MCQ Questions for Class 9 Maths Number Systems Quiz 2

Two rational numbers between

$\dfrac{2}{3}$ and

$\dfrac{5}{3}$ are :

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$\dfrac{1}{6}$ and $\dfrac{2}{6}$
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$\dfrac{1}{2}$ and $\dfrac{2}{1}$
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$\dfrac{5}{6}$ and $\dfrac{7}{6}$
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$\dfrac{2}{3}$ and $\dfrac{4}{3}$

Explanation

Changing the denominators of both numbers to 6, we get

$\dfrac { 2 }{ 3 } =\dfrac { 4 }{ 6 } \quad \& \quad \dfrac { 5 }{ 3 } =\dfrac { 10 }{ 6 }$

Numbers between the given rational numbers from the options are

$\dfrac { 5 }{ 6 } \quad \& \quad \dfrac { 7 }{ 6 }$

So, correct answer is option C.

State true or false:

There are numbers which cannot be written in the form

$\frac{p}{q}$ , where

$q\neq 0$ and both p, q are integers.

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True
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False

Explanation

The statement is true as there are Irrational numbers which dont satisfy the condition of rational numbers i.e irrational number cannot be written in the form of

$\frac { p }{ q }$

$q\neq 0\quad , where \quad p,q\quad$ are integers.
Example,

$\sqrt { 3 } ,\sqrt { 99 } \quad$

The value of

$2\sqrt {3} + \sqrt {3}$ is equal to

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$2\sqrt {6}$
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$3\sqrt {3}$
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$4\sqrt {6}$
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$6$

Explanation

$2\sqrt { 3 } +\sqrt { 3 } =\left( 2+1 \right) \sqrt { 3 } =3\sqrt { 3 }$

Between any two rational numbers,

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there is no rational number
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there is exactly one rational number
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there are infinitely many rational numbers
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there are only rational numbers and no irrational numbers

Explanation

Recall that to find a rational number between

$r$ and

$s,$ you can add

$r$ and

$s$ and divide the sum by

$2,$ that is

$\displaystyle \frac { r+s }{ 2 }$ lies between r and s.

For example,

$\displaystyle \frac { 5 }{ 2 }$ is a number between

$2$ and

$3.$ We can proceed in this manner to find many more rational numbers between

$2$ and

$3.$

Hence, we can conclude that there are infinitely many rational numbers between any two given rational numbers.

Every rational number is

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A natural number
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An integer
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A real number
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A whole number

Explanation

$\textbf{Step-1: Explain property of rational numbers}$

$\text{A real number is a number which can be expressed in the form}$

$\dfrac{p}{q},$

$\text{where}$

$q\neq{0.}$

$\textbf{Step-2: Proving that every rational number is a real number}$

$\text{Real numbers are numbers that include bothe rational and irrational numbers.}$

$\text{Hence, every rational number is a real number.}$

$\textbf{Final Answer: Every rational number is a real number. The correct option is (C).}$

The product of a non - zero rational number with an irrational number is always :

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Irrational number
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Rational number
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Whole number
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Natural number

$\sqrt { 2 } ,\sqrt { 3 }$ are

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Whole numbers
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Rational numbers
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Irrational numbers
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Integers

Explanation

$\sqrt{2}, \sqrt{3}$ are irrational numbers.
Because these two values cannot be written in the form of p/q where p and q both are integers and q should not be equal to zero .

State whether the given statement is True or False.

After rationalising the denominator of

$\dfrac{5}{3\sqrt{2}-2\sqrt{3}}$ , we get its denominator as

$7.$

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True
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False

Explanation

The denominator of

$\dfrac{5}{3\sqrt2-2\sqrt3}$ is

$3\sqrt2-2\sqrt3$
Rationalising the denominator, we get

$(3\sqrt2-2\sqrt3)(3\sqrt2+2\sqrt3)$

$=\left( 3\sqrt { 2 } \right) ^{ 2 }-\left( 2\sqrt { 3 } \right) ^{ 2 }$

$=18-12$

$=6$
So, the denominator is

$6$ .
Thus, answer is false.

Find the product.

$(a^2) (2a^{22}) (4a^{26})$

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$8a^{40}$
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$8a^{50}$
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$8a^{30}$
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$8^{20a}$

Explanation

$(a^2) (2a^{22}) (4a^{26})$

$=(a^2)\times (2a^{22})\times (4a^{26})$

$=8a^{2+22+26}$

$=8a^{50}$

The value of

$(6 + \sqrt{27}) - (3 + \sqrt{3}) + (1 - 2\sqrt{3})$ when simplified is :

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positive and irrational
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negative and rational
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positive and rational
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negative and irrational

Explanation

$6+\sqrt { 27 } -(3+\sqrt { 3 } )+(1-2\sqrt { 3 } )=6+3\sqrt { 3 } -3-\sqrt { 3 } +1-2\sqrt { 3 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =4$

$4$ is a positive rational number

Hence, correct answer is option C.

