RBSE Solutions Class 6 Maths Chapter 2 Relation Among Numbers More Ques

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 factors against the numbers given.
Answer:

In simple words: To find the factors of a number, list all the numbers that can divide it exactly without leaving any remainder.

🎯 Exam Tip: When listing factors, always start with 1 and end with the number itself. Remember that factors come in pairs.

 

Question 2. From the above table, can you say that '1' is the factor of every number?
Answer: Yes, we can confirm that 1 is a factor of every number. This is a fundamental property of numbers, as any number can be divided by 1 without a remainder.
In simple words: Yes, the number 1 can divide any other number perfectly, so it is a factor of every number.

🎯 Exam Tip: Always remember that 1 is a universal factor; it divides all integers.

Pg. No. 24

 

Question 1. Do all the numbers which have 0 and 5 at their units place, have 5 as one of their factors?
Answer: Yes, it is true that all numbers ending with 0 or 5 in their units place have 5 as one of their factors. This is a quick divisibility rule for the number 5.
In simple words: Yes, if a number ends in 0 or 5, you can always divide it by 5.

🎯 Exam Tip: Knowing divisibility rules helps you quickly check if a number can be divided by another without performing long division.

 

Question 2. Are all these numbers divisible by 5?
Answer: Yes, all numbers that end in either 0 or 5 are completely divisible by 5. This is a key divisibility test.
In simple words: If a number ends with 0 or 5, then it can always be divided by 5 without any leftover.

🎯 Exam Tip: Use the divisibility rule for 5 to save time in calculations: simply check the last digit.

 

Question 3. Does any number not having 0 or 5 at its units place, have 5 as a factor?
Answer: No, if a number does not have 0 or 5 in its units place, then 5 will not be a factor of that number. This means it will not be perfectly divisible by 5.
In simple words: No, if a number does not end in 0 or 5, you cannot divide it evenly by 5.

🎯 Exam Tip: Always check the last digit first when testing for divisibility by 5; it's the fastest way to tell.

Pg. No. 25

 

Question 1. In 3672 sum of digits is \( 3 + 6 + 7 + 2 = 18 \), is it divisible by 9? Solve \( 3672 \div 9 \).
Answer: For the number 3672, the sum of its digits is \( 3 + 6 + 7 + 2 = 18 \). Since 18 is divisible by 9, the number 3672 is also divisible by 9. When we divide 3672 by 9, the result is 408.
\( 3672 \div 9 = 408 \)
In simple words: Add up all the digits of the number. If this sum can be divided by 9, then the whole number can also be divided by 9.

🎯 Exam Tip: The divisibility rule for 9 is very useful: sum the digits; if the sum is a multiple of 9, the number is divisible by 9.

 

Question 1. Find out the divisibility by 6 for the numbers 336, 123, 1002, 4236.
Answer: To find divisibility by 6, we check if a number is divisible by both 2 and 3. Here is the test for the given numbers:

In simple words: A number can be divided by 6 if it can be divided by 2 (meaning it's an even number) AND it can also be divided by 3 (meaning its digits add up to a number divisible by 3).

🎯 Exam Tip: For divisibility by 6, both conditions (divisible by 2 and divisible by 3) must be met; one condition alone is not enough.

Pg. No. 29

 

Question 1. Raju's cow gives 15 litres and buffalo gives 20 litres milk. Find out the maximum measurement for measuring pot for both type of milk exactly.
Answer: To find the maximum measurement for a pot that can measure both quantities of milk exactly, we need to find the Highest Common Factor (H.C.F.) of 15 and 20.
Factors of 15 are: 1, 3, 5, 15.
Factors of 20 are: 1, 2, 4, 5, 10, 20.
The common factors are 1 and 5.
The highest common factor is 5.
Therefore, the required measurement of the pot will be 5 litres. This pot can measure 15 litres (three times) and 20 litres (four times) exactly.
In simple words: We need to find the biggest number that can divide both 15 and 20 without any leftovers. This number is 5, so a 5-litre pot is needed.

🎯 Exam Tip: When a question asks for the 'maximum' or 'largest' quantity that can measure given items exactly, it's a strong indicator that you need to find the Highest Common Factor (H.C.F.).

