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Olympiad Test: Whole Numbers - Class 6 MCQ


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20 Questions MCQ Test Mathematics (Maths) Class 6 - Olympiad Test: Whole Numbers

Olympiad Test: Whole Numbers for Class 6 2024 is part of Mathematics (Maths) Class 6 preparation. The Olympiad Test: Whole Numbers questions and answers have been prepared according to the Class 6 exam syllabus.The Olympiad Test: Whole Numbers MCQs are made for Class 6 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: Whole Numbers below.
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Olympiad Test: Whole Numbers - Question 1

Which of the following numbers is a prime number?

Detailed Solution for Olympiad Test: Whole Numbers - Question 1

97 is a prime number because it has no divisors other than 1 and itself.To determine which number is a prime number among the options given:

- 123: Not a prime number because it is divisible by 3 (3 x 41).
- 97: This is a prime number because it has exactly two factors: 1 and 97.
- 121: Not a prime number as it is divisible by 11 (11 x 11).
- 105: Not a prime number as it is divisible by 3 and 5 (3 x 5 x 7).

Therefore, the prime number among the options is 97.

Olympiad Test: Whole Numbers - Question 2

Which of the following is the largest 6-digit number that can be formed using the digits 5, 7, 3, 0, 8, 9 without repetition?

Detailed Solution for Olympiad Test: Whole Numbers - Question 2
The largest 6-digit number formed by arranging the digits in descending order is 987530.
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Olympiad Test: Whole Numbers - Question 3

What is the sum of the smallest 3-digit number and the largest 4-digit number?

Detailed Solution for Olympiad Test: Whole Numbers - Question 3

The largest 4 digit number is 9999 and the smallest 3 digit is 100. The sum is 10099.

Olympiad Test: Whole Numbers - Question 4

How many whole numbers are there between 25 and 75, inclusive?

Detailed Solution for Olympiad Test: Whole Numbers - Question 4

So, the number of whole numbers between 25 and 75, inclusive, is:

75 − 25 + 1 = 51

There are 51 whole numbers between 25 and 75, inclusive.

Olympiad Test: Whole Numbers - Question 5
If the product of two whole numbers is 150 and one of them is 10, what is the other number?
Detailed Solution for Olympiad Test: Whole Numbers - Question 5
The product of 10 and 15 is 150, so the other number is 15.
Olympiad Test: Whole Numbers - Question 6

What is sum of first 10 odd numbers ?

Detailed Solution for Olympiad Test: Whole Numbers - Question 6

To find the sum of the first 10 odd numbers:

- Odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
- Adding these numbers: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100
- The sum of the first 10 odd numbers is 100.
- Therefore, option B (100) is the correct answer.

Olympiad Test: Whole Numbers - Question 7
What is the remainder when 123456789 is divided by 9?
Detailed Solution for Olympiad Test: Whole Numbers - Question 7
The sum of the digits of 123456789 is 45, which is divisible by 9, hence the remainder is 0.
Olympiad Test: Whole Numbers - Question 8
What is the sum of the first 50 whole numbers?
Detailed Solution for Olympiad Test: Whole Numbers - Question 8
The sum of the first n whole numbers is n(n + 1)/2. For the first 50 whole numbers, it is 50*51/2 = 1275.
Olympiad Test: Whole Numbers - Question 9
What is the smallest whole number that is divisible by both 8 and 12?
Detailed Solution for Olympiad Test: Whole Numbers - Question 9
The smallest whole number divisible by both 8 and 12 is their LCM, which is 24.
Olympiad Test: Whole Numbers - Question 10
If 2x - 3 = 7, what is the value of x?
Detailed Solution for Olympiad Test: Whole Numbers - Question 10
Solving the equation 2x - 3 = 7, we get 2x = 10, so x = 5.
Olympiad Test: Whole Numbers - Question 11

What is the smallest number that when divided by 4, 6, and 9 leaves a remainder of 1?