The value of

$\cfrac{3^0+7^0}{5^0}$ is:

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$2$
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$0$
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$\dfrac{9}{5}$
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$\dfrac{1}{5}$

Explanation

$\Rightarrow$

$\dfrac{3^0+7^0}{5^0}$

$\Rightarrow$

$\dfrac{1+1}{1}$

$[\,\because\,x^0=1]$

$\Rightarrow$

$\dfrac{2}{1}$

$\Rightarrow$

$2$

$\therefore$

$\dfrac{3^0+7^0}{5^0}=2$

$\sqrt{5}$ is an irrational number.

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True
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False

Explanation

An irrational number is any real number that cannot be expressed as a ratio

$\dfrac ab,$ where

$a$ and

$b$ are integers and

$b$ is non-zero.

$\sqrt5$ is irrational as it can never be expressed in the form

$\dfrac ab.$

Find the five rational numbers between

$\displaystyle \frac{1}{2}$ and

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

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$0.5 < 0.6 < 0.7 < 0.8 ... < 1.1 < ... < 1.15 < 1.50$
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$0.5 < 0.6 < 1.7 < 3.8 ... < 1.8< ... < 1.15 < 1.50$
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$0.5 < 0.6 < 0.7 < 2.8 ... < 1.1 < ... < 1.15 < 1.50$
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$0.5 < 0.6 < 0.7 < 0.8 ... < 3.1 < ... < 1.15 < 1.50$

Which of the following numbers are rational ?

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$1$
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$-6$
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$3\dfrac{1}{2}$
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All above are rational

Explanation

$\Rightarrow$ A rational number is a type of real numbers which can be expressed in the form of

$\dfrac pq,$ where

$q \neq0.$

$\Rightarrow$ All the numbers are rational as they are in the form of

$\dfrac pq,$ where

$q \neq0.$

i.e,

$\dfrac{1}{1}\ , \dfrac{-6}{1}\ ,\dfrac{7}{2}$

The rationalizing factor of

$(a+\sqrt{b})$ is

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$a-\sqrt{b}$
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$\sqrt{a}-b$
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$\sqrt{a}-\sqrt{b}$
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None of these

Explanation

The rationalizing factor of a+

$\sqrt { b }$ is a-

$\sqrt { b }$ as the product of these two expressions give a rational number.

Simplify:

$3\sqrt{3} + 10\sqrt{3}$

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$13\sqrt{3}$
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$10\sqrt{3}$
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$12\sqrt{3}$
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$11\sqrt{3}$

Explanation

$3 \sqrt{3}+ 10 \sqrt{3} =13 \sqrt{3}$

The value of

$5\sqrt{3} - 3\sqrt{12} + 2\sqrt{75}$ on simplifying is :

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$5\sqrt{3}$
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$6\sqrt{3}$
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$\sqrt{3}$
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$9\sqrt{3}$

Explanation

The given expression is

$5\sqrt { 3 } -3\sqrt { 12 } +2\sqrt { 75 } \\ =5\sqrt { 3 } -6\sqrt { 3 } +10\sqrt { 3 } \\ =9\sqrt { 3 }$ .

Therefore, option

$D$ is correct.

State True or False.

A rational number can always be written in a fraction

$\dfrac{a}{b}$ , where a and b are not integers

$(b \neq 0)$ .

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True
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False

Explanation

A number that can always be written in the form of

$p/q$ , where p is any integer and q is a non-zero integer, is a rational number.

The given statement is false.

Find conjugate of:

$\sqrt{3}+\sqrt{2}$

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$\sqrt{3}-\sqrt{2}$
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$\sqrt{3}+\sqrt{2}$
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$\sqrt{3}\sqrt{2}$
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None of these

Explanation

$\sqrt{3}+\sqrt{2}$
The procedure of multiplying a surd by another surd to get a rational number is called RationalisationThe operands are called rationalizing factor (RF) of the other

$(\sqrt{3}+\sqrt{2}) \times (\sqrt{3}-\sqrt{2})=\sqrt{3}^2-\sqrt{2}^2=9-4=5$
The Rationalizing factor of

$\sqrt{3}+\sqrt{2}$ is

$\sqrt{3}-\sqrt{2}$

Rationalise the denominator of :

$\displaystyle\ \frac{\sqrt{6}}{\sqrt{12}}$

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$\displaystyle\ \frac{\sqrt{2}}{6}$
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$\displaystyle\ \frac{\sqrt{2}}{5}$
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$\displaystyle\ \frac{\sqrt{2}}{2}$
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$\displaystyle\ \frac{\sqrt{2}}{3}$

Explanation

$\cfrac{\sqrt6}{\sqrt{12}}$ =

$\cfrac {\sqrt6\times\sqrt{12}}{\sqrt{12}\times\sqrt{12}}$ =

$\cfrac{6\sqrt2}{12}$ =

$\cfrac{\sqrt2}{2}$

Find conjugate of:
3

$\sqrt{2}$ -1

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3 $\sqrt{2}$ +1
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3 $\sqrt{2}$ -1
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3 $\sqrt{2}$
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None of these

