NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.2

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.2.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 8
Chapter Name  Decimals
Exercise  Ex 8.2
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.2

Question 1.
Complete the table with the help of these boxes and use decimals to write the number.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 17

Ones Tenths Hundredths Number
(a)
(b)
(c)

Solution.

Ones Tenths Hundredths Number
(a) 0 2 6 0.26
(b) 1 3 8 1.38
(c) 1 2 8 1.28

Question 2.
Write the numbers in the following place value table in decimal form:

Hundreds
(100)
Tens
(10)
ones
(1)
Tenths
(Tenths
\(\frac { 1 }{ 10 } \))

Hundredths
(Tenths
\(\frac { 1 }{ 100 } \))
Thousandths
(Tenths
\(\frac { 1 }{ 1000 } \))
(a) 0 0 3 2 5 0
(b) 1 0 2 6 3 0
(c) 0 3 0 0 2 5
(d) 2 1 1 9 0 2
(e) 0 1 2 2 4 1

Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 18
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 19
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 20

Question 3.
Write the following decimals in the place value table:
(a) 0.29
(b) 2.08
(c) 19.60
(d) 148.32
(e) 200.812
Solution.

DecimalNumber Hundreds Tens Ones Tenths Hundredths Thousandths
(a) 0.29 0 0 0 2 9 0
(b) 2.08 0 0 2 0 8 0
(c) 19.60 0 1 9 6 0 0
  (d) 148.32 1 4 8 3 2 0
  (e) 200.812 2 0 0 8 1 2

Question 4.
Write each of the following as decimals: 
(a) 20+9+\(\frac { 4 }{ 10 } \)+\(\frac { 1 }{ 100 } \)
(b)
30+\(\frac { 4 }{ 10 } \) + \(\frac { 8 }{ 100 } \)
(c)
137+ \(\frac { 5 }{ 100 } \)
(d) 
\(\frac { 7 }{ 10 } \) + \(\frac { 6 }{ 100 } \) + \(\frac { 4 }{ 1000 } \)
(e) 23+\(\frac { 2 }{ 10 } \) + \(\frac { 6 }{ 1000 } \)
(f)
700+20+5+\(\frac { 9 }{ 100 } \)
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 21
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 22

Question 5.
Write each of the following decimals in words:
(a) 0.03
(b) 1.20
(c) 108.56
(d) 10.07
(e) 0.032
(f) 5.008.
Solution.
(a) Zero point zero three
(b) One point two zero
(c) One hundred eight point five six
(d) Ten point zero seven
(e) Zero point zero three two
(f) Five point zero eight.

Question 6.
Between which two numbers in tenths place on the number line does each of the given number lie?
(a) 06
(b) 0.45
(c) 19
(d) 0.66
(e) 92
(f) 0.57.
Solution.
Each of the given numbers lies between the two whole numbers 0 and 1 on the number line.
(a) 0 and 0.1
(b) 4 and 0.5
(c) 1 and 0.2
(d)  0.6 and 0.7
(e) 9 and 1.0
(f)  0.5 and 0.6.

Question 7.
Write as fractions in lowest terms:
(a) 0.60
(b) 0.05
(c) 75
(d) 0.18
(e) 25
(f) 0.125.
Solution.
(a) 0.60
\(0.60 =\frac { 60 }{ 100 } =\frac { 60\div 20 }{ 100\div 20 } \)
∵  H.C.F.(60,100) = 20
= \(\frac { 3 }{ 5 } \)

(b) 0.05

\(0.05=\frac { 5 }{ 100 } =\frac { 5\div 5 }{ 100\div 5 } \)
∵  H.C.F.(5,100) = 5
= \(\frac { 1 }{ 20 } \)

(c) 0.75
\(0.75=\frac { 75 }{ 100 } =\frac { 75\div 25 }{ 100\div 25 } \)
∵  H.C.F.(75,100) = 25
= \(\frac { 3 }{ 4 } \)

(d) 0.18
\(0.018 =\frac {18 }{ 100 } =\frac { 18\div 2 }{ 100\div 2 } \)
∵  H.C.F.(18,100) = 2
= \(\frac { 9 }{ 50 } \)

