Playing with Numbers Test

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Playing with Numbers Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

If $$abc$$ is a three digit number, then the number $$abc-a-b-c$$ is divisible by

A
$$9$$

B
$$90$$

C
$$10$$

D
$$11$$

Solution

Expandin $$abc$$, we get,
$$abc=100a+10b+c$$

$$\therefore abc-a-b-c=100a+10b+c-a-b-c\\=99a+9b\\=9(11a+b)$$

Therefore, $$(abc-a-b-c)$$ is divisible by $$9$$
Question 2
1 / -0

If $$5\text{A}+\text{B}3=65$$, then the value of $$A$$ and $$B$$ is

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

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

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

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

Solution

Given,
$$\ \ \ \ \ 5\ \text{A}$$
$$+\ \ \text{B}\ 3$$
$$\overline{\ \ \ \ \ 6\ 5\ \ }$$

In $$1'$$s column,
$$A+3=5$$ or a number whose unit digit is $$5$$
$$\because A$$ is a single digit so the value to be taken is from $$0$$ to $$9$$.
If we take the maximum value i.e., $$9$$,
Then, $$A+3=9+3=12\neq 5$$
Therefore, $$A=5-3=2$$, and there is no carry forward.

In $$10'$$s column,
$$5+B=6$$
$$\Rightarrow B=6-5=1$$

Hence, $$A=2$$ and $$B=1$$.
And option (C) is the correct answer.
Question 3
1 / -0

If $$A3+8B=150$$, then the value of $$A+B$$ is

A
$$13$$

B
$$12$$

C
$$17$$

D
$$15$$

Solution

Given,
$$\ \ \ A\ 3$$
$$+\ 8\ B$$
$$\overline{\ \ 1\ 5\ 0\ \ \ \ }$$

In $$1'$$s column,
$$B+3=10$$
$$\Rightarrow B=7$$ and $$1$$ is carried forward.

In $$10's$$ column,
$$A+8+1=15$$
$$\Rightarrow A=6$$

Therefore, $$A=6$$ and $$B=7$$
And, $$A+B=6+7=13$$

Hence, optiom (A) is correct.
Question 4
1 / -0

If $$6 A\times B=A 8 B$$ , then the value of $$A-B$$ is

A
$$-2$$

B
$$2$$

C
$$-3$$

D
$$3$$

Solution

$$\textbf{Step 1: Finding the relationship between A and B}$$

$$6A\times B=A8B$$

$$\therefore \text{ The unit place of the number when multiplying A and B is B}$$

$$\therefore \text{ The unit place of the number when multiplying 6 and B is 8}$$

$$\text{The only digit when multiplied with 6 that give 8 is 3}$$

$$\Rightarrow B=3$$

$$\text{Since }A\times B\text{ gives B}$$

$$\Rightarrow A=1$$

$$\textbf{Step 2: Finding A-B}$$

$$A-B=1-3=-2$$

$$\textbf{Hence, the value of A-B is -2.}$$
Question 5
1 / -0

The general form of $$456$$ is

A
$$ (4 \times 100)+(5 \times 10)+(6 \times 1) $$

B
$$ (4 \times 100)+(6 \times 10)+(5 \times 1) $$

C
$$ (5 \times 100)+(4 \times 10)+(6 \times 1) $$

D
$$ (6 \times 100)+(5 \times 10)+(4 \times 1) $$

Solution

The given number is $$456$$.

