MCQ Questions for Class 9 Maths Number Systems Quiz 8

Find five rational numbers between

$\displaystyle\frac{-3}{2}$ and

$\displaystyle\frac{5}{3}$ .

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$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-13}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{2}{6}$
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$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{13}{6}$
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$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{11}{6}$
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$\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{2}{6}$

Explanation

Converting the given rational numbers with the same denominators

$\cfrac{-3}{2}=\cfrac{-3\times3}{2\times3}=\cfrac{-9}{6}$ and

$\cfrac{5}{3}=\cfrac{5\times2}{3\times2}=\cfrac{10}{6}$

We know that

$-9\,<\,-8\,<\,-7\,<\,-6\,<\,\dots\,<\,0\,<\,1\,<\,2\,<\,8\,<\,9\,<\,10$

$\Rightarrow\cfrac{-9}{6}\,<\,\cfrac{-8}{6}\,<\,\cfrac{-7}{6}\,<\,\cfrac{-6}{6}\,<\,\dots\,<\,\cfrac{0}{6}\,<\,\cfrac{1}{6}\,<\,\cfrac{2}{6}\,<\,\dots\,<\,\cfrac{8}{6}\,<\,\cfrac{9}{6}\,<\,\cfrac{10}{6}$ .

Thus, we have the following five rational numbers between

$\cfrac{-3}{2}$ and

$\cfrac{5}{3}$

$\Rightarrow \cfrac{-8}{6},\,\cfrac{-7}{6},\,\cfrac{0}{6},\,\cfrac{1}{6}and\cfrac{2}{6}$

Find

$9$ rational numbers between

$-\displaystyle\frac{1}{9}\;and\;\displaystyle\frac{1}{5}$ .

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$\displaystyle\frac{-7}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$
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$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{10}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$
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$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{9}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$
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$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$

Explanation

Convert the rational numbers into equivalent rational numbers with the same denominator.

LCM of

$9$ and

$5$ is

$45$ .

$-\displaystyle\frac{1}{9}=\displaystyle\frac{-1\times5}{9\times5}=\displaystyle\frac{-5}{45}$ and

$\displaystyle\frac{1}{5}=\displaystyle\frac{1\times9}{5\times9}=\displaystyle\frac{9}{45}$

The integers between

$-5$ and

$9$ are

$-4,\,-3,\,-2,\,-1,\,0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8$ .

The corresponding rational numbers are

$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{0}{45},\,\displaystyle\frac{1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{6}{45},\,\displaystyle\frac{7}{45},\,\displaystyle\frac{8}{45}$

On selecting any

$9$ of them, we get

$9$ rational numbers between

$-\displaystyle\frac{1}{9}$ and

$\displaystyle\frac{1}{5}$ as

$\displaystyle\frac{-4}{45},\,\displaystyle\frac{-3}{45},\,\displaystyle\frac{-2}{45},\,\displaystyle\frac{-1}{45},\,\displaystyle\frac{2}{45},\,\displaystyle\frac{3}{45},\,\displaystyle\frac{4}{45},\,\displaystyle\frac{5}{45},\,\displaystyle\frac{8}{45}$

Write any

$10$ rational numbers between

$0\;and\;2$ .

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$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{88}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$
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$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{21}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$
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$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{35}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$
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$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$

Explanation

Let us write

$0$ as

$\dfrac{0}{10}$ and

$2$ as

$\dfrac{20}{10}.$

So, ten rational numbers between these will be,

$\dfrac{1}{10},\;\dfrac{2}{10},\;\dfrac{3}{10},\;\dfrac{4}{10},\;\dfrac{5}{10},\;\dfrac{6}{10},\;\dfrac{7}{10},\ \dfrac{8}{10},\ \dfrac{9}{10},\; 1$

Hence, option

$D$ is correct.

Which of the following is irrational?

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$\displaystyle\frac{1}{3}$
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$\displaystyle\frac{48}{5}$
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$0.7777\dots$
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$1.73202002\dots$

Explanation

$\textbf{Step -1: Definition of irrational numbers.}$

$\text{Numbers that cannot be represented in }\dfrac{p}{q}\text{ form are said to be irrational numbers.}$

$\text{Irrational numbers are non-repeating and non-terminating decimal numbers.}$

$\textbf{Step -2: Using the definition find the answer.}$

$\text{Some examples of irrational numbers are }\sqrt{2}, \sqrt{3}, \pi, 6.9318786..., \text{etc}.$

$\therefore \text{1.73202002... is a non-repeating and non-terminating number, hence is an irrational number.}$

