The number obtained on rationalising the denominator of $$\dfrac{1}{\sqrt{9}-\sqrt{8}}$$ is ?

$$\dfrac{1}{2}\left ( 3-2\sqrt{2} \right )$$

MCQ Questions for Class 9 Maths Number Systems Quiz 1

State whether the statement is true (T) or false (F).
If

\frac{p}{q}

$\dfrac{p}{q}$ is a rational number, then

p

$p$ cannot be equal to zero.

0%
True
0%
False

Explanation

The given statement is false.

p

$p$ can be any integer, including zero.

State whether the statement is true (T) or false (F).
If

\frac{x}{y}

$\dfrac{x}{y}$ is a rational number, then

y

$y$ is always a whole number.

0%
True
0%
False

Explanation

The given statement is false.

y

$y$ should be a non- zero number.

State true or false.

Zero is the smallest rational number.

0%
True
0%
False

5^{0} = 5

$5^{0}=5$

0%
True
0%
False

Explanation

False.
Using law of exponent

= a^{0} = 1

$=a^{0}=1$

\Rightarrow 5^{0} = 1

$\Rightarrow 5^{0}=1$

L H S \neq R H S

$LHS\neq RHS$

x

$x$ be any non-zero integer and

m, n

$m,n$ be positive integers, then

x^{m} \times x^{n}

$x^m \times x^n$ is equal to

0%
$x^m$
0%
$x^{m+n}$
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$x^n$
0%
$x^{m-n}$

Explanation

Using law of exponents,

a^{m} \times a^{n} = (a)^{m + n}

$a^m\times a^n =(a)^{m+n}\quad$ [Where

a

$a$ is non-zero integer]
Similarly

x^{m} \times x^{n} = (x)^{m + n}

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

y

$y$ is any non-zero integer, then

y^{0}

$y^0$ is equal to ....

0%
$1$
0%
$0$
0%
$-1$
0%
Not found

Explanation

Using the law of exponents,

a^{0} = 1

$a^0 =1$

Therefore

y^{0} = 1

$y^0=1$

(- 2)^{0} = 2

$(-2)^{0}=2$

0%
True
0%
False

Explanation

Step -1: Write the equation .

${\textbf{Step -1: Write the equation}}{\textbf{.}}$

{(- 2)}^{0} = 2

${\left( { - 2} \right)^0} = 2$

Step -2: According to law of indices .

${\textbf{Step -2: According to law of indices}}{\textbf{.}}$

{(- 2)}^{0} = 1

${\left( { - 2} \right)^0} = 1$

and

${\text{and }}$

1 \neq 2

${{ 1} } \neq 2$

Hence, the given equation is false .

${\textbf{Hence, the given equation is false}}{\textbf{.}}$

(- 6)^{0} = - 1

$(-6)^{0}=-1$

0%
True
0%
False

Explanation

False.
Using law of exponent

= a^{0} = 1

$=a^{0}=1$

\Rightarrow (- 6)^{0} = 1

$\Rightarrow (-6)^{0}=1$

L H S \neq R H S

$LHS\neq RHS$

Find the positive rational number out of the following.

0%
$\dfrac{-4}{9}$
0%
$\dfrac{-16}{36}$
0%
$\dfrac{-20}{-45}$
0%
$\dfrac{28}{-63}$

Explanation

\frac{- 20}{- 45} = \frac{20}{45}

$\dfrac{-20}{-45}=\dfrac{20}{45}$ is the only positive rational number among all in the above-given questions . Therefore, the option (c) is the odd one.

State whether the following statement is true (T) or false (F):

3^{0} = (1000)^{0}

$3^0 = (1000)^0$ .

0%
True
0%
False

Explanation

3^{0} = 1

$3^0 =1$ and

(1000)^{0} = 1

$(1000)^0 =1$

State true or false:

All terminating decimal numbers are rational numbers.

0%
True
0%
False

Explanation

All terminating decimals, after mathematical manipulations, can be expressed as a rational number.

Hence, the statement is true.

Therefore, option

A

$A$ is correct.

\sqrt{3}

$\sqrt 3$ is

0%
an irrational number
0%
a rational number
0%
a composite number
0%
a prime number

\frac{2}{7}

$\dfrac{2}{7}$ is an irrational number.

0%
True
0%
False

Explanation

False, as

\frac{2}{7}

$\dfrac 27$ is represented in the form of

\frac{p}{q}

$\dfrac pq$ and

7 \neq 0.

$7\neq 0.$ The given fraction is a rational number.

