MCQ Questions for Class 9 Maths Number Systems Quiz 3

Rationalise the denominator of :

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

0%
$\displaystyle\ \frac{\sqrt{6}-\sqrt{2}}{2}$
0%
$\displaystyle\ \frac{\sqrt{6}\sqrt{2}}{2}$
0%
$\displaystyle\ \frac{\sqrt{6}+\sqrt{2}}{2}$
0%
None of these

Explanation

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

Write the simplest rationalization factor of the following surds :

$2\sqrt{2}$

0%
$\sqrt{3}$
0%
$\sqrt{8}$
0%
$\sqrt{2}$
0%
$\sqrt{4}$

Explanation

$2\sqrt 2$

$=2\sqrt2$
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.

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

0%
True
0%
False

Explanation

We

know that,

$\sqrt {2} = 1.414$ and

$\sqrt {3} = 1.732$
Also,

$\sqrt {5} = 2.236$

Now,

$\sqrt {2} + \sqrt {3} = 1.414 + 1.732 = 3.146 \neq \sqrt{5}$

Hence, the given statement is False.

State whether true or false.

$\displaystyle \frac{2}{7}$ is a rational number.

0%
True
0%
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,

$\dfrac {2}{7}$ , is of the form

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

$q \neq 0$
Hence, it is a rational number.

$\sqrt 7$ is

0%
A rational number
0%
An irrational number
0%
Not a real number
0%
Terminating decimal

Explanation

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

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

$q \neq 0$
Numbers which are not rational numbers are called irrational numbers.
Since,

$\sqrt {7}$ cannot be written in

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

$p$ and

$q$ are integers and

$q \neq 0$ ; it is an irrational number.

Are the square roots of all positive integers irrational?

0%
True
0%
False

Explanation

No. Not all square roots of integers are irrational. Examples are

$\sqrt{4}=2$ ,

$\sqrt{1}=1$ ,

$\sqrt{9}=3$ , etc.

State whether the following statement are true or false? Justify your answers.

Every irrational number is a real number.

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True
0%
False

The decimal expansion of the number

$\sqrt {2}$ is

0%
A finite decimal
0%
$1.4121$
0%
Non-Terminating, Recurring
0%
Non-Terminating, Non-Recurring

Explanation

$\sqrt2$ is an irrational number.

$\sqrt{2} = 1.4142136...$
As can be observed from the expansion, the decimal expansion of the number

$\sqrt{2}$ is non-terminating non-recurring.

A number that can be expressed in the form

$\displaystyle\frac{a}{b}$ , where

$a$ and

$b$ are integers and

$b$ is not equal to zero is called

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a fraction
0%
an integer
0%
a rational number
0%
a real number

Explanation

A rational number : A number that can be in the form of

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

$p$ and

$q$ are integers and

$q$ is not equal to zero.

Which of the following statement is false?

0%
Every fraction is a rational number
0%
Every rational number is a fraction
0%
Every integer is a rational number
0%
All the above

Explanation

${\textbf{Step 1: Consider, option A.}}$

${\text{Every fraction is a rational number}}.$

${\text{As we know that,}}$ ${\text{ A rational number is a type of real numbers, }}$

${\text{which is in the form of }}\dfrac{p}{q}{\text{ where }}q{\text{ is not equal to zero}}{\text{. }}$

${\text{Any fraction with non - zero denominators is rational number}}{\text{.}}$

${\text{Thus, option A}}{\text{. is true}}{\text{.}}$

${\textbf{Step 2: Consider, option B}}{\textbf{.}}$

${\text{Every rational number is a fraction}}.$

${\text{As we know that,}}$ ${\text{ A rational number is a type of real numbers, }}$

${\text{which is in the form of }}\dfrac{p}{q}{\text{ where }}q{\text{ is not equal to zero}}{\text{. }}$

${\text{The given statement is not true as 10 is a rational number but it is not a fraction}}{\text{.}}$

${\text{Thus, option B}}{\text{. is false}}{\text{.}}$

${\textbf{Step 3: Consider, option C}}{\textbf{.}}$

${\text{Every integer is a rational number}}.$

${\text{A rational number is a type of real numbers, }}$

${\text{which is in the form of }}\dfrac{p}{q}{\text{ where }}q{\text{ is not equal to zero}}{\text{. }}$

${\text{And we know that an integer can be represented as }}\dfrac{p}{q}$

${\text{form where denominator will be always 1}}{\text{.}}$

${\text{For example: }}10 = \dfrac{{10}}{1}$

${\text{Thus, option C}}{\text{. is true}}{\text{.}}$

${\textbf{Hence, option B is correct answer.}}$

Write two irrational numbers between

$0.2$ and

$0.21$ .

0%
$0.2010010001...$
0%
$0.2020020002...$
0%
$0.2010010001$
0%
$0.2020020002$

Explanation

$0.2$ and

$0.21$ are two rational numbers.

We know, between any two rational numbers, there are infinitely many irrational numbers.

An irrational number has non-terminating non-recurring decimal expansions.