The general form of $$456$$ is $$ (4 \times 100)+(5 \times 10)+(6 \times 1) $$

Hence, $$Op-A$$ is correct.
Question 6
1 / -0

The value of A is :

A
$$9$$

B
$$7$$

C
$$6$$

D
$$1$$

Solution

As the right hand digit of the sum of $$A + B + C = C,$$
it means that $$A+ B = 10$$ so $$A$$ or $$B,$$ none can be $$0.$$
Also, the sum of $$A + B + C$$ can not exceed $$19.$$
So, $$1$$ is carried to the second column.
Now, $$1 +A + B + C$$ gives $$A$$ as the right digit of the sum.
So, $$A= C +1.$$
Further, again $$1$$ is carried over to the third column as A
cannot be 0 and the sum of $$1 + A + B + C$$ can be at most 19.
On adding $$A + B + C + 1$$ we get BA, thus here $$B = 1.$$
Using $$A+ B = 10,A= 10 -1= 9$$
Also, $$A= C + 1$$
$$\Rightarrow C =A-1= 9-1=8.$$
So, $$A= 9, B = 1$$ and $$C = 8$$ satisfy all the conditions.

So, option A is correct.
Question 7
1 / -0

Which of the following is/are true?
(i) $$A = B + C$$
(ii) $$A$$ is the greatest digit integer
(iii) $$C = 0$$

A
i & iii

B
ii & iii

C
i & ii

D
iii only

Solution

As the right hand digit of the sum of $$A + B + C = C,$$
it means that $$A+ B = 10$$ so $$A$$ or $$B,$$ none can be $$0.$$ Also, the sum of $$A + B + C$$ can not exceed $$19$$, As A+B+C=C,
So, $$1$$ is carried to the second column.
Now, $$1 +A + B + C$$ gives $$A$$ as the right digit of the sum.
So, $$A= C +1.$$
Further, again $$1$$ is carried over to the third column as $$A$$ cannot be $$0$$ and the sum of $$1 + A + B + C$$ can be at most $$19$$.
On adding $$A + B + C + 1$$ we get $$BA$$, thus here $$B = 1.$$
Using $$A+ B = 10,A= 10 -1= 9$$
Also, $$A= C + 1$$
$$\Rightarrow C =A-1= 9-1=8$$.
So, $$A= 9, B = 1$$ and $$C = 8$$ satisfy all the conditions.

(i) $$A=9$$
$$B+C = 8+1 = 9$$
$$\therefore A=B+C $$

(ii) $$9>8>1$$
$$\therefore A>C>B$$

(iii) $$C=8 \neq 0$$
$$\therefore$$ this statement is incorrect.
Question 8
1 / -0

If you rearrange the letters 'BARBIT', you would have the name of a/an ____.

A
Ocean

B
Country

C
Animal

D
City

Solution
Question 9
1 / -0

A box contains 3 red balls 4 white balls and 7 black balls one ball is chosen at random what the probability of choosing the black ball is.

A
2/5

B
2/4

C
1/3

D
1/2

Solution
Question 10
1 / -0

$$1, 3, 5, 8, 10, 13$$. How many different sums can be made by adding any two different numbers from the list?

A
$$6$$

B
$$8$$

C
$$10$$

D
$$12$$

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Re-Attempt Weekly Quiz Competition

Playing with Numbers Test - 19

Result

Playing with Numbers Test - 19

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SHARING IS CARING

Question 1

$$9$$

$$90$$

$$10$$

$$11$$

Solution

Question 2

$$A=2, B=3$$

$$A=3, B=2$$

$$A=2, B=1$$

$$A=1, B=2$$

Solution

Question 3

$$13$$

$$12$$

$$17$$

$$15$$

Solution

Question 4

$$-2$$

$$2$$

$$-3$$

$$3$$

Solution

Question 5

$$ (4 \times 100)+(5 \times 10)+(6 \times 1) $$

$$ (4 \times 100)+(6 \times 10)+(5 \times 1) $$

$$ (5 \times 100)+(4 \times 10)+(6 \times 1) $$

$$ (6 \times 100)+(5 \times 10)+(4 \times 1) $$

Solution

Question 6

$$9$$

$$7$$

$$6$$

$$1$$

Solution

Question 7

i & iii

ii & iii

i & ii

iii only

Solution

Question 8

Ocean

Country

Animal

City

Solution

Question 9

2/5

2/4

1/3

1/2

Solution

Question 10

$$6$$

$$8$$

$$10$$

$$12$$

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