MCQ Questions for Class 6 Maths Playing With Numbers Quiz 7

Example for co-prime numbers is

0%
$25, 14$
0%
$18, 16$
0%
$9, 18$
0%
$11, 77$

Explanation

Two numbers are co-prime if their highest common factor (HCF) is

1

$1$ . For example, the numbers

2

$2$ and

3

$3$ are co-primes to each other.

(a) Let us find the HCF of

25

$25$ and

14

$14$ by factorizing these numbers:

$25=5\times 5\\ 14=2\times 7$

So, HCF

$(25,14)=1$ and therefore,

$25$ and

$14$ are co-prime numbers.

(b) Let us find the HCF of

$18$ and

$16$ by factorizing these numbers:

$18=2\times 3\times 3\\ 16=2\times 2\times 2\times 2$

So, HCF

$(18,16)=2$ and therefore,

$18$ and

$16$ are not co-primes.

$9$ and

$18$ by factorizing these numbers:

$18=2\times 3\times 3\\ 9=3\times 3$

So, HCF

$(9,18)=3\times 3=9$ and therefore,

$9$ and

$18$ are not co-primes.

(d) Let us find the HCF of

$11$ and

$77$ by factorizing these numbers:

$11=11\times 1\\ 77=11\times 7$

So, HCF

$(11,77)=11$ and therefore,

$11$ and

$77$ are not co-primes.

Hence,

$25$ and

$14$ are co-prime numbers.

State whether true or false:

If a number is divisible by 9, it must be divisible by 3.

0%
True
0%
False

Which number is divisible by

$6$ ?

0%
$213$
0%
$468$
0%
$621$
0%
None of these

Explanation

${\textbf{Step - 1: Write divisibility test for 6}}{\textbf{.}}$

${\text{The whole number has to be divisible by }}2.{\text{ A number is divisible by }}2{\text{ if the unit place}}$

${\text{digit of the number is even, i}}{\text{.e}}{\text{. }}0,2,4,6,{\text{ and }}8.$

${\text{The whole number has to be divisible by 3}}{\text{. A number is divisible by 3 if the sum}}$

${\text{of all digits of the number is a multiple of 3 or the sum is exactly divisible by 3}}{\text{.}}$

${\textbf{Step - 2: Check the example using the divisibility test}}{\textbf{.}}$

${\text{Option A, as unit digit is not even so not divisible by 2}}{\text{.}}$

${\text{Now sum up all the digits 2 + 1 + 3 = 6}}.$

${\text{As 6 is multiple of 3, but it does not pass divisibility test of 2 so}}$

${\text{it is}}$

${\text{not divisible by 6}}.$

${\text{Option B, as unit digit is divisible by 2 also when summed up 4 + 6 + 8 = 18 is}}$

${\text{divisible by 3}}{\text{.}}$

${\text{Hence, 468 is divisible by 6 as it fulfills both the condition}}{\text{.}}$

${\text{Option C, as unit digit is not divisible by 2}}{\text{. When summed up 6 + 2 + 1 = 9}}{\text{.}}$

${\text{It is divisible by 3 but not divisible by 2 hence not divisible by 6}}{\text{.}}$

${\textbf{Hence, the number (B) 468 is divisible by 6}}{\textbf{.}}$

State whether true or false:

If a number is divisible by 4, it must be divisible by 8.

0%
True
0%
False

Explanation

A number divisible by

$8$ will also be divisible by

$4$ as

$4$ is a factor of

$8,$ but vice versa may not be true.

Which of the following number is divisible by 8?

0%
23, 624
0%
39, 126
0%
1,20,458
0%
None of these

Explanation

To check if a number is divisible by

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

$8$ .

The last three digits of

$23,624$ are

$624$
Now,

$624 = 8 \times 78$

Hence as

$624$ is divisible by

$8$ , the number

$23,624$ is divisible by

$8$ .

Find the number of factors of

$512$ .

