MCQ Questions for Class 6 Maths Playing With Numbers Quiz 11

Which of the following is a composite number?

0%
$23$
0%
$29$
0%
$32$
0%
none of these

Explanation

(a)

$23$
Since it cannot be broken into factors
Hence

$23$ is not a composite number
(b)

$29$
Since it cannot be broken into factors
Hence

$29$ is not a composite number
(c)

$32$
Since it can be broken into factors i.e.

$2 \times 2 \times 2 \times 2 \times 2$
Hence

$32$ is a composite number
Option (c) is the correct answer

If the sum of the digits of a number is divisible by

$3$ , then the number itself is divisible by

$9$ .

0%
True
0%
False

Explanation

According to the divisibility Test for

$9$ , for a number to be divisible by

$9$ the sum of digits of the number should be divisible by

$9$ .

And not all the numbers divisible by

$3$ are divisible by

$9$ .

For example:

$15$ is divisible by

$3$ but not

$9$ .

$\textbf{Therefore the given statement is False}.$

State true of false:
Two prime numbers are always co-prime numbers.

0%
True
0%
False

A number with three or more digits is divisible by

$6$ if the number formed by its last two digits (i.e., ones and tens) is divisible by

$6$ .

0%
True
0%
False

Explanation

We know that according to the divisibility test of

$6$ if a number is divisible by both

$2$ and

$3$ then it is divisible by

$6$ , it does not have to do anything with the last two digits of the number.

For example

$112$ is not divisible by

$6$ although the last two digits of

$112$ are divisible by

$6$ .

$\therefore\ \textbf{The given statement is False}.$

Which of the following are not co-primes?

0%
$8,12$
0%
$9,10$
0%
$6,8$
0%
$15,18$

Explanation

If the H.C.F of two numbers are

$1$ they are said to be co-prime.

(a)

$8, 12$
Here they have a common factor

$4$
Hence

$8, 12$ are not co- primes
(b)

$9, 10$
Here they don’t have a common factor
Hence

$9, 10$ are co- primes
(c)

$6, 8$
Here they have a common factor

$2$
Hence

$6, 8$ are not co- primes
(d)

$15, 18$
Here they have a common factor

$3$
Hence

$15, 18$ are not co- primes

State whether the following statements are true (T) or false (F) :
There are infinitely many prime numbers.

0%
True
0%
False

State whether the following statement is true or false.

HCF of an even number and an odd number is always

$1.$

0%
True
0%
False

Explanation

Let us consider an even number

$6$ and an odd number

$9$

The HCF of

$6$ and

$9$ is

$3$

Hence the statement "HCF of an even number and an odd number is always

$1$ " is false.

Which of the following is an odd composite number ?

0%
$7$
0%
$9$
0%
$11$
0%
$12$

Explanation

$9 = 3 \times 3$ , is an odd composite number.

Which of the following pairs is not co prime ?

0%
$8, 10$
0%
$11, 12$
0%
$1, 3$
0%
$31, 33$

Explanation

We know that Numbers which have only

$1$ as a common factor are called co-prime.
In the given options,
Common factors of

$8$ and

$10$ are

$1, 2$ .
Therefore

$8$ and

$10$ are not co-prime numbers

$\therefore$

$\textbf{Correct option is A}$ .

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

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

$\textbf{Step 1: Recall Odd numbers}$ .

Odd numbers are the numbers which are not divisible by

$2$ .

$\textbf{Step 2: Find the sum of consecutive odd numbers}$

Consecutive odd numbers are

$3$ and

$5, 7$ and

$9, 11$ and

$13, 13$ and

$15$ etc

Sum of consecutive odd numbers

$3$ and

$5 = 3 + 5 = 8$ (Divisible by

$2$ and

$4$ and

$8$ )

Sum of consecutive odd numbers

$7$ and

$9 = 7 + 9 = 16$ (Divisible by

$2, 4$ and

$8$ )

Sum of consecutive odd numbers

$13$ and

$15 = 13 + 15 = 28$ (Divisible by

$2, 4$ and

$7$ )

The sum of consecutive odd numbers is always divisible by

$2, 4$ by taking any consecutive odd numbers
So, we can say that the largest number which always divides the sum of any pair of consecutive odd numbers is

$4$ .

$\therefore$

$\textbf{Option B is correct}$

Any two consecutive numbers are co-prime.

0%
True
0%
False

Explanation

The numbers that have only

$1$ as common factors are called coprime numbers.

HCF of any two consecutive numbers is always

$1$

So, two consecutive numbers are always coprime.

For example,

$99$ and

$100$ are coprime.

$\textbf{Therefore the given statement is True}.$

All factors of

$6$ are

0%
$1, 6$
0%
$2 , 3$
0%
$1, 2 , 3$
0%
$1 , 2 , 3 , 6$

Explanation

The factors of

$6$ are

$1 , 2 , 3 , 6$

Which of the following number is divisible by

$8$ ?

0%
$503786$
0%
$505268$
0%
$305678$
0%
$703568$

Explanation

To check if a number is divisible by

$8$ , we check if the last three digits of the number is divisible by

$8$ .

Since, the number formed by hundreds, tens and ones digit is divisible by

$8$ i.e.

$568 \div 8 = 71$ .

Hence, D will be correct answer.

Which of the following number is divisible by

$6$ ?

0%
$560324$
0%
$650374$
0%
$798653$
0%
$750972$

Explanation

We know that the divisibility test of

$6$ states that a number is divisible by

$6$ if the number is even and the sum of the digits is divisible by

$3$ .

