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

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.

Solution.
Writting factors of numbers in table –

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

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.

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

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.

Pg. No. 32

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

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

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.

Pg. No. 20-21

Question 1.
Look at the factors of the numbers given below :

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  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.

Sol.
Writting numbers stepwise :

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.

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.

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.

Solution.
2 is factor of which numbers, the (RBSESolutions.com) number at their unit place are : 0, 2, 4, 6 and 8.

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

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.

See unit place of all the numbers which have 5 as one factor.
Solution.
Completing the table –

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.
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.

Solution.
Completing the table as given –

Pg. No. 25

Question 1.
Fill the (RBSESolutions.com) table.

Can you find any pattern for the divisibility by 9?
Solution.
Completing the table

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.

Can you see any pattern for divisibility by 6 ?
Solution.
Completing the table

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.

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.

Question 4.
For divisibility by 8, take some numbers and test the pattern in table.

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.

Pg. No. 27

Question 1.
Fill the table and find which numbers are divisible by 11 ?

Solution.
Completing the table.

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.

Thus, multiples (RBSESolutions.com) of 3 and 4 are 12, 24, 36, 48, 60, 72, 84.

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