0%
$8$
0%
$10$
0%
$4$
0%
$14$

Explanation

Factors of

$512=1,2,4,8,16,32,64,128,256,512$
Therefore, number of factors of

$512=10$ .

The number

$10$ has four factors:

$1,2, 5$ and

$10$ . The table below lists the number of factors for some numbers

Numbers	Number of factors
$21$	$4$
$23$	$2$
$25$	$3$
$27$	$4$
$29$	$2$

From this, we can say that the number of prime numbers between

$20$ and

$30$ is:

0%
$0$
0%
$2$
0%
$3$
0%
$4$

Explanation

We know that prime numbers have two factors that is

$1$ and number itself.

In the given table only

$23$ and

$29$ have

$2$ factors so there will be two prime numbers that is

$23$ and

$29$ so correct answer will be option B

State true or false:

If a number is divisible by

$3$ , it must be divisible by

$9$ .

0%
True
0%
False

Explanation

A number divisible by

$9$ will be divisible by

$3$ as

$3$ is a factor of

$9$ , but vice versa may not be true.

For eg.

$6$ is divisible by

$3$ but it is not divisible by

$9$ .

Hence, the given statement is false.

State whether true or false:

If a number is divisible by 8, it must be divisible by 4.

0%
True
0%
False

Explanation

A number divisible by

$8$ will be divisible by

$4$ as

$4$ is a factor of

$8$

(Example :

$24$ is divisible by both

$4$ and

$8$ ),

but vice versa may not be true (Example :

$20$ is divisible by

$4$ but not by

$8$ ).

So, the answer is

$True$

Which of the following number is composite?

0%
$1$
0%
$4$
0%
$2$
0%
$3$

How many numbers from 1 to 100 are there each of which is not only exactly divisible by 4 but also has 4 as a digit ?

0%
7
0%
10
0%
20
0%
12

Explanation

The numbers from

$1$ to

$100$ which are exactly divisible by

$4$ are:-

$4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100$

But each number should have

$4$ as its digit.

Therefore, the required numbers are

$4, 24, 40, 44, 48, 64, 84$ .

Clearly, there are 7 such numbers.

Hence, there are 7 numbers from

$1$ to

$100$ each of which is not only exactly divisible by

$4$ but also has

$4$ as a digit.

A number is divisible by 8 if the number formed by units, tens and hundreds places is divisible by 8

0%
True
0%
False

Explanation

Numbers are divisible by

$8$ if the number formed by the last three individual digits is divisible by

$8$ .

For example, the last three digits of the number

$3624$ is

$624$ , which is divisible by

$8$ .

Therefore,

$3624$ is divisible by

$8$ .

Hence, the statement a number is divisible by

$8$ if the number formed by the digit in units, tens and hundreds places is divisible by

$8$ is true.

Among the following numbers, how many numbers are divisible by

$4$ ?

$124, 118, 280, 314, 512$

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

We know the divisibility rule for

$4$ :
Check the last two digits are a multiple of

$4$ or if the last two digits are

$00$ .
Here,

$124, 280$ and

$512$ is divisible by

$4$ because the last two digits are a multiple of

$4$ and

$0$ .

So, option B is correct.

Number which is divisible by 5 is

0%
$75,620$
0%
$8,319$
0%
$724$
0%
$89,652$

Explanation

${\textbf{Step - 1 : Describe the rules of divisibility test of 5}}{\text{.}}$

$\text{As per the rule , any number can divisible by 5 if and only if the last}$

$\text{digit of that number is 0 or 5}{\text{.}}$

${\textbf{Step - 2 : Check with divisibility test for the pairs}}{\text{. }}$

${\text{For option B ,Here neither 8 nor the last digit of 139 is 0 or 5.}}{\text{.}}$

${\text{For option C ,the last digit of 124 ,is 0 or 5}}{\text{.}}$

${\text{For option D ,Here neither 89 nor the last digit of 652 is 0 or 5.}}{\text{.}}$

${\text{But for option A ,75 and 620 have last digit of 5 and 0}}{\text{.So both can be divided by 5}}{\text{.}}$

