MCQ Questions for Class 6 Maths Playing With Numbers Quiz 6

Which of the following numbers is divisible by

$3$ ?

0%
$50762$
0%
$42063$
0%
$52871$
0%
$37036$

Explanation

We use a rule to check whether the number is divisible by 3 or not.

A number is divisible by 3 if the sum of the digits is evenly divisible by 3.

Sum of digits is

$= 4 + 2 + 0 + 6 + 3 = 15$ which is divisible by

$3$

A four-digit number

$aabb$ is divisible by

$55$ . Then possible value(s) of

$b$ is/are

0%
$0$ and $2$
0%
$2$ and $5$
0%
$0$ and $5$
0%
$7$

Explanation

If the number is divisible by

$55$ ,

It must also be divisible by

$5$ .

Therefore the number ends with

$0$ or

$5$ .

If the sum of digits of a number is divisible by three then the number is always divisible by

0%
$2$
0%
$3$
0%
$6$
0%
$9$

Explanation

If the sum of digits of a number is divisible by three, then the number is always divisible by

$3$ .

This is the test for divisibility by 3.

Hence, (B) is the correct option.

What are the common factors of numbers

$10$ and

$12$ ?

0%
1,2,3,5
0%
1,4,6
0%
1,2,5,8
0%
1,2

Explanation

Factors of

$10$ are

$1, 2, 5,10$
Factors of

$12$ are

$1, 2, 3,4, 6, 12,$

$\therefore$ common factors are

$1, 2$ .

The number that should be added to

$345670$ in order to make it divisible by

$6$ is

0%
$2$
0%
$4$
0%
$5$
0%
$6$

Explanation

For a number to be divisible by

$6$ , it has to be divisible by both

$2$ and

$3$ .
A number is divisible by

$2$ if it ends with any of

$0,2,4,6,8$ hence,

$345670$ is divisible by

$2$ .

A number is divisible by

$3$ if the sum of its digit is divisible by

$3$ . So,

$3+4+5+6+7+0=25$

If we add

$2$ to the given number then the sum of its digits will become

$27$ and the number becomes

$345672$ which is even as well.

Hence, option

$A$ is correct.

How many numbers between

$9$ and

$54$ are exactly divisible by

$9$ but not by

$3$

0%
$8$
0%
$6$
0%
$5$
0%
None of these

Explanation

Any number which is divisible by

$9$ is also divisible by

$3$ . So, no number between 9 and 54, which is exactly divisible by 9 and not by 3, cant be found.

State true or false:

The common factors of

$18$ and

$24$ are

$1,2,3,6$ .

0%
True
0%
False

Explanation

Factors of

$18 = 1, 2, 3, 6, 8, 9$ and

$18$

Factors of

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

$24$

Common factors are

$1, 2, 3, 6$

Write all the composite numbers between

$10$ and

$35.$

0%
$7, 12, 14, 15, 16, 17, 20, 21, 22, 24, 25, 26, 27, 28, 30,33$ and $34$
0%
$7, 12, 14, 15, 16, 18, 20, 21, 21, 24, 25, 26, 27, 28, 32, 33$ and $34$
0%
$7, 12, 14, 15, 16, 19, 20, 21, 22, 24, 25, 26, 27, 28, 32, 33$ and $34$
0%
None of these

Explanation

All composite numbers between

$10$ and

$35$ are

$12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33$ and

$34.$

State the following statement is True or False

The common factors of

$75$ and

$50$ are

$1,5,25$

0%
True
0%
False

Explanation

Factors of

$50 = 1, 2, 5, 10, 25$ and

$50$

Factors of

$75 = 1,3, 5, 15, 25$ and

$75$

Common factors are

$1, 5, 25$

State the following statement is True or False

The H.C.F of

$90$ and

$162$ is

$18$ .

0%
True
0%
False

Explanation

Factors of

$90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90$
Factors of

$162 = 1, 2, 3, 6, 9, 18, 27, 54, 81, 162$

Common factors are

$= 1, 2, 3, 6, 9, 18$

=> HCF

$= 18$

Say true or false:

$12$ is factor of

$36$ .

0%
True
0%
False

Explanation

$12$ is a factor of

$36$ , if

$12$ divides

$36$ completely.

