Real Numbers Test

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Real Numbers Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

When the HCF of $$468$$ and $$222$$ is written in the form of $$ 468 x + 222y$$ then the value of $$ x$$ and $$y$$ is

A
$$x =-9 \ and \ y =19$$

B
$$x =9 \ and \ y = -19$$

C
$$x =9\ and \ y = 19$$

D
$$x =-9 \ and \ y =- 19$$

Solution

According to the definition of Euclid's theorem,
$$a = b \times q + r$$ where $$0 \leq r < b .$$
HCF of $$468$$ and $$222$$
$$468 = \left(222 \times 2\right) + 24$$
$$222 = \left(24\times\ 9\right) + 6$$
$$24 = \left(6\times\ 4\right) + 0$$
$$\therefore HCF = 6$$
$$6 = 222 - \left(24\times\ 9\right)$$
$$ = 222 - \left[\left(468 -222 \times 2\right) \times\ 9\right] $$ [where $$468 = 222 \times 2 + 24$$]
$$ = 222 - \left[468 \times 9 -222 \times 2 \times 9\right]$$
$$= 222 - \left(468 \times9\right) - \left(222\times 18\right)$$
$$ = 222 + \left(222 \times 18\right) - \left(468 \times9\right)$$
$$= 222\left[1 + 18\right]- 468 \times 9$$
$$= 222 \times19- 468 \times 9$$
$$ = 468 \times -9 + 222\times 19$$
$$\therefore x=-9$$ and $$y=19$$.
Question 2
1 / -0

One and only one out of $$n, n + 4, n + 8, n + 12\ and \ n + 16 $$ is ......(where n is any positive integer)

A
Divisible by 5

B
Divisible by 4

C
Divisible by 10

D
Divisible by 12

Solution

We know that any positive integer is of the form 5q, 5q + 1 or 5q + 2, 5q + 3 or 5q + 4 for some integer q and one and onlyone of these possibilities can occur. So, we have the following cases:
Case-I: When $$n=5q$$
In this case, we have
$$n=5q$$, which is divisible by 5
Now, $$n=5q$$
$$\Rightarrow n+4=5q+4$$
$$\Rightarrow n+4$$ leaves remainder 4 when divided by 5
$$\Rightarrow n+4$$ is not divisible by 5.
Now $$n+8=5q+8=5(q+1)+3=5m+3$$, m is an integer.
Clearly, n+8 is not divisible by 5.
Again, $$n+12=5q+12=5(q+2)+2=5m+2$$, m in an integer.
Clearly n+12 is not divisible by 5.
Now $$n+16=5q+16=5(q+13)+1=5m+1$$, m is an integer
$$\Rightarrow n+16$$ is not divisible by 5
Thus, if n = 5q only one out of n, n + 4, n + 8, n +
12 and n + 16 is divlsible by 5,
Similarly, this result can be proved for the rest of .
the cases.
Question 3
1 / -0

If any positive even integer is of the form $$4q$$ or $$4q + 2$$, then $$q$$ belongs to:

A
whole number

B
irrational number

C
Both A & B

D
none of these

Solution

Let $$a$$ be any positive even integer and $$b=4$$.Then by division algorithm,
$$a=4q+r$$ for some integer $$q \ge 0$$ and $$r=0,1,2,3$$
So,
$$a=4q$$ or,
$$4q+1$$,
$$4q+2$$
$$4q+3$$
Because $$0 \ge r \ge 4$$
Now,
$$4q$$ i.e $$2(2q)$$ is an even number
$$\therefore$$ $$4q+1$$ is an odd number
$$4q+2$$ i.e. $$2(2q+1)$$ is an even number
$$\therefore (4q+2)+1=4q+3$$ is an odd number
Thus, We can say that any positive even integer can be written as in the form of $$4q, 4q+2$$ where $$q$$ is the whole number
Question 4
1 / -0

The number $$\displaystyle\frac{3-\sqrt{3}}{3+\sqrt{3}}$$ is

A
Rational

B
Irrational

C
Both

D
Can't say

Solution

$$Here,\quad we\quad will\quad carry\quad out\quad rationalization.\quad \\ \frac { 3-\sqrt { 3 } }{ 3+\sqrt { 3 } } =\frac { 3-\sqrt { 3 } }{ 3+\sqrt { 3 } } x\frac { 3-\sqrt { 3 } }{ 3-\sqrt { 3 } } =\frac { { (3-\sqrt { 3 } ) }^{ 2 } }{ (3+\sqrt { 3) } (3-\sqrt { 3 } ) } =\frac { 9+3-6\sqrt { 3 } }{ 9-3 } =\frac { 12-6\sqrt { 3 } }{ 6 } =\frac { 2-\sqrt { 3 } }{ 1 } \\ Since\quad \sqrt { 3 } is\quad irrational\quad number\quad and\quad subtraction\quad of\quad rational\quad and\quad irrational\quad is\quad irrational.\\ The\quad given\quad expression\quad is\quad irrational.\\ \quad $$
Question 5
1 / -0

