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Class 6

RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1

June 3, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1

RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1 is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 6 Decimal Numbers Exercise 6.1.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 6
Subject Maths
Chapter Chapter 6
Chapter Name Decimal Numbers
Exercise Ex 6.1
Number of Questions 10
Category RBSE Solutions

Rajasthan Board RBSE Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1

Question 1.
Write the numbers for the following (RBSESolutions.com) in the table given below :
(i) 1 tens 2 unit 3 one-tenths
(ii) 1 hundred 3 tens 7 one-tenths
(iii) 2 hundred 5 tens 1 unit 2 one-tenths
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1 image 1
Solution.
On writting numbers in the table :
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1 image 2

RBSE Solutions

Question 2.
Write the place value of following (RBSESolutions.com) decimal numbers in a table.
(i) 19.4
(ii) 0.5
(iii) 10.9
(iv) 205.9
Solution.
On writing the place value of decimal numbers in a table :
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1 image 3

What is 11/16 as a decimal you ask? Converting the fraction 11/16 into a decimal is very easy.

Question 3.
Write each of the following (RBSESolutions.com) in the form of decimal :
(i) 7 one-tenths
(ii) 2 tens 4 one-tenths
(iii) Fourteen decimal nine
(iv) Six hundred point three
Solution.
On writing each in decimal form :
(i) 7 one-tenths = \(\frac { 7 }{ 10 } \) = 0.7
(ii) 2 tens 4 one-tenths 2 tens = 2 × 10 = 20
4 one-tenths = \(\frac { 4 }{ 10 } \) = 0.4
∴ 2 tens 4 one-tenths = 20.4
(iii) Fourteen point Nine = 14.9
(iv) Six hundred point three = 600.3

Question 4.
Represent the following in (RBSESolutions.com) form of decimal fractions :
(i) \(\frac { 3 }{ 10 } \)
(ii) 4 + \(\frac { 8 }{ 10 } \)
(iii) 300 + 50 + 8 + \(\frac { 1 }{ 10 } \)
(iv) 90 + \(\frac { 3 }{ 10 } \)
(v) \(\frac { 3 }{ 2 } \)
(vi) \(\frac { 2 }{ 5 } \)
(vii) 4 \(\frac { 1 }{ 2 } \)
(viii) 3 \(\frac { 3 }{ 5 } \)
Solution.
On representing each in form (RBSESolutions.com) of decimal fractions :
(i) \(\frac { 3 }{ 10 } \) = 0.3
(ii) 4 + \(\frac { 8 }{ 10 } \) = 4 + 0.8 = 4.8
(iii) 300 + 50 + 8 + \(\frac { 1 }{ 10 } \) = 358 + 0.1 = 358.1
(iv) 90 + \(\frac { 3 }{ 10 } \) = 90 + 0.3 = 90.3
(v) \(\frac { 3 }{ 2 } \) for making its denominator 10, multiply its numerator and denominator by 5 :
\(\frac { 3 }{ 2 } \) × \(\frac { 5 }{ 5 } \) = \(\frac { 15 }{ 10 } \) = 1.5
(vi) \(\frac { 2 }{ 5 } \) for making its denominator 10, multiply its numerator and denominator by 2 :
\(\frac { 2 }{ 5 } \) × \(\frac { 2 }{ 2 } \) = \(\frac { 4 }{ 10 } \) = 0.4
(vii) 4 \(\frac { 1 }{ 2 } \) = \(\frac { 9 }{ 2 } \)
for making its denominator 10, multiply its numerator and denominator by 5 :
\(\frac { 9 }{ 2 } \) × \(\frac { 5 }{ 5 } \) = \(\frac { 45 }{ 10 } \) = 4.5
(viii) 3 \(\frac { 3 }{ 5 } \) = \(\frac { 18 }{ 5 } \)
for making its denominator 10, multiply its numerator and denominator by 2 :
\(\frac { 18 }{ 5 } \) × \(\frac { 2 }{ 2 } \) = \(\frac { 36 }{ 10 } \) = 3.6

Question 5.
Write fractions for the following decimal (RBSESolutions.com) numbers and convert in the simplest form.
(i) 0.6
(ii) 2.5
(iii) 2.8
(iv) 13.7
(v) 21.2
(vi) 1.0
(vii) 6.4
Solution.
On writing decimal numbers as fractions and (RBSESolutions.com) converting them in simplest form :
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1 image 4

