angle is 90°

Solution:

Quadrilateral Properties	Rectangle	Parallelogram	Rhombus	Trapezium	Square
All sides are equal	×	×	√	×	√
Only opposite sides are equal	√	√	×	×	×
Diagonals are equal	√	×	×	×	√
Diagonals bisect each other	√	√	√	×	√
Diagonals are perpendicular to each other	×	×	√	×	√
Each angle is 90°	√	×	X	×	√

4. Fill in the blanks:

Question (i)
…………… is a quadrilateral with only one pair of opposite sides parallel.
Solution:
Trapezium

Question (ii)
…………….. is a quadrilateral with all sides equal and diagonals of equal length.
Solution:
Square

Question (iii)
A polygon with atleast one angle is reflex is called ……………….. .
Solution:
Concave polygon

Question (iv)
………….. is a regular quadrilateral.
Solution:
Square

Question (v)
…………… is a quadrilateral with opposite sides equal and diagonals of unequal length.
Solution:
Parallelogram.

5. State True or False:

Question (i)
A rectangle is always a rhombus.
Solution:
False

Question (ii)
The diagonals of a rectangle are perpendicular to each other.
Solution:
False

Question (iii)
A square is a parallelogram.
Solution:
True

Question (iv)
A trapezium is a parallelogram.
Solution:
False

Question (v)
Opposite sides of a parallelogram are parallel.
Solution:
True.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.4

1. Classify each of the following triangles as scalene, isosceles or equilateral:

Solution:
(i) Here, two sides of triangle are equal in length.
∴ It is an isosceles triangle.
(ii) Here, all the three sides of the triangle are equal in length.
∴ It is an equilateral triangle.
(iii) Here, no two sides are equal in length.
∴ It is scalene triangle.
(iv) Here, two sides of triangle are equal in length.
∴ It is an isosceles triangle.
(v) Here, no two sides are equal in length.
∴ It is scalene triangle.
(vi) Here, all the three sides of the triangle are equal in length.
∴ It is equilateral triangle.

2. Classify each of the following triangles as acute, obtuse or right triangle:

Solution:
(i) Here, one angle is 120°, which is obtuse angle.
∴ It is an obtuse-angled triangle.
(ii) Here, one angle is 90°, which is right angle.
∴ It is an right-angled triangle.
(iii) Here, each angle is acute angle.
∴ It is an acute-angled triangle.
(iv) Here, one angle is 90°, which is right angle.
∴ It is an right-angled triangle.
(v) Here, one angle is 120°, which is obtuse angle.
∴ It is an obtuse-angled triangle.
(vi) Here, each angle is 60°, which is actute angle.
∴ It is an actute angled triangle.

3. Which of the following triangles are possible with the given angles?

Question (i)
60°, 60°, 60°
Solution:
In a triangle sum of the three angles of a triangle is equal to 180°.
Here, sum of the three angles of triangle is:
60° + 60° + 60° = 180°
∴ This triangle is possible.

Question (ii)
110°, 50°, 30°
Solution:
Here, sum of the three angles of triangle is:
110° + 50° + 30°= 190° ≠ 180°
∴ This triangle is not possible.

Question (iii)
65°, 55°, 60°
Solution:
Here, sum of the three angles of triangle is:
65°+ 55°+ 60°= 180°
∴ This triangle is possible.

Question (iv)
90°, 40°, 50°
Solution:
Here, sum of the three angles of triangle is:
90°+ 40°+ 50°= 180°
∴ This triangle is possible.

Question (v)
48°, 62°, 50°
Solution:
Here, sum of the three angles of triangle is:
48°+ 62°+ 50°= 160° ≠ 180°
∴ This triangle is not possible.

Question (vi)
90°, 95°, 30°.
Solution:
Here, sum of the three angles of triangle is:
90°+ 95°+ 30° =215° ≠ 180°
∴ This triangle is not possible.

4. Classify each of the following triangles as scalene, isosceles or equilateral triangle:

Question (i)
4 cm, 5 cm, 6 cm
Solution:
The sides of triangle are 4 cm, 5 cm, 6 cm
No, two sides of this triangle are equal.
∴ This is a scalene triangle.

Question (ii)
5 cm, 7 cm, 5 cm
Solution:
The sides of triangle are 5 cm, 7 cm, 5 cm
Here, two sides are equal each of 5 cm in length.
∴ This is an isosceles triangle.

Question (iii)
4.2 m, S3 m, 6.1 m
Solution:
The sides of triangle are 4.2 m, 5.3 m, 6.1 m
Here, all sides are of different length.
∴ This is a scalene triangle.

Question (iv)
3.5 cm, 3.5 cm, 33 cm
Solution:
The sides of triangle are 3.5 cm, 3.5 cm, 3.5 cm
All the sides of triangle are of equal length.
∴ This is an equilateal triangle.

Question (v)
8 cm, 4.2 cm, 4.2 cm
Solution:
The sides of triangle are 8 cm, 4.2 cm, 4.2 cm
Here, two sides of the triangle are of equal length.
∴ This is an isosceles triangle.

Question (vi)
2 cm, 3 cm, 4 cm.
Solution:
The sides of triangle are 2 cm, 3 cm, 4 cm
All the sides of the triangle are of different lengths
∴ This is a scalene triangle.

5. Name the following triangles in both ways: (Based on sides and angles)

Solution:
(i) Based on sides: In this triangle, no two sides of the triangle are equal.
∴ This is a scalene triangle.
Based on angles: All the three angles of the triangle are acute.
∴ This is an acute-angled triangle.

(ii) Based on sides: In this triangle, two sides are of equal length each is 4 cm.
∴ This is an isosceles triangle.
Based on angles: In this triangle, one angle is of 90° which is a right angle.
∴ This is a right-angled triangle.

(iii) Based on sides: In this triangle, two sides are of equal length.
∴ This is an isosceles triangle.
Based on angles: In this triangle one angle is of 110°, which is obtuse angle.
∴ This is an obtuse-angled triangle.

(iv) Based on sides: In this triangle, all the sides are of equal length i.e. each = 4 cm.
∴ This is an equilateral triangle.
Based on angles: In this triangle, all the angles are acute angles.
∴ This is an acute-angled triangle.

(v) Based on sides: In this triangle, all the three sides are of different lengths.
∴ This is a scalene triangle.
Based on angles: In this triangle, one angle is 105°, which is obtuse angle.
∴ This is an obtuse-angled triangle.

6. Fill in the blanks:

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

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

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

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

Question (v)
A triangle whose all sides are different is known as ………………. .
Solution:
Scalene triangle

Question (vi)
A triangle whose all angles are acute is known as ……………….. .
Solution:
Acute angled triangle

Question (vii)
A triangle whose two sides are equal is known as ……………….. .
Solution:
Isosceles triangle

Question (viii)
A triangle whose one angle is obtuse is known as ……………….. .
Solution:
obtuse-angled triangle

Question (ix)
A triangle whose all sides are equal is known as ……………….. .
Solution:
Equilateral triangle

Question (x)
A triangle whose one angle is right angle is known as ……………….. .
Solution:
Right-angled triangle

7. State True or False:

Question (i)
Each equilateral triangle is an isosceles triangle.
Solution:
True

Question (ii)
Each acute-angled triangle is a scalene triangle.
Solution:
False

Question (iii)
Each isosceles triangle is an equilateral triangle.
Solution:
False

Question (iv)
There are two obtuse angles in an obtuse triangle.
Solution:
False

Question (v)
In right triangle, there is only one right angle.
Solution:
True

Question (vi)
Right triangle can never be isosceles.
Solution:
False.