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This mock test of Olympiad Test: Whole Numbers for Class 6 helps you for every Class 6 entrance exam.
This contains 20 Multiple Choice Questions for Class 6 Olympiad Test: Whole Numbers (mcq) to study with solutions a complete question bank.
The solved questions answers in this Olympiad Test: Whole Numbers quiz give you a good mix of easy questions and tough questions. Class 6
students definitely take this Olympiad Test: Whole Numbers exercise for a better result in the exam. You can find other Olympiad Test: Whole Numbers extra questions,
long questions & short questions for Class 6 on EduRev as well by searching above.

QUESTION: 1

Which of the following is not true

Solution:

According to BODMAS rule, multiplication is done forst then addition.

So according to this, LHS = 7+ 8 x 9 = 7+72 = 79

But RHS = (7 + 8) x(7+ 9) = (15) x (16) = 240

Hence Option C is incorrect

QUESTION: 2

By using dot (•) patterns, which of the following numbers can be arranged in all the three ways namely a line, a triangle and a rectangle?

Solution:

A triangle can’t be made by 3 dots, 11 dots or 12 dots and a rectangle can’t be made by 11 dots, hence option A, C and D are incorrect.

A line, a triangle and a rectangle can be drawn by 10 dots as shown,

QUESTION: 3

Which of the following statements is not true?

Solution:

QUESTION: 4

Which of the following statements is not true?

Solution:

0 can not be divided by any number and 0 can not divide any number.

so, 0÷0=0 is not true

QUESTION: 5

The predecessor of 1 lakh is

Solution:

Predecessor is the term just before the given term

So 100000 - 1 = 99999

QUESTION: 6

The successor of 1 million is

Solution:

QUESTION: 7

Number of even numbers between 58 and 80 is

Solution:

Starting from 60, we have 60, 62, 64 .... upto 78 a total of 10 even numbers

QUESTION: 8

Sum of the number of primes between 16 to 80 and 90 to 100 is

Solution:

Prime numbers between 16 and 80 are

17 , 19 , 23 ,29 , 31 , 37 , 41 , 43 , 47 , 53 ,

59 , 61 , 67 , 71 , 73 , 79 ,

Number of primes between 16 and 80 = 16

Prime number between 90 and 100 is 97

Number of prime between 90 and 100 = 1

Therefore ,

Sum of the number of prime between 16 to 80

and 90 to 100 = 16 + 1 = 17

QUESTION: 9

Which of the following statements is not true?

Solution:

QUESTION: 10

The number of distinct prime factors of the largest 4-digit number is

Solution:

Largest 4 digit number = 9999

Prime Factorization of 9999 = 3²*11*101

Distinct prime factors of 9999 = 3, 11 and 101

Therefore, the total number of distinct prime factors of 9999 is 3 (i.e. 3, 11 and 101).

QUESTION: 11

The number of distinct prime factors of the smallest 5-digit number is

Solution:

Option A is right

The smallest 5− digit number =1000.

Now, we find the factor of 10000

∴10000=2×2×2×2×5×5×5×5

Therefore distinct prime factors of 10000 are 2 and 5,

Hence, 2 distinct prime factors.

QUESTION: 12

If the number 7254*98 is divisible by 22, the digit at * is

Solution:

7254698÷22=329759

Therefore 6 is the answer

QUESTION: 13

The largest number which always divides the sum of any pair of consecutive odd numbers is

Solution:

QUESTION: 14

A number is divisible by 5 and 6. It may not be divisible by

Solution:

Take the LCM of 5 and 6 is 30.

5 = 5×1

6 = 3×2×1

LCM of 5 and 6 is

5×1×3×2×1 = 30

And 30 is divisible by 10,15 and 30 but not by 60.

QUESTION: 15

The sum of the prime factors of 1729 is

Solution:

**Prime factorization** calculator of **1729**

Positive Integer **factors of 1729** = 7, 13, 91, 19, **1729** divided by 7, 13, 19, gives no remainder. They are integers and **prime** numbers of **1729**, they are also called composite number.

So, on adding 7, 13, 19, we get the sum = 39.

QUESTION: 16

The greatest number which always divides the product of the predecessor and successor of an odd natural number other than 1, Is

Solution:

3,5,7,9 and so one are other than 1 odd natural number.

Now, the predecessor and successor of 3 are 2 and 4 respectively. The product is ⇒2×4=8

Similarly, the predecessor and successor of 5 are 4 and 6 respectively. The product is ⇒4×6=24 and so on.

Thus, the above shows that greatest number which always divides the product of the predecessor and successor of an odd natural other than 1 is 4

QUESTION: 17

The number of common prime factors of 75, 60, 105 is

Solution:

75 = 3 × 5 × 5

60 = 2 × 2 × 3 × 5

105 = 3 × 5 × 7

Therefore, common prime factors of 75, 60 and 105 are 3 and 5 that is 2 numbers are common.

QUESTION: 18

Which of the following pairs is not coprime?

Solution:

QUESTION: 19

Which of the following numbers is divisible by 11?

Solution:

QUESTION: 20

LCM of 10, 15 and 20 is

Solution:

LCM of 5 ,10, 15, 20 :-

Hence, LCM of 5, 10, 15 and 20 is :-

5 x 2 x 2 x 3 = 60.

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