(e) 0.25
\(0.25 =\frac { 25 }{ 100 } =\frac { 125\div 125 }{ 100\div 25 } \)
∵  H.C.F.(25,100) = 25
= \(\frac { 1 }{ 4 } \)

(f) 0.125
\(0.125 =\frac { 125 }{ 1000 } =\frac { 125\div 125 }{ 1000\div 125 } \)
∵  H.C.F.(125,1000) =125
= \(\frac { 1 }{ 8 } \)

 

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NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.6

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.6 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.6.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 7
Chapter Name Fractions
Exercise  Ex 7.6
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.6

Question 1.
Solve:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 96
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 97
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 98
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 99
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 100
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 101
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 102
Question 2.
Sarita bought \(\frac { 2 }{ 5 }\) metre of ribbon and Lalita bought \(\frac { 3 }{ 4 }\) metre of ribbon. What was the total 4 length of the ribbon they bought?
Solution :
Ribbon bought by Sarita = \(\frac { 2 }{ 5 }\) m
Ribbon bought by Lalita = \(\frac { 3 }{ 4 }\) m
∴ Total length of the ribbon they bought
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 103

Question 3.
Naina was given \(1\frac { 1 }{ 2 }\) piece of cake and Najma was given \(1\frac { 1 }{ 3 }\) piece of cake. Find the total amount of care given to both of them.
Solution :
Cake given to naina = \(1\frac { 1 }{ 2 }\) piece = \(\frac { \left( 1\times 2 \right) +1 }{ 2 }\) piece
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 104
Question 4.
Fill in the boxes:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 105
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 106
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 107

Question 5.
Complete the addition-subtraction box.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 108
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 109
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 110
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 111

Question 6.
A piece of wire \(\frac { 7 }{ 8 }\) meter long broke into two pieces. One piece was \(\frac { 1 }{ 4 }\) metre long. How long is the other piece?
Solution :
Length of the original piece of wire = \(\frac { 7 }{ 8 }\) metre
∴ Length of one piece = \(\frac { 1 }{ 4 }\) metre
Length of the other piece = \(\frac { 7 }{ 8 }\) metre – \(\frac { 1 }{ 4 }\) metre = \(\left( \frac { 7 }{ 8 } -\frac { 1 }{ 4 } \right)\) metre
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 112
Question 7.
Nandini’s house is ~ km from her school. She walked some distance and then took a bus for km to reach the school. How far did she walk?
Solution :
The distance of Nandini’s house from school
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 113
Question 8.
Asha and Samuel have bookshelves of the 5 same size. Asha’s shelf is \(\frac { 5 }{ 6 }\) the full of book and Samuel’s shelf is \(\frac { 2 }{ 5 }\)th full. Whose bookshelf is more full? By what fraction?
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 114
Question 9.
Jaidev takes \(2\frac { 1 }{ 5 }\) minutes to walk across 5 the school ground. Rahul takes \(\frac { 7 }{ 4 }\) minutes to do the same. Who takes less time and by what fraction?
Solution :
Time taken by Jaidev to walk across the school ground = \(2\frac { 1 }{ 5 }\) minutes
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 115

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NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.5

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.5 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.5.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 7
Chapter Name Fractions
Exercise  Ex 7.5
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.5

Question 1.
Write these fractions appropriately as additions or subtractions:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 84
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 85
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 86
Question 2.
Solve
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 87
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 88
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 89

Question 3.
Shubham painted \(\frac { 2 }{ 3 }\) of the wall space in his room. Her sister Madhavi helped and painted \(\frac { 1 }{ 3 }\) of the wall space. How much did they paint together?
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 90
Hence, they painted together with the complete wall space. ,

Question 4.
Fill in the missing fractions :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 91
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 92
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 93
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 94
Question 5.
Javed was given \(\frac { 5 }{ 7 }\) of a basket of oranges. What fraction of oranges was left in the basket?
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 95
Hence, fraction \(\frac { 2 }{ 7 }\) of oranges was left in the basket.

