Playing with Numbers Test

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Playing with Numbers Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

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:

A
$$0$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

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
Question 2
1 / -0

Which of the following number is composite?

A
$$1$$

B
$$4$$

C
$$2$$

D
$$3$$

Solution

A composite number is a positive nteger that has at least one positive divisor other than one and number itself.
Also all the even numbers are composite.
Hence, answer will be B.
Question 3
1 / -0

Number which is divisible by 5 is

A
$$75,620$$

B
$$8,319$$

C
$$724$$

D
$$89,652$$

Solution

$${\textbf{Step - 1 : Describe the rules of divisibility test of 5}}{\text{.}}$$
$$\text{As per the rule , any number can divisible by 5 if and only if the last}$$
$$\text{digit of that number is 0 or 5}{\text{.}}$$
$${\textbf{Step - 2 : Check with divisibility test for the pairs}}{\text{. }}$$
$${\text{For option B ,Here neither 8 nor the last digit of 139 is 0 or 5.}}{\text{.}}$$
$${\text{For option C ,the last digit of 124 ,is 0 or 5}}{\text{.}}$$
$${\text{For option D ,Here neither 89 nor the last digit of 652 is 0 or 5.}}{\text{.}}$$
$${\text{But for option A ,75 and 620 have last digit of 5 and 0}}{\text{.So both can be divided by 5}}{\text{.}}$$
$${\textbf{Hence, the number which is divisible by 5 is 75, 620}}{\text{.}}$$
Question 4
1 / -0

The number of composite number between $$101$$ and $$120$$ are

A
$$11$$

B
$$12$$

C
$$13$$

D
$$None$$

Solution

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 : $$14.$$
Question 5
1 / -0

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

A
units and tens

B
tens and hundreds

C
units

D
units, tens and hundreds

Solution

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 $$4$$.
Question 6
1 / -0

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

A
units and tens

B
tens and hundreds

C
units

D
none of these

Solution

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.
Question 7
1 / -0

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

A
$$3$$

B
$$4$$

C
$$6$$

D
$$9$$

Solution

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 $$9$$.
Question 8
1 / -0

Number which is divisible by 2 is

A
$$7,907$$

B
$$63,195$$

C
$$72,028$$

D
$$3,451$$
Question 9
1 / -0

7248 is divisible by __________.

A
$$2$$

B
$$4$$

C
Both $$(A)$$ and $$(B)$$

D
None of these

Solution

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 $$4$$.
Question 10
1 / -0

HCF of $$36$$ and $$144$$ is ______.

A
$$36$$

B
$$144$$

C
$$4$$

D
$$2$$

Solution

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 $$36$$.

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Re-Attempt Weekly Quiz Competition

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

Playing with Numbers Test - 23

Result

Playing with Numbers Test - 23

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Question 1

$$0$$

$$2$$

$$3$$

$$4$$

Solution

Question 2

$$1$$

$$4$$

$$2$$

$$3$$

Solution

Question 3

$$75,620$$

$$8,319$$

$$724$$

$$89,652$$

Solution

Question 4

$$11$$

$$12$$

$$13$$

$$None$$

Solution

Question 5

units and tens

tens and hundreds

units

units, tens and hundreds

Solution

Question 6

units and tens

tens and hundreds

units

none of these

Solution

Question 7

$$3$$

$$4$$

$$6$$

$$9$$

Solution

Question 8

$$7,907$$

$$63,195$$

$$72,028$$

$$3,451$$

Question 9

$$2$$

$$4$$

Both $$(A)$$ and $$(B)$$

None of these

Solution

Question 10

$$36$$

$$144$$

$$4$$

$$2$$

Solution

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