Playing with Numbers Test

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Playing with Numbers Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

Which number is divisible by $$9$$?

A
$$863$$

B
$$932$$

C
$$752$$

D
$$837$$

Solution

$${\textbf{Step -1: Sum of digits of 863 = 17}}$$

$${\text{17 is not divisble by 9}}$$

$${\textbf{Step -2: Sum of digits of 932 = 14}}$$

$${\text{14 is not divisible by 9}}$$

$${\textbf{Step -3:Sum of digits of 752 = 14}}$$

$${\text{14 is not divisible by 9}}$$

$${\textbf{Step -4: Sum of digits of 837 = 18}}$$

$${\text{18 is divisible by 9}}$$

$${\textbf{Hence, 837 is divisible by 9 .}}$$
Question 2
1 / -0

Find the values of the letters:
$$ \ 3\ 2$$
+ $$\ 7 \ A$$
_______
$$C\ B\ 6$$

A
$$A = 4, B = 0, C = 1$$

B
$$A = 3, B = 0, C = 1$$

C
$$A = 4, B = 1, C = 0$$

D
$$A = 4, B = 1, C = 1$$

Solution

$$ \ 3\ 2$$
+$$\ 7\ A$$
_______
$$C\ B\ 6$$
The addition of $$2$$ and $$A$$ is giving $$6$$ i.e., $$2 + A = 6$$
$$A = 6 - 2$$
$$A = 4$$
Then, $$3 + 7 = B$$
$$B = 10$$, so $$1$$ will be carry for the next step.
Therefore, the addition is as follows:
$$ 3 2$$
+ $$7 4$$
_______
$$1 0 6$$
Hence, $$A, B$$ and $$C$$ are $$4, 0$$ and $$1$$ respectively.
Question 3
1 / -0

Find the values of the letters:
$$ 8 A$$
x $$A$$
_____
$$B B 6$$

A
$$A = 4, B = 3$$

B
$$A = 3, B = 4$$

C
$$A = 0, B = 2$$

D
$$A = 4, B = 0$$

Solution

$$ 8 A$$
$$\times A$$
_____
$$B B 6$$
The multiplication of $$A$$ and $$A$$ is giving $$6$$ i.e., $$A \times A = 6$$
The above condition is possible only when digit $$A$$ is $$4$$, because $$4 \times 4 = 16$$.
Ones digit will be $$6$$. $$1$$ will be carry for the next step to get the value as $$33$$.
Therefore, the multiplication is as follows:
$$8 4$$
X $$ 4$$
____
$$33 6$$
Question 4
1 / -0

Find the value of $$C$$.
$$C C$$
x $$6$$
____
$$1 3 C$$

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$C C$$
x $$6$$
____
$$1 3 C$$
The multiplication of $$C$$ and $$6$$ is giving $$C$$ i.e., $$C \times 6 = C$$.
The above condition is possible only when digit $$C$$ is $$2$$, because $$2 \times 6 = 12.$$
Ones digit will be $$2, 1$$ will be carry for the next step to get the value as $$13$$.
Therefore, the multiplication is as follows:
1
$$ 2 2$$
X $$ 6$$
____
$$13 2$$
The value of $$C$$ is $$2$$.

So, option A is correct.
Question 5
1 / -0

What are the value of $$A$$ and $$B$$?
$$\ A\ B$$
+$$\ 7\ \ 9$$
    ___
$$\ \ 9\ A$$

A
$$A = 1, B = 1$$

B
$$A = 2, B = 1$$

C
$$A = 1, B = 2$$

D
$$A = 2, B = 0$$

Solution

$$A\ B$$
+$$ 7\ 9$$
    ___
$$ 9\ A$$

The addition of $$B$$ and $$9$$ gives $$A$$ i.e., $$B + 9 = A$$
The above condition is possible only when digit $$A$$ is $$1$$, because $$A + 7 = 9$$.
$$1$$ will be the carry for the next step to get the value $$9$$. i.e.,$$1 + 1 + 7 = 9$$
So, the value of $$A = 1$$.

The value of $$B$$ must be $$2$$, because the value of $$A$$ is $$1$$.
$$2 + 9 = 11$$, $$1$$ will be the carry for the next step.
Therefore, the addition is as follows:
$$ 1 2$$
+ $$7 9$$
    ___
$$9 1$$
The value of $$A = 1, B = 2$$.
Question 6
1 / -0

Find $$A$$ in the addition:
$$ \ \ 7\ \ 0 \ \ A$$
+ $$\ \ A\ \ 1\ \ 5$$
__________
$$1\ 5 \ 2\ \ 3$$

A
$$4$$

B
$$1$$

C
$$2$$

D
$$8$$

Solution

Given the expression that $$70A+A15=1523$$
Now here $$A+5=3$$ but $$0+1=2$$ it means there a carry from $$A+5$$
So that if $$A=8$$ add $$5$$ to it we get $$13$$ then $$3$$ will be written and 1 will be carried.
So, $$A+7=15$$
$$\implies A=15-7$$
$$\implies A=8$$
Question 7
1 / -0