 

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NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.4

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.4 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.4.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 7
Chapter Name Fractions
Exercise  Ex 7.4
Number of Questions Solved 10
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.4

Question 1.
Write shaded portion as a fraction. Arrange them in ascending and descending order using correct sign ‘<‘, ‘=’, ‘>’ between the fraction:
(a)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 53
(b)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 54
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 55
appropriate signs between the fractions  given
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 56
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 57
(i) In ascending order, these are
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 58
(ii) In descending order, these are
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 59
(b)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 60
(i) In ascending order, these are
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 61
(ii) In descending order, these are
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 62
(c)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 63
Question 2.
Compare the fractions and put an appropriate sign.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 64
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 65

NCERT Solutions for Class 6 Maths Chapter 7 Fractions 66
Question 3.
Make five more such pairs and make appropriate signs.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 67

Question 4.
Look at the figures and write ‘<’ or ‘>’, ‘=’ between the pairs of fractions.
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 68
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 69
Make five more such problems and solve them with your friends.
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 70
For the remaining part, please try yourself.

Question 5.
How quickly can you do this? Fill appropriate sign (<, =,>)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 71
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 72
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 73
Question 6.
The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 74
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 75
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 76
Question 7.
Find answers to the following. Write and indicate how you solved them.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 77
Solution :
(a) Equivalent fraction of \(\frac { 5 }{ 9 }\) are
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 78
Equivalent fraction of \(\frac { 4 }{ 5 }\) are
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 79
(b) Equivalent fraction of \(\frac { 9 }{ 16 }\) are
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 80
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 81

Question 8.
Ila reads 25 pages of a book containing 100 pages. Lalita reads \(\frac { 1 }{ 2 }\) of the same book. Who read less?
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 82
Question 9.
Rafiq exercised for \(\frac { 3 }{ 6 }\) of an hour, while 6 Rohit exercised for \(\frac { 3 }{ 4 }\) of an hour. Who exercised for a longer time?
Solution :
∴ \(\frac { 3 }{ 4 }\) > \(\frac { 3 }{ 6 }\)
∴ Rohit exercised for a longer time.

Question 10.
In class A of 25 students, 20 passed in first class; in another class B of 30 students, 24 passed in first class. In which class was a greater fraction of students getting first class?
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 83
Hence, in both the classes the same fraction \(\left( \frac { 4 }{ 5 } \right)\)of total students got first class.

 

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NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.3

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.3.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 7
Chapter Name Fractions
Exercise  Ex 7.3
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.3

Question 1.
Write the fractions. Are all these fractions equivalent?
(a)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 18
(b)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 19
Solution :
(a)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 20
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 21
(b)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 22
No ! all these fractions ae not equivalent.

Question 2.
Write the fractions and pair up the equivalent fractions from each row.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 23
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 24
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 25
Equivalent fractions are
(a), (ii);
(b), (,iv);
(c), (i);
(d), (v);
(e), (iii).

Question 3.
Replace in each of the following by the correct number:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 26
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 27
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 28
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 29
Question 4.
Find the equivalent fraction of \(\frac { 3 }{ 5 }\) having
(a) denominator 20
(b) numerator 9
(d) numerator 27
(c) denominator 30
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 30
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 31
Question 5.
Find the equivalent fraction of \(\frac { 36 }{ 48 }\) with
(a) numerator 9
(b) denominator 4.
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 32
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 33
Question 6.
Check whether the given fractions are equivalent :
(a) \(\frac { 5 }{ 9 }\), \(\frac { 30 }{ 54 }\)
(b) \(\frac { 3 }{ 10 }\), \(\frac { 12 }{ 50 }\)
(c) \(\frac { 7 }{ 13 }\), \(\frac { 5 }{ 11 }\)
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 34
∴ The given fractions \(\frac { 5 }{ 9 }\) and \(\frac { 30 }{ 54 }\) are equivalent.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 35
∴ The given fractions \(\frac { 3 }{ 10 }\) and \(\frac { 12 }{ 50 }\) are not equivalent.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 36
∴ The given fractions \(\frac { 7 }{ 3 }\) and \(\frac { 5 }{ 11 }\) are not equivalent.