 

Question 2. Find out H.C.F. by Vedic method.
(i) 8, 12
(ii) 38, 57
(iii) 117, 195
(iv) 99, 165, 231
Answer:
(i) For 8 and 12:
First, find the difference between the numbers: \( 12 - 8 = 4 \).
The H.C.F. of 8 and 12 is 4 because 4 divides both 8 and 12.
(ii) For 38 and 57:
First, find the difference between the numbers: \( 57 - 38 = 19 \).
Next, check if 19 divides both 38 and 57. \( 38 \div 19 = 2 \) and \( 57 \div 19 = 3 \).
Since 19 divides both numbers, the H.C.F. of 38 and 57 is 19.
(iii) For 117 and 195:
First, find the difference between the numbers: \( 195 - 117 = 78 \).
Next, find the difference between the smaller number and the first difference: \( 117 - 78 = 39 \).
Then, find the difference between the first difference and the second difference: \( 78 - 39 = 39 \).
Since the differences repeated as 39, the H.C.F. of 117 and 195 is 39.
(iv) For 99, 165, and 231:
First, find the differences between the numbers:
Difference 1: \( 165 - 99 = 66 \)
Difference 2: \( 231 - 165 = 66 \)
Now, find the difference between the smallest number and one of the differences: \( 99 - 66 = 33 \).
The H.C.F. of 99, 165, and 231 is 33. This is because 33 divides 99, 165, and 231 exactly.
In simple words: The Vedic method for H.C.F. uses differences between numbers. You keep finding differences until you get a number that divides all the original numbers. That number is the H.C.F.

🎯 Exam Tip: The Vedic method for H.C.F. relies on finding successive differences. The H.C.F. is the greatest common factor of these differences, or the difference itself if it divides all original numbers.

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?
Answer: To find out when both bells will ring together again, we need to find the Least Common Multiple (L.C.M.) of their ringing intervals, which are 3 minutes and 5 minutes. Since 3 and 5 are prime numbers, their L.C.M. is simply their product.
L.C.M. of 3 and 5 = \( 3 \times 5 = 15 \)
So, both bells will ring together after every 15 minutes. This is the shortest time after which their ringing schedules will coincide.
In simple words: We need to find the smallest number that both 3 and 5 can divide evenly. This number is 15, so the bells will ring together again in 15 minutes.

🎯 Exam Tip: When a question asks when two or more events will happen 'together again' or 'at the same time' in the future, it usually requires finding the Least Common Multiple (L.C.M.).

Pg. No. 32

 

Question 1. Find out the LCM of 48, 64 and 80 by division method.
Answer: We will find the L.C.M. of 48, 64, and 80 using the division method by continuously dividing by prime numbers until all numbers become 1.

L.C.M. of 48, 64, and 80 \( = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5 = 960 \).
Therefore, the least common multiple is 960.
In simple words: To find the L.C.M., divide all numbers by common prime factors until they become 1. Then, multiply all the divisors to get the final L.C.M.

🎯 Exam Tip: Always use prime numbers for division when finding L.C.M. by the division method, and ensure all numbers are reduced to 1 at the end.

 

Question 2. Find out the L.C.M of 24 and 30 by Vedic method.
Answer: To find the L.C.M. of 24 and 30 using the Vedic method, we can follow these steps:
Step 1. Write the numbers as a fraction: \( \frac { 24 }{ 30 } \).
Step 2. Prime factorise the numerator and the denominator:
\( \frac { 24 }{ 30 } = \frac { 2 \times 2 \times 2 \times 3 }{ 2 \times 3 \times 5 } \)
Step 3. Identify and cancel out the common factors from the numerator and denominator:
\( \frac { 24 }{ 30 } = \frac { (2 \times 3) \times 2 \times 2 }{ (2 \times 3) \times 5 } = \frac { 2 \times 2 }{ 5 } = \frac { 4 }{ 5 } \)
Step 4. Now, perform cross-multiplication with the original numbers and the simplified fraction:
\( 24 \times 5 = 120 \) or \( 30 \times 4 = 120 \).
Thus, the required L.C.M. is 120. This method simplifies the numbers before finding the L.C.M.
In simple words: Write the numbers as a fraction and simplify it. Then, multiply the numerator of one original number with the denominator of the simplified fraction. The result is the L.C.M.