${\textbf{Hence, the number which is divisible by 5 is 75, 620}}{\text{.}}$

The number of composite number between

$101$ and

$120$ are

0%
$11$
0%
$12$
0%
$13$
0%
$None$

Explanation

A composite number is a positive integer that has atleast one positive divisor other than one or the number itself. In other words, a composite number is any integer greater than one that is not a prime number

Now between

$101$ and

$120,$

Numbers are

$102,104,105,106,108,110,111,112,114,115,116,117,118,119$

Total :

$14.$

Fill in the blanks.
Every composite number can be expressed as a product of ____, and this factorization is unique except for the order in which the prime factors occur.

0%
Co-primes
0%
Twin primes
0%
Primes
0%
All of these

A number is divisible by 4 if the number formed by the digits in ............. places is divisible by 4

0%
units and tens
0%
tens and hundreds
0%
units
0%
units, tens and hundreds

Explanation

A number is divisible by

$4$ if the number is formed by its digits in ten’s place and unit’s place (i.e. the last two digits on its extreme right side) is divisible by

$4$ .

For example, consider the number

$124$ ,

The last two digits on its extreme right side of the number

$124$ is

$24$ which is divisible by

$4$ .

Therefore,

$124$ is divisible by

$4$ .

Hence, a number is divisible by

$4$ if the number formed by the digits in the units and tens places is divisible by

$4$ .

Number which is divisible by 2 is

0%
$7,907$
0%
$63,195$
0%
$72,028$
0%
$3,451$

A number is divisible by 2 if it has 0 or 2 or 4 or 6 or 8 in its units place

0%
True
0%
False

Explanation

A number is divisible by

$2$ if the digit at unit place is either

$0$ or multiple of

$2$ .

So a number is divisible by

$2$ if digit at its units place is

$0, 2, 4, 6$ or

$8$ .

Numbers divisible by

$2$ are called even numbers.

Numbers not divisible by

$2$ are called odd numbers.

Let us consider some of the following numbers which are divisible by

$2$ using the divisibility test.

$346, 218, 106, 100, 194, 152$ here all these numbers is divisible by

$2$ because their units place is either

$0$ or multiple of

$2$ .

Whereas,

$317, 125, 103, 159$ , etc. here all these number is not divisible by

$2$ because their units place is not multiple of

$2$ .

Hence, the statement a number is divisible by

$2$ if it has

$0$ or

$2$ or

$4$ or

$6$ or

$8$ in its units place is true.

A number is divisible by 4 if the number formed by units and tens places is divisible by 4

0%
True
0%
False

Explanation

$\Rightarrow$ A number is divisible by

$4$ if the number formed by its digits in tens and units place is divisible by

$4$ . It is the divisibility rule of

$4$

For example:

$\Rightarrow$ A number

$88312$ is divisible by

$4$ because the number formed by its last two digits i.e.,

$12$ is divisible by

$4$ .

$\Rightarrow$ Hence, the statement a number is divisible by

$4$ if the number formed by units and tens places is divisible by

$4$ is True.

A number is divisible by 3 if the sum of the digits in the number are divisible by 3

0%
True
0%
False

Explanation

Numbers are divisible by

$3$ if the sum of all the individual digits is divisible by

$3$ .

For example, the last sum of the digits of the number

$3627$ is

$3+6+2+7=18$ , and

$18$ is divisible by

$3$ .

Therefore,

$3627$ is divisible by

$3$ .

Hence, the statement a number is divisible by

$3$ if the sum of the digits in the number are divisible by

$3$ is true.

7248 is divisible by __________.

0%
$2$
0%
$4$
0%
Both $(A)$ and $(B)$
0%
None of these

Explanation

The number

$7248$ has

$48$ on its extreme right side which is exactly divisible by

$4$ . When we divide

$48$ by

$4$ we get

$12$ .

Therefore,

$7248$ is divisible by

$4$ .

