Dattu

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4.

• Decimals Class 6 Ex 8.1
• Decimals Class 6 Ex 8.2
• Decimals Class 6 Ex 8.3
• Decimals Class 6 Ex 8.5
• Decimals Class 6 Ex 8.6

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 8
Chapter Name	Decimals
Exercise	Ex 8.4
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4

Question 1.
Express as rupees using decimals.
(a) 5 paise
(b) 75 paise
(c) 20 paise
(d) 50 rupees 90paise
(e) 725 paise.
Solution.

Question 2.
Express as meters using decimals.
(a) 15 cm
(b) 6 cm
(c) 2 m 45 cm
(d) 9 m 7cm
(e) 419 cm.
Solution.

Question 3.
Express as cm using decimals.
(a) 5 mm
(b) 60 mm
(c) 164 mm
(d) 9 cm 8 mm
(e) 93mm.
Solution.

Question 4.
Expresc as km using decimals.
(a) 8m
(b) 88m
(c) 8888 m
(d) 70 km 5 m.
Solution.

Question 5.
Express as kg using decimals.
(a) 2g
(b) 100g
(c) 3750g
(d) 5kg8g
(e) 26 kg 50g
Solution.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4 help you. If you have any query regarding. NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3.

• Decimals Class 6 Ex 8.1
• Decimals Class 6 Ex 8.2
• Decimals Class 6 Ex 8.4
• Decimals Class 6 Ex 8.5
• Decimals Class 6 Ex 8.6

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 8
Chapter Name	Decimals
Exercise	Ex 8.3
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3

Question 1.
Which is greater?
(a) 0.3 or 0.4
(b) 0.07 or 0.02
(c) 3 or 0.8
(d) 0.5 or 0.05
(e) 1.23 or 1.2
(f) 0.099 or 0.19
(g) 1.5 or 1.50
(h) 1.431 or 1.490
(i) 3.3 or 3.300
(j) 5.64 or 5.603
Solution.
(a) 0.3 or 0.4
Whole part for both is 0.
Tenth part for 0.3 is 3 and for 0.4 is 4
∵ 4>3
∴ 0.4 > 0.3

(b) 07 or 0.02
Whole part for both is 0.
Tenth part for both is 0.
Hundredth part for 0.07 is 7 and for 0.02 is 2.
∵ 7>2
∴  0.07 > 0.02

(c) 3 or 0.8
Whole part for 3 is 3 and for 0.8 is 0.
∵ 3 >0
∴ 3 >0.8

(d) 5 or 0.05
Whole part for both is 0.
Tenth part for 0.5 is 5 and for 0.05 is 0.
∵ 5 >0
∴ 0.5 > 0.05

(e) 23 or 1.2
Whole part for both is 1.
Tenth part for both is 2.
Hundredth part for 1.23 is 3 and for 1.2 is 0 (as 1.2= 1.20)

(f) 0.099 or 0.19
Whole part for both is 0.
Tenth part for 0.099 is 0 and for 0.19 is 1.
∵ 1>0
∴ 0.19 >0.099

(g)1.5 or 1.50
Whole part for both is 1.
Tenth part for both is 5.
Hundredth part for both is 0 (as 1.5 = 1.50).
∴ 1.5=1.50

(h) 1431 or 490
Whole part for both is 1.
Tenth part for both is 4.
Hundredth part for 1.431 is 3 and for 1.490 is 9
∵ 9>3
∴ 1.490 >1.431

(i) 3 or 3.300
Whole part for both is 3.
Tenth part for both is 3.
Hundredth part for both is 0 (as 3.3 = 3.30).
Thousandth part for both is 0 (as 3.3 = 3.300).
∴ 3.3 = 3.300

(J) 5.64 or 5.603 
Whole part for both is 5.
Tenth part for both is 6.
Hundredth part for 5.64 is 4 and for 5.603 is 0.
∵ 4>0
∴ 5.64 > 5.603