Here,

$0.2010010001...$ and

$0.2020020002...$ are non-terminating non-recurring,

while

$0.2010010001$ and

$0.2020020002$ are terminating.

Hence, the two irrational numbers are

$0.2010010001...$ and

$0.2020020002...$ .

Thus, options

$A$ and

$B$ are the correct answers.

If we divide a positive integer by another positive integer, what is the resulting number ?

0%
Always a natural number
0%
Always an integer
0%
A rational number
0%
An irrational number

Explanation

If we divide a positive integer by another positive integer, the resulting number is always a rational number.

Though it can be a natural number and an integer only if the denominator is

$1.$

Find two irrational numbers between

$0.5$ and

$0.55$ .

0%
$0.5010010001...$
0%
$0.5020020002...$
0%
$0.50101010101...$
0%
$0.50202020202...$

Explanation

${\textbf{Step-1: Which are rational numbers.}}$

${\text{Given numbers are 0.5 and 0.55 which are rational numbers.}}$

${\text{Irrational numbers are non terminating and non-recurring decimals}}$

$\rightarrow$

${\text{point to consider.}}$

${\text{In between 0.5 and 0.55, we can find 0.51 and 0.52.}}$

${\textbf{Step-2: Reference image.}}$

${\text{And these lie in between 0.5 and 0.55 and these are irrational too.}}$

${\text{So,}}$

$0.5101001000100001...$

${\text{and}}$

$0.5202002000200002...$

${\text{are two irrational numbers between 0.5 and 0.55.}}$

${\textbf{ Hence, the option (A) and (B) is correct.}}$

Conjugate surd of

$\displaystyle a-\sqrt{b}$ is:

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$\displaystyle a+\sqrt{b}$
0%
$\displaystyle \sqrt{a}-b$
0%
$\displaystyle \sqrt{a}-\sqrt{b}$
0%
$\displaystyle \sqrt{a}+b$

Explanation

The sum and difference of two simple quadratic surds are said to be conjugate surds to each other.
So, conjugate surd of

$a-\sqrt{b}$ is

$a+ \sqrt{b}$ .

For example, conjugate surd of

$2+\sqrt{3}$ is

$2- \sqrt{3}$ .

The rationalising factor of

$\displaystyle 5+2\sqrt{6}$ is

0%
$\displaystyle -2\sqrt{6}-5$
0%
$\displaystyle \sqrt{6}-10$
0%
$\displaystyle 6-2\sqrt{5}$
0%
$\displaystyle 5-2\sqrt{6}$

Explanation

$\displaystyle 5-2\sqrt{6}$

State true or false.

The three rational number between

$5$ and

$6$ are

$\displaystyle\frac{21}{4},\frac{22}{4},\frac{23}{4}$ .

0%
True
0%
False

Explanation

$5=\frac { 20 }{ 4 } \quad and\quad 6=\frac { 24 }{ 4 } ,\\ \frac { 21 }{ 4 } ,\frac { 22 }{ 4 } \quad and \quad \frac { 23 }{ 4 } \quad are\quad three\quad rational\quad numbers\quad lying\quad between\quad 5\quad and\quad 6.\quad \quad \\$

Simplify

$(32)^{\displaystyle \frac {-2}{5}}\, \div\, (125)^{\displaystyle \frac {-2}{3}}$

0%
$\displaystyle \frac {4}{25}$
0%
$\displaystyle \frac {25}{4}$
0%
$\displaystyle \frac {2}{5}$
0%
$\displaystyle \frac {5}{2}$

Explanation

$\cfrac {(32)^{ \tfrac {-2}{5}}}{(125)^{ \tfrac {-2}{3}}} = \cfrac { \cfrac {1}{4}}{ \cfrac {1}{25}} = \cfrac {25}{4}$

A rational number between

$\displaystyle \frac{1}{4}$ and

$\displaystyle \frac{1}{3}$ is

0%
$\displaystyle \frac{7}{24}$
0%
$\displaystyle \frac{16}{48}$
0%
$\displaystyle \frac{13}{48}$
0%
All the above

Explanation

$\dfrac{1}{4} = \dfrac{6}{24} = \dfrac{12}{48}$

$\dfrac{1}{3} = \dfrac{8}{24} = \dfrac{16}{48}$

From this, we can see

$\dfrac{7}{24} , \dfrac{13}{48}, \dfrac{16}{48}$ all lie between

$\dfrac{1}{4}$ and

$\dfrac{1}{3}$

$\left ( \displaystyle \frac {16}{81} \right )^{\displaystyle \frac {3}{4}}$ = ..........

0%
$\displaystyle \frac {9}{2}$
0%
$\displaystyle \frac {2}{9}$
0%
$\displaystyle \frac {8}{27}$
0%
$\displaystyle \frac {27}{8}$

Explanation

$\left ( \cfrac {16}{81} \right )^{ \tfrac {3}{4}} = \left [ \left ( \cfrac {2}{3} \right )^4 \right ]^{\tfrac {3}{4}} = \cfrac {8}{27}$

If p: every fraction is a rational number

q: every rational number is a fraction

then which of the following is correct?