RBSE Solutions

Question 6.
Using centimeters convert the (RBSESolutions.com) following into decimal form :
(i) 2 mm
(ii) 30 mm
(iii) 116 mm
(iv) 5 cm 2 mm
(v) 95 mm
(vi) 19 cm 1 mm
Solution.
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1 image 5

Question 7.
On the number line between which two whole (RBSESolutions.com) numbers, the following numbers are marked? Which whole number of these is closer to the decimal number?
(i) 0.5
(ii) 5.3
(iii) 9.0
(iv) 4.9
(v) 3.8
Solution.
On checking given each decimal number on number line :
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1 image 6
∵ 0.5 is equally near from 0 and 1, but we let it closer to 1, because (RBSESolutions.com) when last point of number is 5 or more than 5, then it will be closer to next number.
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1 image 7

Questions 8.
Show the following (RBSESolutions.com) on number line.
(i) 0.3
(ii) 1.7
(iii) 3.4
(iv) 2.5
Solution.
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1 image 8

RBSE Solutions

Question 9.
The length of Tulsi’s hand grip is 95 m.m. Write (RBSESolutions.com) it in cm form.
Solution.
The length of Tulsi’s hand grip = 95 mm
∵ 10 mm = 1 cm
∴ 95 mm = \(\frac { 1 }{ 10 } \) × 95 cm = \(\frac { 95 }{ 10 } \) = 9.5 cm
Thus, length of Tulsi’s hand grip is 9.5 cm

Question 10.
Deepu has a 6 cm. scale. It has been broken at 4.4 cm. What is the length of the left piece of scale?
Solution.
Total length of scale = 6 cm
Length of broken scale = 4.4 cm Thus, length (RBSESolutions.com) of remaining scale = (6 – 4.4) cm = (6.0 – 4.4) cm = 1.6 cm

We hope the RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Ex 6.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 6 Maths Chapter 6 Decimal Numbers Exercise 6.1, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise

June 3, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 6
Subject Maths
Chapter Chapter 6
Chapter Name Decimal Numbers
Exercise In Text Exercise
Number of Questions 11
Category RBSE Solutions

Rajasthan Board RBSE Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise

Page No. 75

Question 1.
You too measure pencil, rubber and other things (RBSESolutions.com) from your bag by scale and fill in the table.
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 1
Solution.
Fill the table as below –
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 2
Note : Students measure their own objects.

(Page No. 76)

RBSE Solutions

Question 2.
Write the place value of digits (RBSESolutions.com) in the following number.
(i) 123.4
(ii) 111
Solution.
(i) Place value of 1 = 1 × 100 = 100
Place value of 2 = 2 × 10 = 20
Place value of 3 = 3 × 1 = 3
Place value of 4 = 4 × \(\frac { 1 }{ 10 } \) = 0.4

(ii) Place value of 1 = 1 × 100 = 100
Place value of 1 = 1 × 10 = 10
Place value of 1 = 1 × 1 = 1

ROUND returns n rounded to integer places to the right of the decimal point. … The following example rounds a number to one decimal point.

Page No. 76

Question 1.
Show the decimal (RBSESolutions.com) numbers 0.6, 1.3 and 2.5 on number line.
Solution.
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 3

Page No. 81

Question 1
Which number is greater in the following?
(i) 3.07 and 3.89
(ii) 0.57 and 0.05
(iii) 147.8 and 147.08
(iv) 9.5 and 5.92
Solution.
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 4
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 5
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 6

Page No. 82

Question 1.
Add the (RBSESolutions.com) following.
(i) 1.54 + 1.80
(ii) 2.75 + 0.08
Solution.
(i) 1.54 + 1.80
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 7
(ii) 2.75 + 0.08
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 8

RBSE Solutions

Here we will write the mixed fraction 1 1/2 as a decimal.

