Important Questions Test: Whole Numbers - Class 6 MCQ

 0 does not have a predecessor.

just multiply respective terms  and add respective terms

The option (b) is the correct answer.

The smallest whole number is 0.

The option (c) is the correct answer.

We know that

1 million = 10, 00,000

So the successor of 1 million = 10, 00,000 + 1 = 10, 00,001

The option (a) is the correct answer.

For Example-

Consider whole number = 1

So the successor of 1 = 1 + 1 = 2

Product between them = 1 × 2 = 2

Hence, 2 is an even number.

The option (d) is the correct answer.

For example-

Consider an even natural number = 2

So the predecessor of 2 = 2 − 1 = 1

We get the successor of 2 = 2 + 1 = 3

Product between them = 1 × 3 = 3

Hence, the product is an odd number.

The option (c) is the correct answer.

We know that the set of numbers which has common factor as 1 are called co-prime.

So the LCM of two co-prime numbers and their product are equal.

The option (b) is the correct answer.

The smallest number which is neither prime nor composite is 1

 There is no predecessor for natural number 1. As natural number starts from 1

The option (b) is the correct answer.

We know that we get multiples by multiplying the number by any number.

Hence, every counting number has an infinite number of multiples

The option (b) is the correct answer.

We know that the LCM of 15, 20, 24 and 32 is

15 = 3 × 5 = 3¹ × 5¹

20 = 2 × 2 × 5 = 2² × 5¹

24 = 2 × 2 × 2 × 3 = 2³ × 3¹

32 = 2 × 2 × 2 × 2 × 2 = 2⁵

So the LCM = 2⁵ × 3¹ × 5¹ = 480

Concept:

Counting the whole numbers within the given range.

Solution:

⇒ The sequence of whole numbers between 0 and 10 is: 1, 2, 3, 4, 5, 6, 7, 8, 9

⇒ We simply need to count these numbers.

⇒ There is 1 one. ⇒ There is 1 two. ⇒ There is 1 three. ⇒ There is 1 four. ⇒ There is 1 five. ⇒ There is 1 six. ⇒ There is 1 seven. ⇒ There is 1 eight. ⇒ There is 1 nine.

by adding all the counts, we get 9 whole numbers between 0 and 10.

Hence, the correct answer is "9".

We know that the factors of

91 = 1 × 7 × 13

81 = 1 × 3 × 3 × 3 × 3

87 = 1 × 3 × 29

97 = 1 × 97

Hence, 81, 87 and 91 are not prime numbers.

We know that the set of numbers which has common factor as 1 are called co-prime.

LCM of two co-prime numbers and their product is equal.

So the HCF of two co-prime numbers is 1.

There 29 numbers between 38 and 68 . Hence the number of whole numbers between 38 and 68 are 29
use m - n - 1 for finding no of terms
= 68 - 38 - 1 = 29

Subtract the required remainder from 134 and 167.

So we get 132 and 165.

132 = 2 × 2 × 3 × 11 = 22 × 3 × 11

165 = 3 × 5 × 11 = 31 × 5 × 11

HCF = 3 × 11 = 33

Hence, the greatest number which divides 134 and 167 leaving 2 as remainder in each case is 33

Successor=1000
Predecessor=998
Product= 1000×998
= 998000

We know that the product of two numbers = HCF of two numbers × LCM of two numbers

By substituting the values

1530 = 15 × LCM of two numbers

We get

LCM of two numbers = 1530/15 = 102

Let the whole number be x
According to the question
= x + 25 + 25 - x
= 50

We know that the predecessor of an odd number is an even number and the successor of an odd number is also an even number.

So the two even numbers and their product are two consecutive even numbers which is always divisible by 8.