Class 6 Maths Chapter 3 Exercise 3.5 NCERT Solution – Playing With Numbers

NCERT Class 6 Math Exercise 3.5 Solution of Chapter 3 Playing with Numbers with explanation. Here we provide Class 6 Maths all Chapters in Hindi for cbse, HBSE, Mp Board, UP Board and some other boards.

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NCERT Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.5 Solution in english Medium.

Class 6 Maths Chapter 3 Exercise 3.5 Solution

Question 1
Which of the following statements are true:
A. If a number is divisible by 3, it must be divisible by 9.
B. If a number is divisible by 9 , it must be divisible by 3.
C. If a number is divisible by 18, it must be divisible by both 3 and 6.
D. If a number is divisible by 9 and 10 both , then it must be divisible by 90.
E. If two numbers are co primes , at least one of thme must be prime.
F. All numbers which are divisible by 4 must be divisible by 8.
G. All numbers which are divisible by 8 must also be divisible by 4.
H. If a number exactly divides two numbers separately, it must exactly divide their sum.
I. If a number exactly divides the sum of two numbers , it must exactly divides the two numbers separately.
Answer.
statements B, C, D, G, and H aer true.

Question 2.
Here are two different factor trees for 60. Write the missing numbers.

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Question 3.
Which factors are not included in the prime factorisation of a composite number?
Answer.
1 and the number itself are the factor which is not included in the prime factorisation of a composite number.

Question 4.
Write the greatest 4 digit number and express it in terms of its prime factors.
Answer.
The greatest 4 digit number = 9999
The prime factors of 9999 are 3x3x11x101.

Question 5.
write the smallest 5 digit number and express it in terms of its prime factors.
Answer.
The smallest five digit number is 10000.
The prime factors of 10000 are 2x2x2x2x5x5x5x5.

Question 6.
Find all the prime factors of 1729 and arrange them in ascending order.
Now state the relation, if any between , two consecutive prime numbers.
Answer.
Prime factors of 1729 are 7x13x19.
The difference of two consecutive prime factors is 6.

Question 7.
The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Answer.
Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6.
Example: 2x3x4 = 24.
4x5x6 = 120.

Question 8.
The sum of two consecutive odd numbers is always divisible by 4. Verify this statement with the help of some examples.
Answer.
3 + 5 = 8, is divisible by 4.
5 + 7 = 12, is divisible by 4.
7 + 9 = 16, is divisible by 4.

Question 9.
In which of the following expressions, prime factorisation has been done:
A. 24 = 2x3x4
B. 56 = 7x2x2x2
C. 70 = 2x5x7
D. 54 = 2x3x9
Answer.
In expressions B and C, prime factorisation has been done.

Question 10.
Determine if 25110 is divisible by 45.
Answer.
The prime factorisation of 45 = 5×9
25110 is divisible by 5 as 0 is at its unit place.
25110 is divisible by 9 as sum of digits is divisible by 9.
therefore, the number must be divisible by 5×9 = 45.

Question 11.
18 is divisible by both 2 and 3. It is also divisible by 2×3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4×6 = 24? If not , give an example to justify your answer.
Answer.
No.
Number 12 is divisible by both 6 and 4 but is 12 is not divisible by 24.

Question 12.
I am the smallest number, having four different prime factors. Can you find me?
Answer.
The smallest four prime numbers are 2, 3, 5 and 7.
Hence, the required number is 2x3x5x7 = 210.

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