Question 2.
Make five more examples and find the greater number from them.
Solution.
Five more examples are as follows:
Which is greater?
Ex. 1. 0.4 or 0.5
Ex. 2. 0.06 or 0.04
Ex. 3. 4 ox 0.9
Ex. 4. 0.6 or 0.06
Ex. 5. 0.063 or 0.22
Solution.
1.
0.4 or 0.5
Whole part for both = 0
Tenth part for 0.4 = 4
Tenth part for 0.5 = 5
∵ 5 >4
∴ 0.5 > 0.4

2.
0.06 or 0.04
Whole part for both = 0
Tenth part for both = 0
Hundredth part for 0.06 = 6
Hundredth part for 0.04 = 4
∵ 6 >4
∴ 0.06 > 0.04

3.
4 or 0.9
Whole part for 4 = 4
Whole part for 0.9 = 0
∵ 4 > 0
∴ 4 > 0.9

4.
0.6 or 0.06
Whole part for both = 0
Tenth part for 0.6 = 6
Tenth part for 0.06 = 0
∵  6 > 0
∴ 0.6 > 0.06

5.
0.063 or 0.22
Whole part for both = 0
Tenth part for 0.063 = 0
Tenth part for 0.22 = 2
∵  2 > 0
∴ 0.22 > 0.063.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3 help you. If you have any query regarding. NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3, drop a comment below and we will get back to you at the earliest.

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.

• Decimals Class 6 Ex 8.1
• Decimals Class 6 Ex 8.3
• Decimals Class 6 Ex 8.4
• Decimals Class 6 Ex 8.5
• Decimals Class 6 Ex 8.6

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.

	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.

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.

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
= \(\frac { 1 }{ 8 } \)

 

We hope the NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.2 help you. If you have any query regarding. NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.2, drop a comment below and we will get back to you at the earliest.

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.

• Fractions Class 6 Ex 7.1
• Fractions Class 6 Ex 7.2
• Fractions Class 6 Ex 7.3
• Fractions Class 6 Ex 7.4
• Fractions Class 6 Ex 7.5

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:

Solution :






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

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

Question 4.
Fill in the boxes:

Solution :

Question 5.
Complete the addition-subtraction box.


Solution :

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

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

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 :

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

We hope the NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.6 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.6, drop a comment below and we will get back to you at the earliest.

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.

• Fractions Class 6 Ex 7.1
• Fractions Class 6 Ex 7.2
• Fractions Class 6 Ex 7.3
• Fractions Class 6 Ex 7.4
• Fractions Class 6 Ex 7.6

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:

Solution :


Question 2.
Solve

Solution :

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 :

Hence, they painted together with the complete wall space. ,

Question 4.
Fill in the missing fractions :


Solution :


Question 5.
Javed was given \(\frac { 5 }{ 7 }\) of a basket of oranges. What fraction of oranges was left in the basket?
Solution :

Hence, fraction \(\frac { 2 }{ 7 }\) of oranges was left in the basket.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.5 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.5, drop a comment below and we will get back to you at the earliest.

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.

• Fractions Class 6 Ex 7.1
• Fractions Class 6 Ex 7.2
• Fractions Class 6 Ex 7.3
• Fractions Class 6 Ex 7.5
• Fractions Class 6 Ex 7.6

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)

(b)


appropriate signs between the fractions  given

Solution:

(i) In ascending order, these are

(ii) In descending order, these are

(b)

(i) In ascending order, these are

(ii) In descending order, these are

(c)

Question 2.
Compare the fractions and put an appropriate sign.

Solution :

Question 3.
Make five more such pairs and make appropriate signs.

Question 4.
Look at the figures and write ‘<’ or ‘>’, ‘=’ between the pairs of fractions.
Solution :


Make five more such problems and solve them with your friends.
Solution:

For the remaining part, please try yourself.

Question 5.
How quickly can you do this? Fill appropriate sign (<, =,>)

Solution :


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.

Solution :


Question 7.
Find answers to the following. Write and indicate how you solved them.

Solution :
(a) Equivalent fraction of \(\frac { 5 }{ 9 }\) are

Equivalent fraction of \(\frac { 4 }{ 5 }\) are

(b) Equivalent fraction of \(\frac { 9 }{ 16 }\) are

Question 8.
Ila reads 25 pages of a book containing 100 pages. Lalita reads \(\frac { 1 }{ 2 }\) of the same book. Who read less?