$\textbf{Hence, among the following options, (D) 1.73202002... is an irrational number.}$

Choose the rational number which does not lie between rational numbers

$\displaystyle\frac{3}{5}$ and

$\displaystyle\frac{2}{3}$

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$\displaystyle\frac{46}{75}$
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$\displaystyle\frac{47}{75}$
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$\displaystyle\frac{49}{75}$
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$\displaystyle\frac{50}{75}$

Explanation

Consider the given rational numbers,

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

Now,

$\dfrac{3\times 15}{5\times 15}$ and $\dfrac{2\times 25}{3\times 25}$

$\dfrac{45}{75}$ and $\dfrac{50}{75}$

Now, rational number which does not lie between thes rational numbers is,

$\dfrac{50}{75}$

Hence, this is the answer.

Which of the following rational numbers lies between

$\dfrac {-4}{5}$ and

$\dfrac {-7}{5}$ ?

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$\dfrac {-36}{45}$
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$\dfrac {-35}{45}$
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$\dfrac {-39}{45}$
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$\dfrac {-63}{45}$

Explanation

As per the given options, let us multiply both numbers with

$\dfrac {9}{9}$ .

$\therefore \dfrac {-4}{5}\times \dfrac {9}{9} = \dfrac {-36}{45}$ and

$\dfrac {-7}{5}\times \dfrac {9}{9} = \dfrac {-63}{45}$

$\therefore$ Clearly,

$\dfrac {-39}{45}$ lies between

$\dfrac {-4}{5}$ and

$\dfrac {-7}{5}$ .

The product of

$4\sqrt 6$ and

$3\sqrt {24}$ is:

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$124$
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$134$
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$144$
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$154$

Explanation

$4\sqrt6\times3\sqrt{24}$

$=12\sqrt{6\times24}$

$=12\sqrt{144}$

$=12\times12$

$=144$ .

Therefore, option

$C$ is correct.

$x$ be any integer different from zero and

$m,n$ be any integers then

$({x^m})^n$ is equal

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$x^{m+n}$
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$x^{mn}$
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$\dfrac {m}{x^n}$
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$x^{m-n}$

Explanation

$(a^m)^n =(a)^{m+n}$

[Where,

$a$ is non-zero integer]

Similarly,

$(x^m)^n =(x)^{m\times n}=(x)^{mn}$

Find five rational numbers between

$1$ and

$2.$

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$\dfrac {1}{10}, \dfrac {2}{10}, \dfrac {3}{10}, \dfrac {4}{10}, \dfrac {5}{10}$
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$\dfrac {1}{5}, \dfrac {2}{5}, \dfrac {3}{5}, \dfrac {4}{5}, \dfrac {5}{5}$
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$\dfrac {1}{2}, \dfrac {1}{3}, \dfrac {1}{4}, \dfrac {1}{5}, \dfrac {1}{6}$
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$\dfrac {8}{7}, \dfrac {9}{7}, \dfrac {10}{7}, \dfrac {11}{7}, \dfrac {12}{7}$

Explanation

$\textbf{Step -1: Find required rational numbers.}$

$\text{We need five rational numbers between }1\text{ and }2.$

$\text{So, multiply and divide the numbers with a natural number greater or equal to } 5+1=6.$

$\text{Lets take }7.$

$1=1\times\dfrac{7}{7}=\dfrac{7}{7}$

$\text{and }2=2\times\dfrac{7}{7}=\dfrac{14}{7}$

$\text{Thus, }1\text{ and }2\text{ becomes }\dfrac{7}{7}\text{ and }\dfrac{14}{7}\text{ respectively.}$

$\text{Since, }7<8<9<10<11<12<13<14$

$\therefore 1=\dfrac{7}{7}<\dfrac{8}{7}<\dfrac{9}{7}<\dfrac{10}{7}<\dfrac{11}{7}<\dfrac{12}{7}<\dfrac{13}{7}<\dfrac{14}{7}=2$

$\text{Hence, five rational numbers between }1\text{ and }2\text{ are }\dfrac{8}{7},\dfrac{9}{7},\dfrac{10}{7},\dfrac{11}{7},\text{ and }\dfrac{12}{7}.$

$\textbf{Hence , the correct option is D.}$

The square root of any prime number is

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rational
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irrational
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co-prime
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composite

Explanation

The square root of any prime number is irrational.
Example:

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

The value of

$\sqrt{32}\times \sqrt{64}$ is equal to:

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$32$
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$32\sqrt{2}$
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$64$
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$64\sqrt{2}$

Explanation

$\sqrt{32}\times \sqrt{64}$

$= \sqrt{16\times 2}\times \sqrt{64}$

$=4\times \sqrt{2}\times 8 = 32\sqrt{2}$ .