Question 2.
(i) Subtract (RBSESolutions.com) 1.67 from 5.47.
(ii) Subtract 4.07 from 8.90.
Solution.
(i) 5.47 – 1.65
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 9
(ii) 8.90 – 4.07
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 10

Page No. 74

Question 1.
Say how much is length (RBSESolutions.com) of the Ram’s pencil?
Solution.
Length of pencil in figure is more than 6 cm and less than 7 cm. Now, on making 10 equal parts of 1 cm, we find that length of pencil is 1 part more them 6 cm. Thus, length of pencil is 6.1 cm.
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 11

Question 2.
How much is the length of Rehman’s compass?
Solution.
Length of compass in figure is more than 5 cm and less than 6 cm. Now, on making 10 equal parts of 1 cm, we find that length of compass is 4 part more than 5 cm. Thus, length of compass is 5.4 cm.
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 12

Page No. 75

Question 1.
Find the place value of (RBSESolutions.com) digits of number 523.
Place value of 5 = …………………………….
Place value of 2 = …………………………….
Place value of 3 = …………………………….
Solution.
Following are the place values of digits of number 523.
Place value of 5 = 5 × 100 = 500
Place value of 2 = 2 × 10 = 20
Place value of 3 = 3 × 1 = 3

Question 2.
Fill the following (RBSESolutions.com) table.
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 13
Solution.
Completing the table :
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise image 14

RBSE Solutions

Page No. 79

Question 1.
Write the following (RBSESolutions.com) numbers in words :
1. 45.36 cm =
2. ₹ 325.25 =
Solution.
On writing the given decimal number in words :
1. 45.36 cm – Forty five decimal three six c.m.
2. ₹ 325.25 – Three hundred twenty five decimal two five

We hope the RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 6 Maths Chapter 6 Decimal Numbers In Text Exercise, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 6 Maths Chapter 1 Know the Numbers Ex 1.3

June 2, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 6 Maths Chapter 1 Know the Numbers Ex 1.3 is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 1 Know the Numbers Exercise 1.3.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 6
Subject Maths
Chapter Chapter 1
Chapter Name Know the Numbers
Exercise Ex 1.3
Number of Questions 5
Category RBSE Solutions

Rajasthan Board RBSE Class 6 Maths Chapter 1 Know the Numbers Ex 1.3

Question 1.
Replace each number to the (RBSESolutions.com) nearest hundred of the following and calculate the answer again in the nearest hundred.
(i) 247 + 691
(ii) 4316 + 1567
(iii) 7122 – 3565
(iv) 4543 – 2036
Solution.
(i) 247 + 691
Tens place digit in 247 = 4 < 5 Rounding off 247 to nearest hundred = 200
Tens place digit in number 691 = 9 > 5
Digit of hundred =6 + 1 = 7
Roudning off 691 to nearest = 700
Thus, 200 + 700 = 900

RBSE Solutions

(ii) 4316 + 1567
Tens place digit in 4316 = 1 < 5 Rounding off 4316 to (RBSESolutions.com) nearest hundred = 4300
Tens place digit in 1567 = 6 > 5
Digit of hundred =5 + 1=6
Rounding off 1567 to nearest hundred = 1600
Thus, 4300 + 1600 = 5900

(iii) 7122 – 3565
Tens digit in 7122 = 2 < 5 Rounding off 7122 to nearest hundred = 7100
Tens place digit in 3565 = 6 > 5
Digit of hundred =5 + 1 = 6 Rounding off 3565 to nearest hundred = 3600
Thus, 7100 – 3600 = 3500

(iv) 4543 – 2036
Tens place digit in 4543 = 4 < 5 Rounding off 4543 to nearest (RBSESolutions.com) hundred = 4500
Rounding off 2036 to nearest hundred = 3 < 5
Rounding off 2036 to nearest hundred = 2000
Thus, 4500 – 2000 = 2500

Question 2.
Multiply the nearest tens numbers of the following :
(i) 34 × 57
(ii) 294 × 72
(iii) 869 × 675
Solution.
(i) 34 × 57
Rounding off 34 to (RBSESolutions.com) nearest tens = 30
Rounding off 57 to nearest tens = 60
Thus, 30 × 60 = 1,800

(ii) 294 x 72
Rounding off 294 to nearest tens = 290
Rounding off 72 to nearest tens = 70
Thus, 290 × 70 = 20,300

(iii) 869 × 675
Rouding off 869 to nearest tens = 870
Rounding off 675 to nearest tens = 680
Thus, 870 × 680 = 5,91,600

Rounding to the Nearest Hundredth … The hundredth place is the second number after the decimal.