Using Euclid's division algorithm, find H.C.F.of $$56, 96$$ and $$404.$$

A
$$8$$

B
$$16$$

C
$$12$$

D
$$4$$

Solution

Let us begin by choosing any two number out of any three number.
Say $$56$$ and $$96$$
As $$96>56$$, by applying Euclid's division lemma to $$56$$ and $$96$$ we have,
$$96=56\times 1+40$$
Since remainder $$40 \neq 0.$$
So,applying Euclid's division lemma to $$56$$ and $$40$$ we have,
$$56=40\times 1+16$$
Since remainder $$16 \neq 0$$
So,applying Euclid's division lemma to $$40$$ and $$16$$ we have,
$$40=16\times 2+8$$
Since remainder $$8 \neq 0$$
So, applying Euclid's division lemma to $$16$$ and $$8$$ we have,
$$16=8\times 2+0$$
Since remainder is zero. Hence,divisor $$8$$ is the H.C.F of $$56$$ and $$96.$$
Now,
Again, applying Euclid's division lemma on the H.C.F of the two number and remaining number.
Since, the H.C.F of $$56$$ and $$96$$ is $$8$$ and the remaining number is $$404.$$
So, by applying Euclid's division lemma on $$8$$ and $$404$$ we have,
$$404=8\times 50+4$$
Since remainder $$4 \neq 0$$ So,applying Euclid's division lemma to $$8$$ and $$4$$ we have,
$$8=4\times 2+0$$
Hence, remainder is zero.
Hence, remainder $$4$$ is the H.C.F of $$8$$ and $$404$$
Hence, H.C.F. of $$404, 96$$ and $$56$$ is $$4.$$
Question 6
1 / -0

If the square of an odd positive integer can be of the form $$6q + 1 $$ or $$6q + 3$$ for some $$ q$$ then q belongs to:

A
integers

B
irrational

C
real

D
none of these

Solution

We know that any positive integer can be of the form 6m,6m+1,6m+2,6m+3,6m+4 belongs to integers.
Question 7
1 / -0

Consider the following statements :
1. $$\displaystyle \frac{1}{22}$$ can not be written as terminating decimal

2. $$\displaystyle \frac{2}{15}$$ can be written as a terminating decimal

3. $$\displaystyle \frac{1}{16}$$ can be written as a terminating decimal

Which of the statements given above is/are correct ?

A
$$1$$ only

B
$$3$$ only

C
$$1$$ and $$3$$

D
$$2$$ and $$3$$

Solution

$$\displaystyle \frac{1}{22} = 0.04545454545$$ is not a terminating decimal.

$$\displaystyle \frac{2}{15} = 0.133333333$$ is not a terminating decimal.

$$\displaystyle \frac{1}{16} = 0.0625$$ is a terminating decimal.

Hence, statement $$1$$ and $$3$$ are correct.
Question 8
1 / -0

A rational number can be expressed as a terminating decimal if the denominator has factors:

A
$$2$$ or $$5$$

B
$$2, 3$$ or $$5$$

C
$$3$$ or $$5$$

D
None of these

Solution

Any rational number with its denominator is in the form of $$2^m\times 5^n$$, where $$m,n$$ are positive integers and are terminating decimals. The examples are shown below:

$$(i)\quad \dfrac { 15 }{ 8 } =\dfrac { 15 }{ 2\times 2\times 2 } =\dfrac { 15 }{ 2^{ 3 } } =\dfrac { 15\times 5^{ 3 } }{ 2^{ 3 }\times 5^{ 3 } } =\dfrac { 15\times 125 }{ 8\times 125 } =\dfrac { 1875 }{ 1000 } =1.875\\ (ii)\dfrac { 31 }{ 20 } =\dfrac { 31 }{ 2\times 2\times 5 } =\dfrac { 31\times 5 }{ 2\times 2\times 5\times 5 } =\dfrac { 155 }{ 100 } =1.55$$

In both the examples, the result is a terminating decimal because terminating decimal is a decimal which can be expressed in a finite number of figures.

Hence, a rational number can be expressed as a terminating decimal if the denominator has factors $$2$$ or $$5$$.
Question 9
1 / -0

What is the H.C.F. of two co-prime numbers ?

A
$$1$$

B
$$0$$

C
$$2$$

D
none of these

Solution

$$ {\textbf{Step - 1: Find the required answer using definition of co-prime numbers.}} $$

$$ {\text{By definition of co-prime numbers,}} $$

$$ {\text{Two numbers having only 1 as their common factor are called co-prime numbers.}} $$
$$ {\text{So, HCF of two co-prime numbers is 1.}} $$

$$ {\textbf{Hence, the correct option is A.}} $$
Question 10
1 / -0

The HCF of 136, 170 and 255 is:

A
$$13$$

B
$$15$$

C
$$17$$

D
$$1$$

Solution

Given numbers are $$136,\ 170\text{ and }255$$.
Representing the given numbers as the product of their prime factors:
$$136=2\times 2\times 2\times 17=2^3\times 17$$
$$170=2\times 5\times 17$$
$$255=3\times 5\times 17$$

We can see that $$17$$ is the only factor common between the three numbers.

Hence, HCF of $$136,\ 170\text { and }255$$ is $$17$$

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Real Numbers Test - 27

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Real Numbers Test - 27

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Question 1

$$x =-9 \ and \ y =19$$

$$x =9 \ and \ y = -19$$

$$x =9\ and \ y = 19$$

$$x =-9 \ and \ y =- 19$$

Solution

Question 2

Divisible by 5

Divisible by 4

Divisible by 10

Divisible by 12

Solution

Question 3

whole number

irrational number

Both A & B

none of these

Solution

Question 4

Rational

Irrational

Both

Can't say

Solution

Question 5

$$8$$

$$16$$

$$12$$

$$4$$

Solution

Question 6

integers

irrational

real

none of these

Solution

Question 7

$$1$$ only

$$3$$ only

$$1$$ and $$3$$

$$2$$ and $$3$$

Solution

Question 8

$$2$$ or $$5$$

$$2, 3$$ or $$5$$

$$3$$ or $$5$$

None of these

Solution

Question 9

$$1$$

$$0$$

$$2$$

none of these

Solution

Question 10

$$13$$

$$15$$

$$17$$

$$1$$

Solution

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