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p is true and q is false.
0%
p is false and q is true.
0%
Both p and q are true.
0%
Both p and q are false.

Explanation

Fraction is defined as a part of a whole thing. Every fraction is of the form

$\dfrac{m}{n}$ where

$m$ is a whole number and

$n$ is a natural number.

Rational number is defined as the number of the form

$\dfrac{a}{b}$ where

$a$ and

$b$ are integers and

$b\neq 0$ .

Since all whole numbers and natural numbers are present in set of integers every fraction is a rational number. Examples :

$\dfrac{4}{5}, \dfrac{4}{7}$

So, the statement

$p$ is true.

But every rational number is not a fraction. Examples :

$\dfrac{3}{-4}, \dfrac{17}{-16}$ . Here the denominators are not natural numbers and hence they are not fractions.

So, the statement

$q$ is false.

$\displaystyle \frac{-2}{-19}$ is a

0%
Positive rational number.
0%
Negative rational number.
0%
Either positive or negative number.
0%
Has neither a positive numerator nor a negative number

Explanation

$\because$ Both numerator and denominator are negative (i.e., same sign)

The value of

$(3^{0} - 4^{0})\, \times\, 5^2$ is

0%
$25$
0%
$0$
0%

$-25$
0%
none of these

Explanation

To find the value of

$(3^{0} - 4^{0})\, \times\, 5^2$

$(3^{0} - 4^{0}) \times 5^2$

$= (1 - 1) \times 25$

$= 0\times25$

$= 0$

Hence, the value is

$0$

Evaluate:

$(0.04)^{-1.5}$

0%
$25$
0%
$125$
0%
$250$
0%
$625$

Explanation

$(0.04)^{-1.5} = \left ( \cfrac {4}{100} \right )^{-1.5}$

$=\,\left (\cfrac {1}{25} \right )^{-\tfrac 32}$

$= (25)^{\tfrac 32}$

$= (5^2)^{\tfrac 32}$

$= 5^3 = 125$

$\displaystyle \frac{-5}{0}$ is a ........

0%
Positive rational number.
0%
Negative rational number.
0%
Either positive or negative rational number.
0%
Neither positive nor negative rational number.

Explanation

$\because$ Denominator is '0', it is not a rational number as it makes number undefined.

The value of

$[ (-2)^{(-2)} ]^{(-3)}$ is

0%
64
0%
32
0%
cannot be determined
0%
none of these

Explanation

$[ (-2)^{(-2)} ]^{(-3)}\, =\, (-2)^6\, =\, 64$

Which of the following is a rational number (s)?

0%
$\displaystyle \frac{-2}{9}$
0%
$\displaystyle \frac{4}{-7}$
0%
$\displaystyle \frac{-3}{17}$
0%
All the three given numbers

Explanation

A rational number is a number that can be expressed as a fraction

$\dfrac{p}{q}$ where

$p$ and

$q$ are integers and

$q\neq0$ .

A rational number

$\dfrac{p}{q}$ is said to have numerator

$p$ and denominator

$q.$

In the given options, both numerator and denominator are integers.

Hence, all the three given numbers are rational number.

A rational number is defined as a number that can be expressed in the form

$\dfrac{p}{q}$ where

$p$ and

$q$ are integers and

0%
$q=0$
0%
$q=1$
0%
$q\neq 1$
0%
$q\neq 0$

Explanation

According to the definition of a rational number, it can be expressed in the form of

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

$p$ and

$q$ are integers and

$q\neq$ 0

Evaluate:

$(256)^{0.16}\, \times\, (256)^{0.09}$

0%
$4$
0%
$16$
0%
$64$
0%
$256.25$

Explanation

${ \left( 256 \right) }^{ 0.16 }\times { \left( 256 \right) }^{ 0.09 }$

$={ \left( 256 \right) }^{ 0.16+0.09 }$

$={ \left( 256 \right) }^{ 0.25 }$

$={ \left( 256 \right) }^{ \tfrac { 25 }{ 100 } }$

$={ \left( 256 \right) }^{ \tfrac { 1 }{ 4 } }$

$={ 2 }^{ 8\times \tfrac { 1 }{ 4 } }$

$={ 4 }$

The value of

$(8^{-25}\, -\, 8^{-26})$ is

0%
$7\, \times\, 8^{-25}$
0%
$7\, \times\, 8^{-26}$
0%
$8\, \times\, 8^{-26}$
0%
None of these

Explanation

$(8^{-25} - 8^{-26})$

$=\left ( { \dfrac {1}{8^{25}}} - { \dfrac {1}{8^{26}}} \right )$

$=\dfrac{8^{25}(8-1)}{8^{25}\times 8^{26}}$

$= \cfrac {(8-1)}{8^{26}}$

$=7 \times 8^{-26}$

Thus option

$(B)$ is correct.

The reciprocal of a positive rational number is positive.

0%
True
0%
False
0%
Cannot be determined
0%
None

CBSE Questions for Class 9 Maths Number Systems Quiz 3 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Number Systems Quiz 3 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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