Playing with Numbers Test

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Playing with Numbers Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

If the sum of digits of a number is divisible by three then the number is always divisible by

A
$$2$$

B
$$3$$

C
$$6$$

D
$$9$$

Solution

If the sum of digits of a number is divisible by three, then the number is always divisible by $$3$$.
This is the test for divisibility by 3.
Hence, (B) is the correct option.
Question 2
1 / -0

Directions For Questions

Given:
$$\quad\quad\quad$$ B A N A
$$\quad\quad+$$ N A N A
$$\quad\quad\quad -----$$
$$\quad\quad$$ Q B P B
$$\quad\quad\quad -----$$

where, $$A = 3$$ and $$Q = 8$$.

...view full instructions

Find the value of $$6P$$.

A
$$48$$

B
$$36$$

C
$$24$$

D
$$12$$

Solution

$$\quad\quad\quad$$ B 3 N 3
$$\quad\quad+$$ N 3 N 3
$$\quad\quad\quad ----$$
$$\quad\quad$$ 8 B P B
$$\quad\quad\quad ----$$
Here
A $$= 3$$ and Q $$= 8$$
As, $$3 + 3 = $$ B $$ = 6$$,
N $$= 8 -$$ B $$= 8 - 6 = 2$$,
So, P $$=$$ N $$+$$ N $$= 4$$
Thus, $$6P=6 \times 4=24$$.
Question 3
1 / -0

How many numbers between $$9$$ and $$54$$ are exactly divisible by $$9$$ but not by $$3$$

A
$$8$$

B
$$6$$

C
$$5$$

D
None of these

Solution

Any number which is divisible by $$9$$ is also divisible by $$3$$. So, no number between 9 and 54, which is exactly divisible by 9 and not by 3, cant be found.
Question 4
1 / -0

The value of $$A + B + C$$ is

A
$$15$$

B
$$18$$

C
$$16$$

D
$$12$$

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.
$$\therefore A+B+C = 9+1+8 = 18$$
So, option B is correct.
Question 5
1 / -0

Directions For Questions

Given:
$$\quad\quad\quad$$ B A N A
$$\quad\quad+$$ N A N A
$$\quad\quad\quad -----$$
$$\quad\quad$$ Q B P B
$$\quad\quad\quad -----$$

where, $$A = 3$$ and $$Q = 8$$.

...view full instructions

$$N^2$$ = ?

A
$$16$$

B
$$4$$

C
$$1$$

D
$$9$$

Solution

Given, $$A=3$$ and $$Q=8$$
As per the given condition, we have
$$A+A = B$$ and $$B+N = Q$$
$$\therefore 2A = B$$ and $$B+N=8$$
$$\therefore B=6$$ and $$6+N= 8$$
$$\therefore N=2$$
$$\therefore N^2 = 2^2 = 4$$
Question 6
1 / -0

Directions For Questions

Given:
$$\quad\quad\quad$$ B A N A
$$\quad\quad+$$ N A N A
$$\quad\quad\quad -----$$
$$\quad\quad$$ Q B P B
$$\quad\quad\quad -----$$

where, $$A = 3$$ and $$Q = 8$$.

...view full instructions

The value of P $$\times$$ Q is:

A
$$42$$

B
$$64$$

C
$$4$$

D
$$32$$

Solution

$$\quad\quad\quad$$ B A N A
$$\quad\quad+$$ N A N A
$$\quad\quad\quad ----$$
$$\quad\quad$$ Q B P B
$$\quad\quad\quad ----$$
Here
A $$= 3$$ and Q $$= 8$$
As, $$3 + 3 = $$ B $$ = 6$$,
N $$= 8 -$$ B $$= 8 - 6 = 2$$,
So, P $$=$$ N $$+$$ N $$= 4$$.

