Prime Time Class 6 Extra Questions Maths Chapter 5

Class 6 Maths Chapter 5 Extra Questions Prime Time

Class 6 Maths Prime Time Extra Questions

NCERT Class 6 Maths Chapter 5 Prime Time Extra Questions and Answers

Question 1.
Find the common factors of:
(a) 20 and 50
(b) 56 and 120
Solution:
(a) 20 and 28
We have: 20 = 1 × 20; 20 = 2 × 10; 20 = 4 × 5
All the factors of 20 are: 1, 2, 4, 5, 10 and 20 …..(i)
Again, 50 = 1 × 50; 50 = 2 × 25; 50 = 5 × 10
All the factors of 50 are: 1, 2, 5, 10, 25, 50 ….. (ii)
From (i) and (ii), common factors of 20 and 50 are: 1, 2, 5 and 10.

(b) 56 and 120
Since 56 = 1 × 56; 56 = 2 × 28; 56 = 4 × 14; 56 = 7 × 8
All the factors of 56 are: 1, 2, 4, 7, 8, 14, 28 and 56 …..(i)
Again, 120 = 1 × 120; 120 = 2 × 60; 120 = 3 × 40; 120 = 4 × 30; 120 = 5 × 24; 120 = 6 × 20; 120 = 8 × 15; 120 = 10 × 12
All the factors of 120 are: 1, 2, 3, 4, 5, 6, 8, 10, 12,
15,20, 24, 30, 40, 60 and 120
The common factors of 56 and 120 are: 1, 2, 4 and 8.

Question 2.
Find the product of the common prime factors of 180, 144 and 108.
Solution:
The given numbers are 180,144 and 108. Resolving each of the given numbers into prime factors, we have:

∴ 180 = 2 × 2 × 3 × 3 × 5
144 = 2 × 2 × 2 × 2 × 3 × 3
108 = 2 × 2 × 3 × 3 × 3
The product of common prime factors of 180, 144, 108 = 2 × 2 × 3 × 3 = 36
Thus, the required product is 36.

Question 3.
Find first three common multiples of:
(a) 6 and 8
(b) 12 and 18
Solution:
(a) 6 and 8
Since multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, …
Multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, …
The first three common multiples of 6 and 8 are: 24, 48 and 72.

(b) 12 and 18
Multiples of 12 are: 12, 24, 36, 48, 60, 72, 84, 96, 108 , 120, …
Multiples of 18 are: 18, 36, 54, 72, 90, 108, 126,…
First three common multiples of 12 and 18 are: 36, 72 and 108.

Question 4.
A number is divisible by both 5 and 12. By which other number will that number be always divisible?
Solution:
The given number will be divisible by the product of 5 and 12.
The number will be divisible by 5 × 12 or 60.

Question 5.
A number is divisible by 12. By what other numbers will that number be divisible?
Solution:
The number divisible by 12 will also be divisible by the factors of 12.

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 factors.
Solution:
We have:
The prime factorisation of 1729 = 7 × 13 × 19

Now, ascending order of the prime factors of 1729 is: 7, 13, 19 .
Now, we have: 19 – 13 = 6
13 – 7 = 6
The relation is the difference of two consecutive prime factors is 6.

Question 7.
Find the smallest natural number that has six factors in all.
Solution:
Let us find the number by finding factors of numbers in sequence to find the smallest number.

It is clear from the table that 12 is the smallest natural number that has exactly six factors, which are 1, 2, 3, 4, 6 and 12.

Question 8.
How many multiples of 12 are there between 1 and 100?
Solution:
The multiples of 12 between 1 and 100 are 12, 24, 36, 48, 60, 72, 84, 96. Thus, there are 8 multiples of 12 between 1 and 100.

Question 9.
In the diagram below, Tukku has erased all the numbers except the common factors. Find out what those numbers could be and fill in the missing numbers in the empty regions.

Solution:
A pair of possible numbers whose common factors are 1, 2, 3 and 6 would be 18 and 48.
Factors of 18 = 1, 2, 3, 6, 9 and 18
Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48
Common Factors = 1, 2, 3, and 6

Question 10.
Goofy and Mickey are playing the treasure hunting game. Mickey has kept treasures on numbers 32 and 80 on a line. What sizes of jumps will Goofy take to land on both numbers?
Solution:
In the treasure hunting game, if Mickey has kept treasures on numbers 32 and 80, then the common factors of 32 and 80 will be the jump sizes taken by Goofy to land on both the numbers.
Therefore,
Factors of 32 are: 1, 2, 4, 8, 16, and 32.
Factors of 80 are: 1, 2, 4, 5, 8, 10, 16, 20, 40, and 80.
The common factors of 32 and 80 are: 1, 2, 4, 8 and 16. Thus, the possible jump sizes that Goofy can take to land on both the numbers will be 1, 2, 4, 8 and 16.