Since, the sum of its digit

$= 7 + 5 + 0 + 9 + 7 + 2 = 30$ which is divisible by

$3$ .

and it ends in

$2.$

Hence it is divisible by

$6$

Which of the following is a pair of co-prime numbers ?

0%
$8 , 15$
0%
$3 , 18$
0%
$5 , 35$
0%
$6 , 39$

Explanation

If the H.C.F of two numbers are

$1$ they are said to be co-prime.

$\therefore$

$(14,35)$ is not pair of co-prime.

The factors of

$8$ are

$1 , 2 , 4 , 8$ .

The factor of

$15$ are

$1 , 3 , 5 , 15$ .

The common factor of

$8$ and

$15$ is

$1$

They are co-prime.

Which of the following numbers is divisible by

$4$ ?

0%
$308594$
0%
$506784$
0%
$732106$
0%
$9301538$

Explanation

A number is divisible by

$4$ if the number formed

by the digits in

$\text{units and tens}$ places

is divisible by

$4$ .

Here the number formed by tens and ones digits is

$84$ and divisible by

$4$ i.e.,

$84 \div 4 = 21$ .

A two-digit number

$ab$ is always divisible by

$2$ if

$b$ is an even number.

0%
True
0%
False

Explanation

Rule for divisible by 2.

Any number with the last digit of

$0,2,4,6,8$ are divisible by 2.

A two-digit number is divisible by

$2$ if the unit digit of the number is even.

Hence, given statement is true.

Number of the form

$3N+2$ will leave remainder

$2$ when divided by

$3$ .

0%
True
0%
False

Explanation

True

$3N+2$ is a number whose sum of its digit is

$2$ more than multiple of

$3$ .

$3N$ is completly divisible by 3. hence, remainder=2

$213x27$ is divisible by

$9$ , then the value of

$x$ is

$0$ .

0%
True
0%
False

Explanation

A number is divisible by

$9$ if the sum of its digits is also divisible by

$9$ .

$2+1+3+x+2+7=0,9,18,27,...$

$2+1+3+x+2+7=15+x$
According to the question

$x$ is a single digit.
Hence,

$18=15+x$

$\Rightarrow x=18-15=3$

A three-digit number

$abc$ is divisible by

$5$ if

$c$ is an even number.

0%
True
0%
False

Explanation

False
A three-digit number is divisible by

$5$ if the unit digit of the number is

$0$ or

$5$ .

A four digit number

$abcd$ is divisible by

$4$ if

$ab$ is divisible by

$4$ .

0%
True
0%
False

Explanation

False
A four-digit number is divisible by

$4$ if the number formed by its digit units and

$10's$ place is divisible by

$4$ .

$abc$ is a three digit number, then the number

$abc-a-b-c$ is divisible by

0%
$9$
0%
$90$
0%
$10$
0%
$11$

Explanation

Expandin

$abc$ , we get,

$abc=100a+10b+c$

$\therefore abc-a-b-c=100a+10b+c-a-b-c\\=99a+9b\\=9(11a+b)$

Therefore,

$(abc-a-b-c)$ is divisible by

$9$

A three-digit number

$abc$ is divisible by

$6$ if

$c$ is an even number and

$a+b+c$ is a multiple of

$3$ .

0%
True
0%
False

Explanation

True
If a number is divisible by

$6$ then it is also divisible by

$2$ and

$3$ .

A number is divisible by

$3$ if its unit digit is divisible by

$2$ .
A number is divisible by

$3$ if the sum of its digits is also divisible by

$3$ .

State True or False :
If a number is a factor

$16$ and

$24$ , it is a factor of

$48$

0%
True
0%
False

Explanation

Factor of

$48$ is

$1,2,3,4,6,8,12,16,24$ and

$48$
As

$16$ and

$24$ are factors of

$48$

All even numbers are prime numbers.

0%
True
0%
False

Explanation

False,

$4$ is an even number but not prime numbers.

Write all the factors of 24 and 36 and find out the common factors. Which is the highest common factor of these two numbers?

0%
6
0%
4
0%
12
0%
24

Write all the factors of 16 in circle I and all the factors of 20 in circle II. Write all the common factors in common part of both the circles. Which option shows it correctly?

0%
0%
0%
0%
None of these

Explanation

Factors of

$16$ are

$1, 2,4 , 8 , 16$
Factors of

$20$ are

$1, 2, 4, 5, 10, 20$

$\therefore$ common factors are

$1, 2, 4$ .

Write all the factors of 5, 8 andFind out all the common factors of these numbers. Which is highest common factor?

0%
5
0%
8
0%
10
0%
1

HCF of

$95$ and

$152$ is:

0%
$1$
0%
$19$
0%
$57$
0%
$38$

Explanation

$95 = 5 \times 19$
and

$152 = 2^{3} \times 19$
H.C.F

$= 19$

Using multiplication chart, find all the factors of

$36$ .

0%
1,2,3,5,7,9,12,15,18,36
0%
1,2,3,4,6,9,12,18,36
0%
1,4,7,12,18,36
0%
1,6,9,18,36

CBSE Questions for Class 6 Maths Playing With Numbers Quiz 11 - MCQExams.com

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Practice Class 6 Maths Quiz Questions and Answers

CBSE Questions for Class 6 Maths Playing With Numbers Quiz 11 - MCQExams.com

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Practice Class 6 Maths Quiz Questions and Answers

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