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 :

Hence, in both the classes the same fraction \(\left( \frac { 4 }{ 5 } \right)\)of total students got first class.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.4 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.4, drop a comment below and we will get back to you at the earliest.

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.

• Fractions Class 6 Ex 7.1
• Fractions Class 6 Ex 7.2
• Fractions Class 6 Ex 7.4
• Fractions Class 6 Ex 7.5
• Fractions Class 6 Ex 7.6

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)

(b)

Solution :
(a)


(b)

No ! all these fractions ae not equivalent.

Question 2.
Write the fractions and pair up the equivalent fractions from each row.

Solution:


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:

Solution :



Question 4.
Find the equivalent fraction of \(\frac { 3 }{ 5 }\) having
(a) denominator 20
(b) numerator 9
(d) numerator 27
(c) denominator 30
Solution :


Question 5.
Find the equivalent fraction of \(\frac { 36 }{ 48 }\) with
(a) numerator 9
(b) denominator 4.
Solution :


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 :

∴ The given fractions \(\frac { 5 }{ 9 }\) and \(\frac { 30 }{ 54 }\) are equivalent.

∴ The given fractions \(\frac { 3 }{ 10 }\) and \(\frac { 12 }{ 50 }\) are not equivalent.

∴ The given fractions \(\frac { 7 }{ 3 }\) and \(\frac { 5 }{ 11 }\) are not equivalent.

Question 7.
Reduce the following fractions to simplest form:

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.

(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.

(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.

(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.

(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.

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

For Sheelu
Number of pencils she had = 50 Number of pencils used by her = 25

For Jamaal
Number of pencils he had = 80 Number of pencils used by him = 40

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 :

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.

(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.

(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.

(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.

(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.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.3, drop a comment below and we will get back to you at the earliest.

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.

• Fractions Class 6 Ex 7.1
• Fractions Class 6 Ex 7.3
• Fractions Class 6 Ex 7.4
• Fractions Class 6 Ex 7.5
• Fractions Class 6 Ex 7.6

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:

Solution :
(a)

(b)

(c)

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 :

Question 3.
Express the following as improper fractions:

Solution :

 

We hope the NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3 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.

• Integers Class 6 Ex 6.1
• Integers Class 6 Ex 6.2

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 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.

• Integers Class 6 Ex 6.1
• Integers Class 6 Ex 6.3

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:

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:

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:

(d) 3 less than – 2
We start from – 2 and move to the left by 3 steps and obtain – 5 as shown below:

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.

(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.

(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.

(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.

(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.

(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.

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 8 Maths Chapter 16 Playing with Numbers Ex 16.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2.

• Playing with Numbers Class 8 Ex 16.1

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 16
Chapter Name	Playing with Numbers
Exercise	Ex 16.2
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2

Question 1.
If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Solution.
Since 21y5 is a multiple of 9, its sum of digits 2 + 1+ y + 5 = 8+ y isa multiple of 9; so 8 + y is one of these numbers: 0, 9, 18, 27, 36, 45,… .
But since y is a digit, it can only be possible that 8 + y = 9. Therefore, y = 1.

Question 2.
If 31z5 is a multiple of 9, where z is a digit, what is the value of z?
You will find that there are two answers to the last problem. Why is this so?
Solution.
Since 31z5 is a multiple of 9, its sum of digits 3 + 1 + z + 5 = 9 + z isa multiple of 9; so 9 + z is one of these numbers: 0, 9, 18, 27, 36, 45, … .
But since z is a digit, it can only be possible that 9 + z = 9 or 18. Therefore, z = 0 or 9.

Question 3.
If 24x is a multiple of 3, where x is a digit, what is the value of x?
(Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18, … .
But since xis a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.)
Solution.
The solution is given with question.

Question 4.
If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution.
Since 31z5 is a multiple of 3, its sum of digits 3 + 1 + z + 5 = 9 + z is a multiple of 3; so 9 + z is one of these numbers: 0, 3, 6, 9, 12, 15, 18, … .
But since z is a digit, it can only be possible that 9 + z = 9 or 12 or 15 or 18. Therefore, z = 0 or 3 or 6 or 9.

We hope the NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2, drop a comment below and we will get back to you at the earliest.