Tag: number system

Questions Related to number system

Which of the following numbers lie between $1$ and $3$?

  1. $\dfrac13$

  2. $\sqrt{2}$

  3. $\sqrt{10}$

  4. $\dfrac{8}3$


Correct Option: B,D
Explanation:

Converting all option in decimal format (approximately)

A. $\frac{1}{3}=0.33$
B.$\sqrt {2  } =1.41$
C.$ \sqrt { 10 }=3.16 $
D. $\frac{8}{3}=2.67$
It is clear that option B and D lies between 1 and 3
So correct answer will be option B and D

Between any $2$ real numbers, __________ can always be represented on a number line.

  1. an integer

  2. an irrational number

  3. a natural number

  4. a rational number


Correct Option: B,D
Explanation:

Between any two real number, there is always many rational number.

And between any two rationals there is always an irrational numbers.
Thus, we can represent rationals and irrationals between any two reals.
Hence, option B and D is correct.

Following are the steps to represent $\sqrt5$  on number line.
Arrange them in order.
1) Draw OC on line with $l(OC)=l(OB)$,
2) Draw $AB \perp OA\ and\ l(AB) =1$
3) Take $l(OA)=2$
4) $l(OC)=\sqrt5$, C is required point on real line.

  1. $1,2,3,4$

  2. $2,4,1,3$

  3. $3,2,4,1$

  4. $3,2,1,4$


Correct Option: D
Explanation:

The correct order of representing $\sqrt { 5 } $ on number line is 

Step 1. Take $l(OA) = 2$.
Step 2. Draw $AB$$\perp $$OA\ and\  l(AB)$ $= 1$
Step 3. Draw $OC$ on line with $l(OC) = l(OB)$
Step 4. $l(OC) =$ $\sqrt { 5 } $, $C$ is the required point on real line
Therefore, option(D) is correct.

Which of the following irrational numbers lie between $6$ and $8$?

  1. $\sqrt{49}$

  2. $\sqrt{19}$

  3. $\sqrt{47}$

  4. $\sqrt{62}$


Correct Option: C,D
Explanation:

$6^{2} = 36$

$7^{2} = 49$
$8^{2} = 64$

$\Rightarrow \sqrt47$ and $\sqrt62$ are only irrational numbers from the stated that lie in between $6$ and $8$.

$\sqrt49 = 7$ which isn't irrational.

The number $\sqrt{10}$ lies between $2$ integers $a$ and $b$ such that $b-a = 1$. Then $b+a = \, ?$

  1. $4$

  2. $5$

  3. $7$

  4. None of these


Correct Option: C
Explanation:

$1^{2} = 1$

$2^{2} = 4$
$3^{2} = 9$
$4^{2} = 16$

$\Rightarrow \sqrt10$ lies between $3$ and $4$

$4 = a , 3 = b$ 

$\therefore a+b = 7$

Which one of the following is not true?

  1. $\sqrt{2}$ is an irrational number

  2. $\sqrt{17}$ is a irrational number

  3. $0.10110011100011110...$ is an irrational number

  4. $\sqrt[4]{16}$ is an irrational number


Correct Option: D
Explanation:

$2^{4} =16$


$\because \sqrt[4]{16} = 2$, which is a rational number.

The greater number between $\sqrt{17}-\sqrt{12}$ and $\sqrt{11}-\sqrt{6}$ is ____.

  1. $\sqrt{17}-\sqrt{12}$

  2. $\sqrt{11}-\sqrt{6}$

  3. Both are equal

  4. Cannot comare


Correct Option: B
Explanation:

$\sqrt{17}=4.12\ \sqrt {12}=3.46\ \therefore\sqrt{17}-\sqrt{12}=0.66$

$\sqrt{11}=3.32\ \sqrt6=2.45\ \therefore\sqrt{11}-\sqrt6=0.87$

$\sqrt{11}-\sqrt6>\sqrt{17}-\sqrt{12}$

The value of $0.\overline{2}$ in the form $\frac{p}{q}$ , where p and q are integers and $q\ne 0$ is :

  1. $\dfrac{1}{5}$

  2. $\dfrac{2}{9}$

  3. $\dfrac{2}{5}$

  4. $\dfrac{1}{8}$


Correct Option: B
Explanation:

$Let\quad x=.222....\ On\quad multiplying\quad by\quad 10\quad on\quad both\quad sides\quad \ 10x=2.222....\ On\quad subtracting\quad both\quad equations\quad \ 9x=2\ x=\dfrac { 2 }{ 9 } $

Hence,correct answer is option B.

Which of the following is/are correct?

  1. Every integer is a rational number.

  2. The sum of a rational number and an irrational number is an irrational number.

  3. Every real number is rational.

  4. Every point on the number line is associated with a real number


Correct Option: A,B,D
Explanation:

Yes every integer can be represented in the form of $p/q$ where q is 1 for integers, so every integer is a rational number.
The sum of rational and irrational numbers is always irrational.
No, every real number is not rational. Real numbers are classified as rational numbers and irrational numbers.
Any number on the number line is a real number because that number can be either rational or irrational. Rational and irrational numbers together form real numbers.

To represent a rational number $\sqrt{2}$ on number line, take sides of right triangle as:

  1. $1$ and $1$

  2. $1$ and $2$

  3. $2$ and $0$

  4. $-1$ and $-1$


Correct Option: A
Explanation:
Notice that $\sqrt{2}= \sqrt{(1^2 + 1^2)}$. So, we can form a length of $\sqrt{2}$ units using two mutually perpendicular sides of length $1$ unit each. (Since, $1,1,\sqrt{2}$ form sides of a right angled triangle by Pythagoras theorem).