Question 7.
Reduce the following fractions to simplest form:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 37
Solution :
(a) \(\frac { 48 }{ 60 }\)
Factors of 48 are 1, 2, 3,4, 6, 8, 12, 16, 24 and 48.
Factors of 60 are 1, 2, 3,4, 5, 6, 10,12, 15, 20, 30 and 60.
∴ Common factors of 48 and 60 are 1, 2, 3, 4, 6 and 12. Highest of these common factors is 12.
∴ H.C.F. of 48 and 60 is 12.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 38
(b) \(\frac { 150 }{ 60 }\)
Factors of 150 are 1,2,3,5,6.10,15,25,30,50, 75 and 150.
Factors of 60 are 1, 2. 3, 4. 5, 6, 10, 12, 15, 20, 30 and 60.
∴ Common factors of 150 and 60 are 1, 2, 3, 5,6, 10, 15 and 30.
Highest of these common factors is 30.
∴ H.C.F. of 150 and 60 is 30.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 39
(c) \(\frac { 84 }{ 98 }\)
Factors of 84 are 1, 2, 3,4, 6, 7, 12, 14, 21, 28, 42 and 84.
Factors of 98 are 1, 2, 7, 14,49 and 98.
∴ Common factors of 84 and 98 are 1,7 and 14. Highest of these common factors is 14.
∴ H.C.F. of 84 and 98 is 14.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 40
(d) \(\frac { 12 }{ 52 }\)
Factors of 12 are 1, 2, 3,4, 6 and 12.
Factors of 52 are 1, 2, 4, 13, 26 and 52.
∴ Common factors of 12 and 52 are 1,2 and 4. Highest of these common factors is 4.
∴ H.C.F. of 12 and 52 is 4.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 41
(e) \(\frac { 7 }{ 28 }\)
Factors of 7 are 1 and 7.
Factors of 28 are 1, 2,4, 7, 14 and 28.
∴ Common factors of 7 and 28 are 1 and 7. Highest of these common factors is 7.
∴ H.C.F. of 7 and 28 is 7.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 42
Question 8.
Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of his/her pencils?
Solution :
For Ramesh
Number of pencils he had = 20 Number of pencils used by him =10
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 43
For Sheelu
Number of pencils she had = 50 Number of pencils used by her = 25
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 44

For Jamaal
Number of pencils he had = 80 Number of pencils used by him = 40
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 45
Yes ! each has used up an equal fraction of his/her pencils.

Question 9.
Match the equivalent fractions and write another two more for each :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 46
Solution :
(i) \(\frac { 250 }{ 400 }\)
Factors of 250 are 1, 2, 5, 10, 25, 50, 125 and 250.
Factors of 400 are 1,2,4,5,8,10,16,20,25,40, 80, 100, 200 and 400.
∴ Common factors of 250 and 400 are 1,2,5, 10, 25 and 50.
Highest of these common factors is 50.
∴ H.C.F. of 250 and 400 is 50.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 47

(ii) \(\frac { 180 }{ 200 }\)
Factors of 180 are 1,2, 3,4, 5, 6,9,10,12,15, 18, 20, 30, 36, 45, 60, 90 and 180.
Factors of 200 are 1,2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200.
∴ Common factors of 180 and 200 are 1,2, 4, 5, 10 and 20.
Highest of these common factors is 20.
∴ H.C.F. of 180 and 200 is 20.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 48

(iii) \(\frac { 600 }{ 990 }\)
Factors of 660 are 1,2,3,4,5,6,10,12,22,30, 66, 110, 132, 165, 220, 330 amd 660.
Factors of 990 are 1,2,3,5,6,9,10,11,30,33, 90, 99, 110, 165, 198, 330, 495 and 990.
∴ Common factors of 660 and 990 are 1,2,3, 5,6, 10, 30, 110 and 330.
Highest of these common factors is 330.
∴ H.C.F. of 660 and 990 is 330.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 49