Question 11.
Find the least number that is divisible by all the numbers from 3 to 10 (both inclusive).
Solution:
The least number divisible by all the numbers from 1 to 10 will be the smallest common multiple of the numbers from 3 to 10 (both inclusive).
We have,

Therefore, 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520
Thus, the least number that is divisible by all the numbers from 3 to 10 (both inclusive) will be 2520.

Question 12.
Jatin has to verify that the prime factorization of 40 will always remain the same. Can you help him verify it?
Solution:
Since 40 can be factored into 5 and 8, i.e., 40 = 5 × 8
8 can further be broken down as a product of 2 and 4, and similarly, 4 can be broken down as a product of 2 and 2. 5 is a prime number already.
Therefore, 2 × 2 × 2 × 5 = 40
Hence, she will show the division method and factor tree method in the following way:

∴ This shows that by any method of factorisation, the prime factorisation remains the same.

Question 13.
Verify: Prime factorisation of 300 × Prime factorisation of 500 = Prime factorisation of (300 × 500)
Solution:
Prime factorisation of 300 = 2 × 2 × 3 × 5 × 5
Prime factorisation of 500 = 2 × 2 × 5 × 5 × 5
Prime factorisation of 300 × Prime factorisation of 500 = (2 × 2 × 3 × 5 × 5) × (2 × 2 × 5 × 5 × 5) = 2 × 2 × 2 × 2 × 3 × 5 × 5 × 5 × 5 × 5
Prime factorisation of (300 × 500) = prime factorisation of 150000 = 2 × 2 × 2 × 2 × 3 × 5 × 5 × 5 × 5 × 5
Clearly, Prime factorisation of 300 × Prime factorisation of 500 = Prime factorisation of (300 × 500).

Question 14.
Which of the following numbers is divisible by all of 2, 4, 5, 8 and 10? ‘
(a) 472
(b) 3522
(c) 9000
(d) 8829
(e) 9996
Solution:

Thus, the number 9000 is divisible by all 2, 4, 5, 8 and lo.

Question 15.
Find the product of the common prime factors of the following:
(a) 30, 70, and 60
(b) 140 and 360
(c) 648 and 1440
Solution:

Prime factorisation of 30 = 2 × 3 × 5
Prime factorisation of 70 = 2 × 5 × 7
Prime factorisation of 60 = 2 × 2 × 3 × 5
Common prime factors of 30, 70 and 60=2 and 5
∴ Product of common prime factors of 30, 70 and 60 = 2 × 5 = 10

(b)

Prime factorisation of 140 = 2 × 2 × 5 × 7
Prime factorisation of 360 = 2 × 2 × 2 × 3 × 3 × 5
Common prime factors of 140 and 360 = 2, 2 and 5
∴ Product of common prime factors of 140 and 360 = 2 × 2 × 5 = 20

(c)

Prime factorisation of 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3
Prime factorisation of 1440 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5
Common prime factors of 648 and 1440 = 2, 2, 2, 3 and 3. ‘
∴ Product of common prime factors of 648 and 1440
= 2 × 2 × 2 × 3 × 3 = 72.

Question 16.
What could be only two numbers, so that if 28720 is divisible by those two numbers, then we could say it will be divisible by all of 2, 4, 5, 8 and 10?
Solution:
Since 28720 is an even number ending with 0, the number formed by the last 3 digits (720) is divisible by 8 as 720 ÷ 8 = 90.
Thus, 28720 is divisible by 8.
Also, 28720 ends with 0. So, it is also divisible by 5 and 10 as well.
Thus, if a number is divisible by 8 and 10, then it must be divisible by 2, 4 and 5 also.
Hence, 8 and 10 are two possible numbers, so if 28720 is divisible by 8 and 10, then we could say it will be divisible by all of 2, 4, 5, 8 and 10.

Question 17.
In a school library, there are 780 books of English and 364 books of Science. Ms. Sharma, the librarian of the school, wants to store these books on shelves such that each shelf should have the same number of books of each subject. What should be the maximum number of books on each shelf?
Solution:
Given, number of English books = 780
and number of Science books = 364
The maximum number of books on each shelf, will be the product of the common factors of 780 and 364.

Prime factorization of 780 = 2 × 2 × 3 × 5 × 13
Prime factorization of 3 64 = 2 × 2 × 7 × 13
Therefore, common factors of 780 and 364 are 2, 2 and 13.
Thus, the product of the common factors of 780 and 364 is 2 × 2 × 13 = 52
Hence, the maximum number of books in each shelf = HCF of 780 and 364 = 52.

Question 18.
What number should be added to 63781 to make it exactly divisible by 4?
Solution:
We know that a number is divisible by 4 if the last two digits of a number is divisible by 4. Let’s check.
Here, in 63781, the last two digits are 81, which is not exactly divisible by 4.
If we add 3 to 81, we get 84, which is exactly divisible by 4.
Thus, we can add 3 to 63781 and it will be 63784 which is exactly divisible by 4.

Class 6 Maths Extra Questions