Question 3.
In the school library, there are 2541 books of stories, 1017 subject books (RBSESolutions.com) and other books are 857. Find out the approximate number of books in the school. (Rounding the number to 100).
Solution.
Number of story books = 2541
Round off 2541 = 2500
Number of subject books = 1017
Round off 1017 = 1000
Other books = 857
Round off 857 to nearest hundred = 900
Thus, total books = 2500 + 1000 + 900 = 4400

Question 4.
8596 cows and 7015 buffaloes are there (RBSESolutions.com) in a village. Find out which cattle is more than the other and how much ? (Rounding the number to 100).
Solution.
Number of cows = 8596
Round off 8596 to nearest hundred = 8600
Number of buffaloes = 7015
Round off 7015 to nearest hundred = 7000
Difference = 8600 – 7000 = 1600
Thus, cows are more than others, approximately 1600.

Question 5.
A car runs 15 kilometer with 1 litre petrol. How (RBSESolutions.com) much petrol does it need to go 100 kms ? (Find out the value rounding the number to 10).
Solution.
Petrol need to ran 15 litre = 1 litre
Petrol need to run 1 km = \(\frac { 1 }{ 15 } \) litre
Petrol need to run 100 km
\(=\frac { 1\times 100 }{ 15 } =6.67=7\) litre (approx)

RBSE Solutions

We hope the RBSE Solutions for Class 6 Maths Chapter 1 Know the Numbers Ex 1.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 6 Maths Chapter 1 Know the Numbers Exercise 1.3, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers Additional Questions

June 1, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers Additional Questions

RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers Additional Questions is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 2 Relation Among Numbers Additional Questions.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 6
Subject Maths
Chapter Chapter 2
Chapter Name Relation Among Numbers
Exercise Additional Questions
Number of Questions 17
Category RBSE Solutions

Rajasthan Board RBSE Class 6 Maths Chapter 2 Relation Among Numbers Additional Questions

Multiple Choice Questions

Question 1.
Factors of number 24 are :
(i) 1, 2, 3
(ii) 4, 6
(iii) 12, 24
(iv) All above

Question 2.
Number of prime (RBSESolutions.com) numbers between 1 to 100 are :
(i) 24
(ii) 25
(iii) 26
(iv) 27

RBSE Solutions

Question 3.
Number which is not divisible by 2 :
(i) 267
(ii) 468
(iii) 192
(iv) 374

Question 4.
First three multiples of 15 are :
(i) 15, 30, 60
(ii) 15, 45, 60
(iii) 15, 30, 45
(iv) 30, 45, 60

Question 5.
H.C.F. of 14 and 24 is :
(i) 7
(ii) 3
(iii) 2
(iv) 6.

Question 6.
L.C.M. of 40, 15, 20 is :
(i) 120
(ii) 100
(iii) 140
(iv) 160

The Postive Factors of 50 are therefore all the numbers we used to divide (divisors) above to get an even number.

Question 7.
Common (RBSESolutions.com) factor of 18, 27 is :
(i) 2
(ii) 4
(iii) 6
(iv) 9

Question 8.
Prime number in the following is :
(i) 6
(ii) 7
(iii) 9
(iv) 10

Ans.
1. (iv)
2. (ii)
3. (i)
4. (iii)
5. (iii)
6. (i)
7. (iv)
8. (ii)

RBSE Solutions

Prime Factorization of 38 with a Factor Tree ·

Fill in the blanks

(i) Reverse process of factors is called ………………..
(ii) The numbers having more than two factors are called ………………..
(iii) ……………….. is even prime number.
(iv) The numbers completely divisible by 2 are called.
(v) Multiples of same multiple numbers are called ………………..
(vi) Smallest odd (RBSESolutions.com) prime number is ………………..

Answer.
(i) expansion
(ii) composite
(iii) 2
(iv) even
(v) common factor
(vi) 3

Very Short Answer Type Questions

Question 1.
Write all factors of 38.
Solution.
Factors of 38
38 = 1 × 38
38 = 2 × 19
38 = 19 × 2
∴ Thus, all factors of 38 are 1,2,19 and 38.

Question 2.
Write first seven (RBSESolutions.com) multiples of 4.
Solution.
First seven multiples of 4 are :
4 × 1 = 4
4 × 2 = 8
4 × 3 = 12
4 × 4 = 16
4 × 5 = 20
4 × 6 = 24
4 × 7 = 28
Thus, first seven multiples of 4 are 4,8,12,16, 20, 24 and 28.