Explanation

$3\sqrt{2}+1$
The procedure of multiplying a surd by another surd to get a rational number is called Rationalisation
The operands are called rationalizing factor (RF) of the other

$(3\sqrt{2}+1) \times (3\sqrt{2}-1)=(3\sqrt{2})^2-1^2=18-1=17$
The Rationalizing factor of

$3\sqrt{2}+1$ is

$3\sqrt{2}-1$

Rationalise the denominator of :

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

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$\displaystyle\ \frac{2\sqrt{7}}{3}$
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$\displaystyle\ \frac{2\sqrt{6}}{3}$
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$\displaystyle\ \frac{2\sqrt{2}}{3}$
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$\displaystyle\ \frac{2\sqrt{6}}{10}$

Explanation

$\cfrac{2\sqrt{2}}{\sqrt{3}}$

$=\cfrac{2\sqrt{2}\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$

$=\cfrac{2\sqrt{6}}{3}$

Rationalise the denominator of :

$\displaystyle\ \frac{6}{\sqrt{10}-2}$

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$\sqrt{10}-2$
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$\sqrt{10}+2$
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$2\sqrt{10}$
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None of these

Explanation

$\quad \\ Multiplying\quad and\quad dividing\quad by\quad \sqrt { 10 } +2:\\ \cfrac{6(\sqrt { 10 } +2)}{(10-4)}=\sqrt { 10 } +2\\$

Rationalise the denominator of:

$\displaystyle\ \frac{2}{\sqrt{5}+\sqrt{3}}$

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$\sqrt{5}-\sqrt{3}$
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$\sqrt{4}-\sqrt{3}$
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$\sqrt{2}-\sqrt{3}$
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$\sqrt{6}-\sqrt{3}$

Explanation

$\quad \\ Multiplying\quad and\quad dividing\quad by\quad \sqrt { 5 } -\sqrt { 3 } :\\\cfrac{ 2(\sqrt { 5 } -\sqrt { 3 } )}{(5-3)}=\sqrt { 5 } -\sqrt { 3 } \\$

Rationalise the denominator

(i)

$\displaystyle\ \frac{22}{2\sqrt{3}+1}$

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$(2\sqrt{3}-1)$
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$3(2\sqrt{3}-1)$
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$2(2\sqrt{3}-1)$
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$4(2\sqrt{3}-1)$

Explanation

$\\ Multiplying\quad and\quad dividing\quad by\quad 2\sqrt { 3 } -1:\quad \\ \cfrac{22(2\sqrt { 3 } -1)}{(12-1)}=2(2\sqrt { 3 } -1)\\$

Write the simplest rationalisation factor of the following surds:

$\sqrt{32}$

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$\sqrt{2}$
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$\sqrt{7}$
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$\sqrt{5}$
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$\sqrt{3}$

Explanation

$\sqrt{32}$

$=\sqrt{2\times2\times2\times2\times2}$

$=4\sqrt2$
The procedure of multiplying a surd by another surd to get a rational number is called RationalisationThe operands are called Rationalizing factor (RF) of the other.
Here,

$\sqrt 2$ is the Rationalisation Factor.

State whether true or false.

$\displaystyle \frac{5}{11}$ is a rational number.

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True
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False

Explanation

True.

Rational numbers are those numbers which can be expressed in the form

$\dfrac {p}{q}$ , where p and q are integers and

$q \neq 0$
Now,

$\dfrac {5}{11}$ , is of the form

$\dfrac {p}{q}$ , where p and q are integers and

$q \neq 0$

Hence, it is a rational number.

Rationalise the denominator of :

$\displaystyle\ \frac{5}{\sqrt{7}+\sqrt{2}}$

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$\sqrt{7}+\sqrt{2}$
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$\sqrt{7}-\sqrt{2}$
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$\sqrt{7}\sqrt{2}$
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None of these

Explanation

$\quad \\ Multiplying\quad and\quad dividing\quad by\quad \sqrt { 7 } -\sqrt { 2 } :\\ \cfrac{5(\sqrt { 7 } -\sqrt { 2 } )}{(7-2)}=\sqrt { 7 } -\sqrt { 2 } \\$

State true or false:

There can be a pair of irrational numbers whose sum is irrational such as

$\displaystyle \sqrt{3}+2$ and

$\displaystyle 5+\sqrt{2}$ .

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True
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False

Explanation

To get the sum as irrational, both the numbers need to have an irrational part which are different from each other.

Example, the given pair of numbers

$\sqrt{3} + 2$ and

$5 + \sqrt {2}$ have the sum

$\sqrt{3} + 2 + 5 + \sqrt {2} = 7 + \sqrt {2} + \sqrt {3}$ which is an irrational number too.

Hence, the given statement is true and option

$A$ is correct.

State true or false:

$\sqrt3$ is an irrational number

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True
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False

Explanation

$\sqrt{3}$ is irrational number because it will have non terminating and non recurring decimal

CBSE Questions for Class 9 Maths Number Systems Quiz 2 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Number Systems Quiz 2 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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