Find the value of letter $$B$$:
$$ \ \ 6\ \ B\ \ \ 2$$
+ $$\ \ 4\ \ 2\ \ B$$
______________
$$\ 1\ 0\ \ 2\ \ 2$$

A
$$0$$

B
$$2$$

C
$$1$$

D
$$3$$

Solution

$$ 6\ B\ 2$$
+$$ 4\ 2\ B$$
_____
$$10 \ 2 \ 2$$
From ones column we get, $$2 + B = 2$$
$$B = 2 - 2 = 0$$
So, the value of $$B$$ is $$0$$.
Question 8
1 / -0

Find the values of $$A$$ and $$B$$.
$$A B$$
x $$A B$$
_____
$$5 A\ 9$$

A
$$A = 1, B = 1$$

B
$$A = 1, B = 3$$

C
$$A = 2, B = 3$$

D
$$A = 2, B = 1$$

Solution

$$ A B$$
x$$ A B$$
_____
$$5 A 9$$
The multiplication of $$B$$ and $$B$$ is giving $$9$$ i.e., $$B \times B = 9$$.
The above condition is possible only when digit $$B$$ is $$3$$, because $$3 \times 3 = 9.$$
Ones digit will be $$9$$, $$A \times A = A$$ the next step to get the value as $$5A$$.
If $$A = 1, 13 \times 13 = 169$$. But the product should be more than $$500.$$
If $$A = 2, 23 \times 23 = 529$$
Therefore, the multiplication is as follows:
$$ 2 3$$
X $$2 3$$
____
$$ 52 9$$
The value of $$A = 2, B = 3.$$

So, option C is correct.
Question 9
1 / -0

Find the numbers in place of letters:
$$B A 3$$
- $$A B$$
_____
$$B\ 0\ 2$$

A
$$A = 2, B = 0$$

B
$$A = 0, B = 1$$

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

D
$$A = 2, B = 1$$

Solution

The subtraction of $$3$$ and $$B$$ is giving $$2$$ i.e., $$3 - B = 2$$.
The above condition is possible only when digit $$B$$ is $$1$$, because $$3 - 1 = 2.$$
The next step to get the value of $$A$$ as $$A - A = 0$$. $$A = 2$$.value of A can be any digit so from given options we ake $$A=2$$
Therefore, the subtraction is as follows:
$$ 1 2 3$$
-$$ 2 1$$
______
$$ 1 0 2$$
The numbers are in the place of $$A = 2$$ and $$B = 1$$.

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

Find the digits $$A$$ and $$B$$.
$$ B\ 5$$
x $$ B A$$
_____
$$ 1225$$

A
$$A = 5$$ and $$B = 3$$

B
$$A = 3$$ and $$B = 5$$

C
$$A = 2$$ and $$B = 1$$

D
$$A = 3$$ and $$B = 2$$

Solution

Ones digit is $$5 \times A = 5$$.
So $$A = 3$$ or $$A = 5$$.
If $$B = 2, 25 \times 25 = 625$$. But the product should be more than $$600$$.
If $$B = 3, 35 \times 35 = 1,225$$
Therefore, the multiplication is as follows:
$$ 3 5$$
X$$ 3 5$$
_____
$$1225$$
The digits $$A = 5$$ and $$B = 3$$.

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Re-Attempt Weekly Quiz Competition

Playing with Numbers Test - 16

Result

Playing with Numbers Test - 16

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SHARING IS CARING

Question 1

$$863$$

$$932$$

$$752$$

$$837$$

Solution

Question 2

$$A = 4, B = 0, C = 1$$

$$A = 3, B = 0, C = 1$$

$$A = 4, B = 1, C = 0$$

$$A = 4, B = 1, C = 1$$

Solution

Question 3

$$A = 4, B = 3$$

$$A = 3, B = 4$$

$$A = 0, B = 2$$

$$A = 4, B = 0$$

Solution

Question 4

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 5

$$A = 1, B = 1$$

$$A = 2, B = 1$$

$$A = 1, B = 2$$

$$A = 2, B = 0$$

Solution

Question 6

$$4$$

$$1$$

$$2$$

$$8$$

Solution

Question 7

$$0$$

$$2$$

$$1$$

$$3$$

Solution

Question 8

$$A = 1, B = 1$$

$$A = 1, B = 3$$

$$A = 2, B = 3$$

$$A = 2, B = 1$$

Solution

Question 9

$$A = 2, B = 0$$

$$A = 0, B = 1$$

$$A = 2, B = 2$$

$$A = 2, B = 1$$

Solution

Question 10

$$A = 5$$ and $$B = 3$$

$$A = 3$$ and $$B = 5$$

$$A = 2$$ and $$B = 1$$

$$A = 3$$ and $$B = 2$$

Solution

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