(iv) \(\frac { 180 }{ 360 }\)
Factors of 180 are 1, 2, 3,4, 5, 6, 9, 10, 12,15, 18, 20, 30, 36,45, 60, 90 and 180.
Factors of 360 are 1, 2, 3,4, 5, 6,9,10, 12, 15, 18, 20, 24, 30, 36,40, 60, 72, 90, 120, 180 and 360.
∴ Common factors of 180 and 360 are 1,2, 3,4,5,6,9,10,12,15,18.20, 30, 36,60,90 and 180.
Highest of these common factors is 180.
∴ H.C.F. of 180 and 360 is 180.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 50

(v) \(\frac { 220 }{ 550 }\)
Factors of 220 are 1,2,4, 5,10,11, 20, 22,44, 55, 110 and 220.
Factors of 550 are 1, 2, 5, 10, 22, 25, 55, 110, 275 and 550.
∴ Common factors of 220 and 550 are 1,2,5, 10, 20 and 110.
Highest of these common factors is 110.
∴ H.C.F. of 220 and 550 is 110.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 51
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 52

 

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NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.2

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.2.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 7
Chapter Name Fractions
Exercise  Ex 7.2
Number of Questions Solved 3
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.2

Question 1.
Draw number lines and locate the points on them:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 9
Solution :
(a)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 10
(b)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 11
(c)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 12
Question 2.
Express the following as mixed fractions:
(a) \(\frac { 20 }{ 3 }\)
(b) \(\frac { 11 }{ 5 }\)
(c) \(\frac { 17 }{ 7 }\)
(d) \(\frac { 28 }{ 5 }\)
(e) \(\frac { 19 }{ 6 }\)
(f) \(\frac { 35 }{ 9 }\).
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 13
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 14
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 15

Question 3.
Express the following as improper fractions:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 16
Solution :
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 17

 

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NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 6
Chapter Name Integers
Exercise  Ex 6.3
Number of Questions Solved 4
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3

Question 1.
Find:
(a)
35 – (20)
(b) 72 – (90)
(c) (-15) – (-18)
(d) (-20) – (13)
(e) 23 -(-12)
(f) (-32) -(- 40).
Solution.
(a) 35-(20)
35 – (20)
= 35 + (additive inverse of 20)
= 35 + (- 20)
= 15 + 20 + (- 20)
= 15 + 0= 15

(b) 72-90
72-90
= 12 + (additive inverse of 90)
= 72 + (- 90)
= 72 + (- 72) + (- 18)
= 0 + (- 18) = – 18

(c) (- 15) – (- 18)
(-15)-(-18)
= (- 15) + (additive inverse of -18)
= (- 15) + (18)
= (- 15) + (15) + (3)
= 0 + (3) = 3

(d) (- 20) – (13)
= (-20)-(13)
= (- 20) + (additive inverse of 13)
= (- 20) + (- 13) = – 33

(e) 23 – (- 12)
23 – (- 12)
= 23 + (additive inverse of – 12)
= 23+12 = 35

(f) (- 32) – (- 40)
(-32)-(-40)
= (- 32) + (additive inverse of – 40)
= (-32)+ (+40)
= (- 32) + (+ 32) + (+ 8)
= 0 + (+ 8) = 8.

Question 2.
Fill in the blanks with >, < or = sign :
(a)
(- 3) + (- 6)……… (-3)-(-6)
(b) (- 21) – (- 10)….. (-31)+ (-11)
(c) 45 -(- 11)……. 57 +(-4)
(d) (- 25) – (- 42)……. (-42)-(-25).
Solution.
(a) L.H.S. = (- 3) + (- 6) = – 9
R.H.S. = (- 3) – (- 6)
= (- 3) + (additive inverse of – 6)
= (- 3) + 6
= (- 3) + 3 + 3
=0+3=3
∴ (- 3) + (- 6) < (- 3) – (- 6)

(b) L.H.S. = (- 21) – (- 10)
= (- 21) + (additive inverse of -10)
= (- 21) + 10
= (- ll) + (- 10)+ 10
= (- 11) + 0 = – 11
R.H.S. = (-31)+ (-11)
= -42
∴ (-21)-(-10) >(-31)+ (-11)

(c) L.H.S. =45-(-11)
= 45 + (additive inverse of – 11)
= 45 + 11 = 56
R.H.S.;= 57 + (-4)
= 53 + 4 + (- 4)
= 53 + 0 = 53
∴ 45-(-11) >57 +(-4)

(d) L.H.S. = (-25) – (-42)
= (- 25) + (additive inverse of – 42)
= (-25)+ (+42)
= (- 25) + (+ 25) + (+ 17)
= 0 + (+ 17)= 17
R.H.S. = (- 42) – (- 25)
= (- 42) + (additive inverse of – 25)
= (- 42) + (+ 25)
= (- 17) + (- 25) + (+ 25)
= (- 17) + 0 = – 17
∴ (-25)-(-42) >(-42)-(-25).