Question 3.
Write all prime numbers between 1 and 50.
Solution.
Prime numbers (RBSESolutions.com) between 1 and 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.

Question 4.
Find common factors of 15, 25, 35 and 45.
Solution.
Finding common factor of 15, 25,35 and 45,
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers Additional Questions image 1
∴ Common factors of 15, 25, 35 and 45 = 1, 5

RBSE Solutions

Short/Long Answer Type Questions

Question 1.
Test divisibility (RBSESolutions.com) of 585 by 3, without division.
Solution.
Test divisibility by 3
Sum of digits in number =5 + 8 + 5 = 18 ⇒ 1 + 8 = 9
Which is divisible by 3.
Thus, 585 will be divisible by 3.

Question 2.
Test divisibility of 7640 by 8, without division.
Solution.
Test divisibility by 8
In number. 7640 number formed by digits of hundred tens and ones place = 640
Dividing 640 by 8
640 ÷ 8 = 80
Which is divisible by 8.
Thus, 7640 will be divisible by 8.

Question 3.
Three containers contains 26 l, 65 litre 117 l milk respectively. Find maximum (RBSESolutions.com) measurement of container milk of all three containers.
Solution.
Milk contains in 1st container= 26 l
Milk contains in 2nd container =65 l
Milk contains in 3rd container = 117 l
Factors of 26 = 2 × (13)
Factors of 65 = 5 × (13)
Factors of 117 = 3 × 3 × (13)
∴ H.C.M. of 26, 65 and 117 = 13
Thus, required maximum measurement of container will be 13 l.

Question 4.
Find the smallest number (RBSESolutions.com) which is completely divisible by 18, 36, 50.
Solution.
Finding L.C.M. 18, 36 and 50
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers Additional Questions image 2

RBSE Solutions

We hope the RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 6 Maths Chapter 2 Relation Among Numbers Additional Questions, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise

June 1, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 6
Subject Maths
Chapter Chapter 2
Chapter Name Relation Among Numbers
Exercise In Text Exercise
Number of Questions 31
Category RBSE Solutions

Rajasthan Board RBSE Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise

Pg. No. 20

Question 1.
In the following table, write the (RBSESolutions.com) factors against the numbers given.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 1
Solution.
Writting factors of numbers in table –
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 2

RBSE Solutions

Question 2.
From the above table, (RBSESolutions.com) can you say that ‘1’ is the factor of every number?
Solution.
Yes, we can say that ‘1’ is the factor of every number.

Pg. No. 22

Question 1.
Write even and odd numbers separately.
(i) 357
(ii) 436
(iii) 77
(iv) 1900
(v) 5001
Even numbrs …………….. Odd numbers ……………..
Solution.
Even numbers are (ii) 436, (iv) 1900 Odd numbers are (i) 357, (iii) 77, (v) 5001

Pg. No. 24

Question 1.
Do all the (RBSESolutions.com) numbers which have 0 and 5 at their units place, have 5 as one of their factors?
Solution.
Yes, numbers which have 0 and 5 at their units place, have 5 as one of their factor.

Question 2.
Are all these numbers divisible by 5?
Solution.
Yes, all these numbers are divisible by 5.

Question 3.
Does any number not having 0 or 5 at its units place, have 5 as a factor?
Solution.
No, if any number not having 0 or 5 at its unit place, they have not 5 as a factor.

Pg. No. 25

Question 1.
In 3672 sum of digits is 3 + 6 + 7 + 2 = 18, is it (RBSESolutions.com) divisible by 9? Solve 3672 + 9.
Solution.
In number 3672 sum of digits 3 + 6 + 7 + 2 = 18,
which is divisible by 9.
Thus, 3672 is divisible by 9.
∴ 3672 ÷ 9 = 408

RBSE Solutions

Pg. No. 26

Question 1.
Find out the divisibility by 6 for the numbers 336, 123, 1002, 4236.
Solution.
Testing for divisibility by 6 for the following numbers 336, 123, 1002, 4236.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 3

Pg. No. 29

Question 1.
Raju’s cow gives 15 litres and buffalo gives 20 litres milk. Find (RBSESolutions.com) out the maximum measurement for measuring pot for both type of milk exactly.
Solution.
Factors of 15 = 3 × (5)
Factors of 20 = 2 × 2 × (5)
∴ H.C.F. of 15 and 20 = 5
Thus, required measurement will be 5 l.