Above addition is reduced to
$$\quad\quad\quad$$ 6 3 2 3
$$\quad\quad+$$ 2 3 2 3
$$\quad\quad\quad ----$$
$$\quad\quad$$ 8 6 4 6
$$\quad\quad\quad ----$$

P $$=4$$ and Q $$=8$$
So, P $$\times$$ Q $$=4\times8 = 32$$

Hence, option D.
Question 7
1 / -0

Directions For Questions

Given:
$$\quad\quad\quad$$ B A N A
$$\quad\quad+$$ N A N A
$$\quad\quad\quad -----$$
$$\quad\quad$$ Q B P B
$$\quad\quad\quad -----$$

where, $$A = 3$$ and $$Q = 8$$.

...view full instructions

Product of all the integers of BANANA is

A
$$846$$

B
$$648$$

C
$$468$$

D
$$688$$

Solution

$$\quad\quad\quad$$ B A N A
$$\quad\quad+$$ N A N A
$$\quad\quad\quad ----$$
$$\quad\quad$$ Q B P B
$$\quad\quad\quad ----$$
Here
A $$= 3$$ and Q $$= 8$$
As, $$3 + 3 = $$ B $$ = 6$$,
N $$= 8 -$$ B $$= 8 - 6 = 2$$,
So, P $$=$$ N $$+$$ N $$= 4$$.

Therefore,
B $$\times$$ A $$\times$$ N $$\times$$ A $$\times$$ N $$\times$$ A $$= 6 \times 3 \times 2 \times 3 \times 2 \times 3 = 648$$

Hence, option B.
Question 8
1 / -0

A number $$\displaystyle a_{1}a_{2}a_{3}a_{4}a_{5}$$ is divisible by 9 if
(i) $$\displaystyle a_{1}+a_{2}+a_{3}+a_{4}+a_{5}$$ is divisible by $$9.$$
(ii) $$\displaystyle a_{1}-a_{2}+a_{3}-a_{4}+a_{5}$$ is divisible by $$9.$$

Which of the above statements is/are correct?

A
only (i)

B
only (ii)

C
both (i) and (ii)

D
neither (i) nor (ii)

Solution

The given number is divisible by $$9$$, if sum of
its digits is divisible by $$9$$.
$$\therefore \displaystyle a_{1}+a_{2}+a_{3}+a_{4}+a_{5}$$ is divisible by 9

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

Among the following numbers, number which is divisible by $$5$$ is

A
$$75$$ and $$620$$

B
$$8$$ and $$319$$

C
$$724$$

D
$$89$$ and $$652$$

Solution

A number is divisible by $$5$$ if the last digit is $$ 5\ \text{ or}\ 0.$$

$$\therefore 75$$ and $$620$$ are divisible by $$5$$.
Option A is correct.
Question 10
1 / -0

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

A
$$3$$

B
$$9$$

C
$$6$$

D
$$2$$

Solution

A number is divisible by 9 if the sum of the digits of the number is divisible by 9.
For example $$ 81 $$ is divisible by $$ 9 $$ as $$ 8 + 1 = 9 $$ is divisible by $$ 9 $$.

So, option B is correct.

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Re-Attempt Weekly Quiz Competition

Playing with Numbers Test - 12

Result

Playing with Numbers Test - 12

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SHARING IS CARING

Question 1

$$2$$

$$3$$

$$6$$

$$9$$

Solution

Question 2

$$48$$

$$36$$

$$24$$

$$12$$

Solution

Question 3

$$8$$

$$6$$

$$5$$

None of these

Solution

Question 4

$$15$$

$$18$$

$$16$$

$$12$$

Solution

Question 5

$$16$$

$$4$$

$$1$$

$$9$$

Solution

Question 6

$$42$$

$$64$$

$$4$$

$$32$$

Solution

Question 7

$$846$$

$$648$$

$$468$$

$$688$$

Solution

Question 8

only (i)

only (ii)

both (i) and (ii)

neither (i) nor (ii)

Solution

Question 9

$$75$$ and $$620$$

$$8$$ and $$319$$

$$724$$

$$89$$ and $$652$$

Solution

Question 10

$$3$$

$$9$$

$$6$$

$$2$$

Solution

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