Playing with Numbers Test

Result

Playing with Numbers Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following numbers are divisible by $$5$$ ?

A
$$125$$

B
$$625$$

C
$$1000$$

D
All of the above

Solution

A number is divisible by $$5$$, if it contains $$0$$ or $$5$$ in the units place.
Since all the numbers have $$0$$ or $$5$$ in the units place,
therefore all are divisible by $$5$$.

So, option D is correct.
Question 2
1 / -0

Which of these numbers are divisible by $$2$$?

A
$$211$$

B
$$303$$

C
$$200$$

D
$$101$$

Solution

We know the divisibility rule for $$2$$:
Always check the last digit end with $$0, 2, 4, 6$$ or $$8$$.
Here, $$200$$ is divisible by $$2$$ because the last digit ends with $$0$$.

So, option C is correct.
Question 3
1 / -0

Standard form of
$$900000000 + 3000 + 7$$ is:

A
$$90003007$$

B
$$900003007$$

C
$$900003000$$

D
$$9000307$$

Solution

The standard form of $$900000000+3000+7$$ is $$900003007$$.
Hence, the answer is $$900003007$$.
Question 4
1 / -0

The expanded form of $$999$$ is

A
$$1000 - 3$$

B
$$1005 - 6$$

C
$$1000 \times 1 -1$$

D
None of the above

Solution

Expanded form of 999 is
$$999 = 100 \times 9 + 9\times10+9$$
Therefore none of the answers are correct.
Question 5
1 / -0

Which of the following is divisible by $$3$$ ?

A
$$18$$

B
$$60$$

C
$$24$$

D
All of the above

Solution

A number is divisible by $$3$$ if the sum of its digits is divisible by $$3$$.
$$18 = 1 + 8 = 9$$
$$60 = 6 + 0 = 6$$
$$24 = 4 + 2 = 6$$
Therefore, all are divisible by $$3$$.
So, option D is correct.
Question 6
1 / -0

Which of the following numbers are divisible by $$3$$ ?

A
$$21$$`

B
$$24$$

C
$$27$$

D
All of the above

Solution

A number is divisible by $$3$$ only when sum of the digits of the number is divisible by $$3$$
Sum of digits of given numbers :
$$ 21 = 2+1=3$$
$$24 = 2+4=6$$
$$27 = 2+7=9$$
$$3,6,9$$ are divisible by $$3$$.
$$\therefore $$ All are divisible by $$3$$.
So, option D is correct.
Question 7
1 / -0

Which of the following numbers are divisible by $$5$$ ?

A
$$60$$

B
$$50$$

C
$$120$$

D
All of the above

Solution

A number is divisible by $$5$$ if it contains a $$0$$ or $$5$$ in the units place.
Since all the numbers have $$0$$ in the units place, therefore all are divisible by $$5$$.
So, option D is correct.
Question 8
1 / -0

Find the largest number divisible by $$10?$$

A
$$86340$$

B
$$98002$$

C
$$12976$$

D
$$12755$$

Solution

We know the divisibility rule for $$10$$: The last digit is $$0$$
Here, $$86340$$ is divisible by $$10$$.
Since the last digit is $$0$$.
So, option A is correct.
Question 9
1 / -0

Find the number divisible by $$3$$.

A
$$602$$

B
$$563$$

C
$$991$$

D
$$753$$

Solution

We know the divisibility rule for $$3$$:
If the sum of the digits of a number is divisible by $$3$$, then the number is also divisible by $$3$$.
Here, $$7 + 5 + 3 = 15$$ is divisible by $$3$$.

So, option D is correct.
Question 10
1 / -0

What is the largest $$4$$ digit number exactly divisible by $$10$$?

A
$$9870$$

B
$$9840$$

C
$$9990$$

D
$$9980$$

Solution

Here, $$9990$$ is the largest number divisible by $$10$$.
According to the divisibility rule, the last digit is $$0$$.
All options are divisible by $$10$$, but the largest number is $$9990$$
So, option C is correct.

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Re-Attempt Weekly Quiz Competition

Playing with Numbers Test - 14

Result

Playing with Numbers Test - 14

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SHARING IS CARING

Question 1

$$125$$

$$625$$

$$1000$$

All of the above

Solution

Question 2

$$211$$

$$303$$

$$200$$

$$101$$

Solution

Question 3

$$90003007$$

$$900003007$$

$$900003000$$

$$9000307$$

Solution

Question 4

$$1000 - 3$$

$$1005 - 6$$

$$1000 \times 1 -1$$

None of the above

Solution

Question 5

$$18$$

$$60$$

$$24$$

All of the above

Solution

Question 6

$$21$$`

$$24$$

$$27$$

All of the above

Solution

Question 7

$$60$$

$$50$$

$$120$$

All of the above

Solution

Question 8

$$86340$$

$$98002$$

$$12976$$

$$12755$$

Solution

Question 9

$$602$$

$$563$$

$$991$$

$$753$$

Solution

Question 10

$$9870$$

$$9840$$

$$9990$$

$$9980$$

Solution

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