Question 2.
Find out H.C.F. by Vedic method.
(i) 8, 12
(ii) 38, 57
(iii) 117, 195
(iv) 99,165, 231
Solution.
Finding H.C.F. by vedic method,

(i) 8, 12
First (RBSESolutions.com) difference = 12 – 8 = 4
Thus, possible H.C.F. = 4
Second difference =8 – 4 = 4
∵ First difference = Second difference
∴ H.C.F. of 8 and 12 =4

(ii) 38, 57
First difference = 57 – 38 = 19
Thus, possible H.C.F. = 19
Second difference = 38 – 19 = 19
∵ First difference = Second difference
∴ H.C.F of 38 and 57 = 19

(iii) 117,195
First difference = 195 – 117 = 78
Thus, possible H.C.F = 78
Second difference = 117 – 78 = 39
Thus, possible H.C.F = 39
Third difference = 78 – 39 = 39
∵ Second difference = Third difference
∴ HCF 117 and 195 = 39

(iv) 99, 165, 231
Addition of two (RBSESolutions.com) numbrs = 99 + 231 = 330
First difference = 99 + 231 – 165 = 165
Thus, possible H.C.F = 165
Second difference = 231 – 165 = 66
Thus, possible H.C.F = 66
Third difference = 99 – 66 = 33
Thus, possible H.C.F = 33
∵ Possible H.C.F is a factor of 66.
∴ HCF of 99, 165 and 231 = 33

RBSE Solutions

Pg. No. 30

Question 1.
Two bells start ringing together. First bell rings after every three minutes and second bell rings after every five minutes then after how much time both bells will ring together?
Solution.
Since, first bell rings after every three minutes and second bell rings after every five minutes.
∴ L.C.M. of 3 and 5 = 3 × 5 = 15 Thus, both bells will ring together, after 15 minutes.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 4

Pg. No. 32

365-838 registered under.

Question 1.
Find out the LCM of 48, 64 and 80 by (RBSESolutions.com) division method.
Solution.
To find L.C.M by division method
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 5

Question 2.
Find out the L.C.M of 24 and 30 by Vedic method.
Solution.
Finding L.C.M of 24 and 30 by Vedic (RBSESolutions.com) method :
Step 1. 24 and 30 is written as \(\frac { 24 }{ 30 } \) in fraction form.
Step 2. Prime factorising of 24 and 30
\(\frac { 24 }{ 30 } \) = \(\frac { 2\times 2\times 2\times 3 }{ 2\times 3\times 5 } \)
Step 3. Common in numerator and denominator
\(\frac { 24 }{ 30 } \) = \(\frac { 2\times 2 }{ 5 } \) = \(\frac { 4 }{ 5 } \)
Step 4. By cross multiplication 24 × 5 = 30 × 4 = 120
Thus required L.C.M. = 120

The lcm of 15 and 20 is 60. … Correct Devanshu.

Pg. No. 19

Question 1.
Can we say that all numbers which (RBSESolutions.com) perfectly divides 16 are factors of 16 ?
Solution.
Yes, it is true. As 1, 2, 4, 8, 16 are factors of 16.

RBSE Solutions

Pg. No. 20-21

Question 1.
Look at the factors of the numbers given below :
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 6
See the table and find out which numbers have only two factors?
Solution.
There are four numbers (2, 3, 5, 7) in table which have only two factors.

Number Game
Let us play a game where we can tell if a (RBSESolutions.com) number is prime or not without factorization. First write the numbers 1-100 as shown below :
Step 1. Make a box RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 7 on the number 1 as it is neither prime nor non-prime.
Step 2. Encircle number 2, and cross all its multiples such as 4, 6, 8 etc. (except 2).
Step 3, The next number not crossed is 3. Encircle 3 and cross all its remaining multiples.
Step 4. Continue this process until all numbers have either been encircled or crossed. All encircled numbers are prime numbers.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 8
Sol.
Writting numbers stepwise :
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 9

Question 2.
How many (RBSESolutions.com) prime numbers did you get between 1 – 100 ?
Solution.
25 prime numbers are obtained.