Question 3.
Fill in the blanks: 

(a) (-8) + =0
(b) 13 + = 0
(c) 12 + (-12) –
(d) (-4) + =-72
(e) – 75 = -10.
Solution.
(a) (-8)+ 8 = 0
(b) 13 + (-13) = 0
(c) 12 + (- 12) = 0.
(d)(-4) + (-8) = -12
(e) (+5)-15 =-10.

Question 4.
Find:
(a) (-7) -8 -(-25)
(b) (-13)+ 32-8-1
(c) (-7) + (-8) + (-90)
(d) 50-(-40)-(-2)
Solution.
(a) (- 7) – 8 – (- 25)
(- 7) – 8 – (- 25)
= (- 7) + (additive inverse of 8) – (- 25)
= (- 7) + (- 8) – (- 25)
= – 15-(-25)
= – 15 + (additive inverse of – 25)
= -15+ (+25)
= – 15 + (+ 15) + (+ 10)
= 0 + (+ 10) = 10

(b) (- 13) + 32 – 8 – 1
(-13)+ 32-8-1
= (-13)+ 32-9
= (- 13) + 32 + (additive inverse of 9)
= (- 13) + 32 + (- 9)
= (- 13) + 23 + 9 + (- 9)
= (-13)+ 23 + 0
= (- 13)+ 23
= (-13)+13+10
= 0+ 10 = 10

(c) (- 7) + (- 8) + (- 90)
(-7) +(-8)+ (-90)
= (-15)+ (-90)
= -105

(d) 50-(-40)-(-2) 
= 50 – (- 40) – (- 2)
= 50 + (additive inverse of – 40) – (- 2)
= 50+ (40)-(-2)
= 90-(-2)
= 90 + (additive inverse of – 2)
= 90 + 2 = 92.

We hope the NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 6
Chapter Name Integers
Exercise  Ex 6.2
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2

Question 1.
Using number line write the integer which is:
(a) 3 more than 5
(b) 5 more than – 5
(c) 6 less than 2
(d) 3 less than – 2.
Solution.
(a) 3 more than 5
We start from 5 and proceed 3 steps to the right of 5 to reach 8 as shown below:
NCERT Solutions for Class 6 Maths Chapter 6 Integers 5
Therefore, 3 more than 5 is 8.

(b) 5 more than – 5
We start from – 5 and move to the right by 5 steps and obtain 0 as shown below:
NCERT Solutions for Class 6 Maths Chapter 6 Integers 6
Therefore, 5 more than – 5 is 0.

(c) 6 less than 2
We start from 2 and proceed 6 steps to the left of 2 to reach – 4 as shown below:
NCERT Solutions for Class 6 Maths Chapter 6 Integers 7

(d) 3 less than – 2
We start from – 2 and move to the left by 3 steps and obtain – 5 as shown below:
NCERT Solutions for Class 6 Maths Chapter 6 Integers 8
Therefore, 3 less than – 2 is – 5.