State, in each case, whether true or false:

\frac{5}{11}

$\dfrac{5}{11}$ is a rational number

0%
True
0%
False

Explanation

True, as

\frac{5}{11}

$\dfrac {5}{11}$ is represented in the form of

\frac{p}{q}

$\dfrac pq$ and

11 \neq 0

$11\neq 0$

State, in each case, whether true or false:
All rational numbers are real numbers.

0%
True
0%
False

The number obtained on rationalizing the denominator of

\frac{1}{\sqrt{7} - 2}

$\dfrac {1}{\sqrt {7} - 2}$ is

0%
$\dfrac {\sqrt {7} + 2}{3}$
0%
$\dfrac {\sqrt {7} - 2}{3}$
0%
$\dfrac {\sqrt {7} + 2}{5}$
0%
$\dfrac {\sqrt {7} + 2}{45}$

Explanation

We use the identity

(a + \sqrt{b}) (a - \sqrt{b}) = a^{2} - b

$\left( a+\sqrt { b } \right) \left( a-\sqrt { b } \right) ={ a }^{ 2 }-b$ .

Multiply and divide

\frac{1}{\sqrt{7} - 2}

$\dfrac { 1 }{ \sqrt { 7 } -2 }$ by

\sqrt{7} + 2

$\sqrt { 7 } +2$ to get

\frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3}

$\dfrac { 1 }{ \sqrt { 7 } -2 } \times \dfrac { \sqrt { 7 } +2 }{ \sqrt { 7 } +2 } =\dfrac { \sqrt { 7 } +2 }{ 7-4 } =\dfrac { \sqrt { 7 } +2 }{ 3 }$

Hence, option

A .

$A.$

After rationalising the denominator of

\frac{7}{3 \sqrt{3} - 2 \sqrt{2}}

$\dfrac{7}{3\sqrt{3}-2\sqrt{2}}$ , we get the denominator of the resultant equivalent expression as:

0%
$5$
0%
$13$
0%
$19$
0%
$35$

Explanation

\frac{7}{3 \sqrt{3} - 2 \sqrt{2}} = \frac{7}{3 \sqrt{3} - 2 \sqrt{2}} \times (\frac{3 \sqrt{3} + 2 \sqrt{2}}{3 \sqrt{3} + 2 \sqrt{2}}) = \frac{7 (3 \sqrt{3} + 2 \sqrt{2})}{27 - 8} = \frac{7 (3 \sqrt{3} + 2 \sqrt{2})}{19}

$\dfrac { 7 }{ 3\sqrt { 3 } -2\sqrt { 2 } } =\dfrac { 7 }{ 3\sqrt { 3 } -2\sqrt { 2 } } \times \left( \dfrac { 3\sqrt { 3 } +2\sqrt { 2 } }{ 3\sqrt { 3 } +2\sqrt { 2 } } \right) =\dfrac { 7\left( 3\sqrt { 3 } +2\sqrt { 2 } \right) }{ 27-8 } =\dfrac { 7\left( 3\sqrt { 3 } +2\sqrt { 2 } \right) }{ 19 }$ . Hence, the denominator of the equivalent expression is

19.

$19.$

The number obtained on rationalising the denominator of

\frac{1}{\sqrt{9} - \sqrt{8}}

$\dfrac{1}{\sqrt{9}-\sqrt{8}}$ is ?

0%
$\dfrac{1}{3+2\sqrt{2}}$
0%
$3-2\sqrt{2}$
0%
$3+2\sqrt{2}$
0%
$\dfrac{1}{2}\left ( 3-2\sqrt{2} \right )$

Explanation

We use the identity

(\sqrt{a} + \sqrt{b}) (\sqrt{a} - \sqrt{b}) = a - b

$\left( \sqrt { a } +\sqrt { b } \right) \left( \sqrt { a } -\sqrt { b } \right) =a-b$ .

So,

$\dfrac { 1 }{ \sqrt { 9 } -\sqrt { 8 } } =\dfrac { 1 }{ \sqrt { 9 } -\sqrt { 8 } } \times \left( \dfrac { \sqrt { 9 } +\sqrt { 8 } }{ \sqrt { 9 } +\sqrt { 8 } } \right) \\$

$=\dfrac { \sqrt { 9 } +\sqrt { 8 } }{ 9-8 }$

$=\dfrac { \sqrt { 9 } +\sqrt { 8 } }{ 1 }$

$=\sqrt { 3^2 } +\sqrt { 4\times 2 }$

$=3+2\sqrt { 2 }$

Which of the following is irrational number?

0%
$\sqrt{\dfrac{8}{18}}$
0%
$\sqrt{\dfrac{12}{3}}$
0%
$\sqrt{\dfrac{28}{8}}$
0%
$\sqrt{81}$

Explanation

All numbers that can be written in the form of

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

$p$ and

$q$ are integers are rational numbers.