🎯 Exam Tip: The Vedic method often simplifies calculations by reducing numbers to their simplest ratio first, making it easier to find the L.C.M. through cross-multiplication.

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?
Answer: Based on the table of factors, the numbers that have only two factors are those with factors (1, 2), (1, 3), (1, 5), and (1, 7). These numbers are 2, 3, 5, and 7. Numbers with exactly two factors (1 and themselves) are called prime numbers.

In simple words: The numbers 2, 3, 5, and 7 from the table have only two factors: 1 and themselves. These are called prime numbers.

🎯 Exam Tip: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Number Game

 

Question 2. How many prime numbers did you get between 1 - 100?
Answer: There are 25 prime numbers between 1 and 100. Identifying them often involves a process like the Sieve of Eratosthenes, where multiples of smaller primes are eliminated.
In simple words: If you list all numbers from 1 to 100 and remove numbers that can be divided by others (except 1 and themselves), you will find 25 prime numbers.

🎯 Exam Tip: Familiarize yourself with prime numbers up to 100, as they frequently appear in various mathematical problems.

 

Question 3. Write these numbers in sequence.
Answer: The prime numbers between 1 and 100, written in increasing order, 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.
In simple words: The prime numbers from 1 to 100 are listed from the smallest to the largest in order.

🎯 Exam Tip: Memorizing the list of prime numbers up to 100 can be helpful for quick identification in number theory problems.

Pg. No. 21

Odd-Even Numbers

 

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?
Answer: Yes, we can establish a rule for Eki and Beki based on the game. Numbers that have 2, 4, 6, 8, or 0 in their units place are called even numbers (Beki). Numbers that have 1, 3, 5, 7, or 9 in their units place are called odd numbers (Eki). This rule consistently identifies whether a number is even or odd.

In simple words: We can call numbers ending in 2, 4, 6, 8, or 0 as 'Beki' (even numbers) and numbers ending in 1, 3, 5, 7, or 9 as 'Eki' (odd numbers).

🎯 Exam Tip: The easiest way to identify an even or odd number is by looking at its last digit (the units place).

Pg. No. 23

 

Question 1. Can you say all even numbers are divisible by 2.
Answer: Yes, all even numbers are divisible by 2 without leaving any remainder. This is the definition of an even number. Even numbers always end in 0, 2, 4, 6, or 8.
In simple words: Yes, every even number can be divided by 2 perfectly.

🎯 Exam Tip: The definition of an even number is any integer that is exactly divisible by 2.

 

Question 2. Write factors of 24, 15, 26, 48, 13, 11.
Answer: Here are the factors for each of the given numbers:
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
In simple words: Factors are numbers that divide another number evenly. For example, the factors of 10 are 1, 2, 5, and 10.

🎯 Exam Tip: To ensure you've found all factors, you can list them in pairs (e.g., for 12: 1x12, 2x6, 3x4).

 

Question 3. Write the unit digit of those numbers whose one factor is 2.
Answer: Numbers that have 2 as one of their factors are even numbers. The unit digit of such numbers will always be 0, 2, 4, 6, or 8. These are the digits that make a number divisible by 2.

In simple words: If a number can be divided by 2, its last digit must be 0, 2, 4, 6, or 8.

🎯 Exam Tip: The divisibility rule for 2 is one of the simplest: a number is divisible by 2 if its last digit is an even number.

 

Question 4. Write some more numbers in the table. Do you find any pattern in the numbers divisible by 10 at its units place?
Answer: Here's the completed table. The pattern observed is that all numbers that have 0 in their units place are divisible by 10.

All numbers that have a 0 at their units place, or numbers for which 10 is one of their factors, are divisible by 10.
In simple words: If a number ends with a zero, then you can always divide it by 10 without any remainder.

🎯 Exam Tip: A number is divisible by 10 if and only if its last digit is 0. This is a straightforward divisibility rule.

Pg. No. 24

 

Question 1. Write all the factors of given table.
Answer: Here is the completed table showing all the factors for the given numbers. Numbers ending in 0 or 5 will always have 5 as a factor.

All the numbers that have unit digits 0 or 5 also have 5 as one of their factors.
In simple words: Look at the last digit of a number. If it is 0 or 5, then 5 is one of its factors.