So, option

$B$ is correct.

The fraction

$\displaystyle \frac{4}{\sqrt{6}}$ is equivalent to

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$\displaystyle \frac{2}{\sqrt{3}}$
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$\displaystyle \frac{2\sqrt{6}}{3}$
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$\displaystyle \frac{4\sqrt{6}}{3}$
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None of the above

Explanation

$\displaystyle \frac{4}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}=\frac{4\sqrt{6}}{6}=\frac{2\sqrt{6}}{3}$

So, option B is correct.

The fraction

$\displaystyle \frac{2}{\sqrt{6}}$ is equal to

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$\displaystyle \frac{\sqrt{6}}{3}$
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$\displaystyle \frac{\sqrt{6}}{\sqrt{3}}$
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$\displaystyle \frac{\sqrt{6}}{2}$
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None of the above

Explanation

$\displaystyle \frac{2}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}$

$=\displaystyle \frac{2\sqrt{6}}{6}=\frac{\sqrt{6}}{3}$

So, option A is correct.

The fraction

$\displaystyle \frac{\sqrt{5}}{2\sqrt{3}}$ is equivalent to

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$\displaystyle \frac{\sqrt{5}}{6}$
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$\displaystyle \frac{\sqrt{15}}{6}$
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$\displaystyle \frac{\sqrt{5}}{3}$
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$\displaystyle \frac{\sqrt{10}}{6}$

Explanation

$\displaystyle \frac{\sqrt{5}}{2\sqrt{3}} = \frac{\sqrt{5}}{2\sqrt{3}}\times \frac{2\sqrt{3}}{2\sqrt{3}}$

$=\displaystyle \frac{2\sqrt {15}}{12}$

$=\displaystyle \frac{\sqrt{15}}{6}$

So, option B is correct

$\dfrac {17}{8}$ can be expressed as

$.....$ . It is a

$........$ decimal.

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$2.125$ , terminating
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$2.321321...$ , non-terminating
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$1.125125124...$ , recurring
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$2.125$ , irrational

Explanation

$\dfrac {17}{8}=\dfrac{17\times125}{8\times 125}= \dfrac{2125}{1000}=2.125$ ,

which is terminating.

Therefore, option

$A$ is the correct answer.

............... numbers have terminating and non- terminating repeating decimals.

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Integers
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Whole
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Rational
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Irrational

Explanation

$\dfrac {1}{4} = 0.25$ is a terminating decimal.

$\dfrac {8}{3} = 2.666666666......$ is a non-terminating repeating decimal.

Both are rational numbers as both can be represented in the form of

$\dfrac{p}{q}$ .

Therefore,

$C$ is the correct answer.

What rational number lies exactly halfway between

$\dfrac{2}{3}$ and

$\dfrac{3}{4}$ ?

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$\dfrac{3}{5}$
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$\dfrac{5}{6}$
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$\dfrac{7}{12}$
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$\dfrac{9}{16}$
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$\dfrac{17}{24}$

Explanation

Consider

$3 \times 4 = 12$ ,

So,

$\dfrac {2}{3} = \dfrac{8}{12}$

$\dfrac 34 = \dfrac{9}{12}$

Multiplying the numerator and denominator by

$2$ :

$\dfrac{16}{24}$ and

$\dfrac{18}{24}$ .

The mid point is

$\dfrac{17}{24}$

Hence option

$E$ is correct.

Identify a rational number between

$\dfrac {1}{3}$ and

$\dfrac {4}{5}$

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$\dfrac {1}{4}$
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$\dfrac {9}{10}$
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$\dfrac {17}{30}$
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$\dfrac {7}{10}$

Explanation

$\frac{1}{3}\,and\,\frac{4}{5}\text{ }$

$To\text{ find rational number between to two number make the denominator }$

$\text{same and then check the numbers in between numerators}$

$\frac{6}{30}\,and\,\frac{24}{30}$

$now\,\text{ the possible numbers between 1 and 4 are 7,8,9,10,11,12,13,14,15...23}$

$\therefore \,\text{ the possible rational number between}\frac{1}{3}and\frac{4}{5\,}is \frac{17}{30}\,and\,\frac{7}{10}$

Calculate the number of irrational numbers between

$1$ and

$8$ .