We know that,

$\dfrac{36}{12}=3$

Hence, the given statement

$12$ is a factor of

$36$ is true.

Which one of the following is the largest prime number of three digits?

0%
$997$
0%
$999$
0%
$991$
0%
$993$

Explanation

There are two numbers

$997$ and

$991$ are prime numbers out of given numbers.

$\therefore$ largest prime no. is

$997$

Option

$A$ is correct.

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 formed

by the digits in

$\text{units and tens}$ places

is divisible by

$4$ .
For example:

$464$ is divisible by

$4$ as

$64$ is divisible by

$4$ .

Hence, option A.

Among the following numbers, number which is divisible by

$5$ is

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

Explanation

A number is divisible by

$5$ if the last digit is

$5\ \text{ or}\ 0.$

$\therefore 75$ and

$620$ are divisible by

$5$ .
Option A is correct.

Which of the following numbers have most number of divisors?

0%
$176$
0%
$182$
0%
$99$
0%
$101$

Explanation

$176 = 2\times 2\times 2\times 2\times 11$

$182 = 2\times 7\times 13$

$99 = 3\times 11\times 3$

$101$ is a prime so has only 1 and itself as factors.
Hence, clearly,

$176$ has the greatest number of divisors.

What is the H.C.F. of

$4 \times 27 \times 3125, 8 \times 9 \times 25\times 7, 16 \times 81 \times 5 \times 11\times 49?$

0%
$180$
0%
$360$
0%
$540$
0%
$1260$

Explanation

We have

$\displaystyle 4\times27\times3125=2\times2\times3\times3\times3\times5\times5\times5\times5\times5$

$\displaystyle 8\times 9\times 25\times 7=2\times2\times2\times3\times3\times 5\times5\times 7$

$\displaystyle 16\times 81\times 5\times 11\times 49=2\times2\times2\times2\times 3\times3\times3\times3\times 5\times 11\times 7\times7$
Therefore, H.C.F.

$=$

$\displaystyle 2\times2\times 3\times3\times 5=4\times 9\times 5=180$

Among the following numbers, number which is divisible by

$2$ is

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

Explanation

A number is divisible by

$2$ if its last digit is an even number (i.e.

$0,2,4,6\ or\ 8$ ).

$\therefore 72,028$ is divisible by

$2$ because its last digit is an even i.e

$8$

Rest all numbers have last digit odd.
So, option C is correct.

(i) Every prime number is odd.

(ii) The product of any two prime numbers is odd.

Which of the above statements(s) is/are correct?

0%
Statement I alone.
0%
Statement II alone.
0%
Both statements I and II.
0%
Neither statement I nor statement II.

Consider the following statements:
A. The sum of two prime numbers is a prime number.
B. The product of two prime numbers is a prime number.
Which of these statements is/are correct ?

0%
Neither A nor B
0%
A alone
0%
B alone
0%
Both A and B

Explanation

Neither the sum nor the product of two prime numbers is a prime number

Eg.

$3+5=8$ not a prime number.

$3\times 5=15$ not a prime number.

HCF of the numbers

$24,36$ and

$92$ is

0%
$24$
0%
$36$
0%
$12$
0%
$4$

Explanation

Factors of $24 =1,2,3,4,6,8,12,24$
Factors of $36= 1,2,3,4,6,9,12,18,36$

Factors of

$92=1,2,4,23,46,92$

Common factors are

$=1,2,4$

$\therefore$ HCF

$=4.$

Choose the co-prime numbers from the following pairs:

0%
$7$ and $63$
0%
$36$ and $25$
0%
$35$ and $21$
0%
$63$ and $81$

Explanation

A set of numbers which do not have any other common factor other than

$1$ are called co-prime.

HCF of

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

$25 = 5 \times 5$ is 1 so they are co prime numbers.

Which of the following statement is true?

0%
1 is the smallest prime number
0%
Every prime number is an odd number
0%
The sum of two prime numbers is always a prime number
0%
None of these

Explanation

${\because}$ All the given statements are false.

Example for co-prime numbers from the below options is ?

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

Explanation

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

Out of the given pair of numbers,
Factors of

$25$ are

$1, 5, 25$
Factors of

$14$ are

$1, 2, 7, 14$
Their common factor is

$1$ .
Hence

$25, 14$ are co-prime numbers.

