Playing with Numbers Test

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Playing with Numbers Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

Among the following numbers, number which is divisible by $$2$$ is

A
$$7,907$$

B
$$63,195$$

C
$$72,028$$

D
$$3,451$$

Solution

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.
Question 2
1 / -0

Find the values of the letters in following, and give reason for the steps involved.
$${\;}A{\;}B$$
$$\underline {\times {\;}6}$$
$$\underline {B{\;}B{\;}B}$$

A
$$A=7, B=4$$

B
$$A=8, B=2$$

C
$$A=7, B=2$$

D
$$A=8, B=4$$

Solution

Given: $${\;}A{\;}B$$
$$\underline {\times {\;}6}$$
$$\underline {B{\;}B{\;}B}$$

$$AB\times6=BBB$$

Numbers in the tens position are multiplied by $$10$$ and in hundreds position are $$x$$ by $$100$$ when we break the number.

$$ (10A+B)\times6=100B+10B+B$$

$$60{ A }+6B=100B+10B+B$$

$$60{ A }=105B$$

$$4{ A }=7B$$

$$\text {Only possible values are }A=7,B=4 \quad (A\& B<10)$$
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

Number which is divisible by 10 is

A
$$9,15,525$$

B
$$61,200$$

C
$$72,831$$

D
$$1,141$$

Solution

We know that the divisibility test of $$10$$ states that a number is divisible by $$10$$ if its last digit is $$0$$.

Now, consider the number $$61200$$, the last digit of the number is $$0$$. Therefore, $$61200$$ is divisible by $$10$$.

Hence, the number $$61200$$ is divisible by $$10$$..
Question 5
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 6
1 / -0

The unit digit of a number of two digits is $$y$$ and the tens digit is $$(y-x)$$, then the number will be

A
$$10(y-x)y$$

B
$$(y-x)y$$

C
$$11y-x$$

D
$$11y-10x$$

Solution

Required Number $$=10(y-x)+y$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=10y-10x+y$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=11y-10x$$
Question 7
1 / -0

Number which is divisible by 2 is

A
$$7,907$$

B
$$63,195$$

C
$$72,028$$

D
$$3,451$$
Question 8
1 / -0

The expanded form of $$67$$ is

A
$$66+1$$

B
$$65+2$$

C
$$60 \times 1 + 7$$

D
None of the above

Solution

The general form of two digits number will be, $$ab= a \times 10 + b $$

Therefore, the expand form of $$67$$ = $$60\times 1+7$$
Question 9
1 / -0

The numbers which are not multiples of 2 are called

A
even

B
odd

C
prime

D
composite

Solution

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.
Question 10
1 / -0

Among the following, which is the largest $$3$$ digit number exactly divisible by $$2$$?

A
$$340$$

B
$$542$$

C
$$848$$

D
$$921$$

Solution

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.

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Re-Attempt Weekly Quiz Competition

Playing with Numbers Test - 13

Result

Playing with Numbers Test - 13

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SHARING IS CARING

Question 1

$$7,907$$

$$63,195$$

$$72,028$$

$$3,451$$

Solution

Question 2

$$A=7, B=4$$

$$A=8, B=2$$

$$A=7, B=2$$

$$A=8, B=4$$

Solution

Question 3

$$75,620$$

$$8,319$$

$$724$$

$$89,652$$

Solution

Question 4

$$9,15,525$$

$$61,200$$

$$72,831$$

$$1,141$$

Solution

Question 5

$$3$$

$$4$$

$$6$$

$$9$$

Solution

Question 6

$$10(y-x)y$$

$$(y-x)y$$

$$11y-x$$

$$11y-10x$$

Solution

Question 7

$$7,907$$

$$63,195$$

$$72,028$$

$$3,451$$

Question 8

$$66+1$$

$$65+2$$

$$60 \times 1 + 7$$

None of the above

Solution

Question 9

even

odd

prime

composite

Solution

Question 10

$$340$$

$$542$$

$$848$$

$$921$$

Solution

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