The number

$7248$ has

$8$ on its unit place which is an even number so,

$7248$ is divisible by

$2$ .

Hence,

$7248$ is divisible by both

$2$ and

$4$ .

The numbers which are not multiples of 2 are called

0%
even
0%
odd
0%
prime
0%
composite

Explanation

Even numbers are those numbers which are divisible by

$2$ .

Odd numbers are those numbers which are not divisible by

$2$ .

Prime numbers are those which have factors

$1$ and itself.

Composite numbers are those which have more than

$2$ factors.

Hence, the number which are not multiples of

$2$ are called odd numbers.

A number is divisible by 5 if it has 0 or 5 in its units place

0%
True
0%
False

Explanation

The divisibility test of

$5$ states that a number is divisible by

$5$ if its last digit is

$0$ or

$5$ . For example:

In the number

$75620$ , the last digit is

$0$ . Therefore,

$75620$ is divisible by

$5$ .

Hence, the statement a number is divisible by

$5$ if it has

$0$ or

$5$ in its units place is true.

Some factor is missing in the factor tree. Fill it up.

0%
$4$
0%
$6$
0%
$3$
0%
$2$

Explanation

$56 = 7\times \underline {8}$

$= 7\times \underline {2\times 4}$

$= 7\times 2\times \underline {2\times 2}$
The missing factor is

$4.$
Therefore

$,$

$A$ is the correct answer.

A number is divisible by 8 if the number formed by the digit in ............. place is divisible by 8.

0%
units and tens
0%
tens and hundreds
0%
units
0%
none of these

Explanation

Numbers are divisible by

$8$ if the number formed by the last three individual digits is divisible by

$8$ .

For example, the last three digits of the number

$3624$ is

$624$ , which is divisible by

$8$ .

Therefore,

$3624$ is divisible by

$8$ .

Hence, a number is divisible by

$8$ if the number formed by the digit in units, tens and hundreds place is divisible by

$8$ . So, none of the given options is correct.

A number is divisible by 9 if the sum of the digits of the number is divisible by ............

0%
$3$
0%
$4$
0%
$6$
0%
$9$

Explanation

Numbers are divisible by

$9$ if the sum of all the individual digits is divisible by

$9$ .

For example, the sum of the digits of the number

$3627$ is

$3+6+2+7=18$ , and

$18$ is divisible by

$9$ .

Therefore,

$3627$ is divisible by

$9$ .

Hence, a number is divisible by

$9$ if the sum of the digits of the number is divisible by

$9$ .

HCF of

$36$ and

$144$ is ______.

0%
$36$
0%
$144$
0%
$4$
0%
$2$

Explanation

To find the HCF of

$36$ and

$144$ , first factorise them:

$36= 2 \times 2 \times 3 \times 3$

$144= 2 \times 2 \times 2 \times 3 \times 3$

Taking common factor, we get

HCF

$= 2 \times 2 \times 3 \times 3$

$=36$

Hence, the answer is

$36$ .

HCF of

$120, 144$ and

$216$ is:

0%
$38$
0%
$24$
0%
$120$
0%
$144$

Explanation

The HCF of

$120,144,216$ is

$120= 2 \times 2 \times 2 \times 3 \times 5$

$144= 2 \times 2 \times 2 \times 2 \times 3 \times 3$

$216= 2 \times 2 \times 2\times 3 \times 3 \times 3$

Since, the Common factors are

$2\times 2\times 2\times 3=24$

Hence, the HCF of

$120,144$ and

$216$ is

$24$ .

The two numbers which have only 1 as their common factor are called ...........

0%
Co-primes
0%
Twin primes
0%
Composite
0%
Even numbers

CBSE Questions for Class 6 Maths Playing With Numbers Quiz 7 - MCQExams.com

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Practice Class 6 Maths Quiz Questions and Answers

CBSE Questions for Class 6 Maths Playing With Numbers Quiz 7 - MCQExams.com

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Practice Class 6 Maths Quiz Questions and Answers

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