Class 6 Maths Chapter 3 Exercise Question 1, �����, �������� ������ NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise , Exercise , Exercise , Exercise , Exercise , Exercise and Exercise in English and Hindi Medium for session The NCERT Solutions For Class 6 Maths Chapter 3 Playing with Numbers Exercise explains the method of solving problems related to finding the HCF of a given set of numbers. These NCERT Solutions are prepared by the subject experts at BYJU�S to help students ace the exam. Sep 10, �� NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers PDF Ex solved by Subject Experts as per NCERT (CBSE) Book guidelines. Playing with Numbers Class 6 Maths Chapeter 3 Exercise Questions with Solutions to help .
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So, it is not divisible by 2. Since the given number is not divisible by both 2 and 3 therefore, it is not divisible by 6. Since, the given number is not divisible by both 2 and 3, it is not divisible by 6.

Since, the given number is not divisible by both 2 and 3 hence, it is not divisible by 6. Using divisibility tests, determine which of the following numbers are divisible by a b c d e We know that a number is divisible by 11 if the difference between the sum of the digits at odd places from the right and the sum of the digits at even places from the right of the number is either 0 or divisible by Hence, the given number is divisible by Hence, the given number is not divisible by Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3.

Hence, the required digits are 0 and 9. Write a digit in the blank space of each of the following numbers so that the numbers formed is divisible by Find the common factors of: a 20 and 28 b 15 and 25 c 35 and 50 d 56 and Solution: a Given numbers are : 20 and 28 Factors of 20 are 1, Class 5 Maths Chapter 4 Question Answer Youtube 2, 4, 5, 10, 20 Factors of 28 are 1, 2, 4, 7, 28 Hence, the common factors are 1, 2 and 4.

Find the common factors of: a 4, 8 and 12 b 5, 15 and 25 6 8 Solution: a Given numbers are: 4, 8 and 12 Factors of 4 are 1, 2, 4 Factors of 8 are 1, 2, 4, 8 Factors of 12 are 1, 2, 3, 4, 6, 12 Hence, the common factors are 1, 2 and 4. Write all the numbers less than which are common multiples of 3 and 4. Solution: Given numbers are 3 and 4. Multiples of 3 less than are: Hence, the common multiples of 3 and 4 less than are: 12, 24, 36, 48, 60, 72, 84 and Which of the following numbers are co-prime?

Hence, 18 and 35 are co-prime. Hence, they are co-prime. A number is divisible by both 5 and By which other will that number be always divisible?

Solution: If the number is divisible by both 5 and 12 this the number will also be divisible by 5 x 12 i. A number is divisible by By what other will that number be divisible? Solution: Factors of 12 are 1, 2, 3, 4, 6, 12 Hence the number which is divisible by 12, will also be divisible by its factors i. Which of the following statements are true? Here are two different factor trees for Write the missing numbers. Which factors are not included in the prime factorisation of a composite number?

Solution: 1 and the number itself are not included in the prime factorisation of a composite number. Write the greatest 4-digit number and express it in terms of its prime factors. Write the smallest 5-digit number and express it in the form of its prime factors. Find all the prime factors of and arrange them in ascending order. Now state the relations, if any, between the two consecutive prime factors.

The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples. Solution: Example 1: Take three consecutive numbers 20, 21 and Here 20 is divisible by 2 and 21 is divisible by 3. Example 2: Take three consecutive numbers 30; 31 and Here 30 is divisible by 3 and 32 is divisible by 2.

Example 3: Take three consecutive numbers 48, 49 and Here, 48 is divisible by 3 and 50 is divisible by 2.

The sum of two consecutive odd numbers is divisible by 4. Solution: Example 1: Let us take two consecutive odd numbers 97 and Hence, the sum of numbers 97 and 99 i. Example 2: Let us take two consecutive odd numbers and Example 3: Let us take two consecutive odd numbers and In which of the following expressions, prime factorisation has been done? Verify this statement with the help of some examples.

Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6. Renu purchases two bags of fertilizer of weights 75 kg and 69 kg.

Find the maximum value of weight which can measure the weight of the fertilizer exact number of times. For finding maximum weight, we have to find H. The sum of two consecutive odd numbers is always divisible by 4.

I am the smallest number, having four different prime factors. Can you find me? The smallest four prime numbers are 2, 3, 5 and 7. The length, breadth and height of a room are cm, cm and cm respectively. Find the longest tape which can measure the three dimensions of the room exactly. The LCM of 12 and 16 is: a 32 b 60 c 48 d none of these. Question 2. What are the numbers which have more than two factors called? Question 3. The sum of two odd and one even numbers is a Even b Prime c Composite d Odd.

Question 4. The number which is divisible by 5 is: a b c d Question 6. Question 7. What is the H. Question 8. The LCM of 12, 15, 45 is: a b c d




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