While other options has common factors other than

$1$ .

For option

$B$ the common factor is

$2$ .

For option

$C$ the common factor is

$3$ .

For option

$D$ the common factor is

$11$ .

HCF of the numbers 120, 144 and 216 is __.

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

Explanation

Given numbers are

$120,\ 144\ and \ 216$ .

To find out: HCF of the given numbers.

On factorising the given numbers, we get:

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

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

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

The common factors amongst the three numbers are:

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

Hence, the HCF of

$120,\ 144\ and \ 216$ is

$24$ .

Example for co-prime numbers is

0%
$27,14$
0%
$18,16$
0%
$9,18$
0%
$11,77$

Explanation

The two numbers which have only

$1$ as their common factor are called co-primes.

Out of the given pair of numbers,
Factors of

$27$ are

$1,3,9,27$

Factors of

$14$ are

$1,2,7,14$
Their common factor is

$1$ .
Hence

$25,14$ are co-prime numbers.

What is the HCF of

$24, 36$ and

$92$ ?

0%
$24$
0%
$36$
0%
$12$
0%
$4$

Explanation

Factors of $24 = 1, 2, 3, 4, 6, 8, 12, 24$
Factors of $36 = 1, 2, 3, 4, 6, 9, 12, 18, 36$

Factors of $92 = 1, 2, 4, 23, 46, 92$

Common factors are $= 1, 2, 4$
$\therefore$ HCF $= 4$

A number which is not a factor of every number is

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

Explanation

${\textbf{Step -1: Solve with example}}$

${\text{As we know Factors are all the numbers that divide a number completely, }}$

${\text{i}}{\text{.e}}{\text{., without leaving any remainder, are called factors of that number}}{\text{.}}$

${\text{Multiples are those numbers which on multiplication give the desired number}}{\text{. }}$

${\text{All these multiples individually can be named as a factor of the given number}}{\text{.}}$

${\textbf{Step -2: Find factors}}$

${\text{To find the factors of any given number, }}$

${\text{we express each number as a product of prime numbers since they do not leave any remainder}}{\text{.}}$

${\text{For example, consider some numbers such as 1, 8, 21, 29}}{\text{.The factor of 1 is 1}}{\text{.}}$

${\text{The factors of 8 are 1, 2, 4 and 8}}{\text{.}}$

${\text{The factors of 21 are 1, 3, 7 and 21}}$

${\text{The factors of 29 are 1 and 29}}{\text{.}}$

${\text{From the above data, we observe that - }}$

${\text{1 is the greatest factor of itself and the smallest factor of any other number}}{\text{.}}$

${\textbf{Hence, the correct answer is option A, C, D}}{\text{. }}$

A number

$\displaystyle a_{1}a_{2}a_{3}a_{4}a_{5}$ is divisible by 9 if
(i)

$\displaystyle a_{1}+a_{2}+a_{3}+a_{4}+a_{5}$ is divisible by

$9.$
(ii)

$\displaystyle a_{1}-a_{2}+a_{3}-a_{4}+a_{5}$ is divisible by

$9.$

Which of the above statements is/are correct?

0%
only (i)
0%
only (ii)
0%
both (i) and (ii)
0%
neither (i) nor (ii)

Explanation

The given number is divisible by

$9$ , if sum of

its digits is divisible by

$9$ .

$\therefore \displaystyle a_{1}+a_{2}+a_{3}+a_{4}+a_{5}$ is divisible by 9

So, option A is correct.

A number is divisible by 9 if the sum of the digits in the number is divisible by 3

0%
True
0%
False

Explanation

Numbers are divisible by

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

$9$ .

For example, the last 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, the statement a number is divisible by

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

$3$ is false.

Among the following, which is the largest

$3$ digit number exactly divisible by

$2$ ?

0%
$340$
0%
$542$
0%
$848$
0%
$921$

Explanation

We know the divisibility rule for

$2$ :
Always check the last digit end with

$0, 2, 4, 6$ or

$8$ .
Here,

$848$ is the largest

$3$ digit number exactly divisible by

$2$ .

So, option C is correct.

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

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Practice Class 6 Maths Quiz Questions and Answers

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

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Practice Class 6 Maths Quiz Questions and Answers

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