🎯 Exam Tip: The divisibility rule for 5 is helpful: if a number ends in 0 or 5, it is divisible by 5.

 

Question 2. Complete the table on the basis of given instructions.
Answer: Here is the completed table following the instructions, which focus on divisibility by 3. A number is divisible by 3 if the sum of its digits is divisible by 3.

In simple words: To check if a number can be divided by 3, add all its digits together. If that sum can be divided by 3, then the original number can also be divided by 3.

🎯 Exam Tip: The divisibility rule for 3 is to sum the digits and see if that sum is a multiple of 3. This is a very handy rule for larger numbers.

Pg.No.26

 

Question 1. Can you see any pattern for divisibility by 6?
Answer: Yes, a clear pattern for divisibility by 6 emerges from the table. A number is divisible by 6 if it is divisible by both 2 and 3 separately. This means the number must be even (divisible by 2) and the sum of its digits must be divisible by 3.

From the table, we observe that if any number is divisible by both 2 and 3 separately, then it is also divisible by 6.
In simple words: A number can be divided by 6 only if it can be divided by 2 (it's even) AND also by 3 (its digits add up to a number that 3 can divide).

🎯 Exam Tip: Remember the compound divisibility rule: for a number to be divisible by a composite number (like 6), it must be divisible by all of its prime factors (2 and 3).

 

Question 2. Take some numbers for divisibility by 4 and test the pattern.
Answer: Here are some numbers and a test for their divisibility by 4. The pattern is clear: a number is divisible by 4 if its last two digits (tens and units place) form a number that is divisible by 4, or if its last two digits are both 0.

If a number's last two digits form a number divisible by 4, or if its tens and units digits are 0, then the number is divisible by 4.
In simple words: To know if a number can be divided by 4, just look at its last two digits. If those two digits make a number that 4 can divide, then the whole number can be divided by 4.

🎯 Exam Tip: The divisibility rule for 4 is to check the number formed by the last two digits. If that two-digit number is divisible by 4, the original number is also 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.
Answer: Meena's observation is correct. The number 9212 has 12 as its last two digits. Since 12 is divisible by 4 \( (12 \div 4 = 3) \), the entire number 9212 is also divisible by 4. Dividing 9212 by 4 gives 2303.
\( 9212 \div 4 = 2303 \)
This demonstrates the divisibility rule for 4 perfectly. It's a quick way to check larger numbers.
In simple words: The number 9212 can be divided by 4 because its last two digits, 12, can be divided by 4.

🎯 Exam Tip: Always remember that the divisibility rule for 4 is based solely on the last two digits of the number.

 

Question 4. For divisibility by 8, take some numbers and test the pattern in table.
Answer: Here are some numbers with their divisibility by 8 tested. The pattern shows that a number is divisible by 8 if the number formed by its last three digits (hundreds, tens, and units) is divisible by 8, or if the last three digits are 0s.

If the number formed by the last three digits (units, tens, and hundreds) is divisible by 8, or if any number has 0 as its units, tens, and hundreds digits, then the number is divisible by 8. This is similar to the rule for 4 but uses three digits instead of two.
In simple words: To check if a number can be divided by 8, look at its last three digits. If these three digits form a number that 8 can divide, or if they are all zeros, then the whole number can be divided by 8.

🎯 Exam Tip: For divisibility by 8, focus on the number formed by the last three digits. This rule extends the pattern seen with divisibility by 2 (one digit) and 4 (two digits).

 

Question 2. What are the multiples of 3 and 4? Circle them.
Answer: Multiples of a number are the results you get when you multiply that number by a whole number. To find the common multiples of 3 and 4, we list the multiples of each number and then identify the numbers that appear in both lists.

The common multiples of 3 and 4 (numbers that appear in both lists) are: 12, 24, 36, 48, 60, 72, 84. These numbers can be divided by both 3 and 4 without leaving a remainder.
In simple words: We are looking for numbers that can be divided evenly by both 3 and 4. These numbers are 12, 24, 36, 48, 60, 72, and 84 within this range.

🎯 Exam Tip: When finding common multiples, you can also first find the Least Common Multiple (LCM). For 3 and 4, the LCM is 12. Then, all subsequent multiples of 12 (24, 36, 48, etc.) will also be common multiples of 3 and 4.