Detailed Solution for Olympiad Test: Whole Numbers - Question 11

To find the smallest number that satisfies the given condition, we can use the concept of the least common multiple (LCM) of 4, 6, and 9.

The LCM of 4, 6, and 9 is the smallest number that is divisible by all three numbers.

The prime factorization of these numbers is:

  • 4 = 2^2
  • 6 = 2 * 3
  • 9 = 3^2

To find the LCM, we take the highest power of each prime factor that appears in any of the factorizations:

LCM(4, 6, 9) = 2^2 * 3^2 = 4 * 9 = 36

So, the smallest number that when divided by 4, 6, and 9 leaves a remainder of 1 is 36 + 1 = 37.

Olympiad Test: Whole Numbers - Question 12
How many whole numbers between 100 and 200 are divisible by 3?
Detailed Solution for Olympiad Test: Whole Numbers - Question 12
Whole numbers divisible by 3 between 100 and 200 are 102, 105, ..., 198. There are 34 such numbers.
Olympiad Test: Whole Numbers - Question 13
If the square of a whole number is 961, what is the number?
Detailed Solution for Olympiad Test: Whole Numbers - Question 13
The square root of 961 is 31, so the number is 31.
Olympiad Test: Whole Numbers - Question 14

If x^3 = 729, what is the value of x?

Detailed Solution for Olympiad Test: Whole Numbers - Question 14

Olympiad Test: Whole Numbers - Question 15
What is the sum of all whole numbers from 1 to 100?
Detailed Solution for Olympiad Test: Whole Numbers - Question 15
The sum of the first 100 whole numbers is 100*101/2 = 5050.
Olympiad Test: Whole Numbers - Question 16
If a number is divisible by both 2 and 3, it must also be divisible by:
Detailed Solution for Olympiad Test: Whole Numbers - Question 16
A number divisible by both 2 and 3 must be divisible by 6.
Olympiad Test: Whole Numbers - Question 17

What is the greatest 5-digit number divisible by 8?

Detailed Solution for Olympiad Test: Whole Numbers - Question 17

Since we know that 99999 is the largest 5-digjt no. So when we divide it by 8 we get 7 as remainder.

When 7 is subtracted from 99999 , 99992 is obtained which is divisible by 8 and is the highest 5-digit no. Divisble by 8.

Another method :-

Since, we know that if a number’s last three digits are divisible by 8 then the whole no. Is divisible by 8 ( for more information search divisibility rule of 8 on google ) .

So on dividing 999 by 8 , 7 is obtained as a remainder and we know that ,

Dividend = Divisor *Quotient + remainder

=> Dividend - remainder = divisor * quotient

=> if remainder is subtracted from the dividend then it is divisible by the divisor.

Therefore ,(999–7) is divisible by 8

=> 992 is divisible by 8 .

As I earlier wrote that if a number's last three digits are divisible by 8 then the whole no. Is divisible by 8 .

992 is divisible by 8

Therefore , 99992 is also divisible by 8 and is the highest 5 digit no. Divisible by 8.

Olympiad Test: Whole Numbers - Question 18

Which of the following is smallest factor of 2310?

Detailed Solution for Olympiad Test: Whole Numbers - Question 18

2310 = 2 * 3 * 5 * 7 * 11, and 15 is a factor since 2310 is divisible by 15. Among the above options 15 is the smallest factor of 2310.

Olympiad Test: Whole Numbers - Question 19

What is the product of the smallest 2-digit prime number and the largest 1-digit prime number?

Detailed Solution for Olympiad Test: Whole Numbers - Question 19

The smallest 2-digit prime number is 11 and the largest 1-digit prime number is 7. Their product is 11 * 7 = 77.

Olympiad Test: Whole Numbers - Question 20
If the LCM of two numbers is 84 and their HCF is 6, what is the product of the two numbers?
Detailed Solution for Olympiad Test: Whole Numbers - Question 20
The product of two numbers is equal to the product of their LCM and HCF. Hence, 84 * 6 = 504.
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