Question 2.
Use number line and add the following integers:
(a)
9 +(-6)
(b) 5+ (-11)
(c) (- 1) +(- 7)
(d) (-5)+ 10
(e) (-1) +(-2) + (-3)
(f) (-2) + 8 + (-4).
Solution.
(a) 9 +(-6)
On the number line, we first move 9 steps to the right from 0 reaching 9 and then we move 6 steps to the left of 9 and reach 3.
Thus, 9 + (- 6) = 3.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 9

(b) 5 + (-11)
On the number line, we first move 5 steps to the right from 0 reaching 5 and then we move 11 steps to the left of 5 and reach – 6.
Thus, 5 + (- 11) = – 6.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 10

(c) (-1) + (-7)
On the number line we first move 1 step to the left of 0 reaching – 1, then we move 7 steps to the left of – 1 and reach – 8.
Thus, (- 1) + (- 7) = – 8.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 11

(d) (- 5) + 10
First, we move 5 steps to the left of 0 reaching – 5, then from this point, we move 10 steps to the right. We reach the point + 5.
Thus, (- 5) + 10 = 5.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 12

(e) (-l) + (- 2) + (-3)
First, we move 1 step to the left of 0 reaching – 1, then from this point, we move 2 steps to the left to reach – 3 and finally from – 3, we move 3 steps to the left. We reach the point – 6.
Thus, (- 1) + (- 2) + (- 3) = – 6.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 13

(f) (- 2) + 8 + (- 4)
First, we move 2 steps to the left of 0 reaching – 2, then from this point, we move 8 steps to the right to reach + 6 and finally from + 6 we move 4 steps to the left. We reach the point 2.
Thus, (- 2) + 8 + (- 4) = 2.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 14

Question 3.
Add without using number line:
(a)
11 + (- 7)
(b) (- 13) + (+ 18)
(c) (-10) + (+ 19)
(d) (-250) + (+ 150)
(e) (- 380) + (- 270)
(f) (-217) + (-100).
Solution.
(a) 11 + (- 7)
11+(-7)
= 4 + 7 + (- 7)
=4+0=4

(b) (- 13) + (+ 18)
(-13)+ (+18)
= (- 13)+ (+.13) +(+5)
= 0 + (+5) = 5

(c) (-10) + (+ 19)
(-10)+ (+19)
= (- 10) + (+ 10) + (+ 9)
= 0 + (+ 9) = 9=

(d) (- 250) + (+ 150)
(- 250) + (+ 150)
= (- 100) + (- 150) + (+ 150)
= (- 100) + 0 = – 100

(e) (- 380) + (- 270)
(- 380) + (- 270)
= -650 (/) (- 217) + (-100)
= -317.

Question 4.
Find the sum of:
(a)
137 and – 354
(b) – 52 and 52
(c) – 312, 39 and 192
(d) – 50, – 200 and 300.
Solution.
(a) 137 and – 354
137+ (-354)
= 137+ (- 137)+ (-217)
= 0 +(-217) =-217

(b) – 52 and 52
– 52 + (52)
= 0

(c) – 312,39 and 192
(-312)+ (39)+ (192)
= (-312)+ (231)
= (-81)+ (-231)+ (231)
= (- 81) + 0 = – 81

(d) – 50, – 200 and 300
(-50)+ (-200)+ (300)
= (- 250) + (300)
= (- 250) + (250) + (50)
= 0 + (50) = 50.

Question 5.
Find the value of:
(a) (-7) +(-9)+ 4 + 16
(b) (37) +(-2) +(-65) +(-8).
Solution.
(a) (- 7) + (- 9) + 4 + 16
(- 7) + (- 9) + 4 + 16
= (-16)+ 20 = (-16) +16 + 4 =0+4=4

(b) (37) +(-2) +(-65) +(-8)
= (37)+ (75)
= (37)+ (-37)+ (-38) = 0 + (- 38)
= -38.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.8

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.8 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.8.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 5
Chapter Name Understanding Elementary Shapes
Exercise  Ex 5.8
Number of Questions Solved 45
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.8

Question 1.
Examine whether the following are polygons. If anyone among them is not, say why?
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 25
Solution :
(a) is not a closed figure and hence is not a polygon.
(b) is a polygon of six sides.
(c) is not a polygon since it is not made of line segments.
(d) is not a polygon since it is not made of line segments.

Question 2.
Name each polygon:
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 26
Make two more examples of each of these.
Solution :
(a) A Quadrilateral
(b) A Triangle
(c) A Pentagon (5-sided)
(d) An Octagon (8-sided).
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 27

Question 3.
Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.
Solution :
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 28
The triangle ABC drawn is an obtuse-angled isosceles triangle.

