MCQ Questions for Class 9 Maths Number Systems Quiz 7

Say true or false:

$87, 54, 0, -13, \sqrt{16}$ are integers

0%
True
0%
False

Explanation

The real value of

$\sqrt { 16 } =4$

All other numbers are integers.

So, the given statement is true.

Simplify:

$7\sqrt{48}-\sqrt{72}-\sqrt{27}+3\sqrt{2}$

0%
$25\sqrt{3} - 13\sqrt{2}$
0%
$5\sqrt{3} - 3\sqrt{2}$
0%
$25\sqrt{3} - 9\sqrt{2}$
0%
$25\sqrt{3} - 3\sqrt{2}$

Explanation

$7\sqrt{48}-\sqrt{72}-\sqrt{27}+3\sqrt{2}$

$=7\times 4\sqrt{3}-6\sqrt{2}-3\sqrt{3}+3\sqrt{2}$

$=(28-3)\sqrt{3}+(3-6)\sqrt{2}$

$=25 \sqrt{3}-3 \sqrt{2}$

$\sqrt 5$ is

0%
an integer
0%
a rational number
0%
an irrational number
0%
none of these

Explanation

An integer is a whole number (not a fractional number) that can be positive, negative, or zero and no digits after the decimal point.

Examples of integers are: -5, 1, 5, 8, 97, and 3,043.

A Rational Number can be made by dividing two integers. (An integer is a number with no fractional part.)

$\sqrt{5}$ is an irrational number as it cant be represented as a fraction.

Answer is Option C

Insert two irrational numbers between

$5$ and

$6$ .

0%
$\sqrt{36},\,\sqrt{25}$
0%
$\sqrt{32},\,\sqrt{33}$
0%
$\sqrt{36},\,\sqrt{49}$
0%
$\sqrt{24},\,\sqrt{33}$

Explanation

$\textbf{Step -1: Calculating the two irrational numbers }$

$\text{We need to find two irrational numbers between 5 and 6}$

$\text{Since 5 and 6 are consecutive numbers the root of every}$

$\text{number between their squares is irrational}$

$\implies 25\leq x^2\leq 36$

$\implies x^2 \text{ can take the values 32 and 33}$

$\implies x=\sqrt{32},\sqrt{33}$

$\textbf{Hence,Two irrational numbers between 5 and 6 are }\mathbf{\sqrt{32},\sqrt{33} .}$

Simplify

$\displaystyle 4\sqrt{12}-\sqrt{75}-7\sqrt{48}$

0%
$25\sqrt{3}$
0%
$-25 \sqrt{3}$
0%
$5 \sqrt{3}$
0%
$-5 \sqrt{3}$

Explanation

$\displaystyle 4\sqrt{12}-\sqrt{75}-7\sqrt{48}$

$=4\sqrt{12}-\sqrt{75}-7\sqrt{48}$

$=4\times 2 \sqrt{3} - 5 \sqrt{3} -7 \times 4\sqrt{3}$

$=\sqrt{3}[8-5-28]$

$=-25 \sqrt{3}$

Which of the following is an irrational number?

0%
$\dfrac {22}{7}$
0%
$3.141592$
0%
$2.78181818$
0%
$0.123223222322223.......$

Explanation

${\textbf{Step 1: Define Irrational number.}}$

${\text{An irrational number is a number that cannot be expressed as a dfration}}{\text{.}}$

${\text{Also, they have decimal expression neither terminate nor become periodic}}{\text{.}}$

${\text{Now, check it for all the four options}}{\text{.}}$

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

$\dfrac{22}{7}$ $=3.\overset{\centerdot }{\mathop{1}}\,4285\overset{\centerdot }{\mathop{7}}\,$

$\text{Where }\centerdot \text{ represent beginning and end of recurring group of numbers}\text{.}$

$\Rightarrow \dfrac{{22}}{7}$ ${\text{is a rational number as as it is a recurring decimal}}{\text{.}}$

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

$3.141592$

${\text{It is a terminating decimal number}}{\text{.}}$

$\Rightarrow {\text{3}}{\text{.141592 is a rational number}}{\text{.}}$

${\textbf{Step 4: Consider, option C.}}$

$2.78181818$

${\text{It is a terminating decimal number}}{\text{.}}$

$\Rightarrow 2.78181818{\text{ is a rational number}}{\text{.}}$

${\textbf{Step 5: Consider, option D.}}$

$0.123223222322223 \ldots$

${\text{It is a non - terminating decimal number and also it is not a recurring decimal}}{\text{.}}$

$\Rightarrow 0.123223222322223 \ldots {\text{ is a irrational number}}{\text{.}}$

${\textbf{Final Answer: Hence, option (D)}}$ $\mathbf{0.123223222322223 \dots} {\textbf{ is correct answer.}}$

Find the product of following with its conjugate pair.

$\displaystyle \sqrt{5} + 1$

0%
conjugate pair $= \sqrt{5}+1$ , product $3$
0%
conjugate pair $= \sqrt{5}-1$ , product $4$
0%
conjugate pair $= \sqrt{5}+1$ , product $4$
0%
conjugate pair $= \sqrt{5}-1$ , product $3$

Explanation

The conjugate of

$\sqrt{5}+1$ is

$\sqrt{5}-1$
Product =

$(\sqrt{5}+1) \times (\sqrt{5}-1)={\sqrt{5}}^2-1^2=5-1=4$

Give an example of two irrational numbers, whose difference is an irrational number.

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

Explanation

Let be the Number are

$4\sqrt{3}$ and

$2\sqrt{3}$ .
Difference of Number

$4\sqrt{3} - 2\sqrt{3} = 2\sqrt{3}$ , which is a irrational number.

The numbers in options

$B$ ,

$C$ and

$D$ , on taking the difference gives value

$=0$ , which is a rational number.