Question 3.
Write these numbers in sequence.
Solution.
In sequence, these numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

Pg. No. 21

Odd-Even Numbers
Kanak and Pritam were playing marbles.
Kanak : Look Pritam, let me teach you a game. (RBSESolutions.com) Take any number of marbles in your hand and close your fist. Now, I have to tell if the marbles in your hand are in pairs or not. This game is also called Eki or Beki. Eki means that when you make pairs of marbles in your hand and if one marble is left without a pair, then it is Eki, if all marbles are in pairs, then it is called Beki. Kanak and Pritam played this game and wrote it in a table.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 10

Question 1.
Play this game with your friends and decide which numbers should be called Eki and which ones Beki? Were you able to frame any rule?
Solution.
Yes, we have played this game with our friends and find a rule. Numbers with 2, 4, 6, 8, 0«in units place are known as even numbers If 1, 3, 5, 7, 9 are in units place, then the numbers are known as odd numbers.

RBSE Solutions

Pg. No. 23

Question 1.
Can you say all even numbers are divisible by 2.
Solution.
Yes, we can say all even numbers are divisible by 2.

Question 2.
Write (RBSESolutions.com) factors of 24, 15, 26, 48,13, 11.
Solution.
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 15 = 1, 3, 5, 15
Factors of 26 = 1, 2, 13, 26 Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Factors of 13 = 1, 13
Factors of 11 = 1, 11

Question 3.
Write the unit digit of those numbers whose one factor is 2.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 11
Solution.
2 is factor of which numbers, the (RBSESolutions.com) number at their unit place are : 0, 2, 4, 6 and 8.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 12

Question 4.
Write some more numbers in the table. Do you find any pattern in the numbers divisible by 10 at its units place?
Solution.
Completing the table
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 13
All those numbers which have 0 at units place or which have 10 as one of their factors are divisible by 10.

Pg. No. 24

Question 1.
Write all the (RBSESolutions.com) factors of given table.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 14
See unit place of all the numbers which have 5 as one factor.
Solution.
Completing the table –
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 15
All the numbers which has unit digits 0 or 5, their one factor is 5.

Question 2.
Complete the table on (RBSESolutions.com) the basis of given instructions.
Teacher will arrange a game in the class.
1. Think about a number.
2. Add the digits of that number.
3. Divide the sum by 3.
4. Was it perfectly divided?
5. Divide the number by 3 directly.
6. Could it be divided perfectly.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 16
Solution.
Completing the table as given –
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 17

RBSE Solutions

Pg. No. 25

Question 1.
Fill the (RBSESolutions.com) table.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 18
Can you find any pattern for the divisibility by 9?
Solution.
Completing the table
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 19
We find a pattern from this table if sum of digits of any number is divisible by 9, then that number is also divisible by 9.

Pg.No.26

Question 1.
Write some more numbers in (RBSESolutions.com) table and complete the table.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 20
Can you see any pattern for divisibility by 6 ?
Solution.
Completing the table
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 21
From table, we see pattern for divisibility by 6. If any number is divisible by 2 and 3 separately, then it is also divisible by 6.

Question 2.
Take some numbers for (RBSESolutions.com) divisibility by 4 and test the pattern.
Solution.
If a number has its last two digits divisible by 4 or if its tens and units digits are 0, then if is divisible by 4.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 22

Question 3.
Meena took one number 9212, its last two digits are 12 which is divisible by 4. You try and divide it.
Solution.
There are 12 tens and unit place digits of (RBSESolutions.com) number 9212 which is divisible by 4. So, given number will be divisible by 4 completely.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 23

Question 4.
For divisibility by 8, take some numbers and test the pattern in table.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 24
Solution.
If the number framed by last three digits i.e., units, (RBSESolutions.com) tens and hundred is divisible by 8 or if any number has 0 as its units, tens and nundreds digits, then the number is divisible by 8.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 25

RBSE Solutions

Pg. No. 27

Question 1.
Fill the table and find which numbers are divisible by 11 ?
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 26
Solution.
Completing the table.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 27
From given table we define a rule for divisibility by 11. AH those numbers which have the difference between sum of it odd place digits and even place digits as zero (O) or multiple of 11 is divisible by 11.

Question 2.
What are the mu1tiple (RBSESolutions.com) of 3 and 4? Circle them.
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 28
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 29
RBSE Solutions for Class 6 Maths Chapter 2 Relation Among Numbers In Text Exercise image 30
Thus, multiples (RBSESolutions.com) of 3 and 4 are 12, 24, 36, 48, 60, 72, 84.

RBSE Solutions

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