Question 4.
Draw a rough sketch of a regular octagon. (Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Solution :
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 29
Question 5.
A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Solution :
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 30

 

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NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.7

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.7 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.7.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 5
Chapter Name Understanding Elementary Shapes
Exercise  Ex 5.7
Number of Questions Solved 3
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.7

Question 1.
Say True or False :
(a) Each angle of a rectangle is a right angle.
(b) The opposite sides of a rectangle are equal in length.
(c) The diagonals of a square are perpendicular to one another.
(d) All the sides of a rhombus are of equal length.
(e) All the sides of a parallelogram are of equal length.
(f) The opposite sides of a trapezium are parallel.
Solution :
(a) True
(b) True
(c) True
(d) True
(e) False
(f) False.

Question 2.
Give reasons for the following :
(a) A square can be thought of as a special rectangle.
(b) A rectangle can be thought of as a special parallelogram.
(c) A square can be thought of as a special rhombus.
(d) Squares, rectangles, parallelograms are all quadrilaterals.
(e) Square is also a parallelogram.
Solution :
(a) A rectangle with all sides equal becomes a square.
(b) A parallelogram with each angle a right angle becomes a rectangle.
(c) A rhombus with each angle a right angle becomes a square.
(d) All these are four-sided polygons made of line segments.
(e) The opposite sides of a square are parallels, so it is a parallelogram.

Question 3.
A figure is said to be regular, if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?
Solution :
A square is a regular quadrilateral.

 

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NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.6

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.6 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.6.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 5
Chapter Name Understanding Elementary Shapes
Exercise  Ex 5.6
Number of Questions Solved 4
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.6

Question 1.
Name the types of following triangles:
(a) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(b) ΔABC with AB = 8.7cm, AC =7cm andBC = 6 cm.
(c) ΔPQR such that PQ = QR = PR = 5 cm.
(d) ΔDEF with m ∠D = 90°.
(e) ΔXYZ with m ∠Y = 90° and XY = YZ.
(f) ΔLMN with m ∠L = 30°, m ∠M =70° and m ∠N = 80°.
Solution :
(a) Scalene triangle
(b) Scalene triangle
(c) Equilateral triangle
(d) Right angled triangle
(e) Isosceles right triangle
(f) Acute angled triangle.

Question 2.
Match the following :

Measures of triangle Type triangle
(i) 3 sides of equal length (a) Scalene
(ii) 2 sides of equal length (b) Isosceles right angled
(iii) All sides are of different length (c) Obtuse angled
(iv) 3 acute angles (d) Right angled
(v) 1 right angle (e) Equilateral
(vi) 1 obtuse angle (f) Acute-angled
(vii) 1 right angle with two sides of equal length (g) Isosceles

Solution :

Measures of triangle Type triangle
(i) 3 sides of equal length (e) Equilateral
(ii) 2 sides of equal length (g) Isosceles
(iii) All sides are of different length (a) Scalene
(iv) 3 acute angles (f) Acute-angled
(v) 1 right angle (d) Right angled
(vi) 1 obtuse angle (c) Obtuse angled
(vii) 1 right angle with two sides of equal length (b) Isosceles right angled

Question 3.
Name each of the following triangles in two different ways : (you may judge the nature of the angle by observation).
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 22
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 23
Solution :
(a) Acute-angled and isosceles.
(b) Right-angled and scalene.
(c) Obtuse-angled and isosceles.
(d) Right-angled and isosceles.
(e) Equilateral and acute-angled.
(f) Obtuse-angled and scalene.

Question 4.
Try to construct triangles using matchsticks. Some are shown here. Can you make a triangle with
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 24
(a) 3 matchsticks?
(b) 4 matchsticks?
(c) 5 matchsticks?
(d) 6 matchsticks?
(Remember you have to use all the available matchsticks in each case). Name the type of triangle in each case. If you cannot make a triangle, think of reasons for it.
Solution :
(a) Yes, an equilateral triangle.
(b) No, since the sum of the lengths of any two sides of a triangle is always greater than the length of the third side.
(c) Yes, an isosceles triangle.
(d) Yes, an equilateral triangle.

 

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