Option

$A :$

$\sqrt{\dfrac{8}{18}}$

$= \sqrt{\dfrac{4}{9}}$

$= \dfrac{2}{3}$
Hence, it is rational.

Option

$B :$

$\sqrt{\dfrac{12}{3}}$

$=\sqrt{4}$

$=2$
Hence, it is a rational number.

Option

$C :$

$\sqrt{\dfrac{28}{8}}$

$=\sqrt{\dfrac{7}{2}}$
This cannot be simplified further.

This is an irrational number.

Option

$D :$

$\sqrt{81}$

$=9$
This is a rational number.

Hence, option

$C$ is correct.

$\sqrt {10}\times \sqrt {15}$ is equal to

0%
$\sqrt {25}$
0%
$5\sqrt {6}$
0%
$6\sqrt {5}$
0%
$10\sqrt {5}$

Explanation

$\sqrt { 10 } \times \sqrt { 15 } =\sqrt { 5\times 2 } \times \sqrt { 5\times 3 } =\sqrt { 5\times 5\times 2\times 3 } =5\sqrt { 6 }$

The product of any two irrational numbers

0%
Is always an irrational number
0%
Is always a rational number
0%
Is always an integer
0%
Can be rational or irrational

Explanation

The product of two irrational numbers can be rational or irrational depending on the two numbers.

For example,

$\sqrt{2}\times\sqrt{2}$ is

${2}$ , which is a rational number whereas

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

$\sqrt{6}$ , which is an irrational number.

Hence, option

$D.$

The decimal expansion of

$\pi$ is :

0%
terminating
0%
non-terminating and non-recurring
0%
non-terminating and recurring
0%
doesnt exist

Explanation

We know that

$\pi$ is irrational number and Irrational numbers have decimal expansions that neither terminate nor become periodic.

So, correct answer is option

$B$ .

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

0%
Irrational number
0%
Rational number
0%
Whole number
0%
Natural number

Rationalise the denominator and find the equivalent of :

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

0%
$\sqrt{12}$
0%
$\sqrt{3}$
0%
$\sqrt{6}$
0%
$\sqrt{2}$

Explanation

$\cfrac{2\sqrt{3}}{\sqrt{2}}$
Rationalising the denominator, we get

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

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

$=\sqrt{6}$

Find conjugate of:

$3-\sqrt{5}$

0%
$3+\sqrt{5}$
0%
$3-\sqrt{5}$
0%
$3\sqrt{5}$
0%
None of these

Explanation

$3-\sqrt { 5 }$
The procedure of multiplying a surd by another surd to get a rational number is called Rationalization. The operands are called rationalizing factor (RF) of the other

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

$3-\sqrt { 5 }$ is

$3+\sqrt { 5 }$

Rationalise the denominator of:
(i)

$\displaystyle\ \frac{1}{2-\sqrt{3}}$

0%
$2+\sqrt{7}$
0%
$2+\sqrt{2}$
0%
$2+\sqrt{3}$
0%
$2+\sqrt{5}$

Explanation

$\quad \\ Multiplying\quad and\quad dividing\quad by\quad 2+\sqrt { 3 } :\\ \cfrac{(2+\sqrt { 3 } )}{(4-3)}=2+\sqrt { 3 } \\$

Is zero a rational number? Can you write it in the form

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

$p$ and

$q$ are integers and

$q\ne 0$ ?

0%
True
0%
False

Classify the result as rational or irrational:

$(3+\sqrt{23})-\sqrt{23}$ .

0%
Rational number
0%
Irrational number
0%
Data Insufficient
0%
None of the above

Explanation

$(3+\sqrt{23})-\sqrt{23}$

$=3+\sqrt{23}-\sqrt{23}$

$=3$ .
Here,

$3$ is a rational number.

Therefore, option

$A$ is correct.

Write the simplest rationalization factor of the following surds :

$4\sqrt[3]{3}$

0%
$\sqrt[3]{9}$
0%
$\sqrt[2]{9}$
0%
$\sqrt{9}$
0%
$\sqrt{3}$

Explanation

$4\sqrt[3]{3}$ ,the simplest rationalization factor is

$\sqrt[3]{3^2}=\sqrt[3]{9}$

A number is an irrational if and only if its decimal representation is:

0%
non terminating
0%
non terminating and repeating
0%
non terminating and non repeating
0%
terminating

CBSE Questions for Class 9 Maths Number Systems Quiz 1 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Number Systems Quiz 1 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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