Regular revision with Class 6 Maths MCQ with Answers and Ganita Prakash Class 6 Maths Chapter 5 Prime Time MCQ improves accuracy in objective exams.
MCQ on Prime Time Class 6
Prime Time MCQ Class 6
Class 6 Maths Prime Time MCQ
Question 1.
Which of the following is a factor of 36?
(a) 5
(b) 14
(c) 9
(d) 15
Solution:
(c) 9
We can write 36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6
Hence, the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.
Question 2.
Which of the following is a multiple of 15?
(a) 125
(b) 135
(c) 145
(d) 155
Solution:
(b) 135
We can write 135 = 15 × 9.
Hence, 135 is a multiple of 15.
Question 3.
Which of the following is a perfect number?
(a) 4
(b) 5
(c) 6
(d) 8
Solution:
(c) 6
We know that a number for which the sum of all its factors is equal to twice of itself, is called a perfect number.
Factors of 6 are 1, 2, 3 and 6.
Sum of factors of 6 = 1 + 2 + 3 + 6= 12, which is double of number 6.
Hence, 6 is a perfect number.
Question 4.
Which of the following is not a multiple of 24?
(a) 72
(b) 120
(c) 168
(d) 206
Solution:
(d) 206
Clearly, 72 = 24 × 3, 120 = 24 × 5 and 168 = 24 × 7
Hence, 72, 120 and 168 are multiples of 24. But 206 is not a multiple of 24.
Question 5.
Which of the following is not a common multiple of 2 and 5?
(a) 30
(b) 40
(c) 45
(d) 50
Solution:
(c) 45
Clearly, 30 = 2 × 15 = 5 × 6, 40 = 2 × 20 = 5 × 8 and 50 = 2 × 25 = 5 × 10
∴ 30, 40 and 50 are common multiples of 2 and 5.
Since 45 = 5 × 9, 45 is a multiple of 5 but not of 2.
Thus, 45 is not a common multiple of 2 and 5.
Question 6.
Which of the following is a composite number?
(a) 47
(b) 67
(c) 57
(d) 89
Solution:
(c) 57
Factors of 47 are 1 and 47.
Factors of 67 are 1 and 67.
Factors of 57 are 1, 3, 19 and 57.
Factors of 89 are 1 and 89.
Since 57 has more than two factors, 57 is a composite number.
Question 7.
Which of the following numbers is the product of exactly three distinct prime numbers?
(a) 135
(b) 145
(c) 155
(d) 165
Solution:
(d) 165
Every number can be expressed as a product of primes.
135 = 3 × 3 × 3 × 5 (only two prime numbers 3 and 5)
145 = 5 × 29 (only two prime numbers 5 and 29)
155 = 5 × 31 (only two prime numbers 5 and 31)
165 = 3 × 5 × 11 (three distinct prime numbers 3, 5 and 11)
Question 8.
The difference between the prime numbers between 80 and 90 is:
(a) 6
(b) 7
(c) 8
(d) 9
Solution:
(a) 6
The prime numbers between 80 and 90 are 83 and 89.
Hence, the required difference = 89 – 83 = 6
Question 9.
Which of the following are co-prime?
(a) 8 and 14
(b) 12 and 15
(c) 14 and 15
(d) 18 and 21
Solution:
(c) 14 and 15
Factors of 14 are 1, 2, 7 and 14.
Factors of 15 are 1, 3, 5 and 15.
Since 1 is the only common factor of 14 and 15, they are co-prime.
Question 10.
Which of the following are twin primes?
(a) 3 and 7
(b) 11 and 13
(c) 13 and 19
(d) 17 and 23
Solution:
(b) 11 and 13
Twin primes are pairs of primes having a difference of 2.
13 – 11 = 2
Hence, 11 and 13 are twin primes.
Question 11.
The smallest number having four different prime factors is:
(a) 210
(b) 215
(c) 230
(d) 240
Solution:
(a) 210
The smallest number, having four different prime factors, will be obtained as the product of the first four prime numbers.
The first four smallest prime numbers are 2, 3, 5 and 7.
Thus, the required number = 2 × 3 × 5 × 7 = 210
Question 12.
The prime factorisation of 44100 is:
(a) 2 × 2 × 2 × 3 × 3 × 5 × 5 × 7
(b) 2 × 2 × 3 × 3 × 5 × 5 × 7 × 7
(c) 2 × 3 × 5 × 5 × 5 × 7 × 7
(d) 2 × 3 × 3 × 5 × 7 × 7 × 7
Solution:
(b) 2 × 2 × 3 × 3 × 5 × 5 × 7 × 7
The prime factorisation of 44100 is
44100 = 2 × 2 × 3 × 3 × 5 × 5 × 7 × 7
Question 13.
In prime factorisation of 2250, number 5 appears:
(a) 1 time
(b) 2 times
(c) 3 times
(d) 4 times
Solution:
(c) 3 times
The prime factorisation of 2250 is 2250 = 2 × 3 × 3 × 5 × 5 × 5.