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Fewer than $3$
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$3$
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$6$
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$7$
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More than $7$

Explanation

Between any two rational numbers there are infinite number of irrational numbers. Hence, the required answer is more than

$7$ .

How many irrational numbers are there between

$2$ and

$6$ ?

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$1$
0%
$3$
0%
$4$
0%
$10$
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Infinitely many

Explanation

There can be infinite number of irrational numbers between any two rational numbers.

Hence,

$Op-E$ is correct.

Consider the following statements:

$\dfrac {1}{22}$ cannot be written as a terminating decimal.
2.

$\dfrac {2}{15}$ can be written as a terminating decimal.
3.

$\dfrac {1}{16}$ can be written as a terminating decimal.
Which of the statements given above is/are correct?

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$1$ only
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$2$ only
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$3$ only
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$2$ and $3$

Explanation

$\dfrac{1}{22}$ is an irrational number hence it is a non terminating number

$\dfrac{2}{15}$ is an irrational number hence it is a non terminating number

$\dfrac{1}{16}$ is an rational number hence it is a terminating number

Identity the rational number that does not lie between

$\cfrac{3}{5}$ and

$\cfrac{2}{3}$ .

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$\cfrac{46}{75}$
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$\cfrac{47}{75}$
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$\cfrac{49}{75}$
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$\cfrac{50}{75}$

Identify the rational numbers from the followings:

$72, \dfrac {5}{27}, \dfrac {62}{0}, 1\dfrac {3}{5}, -7\dfrac {5}{4}, 0$

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All except $\dfrac {62}{0}$
0%
All are rational numbers
0%
All except $\dfrac {62}{0}$ and $0$
0%
All except $1\dfrac {3}{5}, -7\dfrac {5}{4}$

State whether the following statements are true or false.

$\sqrt {n}$ is not irrational if n is a perfect square

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True
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False

Explanation

False ,

$\sqrt{4}=2$ where 2 is a rational number.Here n is perfect square the

$\sqrt{n}$ is rational number

$\sqrt{5}=2.236..$ is not rational number But it is irrational number . here n is not a perfect square the

$\sqrt{n}$ is irrational number

$\sqrt{n}$ is not irrational number if n is perfect square

State true or false.
Zero is a rational number.

0%
True
0%
False

Explanation

Every rational number is written in the form of

$\dfrac{p}{q}$ where

$p , q$ are integers and

$q \neq 0$

$0$ can also be written as above form

i.e

$\dfrac{0}{5} = 0$ or

$\dfrac{0}{100} = 0$ etc

So Zero is a Rational number

The given statement is True

In mathematics, the irrational number e

$= 2.718$ ... Which of the following letters corresponds to the number e on the number line below?

0%
A
0%
B
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C
0%
D

Explanation

Since the part between

$2.70$ and

$2.75$ is divided into

$5$ parts , so, each part measures

$0.01$ .

$e=2.718$ is to be located on the number line.

So, on analysis, we can say that point B is the correct point.

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.

0%
$1,2,3,4$
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$2,4,1,3$
0%
$3,2,4,1$
0%
$3,2,1,4$

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 number is different from others?

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$\sqrt 7$
0%
$\sqrt 6$
0%
$\sqrt {25}$
0%
$\sqrt{10}$

Explanation

$\sqrt{7}$ is an irrational number

$\sqrt{6}$ is an irrational number

$\sqrt{10}$ is an irrational number

$\sqrt{25}=5$ is different from others because others are irrational number but

$\sqrt{25}$ is a rational number

Hence, option C is correct.

In mathematics, the number

$\phi$ is an irrational number that is roughly equal to

$1.618$ ... Which letter corresponds to the correct placement of

$\phi$ on the number line below?

0%
A
0%
B
0%
C
0%
D

Explanation

Since the part between

$1$ and

$2$ is divided into

$10$ parts , so, each part measures

$0.1$ .

$\phi=1.618$ is to be located on the number line.

$1.6<1.618<1.7$

So, on analysis, we can say that point B is the correct point.

$p$ is prime, then

$\sqrt {p}$ is:

0%
Composite number
0%
Rational number
0%
Positive integer
0%
Irrational number

Explanation

SInce, we know that prime numbers are those which are never perfect square and not divisible by any other number except by itself.
which are

$2,3,5,7,...$
Clearly, if

$p$ is prime then

$\sqrt p$ is irrational number.
Option

$D$ is correct.

CBSE Questions for Class 9 Maths Number Systems Quiz 8 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Number Systems Quiz 8 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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