Here, 2 appears once, 3 appears 2 times and 5 appears 3 times.
Question 14.
Which one of the following numbers is divisible by 8?
(a) 7634452
(b) 4078376
(c) 7232854
(d) 5736500
Solution:
(b) 4078376
For a number to be divisible by 8, the number formed by its last three digits should be divisible by 8.
The number formed by the last three digits of 4078376 is 376.
Since 376 = 8 x 47, it is divisible by 8.
Thus, 4078376 is divisible by 8.
Question 15.
Which one of the following numbers is divisible by all of 2, 4, 5, 8 and 10?
(a) 72460
(b) 73360
(c) 78945
(d) 76390
Solution:
(b) 73360
We know that if a number is divisible by 5 and
8, then it is also divisible by 2, 4 and 10.
All four numbers end with either ‘0’ or ‘5’, therefore they are divisible by 5.
The number formed by the last three digits of 73360 is 360.
Since 360 = 8 × 45, it is divisible by 8.
Thus, 73360 is divisible by 8.
So, 73360 is divisible by all of 2, 4, 5, 8 and 10.
Question 16.
Which digit should replace * so that the number 80405*6 is divisible by 4?
(a) 2
(b) 0
(c) 5
(d) 8
Solution:
(c) 5
For a number to be divisible by 4, the number formed by its last two digits should be divisible by 4.
Digits 1, 3, 5, 7 or 9 can replace * because 16, 36, 76 and 96 are divisible by 4.
Thus, 8040516, 8040536, 8040556, 8040576 and 8040596 are divisible by 4.
Question 17.
Which one of the following numbers is divisible by 10?
(a) 101010
(b) 232323
(c) 656565
(d) 727272
Solution:
(a) 101010
Since 101010 ends with digit 0, it is divisible by 10.
Question 18.
Which of the following statements are true?
(i) A number which is divisible by 10 will always be divisible by 2 also.
(ii) Every multiple of 5 ends with 5 only.
(iii) A factor of a number is always equal to or smaller than the number.
(iv) A composite number has more than 2 factors.
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iii) only
(c) (i), (iii) and (iv)
(d) (iii) and (iv) only
Solution:
(c) (ii) and (iv)
We know that a number is divisible by 10 if it ends with 0.
And, if a number ends with 0, it is also even and hence divisible by 2.
So, any number that is divisible by 10 is divisible by 2.
Alternative explanation:
10 = 2 × 5, which means if a number is divisible by 10, it must also be divisible by both 2 and 5, since they are the factors of 10. Thus, statemen (i) is true.
We know that a number is divisible by 5 or a multiple of 5 if it ends with either 0 or 5.
Thus, statement (ii) is false.
We know that a factor divides a number exactly, without leaving any remainder. So, it cannot be greater than the number itself. The largest factor a number can have is the number itself. Thus, statement (iii) is true.
We know that a number is said to be composite if it has more than 2 factors.
Question 19.
Which of the following is/are not the correct prime factorisation?
(i) 144 = 2 × 2 × 2 × 3 × 3
(ii) 225 = 3 × 3 × 5 × 5
(iii) 100 = 2 × 5 × 10
(iv) 160 = 2 × 2 × 2 × 2 × 2 × 5
Choose the correct option from the following:
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
Solution:
(a) (i) and (iii)
Prime facionsation of 1 44 = 2 × 2 × 2 × 2 × 3 × 3
Prime factorisation of 225 = 3 × 3 × 5 × 5
Prime factorisation of 100 = 2 × 2 × 5 × 5
Prime factorisation of 160 = 2 × 2 × 2 × 2 × 2 × 5
Thus, in the given prime factorisations, prime factorisation of 144 and 100 are incorrect.
Question 20.
Which of the following numbets can be expressed as the product of 4 distinct primes?
(i) 462
(ii) 315
(iii) 1155
(iv) 1925
Choose the correct option from the following:
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
Solution:
(b) (i) and (iii)
Prime factorisation of 462 = 2 × 3 × 7 × 11 (Product of 4 distinct primes)
Prime factorisation of 315 = 3 × 3 × 5 × 7 (Product of only 3 distinct primes)
Prime factorisation of 1155 = 3 × 5 × 7 × 11 (Product of 4 distinct primes)
Prime factorisation of 1925 = 5 × 5 × 7 × 11 (Product of only 3 distinct primes)
Thus, the prime factorisation of 462 and 1155 have four distinct primes.
Prime Time Class 6 Assertion and Reason Questions
The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Question 1.
(A): The common factors of two numbers can never be greater than the smaller number.
(R): If a number is divisible by another, the second number is called a factor of the first.
Solution:
(a) (a) Both A and R are true and R is the correct explanation of A.
When a number is divisible by another, the second number is called a factor of the first number. Since second number can never be greater than the first number, the factors of a number are always smaller than or equal to the number. Therefore, the common factors of two numbers can never be greater than the smaller number.
Hence, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Question 2.
(A): 8 is one of the factors of 36.
(R): Every factor of a number N is always smaller than or equal to N.
Solution:
(d) A is false but R is true.
We can write 36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6.
∴ 1,2, 3,4,6, 9,12, 18 and 36 are factors of 36.
Thus, 8 is not a factor of 36.
Every factor of a number is always smaller than or equal to the number.
Hence, Assertion (A) is false, but Reason (R) is true.
Question 3.
(A): 4 is one of the factors of 44.
(R): Every factor of a number N is always smaller than N.
Solution:
(c) A is true but R is false.
We can write 44 = 1 × 44 = 2 × 22 = 4 × 11.
∴ 1,2, 4, 11, 22 and 44 are factors of 44.
Thus, 4 is one of the factors of 44.
44 also is a factor of 44, but 44 is not smaller than 44.
Every factor of a number is either smaller than or equal to the number.
Hence, Assertion (A) is true, but Reason (R) is false.
Question 4.
(A): 148725 is divisible by 5.
(R): If a number ends with either digit ‘O’ or ‘5’, then the number is divisible by 5.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know, if a number ends with either digit ‘0’ or ‘5’, then the number is divisible by 5.
Thus, 148725 is divisible by 5.
Hence, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Question 5.
(A): 425783 is divisible by 2.
(R): If a number ends with either digit ‘0’ or ‘5’, then the number is divisible by 5.
Solution:
(d) A is false but R is true.
We know, if a number ends with digit 0, 2, 4, 6 or 8, then the number is divisible by 2.
But 425783 ends with the digit 3, therefore it is not divisible by 2.
And, if a number ends with either digit ‘0’ or ‘5’, then the number is divisible by 5.
Hence, Assertion (A) is false, but Reason (R) is true.
Prime Time Class 6 Fill in the Blanks
Question 1.
21 is a multiple of _________ and _______ .
Solution: 3, 7
We can write 21 = 3 × 7.
21 is a multiple of 3 and 7.
Question 2.
The smallest number that is a multiple of all the numbers from 1 to 10 is ______.
Solution: 2520
We know that, 8 is a multiple of 1, 2, 4 and 8.
9 is a multiple of 1, 3 and 9.
8 × 9 is a multiple of 6.
5 × 8 × 9 is a multiple of 5 and 10. [As 5 x 8 = 40 is a multiple of 5 and 10.]
5 × 7 × 8 × 9 is a multiple of 7.
∴ The required smallest number is 5 × 7 × 8 × 9 = 2520.
Hence, the smallest number that is a multiple of all the numbers from 1 to 10 is 2520.
Question 3.
The smallest 2-digit composite number is ________.
Solution: 10
The smallest 2-digit composite number is 10.
Question 4.
The number 1 is _______ a prime _______ a composite number.
Solution: neither, nor
The number 1 is neither a prime nor a composite number.
Question 5.
The largest 2-digit prime number is _________.
Solution: 97
The largest 2-digit prime number is 97.
Question 6.
The total number of even prime numbers is ________.
Solution: 1
2 is the only even prime number.
Thus, the total number of even prime numbers is 1.
Question 7.
The smallest number whose prime factorisation has three different prime numbers is _________.
Solution: 30 (2 × 3 × 5)
A smallest number, having three different prime factors, will be obtained as the product of the first three prime numbers. The first three smallest prime numbers are 2, 3 and 5.
Thus, the smallest number whose prime factorisation has three different prime numbers is 30 (2 × 3 × 5).
Question 8.
The prime factorisation of the smallest 4-digit number is ________.
Solution: 2 × 2 × 2 × 5 × 5 × 5
The smallest 4-digit number is 1000.
Prime factorisation of 1000 is 1000 = 10 × 10 × 10
= (2 × 5) × (2 × 5) × (2 × 5)
= 2 × 2 × 2 × 5 × 5 × 5
Hence, the prime factorisation of the smallest 4-digit number is 2 × 2 × 2 × 5 × 5 × 5.
Question 9.
The largest prime number in prime factorisation of the largest 4-digit number is _______.
Solution: 101
The largest 4-digit number is 9999.
Prime factorisation of 9999 is 9999 = 9 × 1111
= (3 × 3) × 11 × 101 = 3 × 3 × 11 × 101
Thus, the largest prime number in prime factorisation of the largest 4-digit number is 101.
Question 10.
The largest 4-digit palindromic number divisible by 4 is __________ .
Solution: 8888
The largest 4-digit palindromic number divisible by 4 is 8888.
Question 11.
The largest 6-digit number divisible by 8 is ________.
Solution: 999992
The largest 6-digit number divisible by 8 is 999992.
Question 12.
Smallest digit to replace * so that the number 2317*4 is divisible by 11 is ________.
Solution: 0
Given, 2317*4 is divisible by 11.
Sum of its digits in odd places = 4 + 7 + 3= 14
Sum of its digits in even places = * + 1 + 2 = * + 3
Difference between the two sums = 14 – (* + 3) = 11 -*
So, smallest digit to replace * is 0.