Hence, option

$A$ is correct.

Consider option

$A$ .

The number are

$4\sqrt{3}$ and

$2\sqrt{3}$ .
Their difference

$4\sqrt{3} - 2\sqrt{3} = 2\sqrt{3}$ , which is an irrational number.

Consider option

$B$ .

The number are

$\sqrt{3}$ and

$\sqrt{3}$ .
Their difference

$\sqrt{3} - \sqrt{3} = 0$ , which is a rational number.

Consider option

$C$ .

The number are

$2\sqrt{3}$ and

$2\sqrt{3}$ .
Their difference

$2\sqrt{3} - 2\sqrt{3} = 0$ , which is an rational number.

Consider option

$D$ .

The number are

$4\sqrt{3}$ and

$4\sqrt{3}$ .
Their difference

$4\sqrt{3} - 4\sqrt{3} = 0$ , which is an rational number.

Therefore, option

$A$ is correct.

Give an example of two irrational numbers, whose difference is a rational number.

0%
$4+\sqrt{2}, 2+\sqrt{2}$
0%
$4-\sqrt{2}, 2+\sqrt{2}$
0%
$4+\sqrt{2}, 2-\sqrt{2}$
0%
$4+\sqrt{2}, -2-\sqrt{2}$

Explanation

A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator and the denominator are whole numbers.

On the other hand, all numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction.

So, let us take two irrational numbers

$a=4+\sqrt { 2 }$ and

$b=2+\sqrt { 2 }$ and we now calculate

$a-b$ as follows:

$a-b=\left( 4+\sqrt { 2 } \right) -\left( 2+\sqrt { 2 } \right) =4+\sqrt { 2 } -2-\sqrt { 2 } =2=\dfrac {2}{1}$

Therefore, the difference of two irrational numbers

$a$ and

$b$ is

$2$ and from the above defination of rational numbers, we get that

$2$ is a rational number.

Hence, the difference of two irrational numbers

$4+\sqrt { 2 }$ and

$2+\sqrt { 2 }$ is a rational number.

Write two irrational numbers between

$0.21$ and

$0.2222...$

0%
$0.21010010001...$
0%
$0.2102020202...$
0%
$0.21020020002...$
0%
$0.210101010101...$

Explanation

$0.21$ and

$0.2222...$ 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 expansion.

Here,

$0.21010010001...$ and

$0.21020020002...$ are non-terminating non-recurring,

while

$0.210101010101...$ and

$0.2102020202...$ are non-terminating recurring.

Hence, the two irrational numbers are

$0.21010010001...$ and

$0.21020020002...$ .

Thus, options

$A$ and

$C$ are the correct answers.

The value of

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

0%
$5\sqrt{3}$
0%
$6\sqrt{3}$
0%
$\sqrt{3}$
0%
$9\sqrt{3}$

Explanation

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

Say true or false:

$0.120 1200 12000 120000$ ....is a rational number

0%
True
0%
False

Explanation

Given,

$0.120 1200 12000 120000 ....$
Since, the decimal expansion is neither terminating nor non-terminating repeating, therefore, the given real number is not rational.
they are not rational, so we can't write of the form

$\displaystyle \frac {p}{q}$ .

The product of

$4\sqrt{6}$ and

$3\sqrt{24}$ is

0%
124
0%
134
0%
144
0%
154

Explanation

$4\sqrt { 6 } \times3\sqrt { 24 }$

$=4\sqrt { 6 } \times 6\sqrt { 6 }$

$=24\sqrt { 6 } \sqrt { 6 }$

$=24*6=144\\ \\ \quad$

The value of

$\displaystyle 2\sqrt{3}-3\sqrt{12}+5\sqrt{75}$ is equal to:

0%
$\displaystyle 4\sqrt{3}$
0%
$\displaystyle 7\sqrt{5}$
0%
$\displaystyle 3\sqrt{5}$
0%
$\displaystyle 21\sqrt{3}$

Explanation

$\displaystyle 2\sqrt{3}-3\sqrt{12}+5\sqrt{75}$

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

$=\displaystyle 2\sqrt{3}-6\sqrt{3}+25\sqrt{3}$

$=\displaystyle \sqrt{3}\left ( 2-6+25 \right )$

$=\displaystyle \sqrt{3}(21)=21\sqrt{3}$ .

Therefore, option

$D$ is correct.

State True or False.

The five rational numbers between

$\dfrac{3}{5}$ and

$\dfrac{4}{5}$ are

$\displaystyle \frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30}$ .

0%
True
0%
False

Explanation

Since we want five numbers we write

$\dfrac{3}{5}$ and

$\dfrac{4}{5}$
So multiply in numerator and denominator by

$5+1 = 6$ we get

$\Rightarrow \dfrac{3}{5}=\dfrac{3\times 6}{5\times 6}=\dfrac{18}{30}$

$\Rightarrow \dfrac{4}{5}=\dfrac{4\times 6}{5\times 6}=\dfrac{24}{30}$

We know that

$18<19<20<21<22<23<24$

$\Rightarrow \displaystyle \frac{18}{30}<\frac{19}{30}<\frac{20}{30}<\frac{21}{30}<\frac{22}{30}<\frac{23}{30}<\frac{24}{30}$

Hence 5 rational number between

$\dfrac{3}{5} and \dfrac{4}{5}$ are

$\displaystyle \frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30},$

State TRUE or FALSE

The three rational number between

$\dfrac{1}{3}$ and

$\dfrac{1}{2}$ are

$\displaystyle\frac{9}{24},\frac{10}{24},\frac{11}{24}$ .

0%
True
0%
False

Explanation

$x=\frac { 1 }{ 3 } =\frac { 8 }{ 24 } \quad and\quad y=\frac { 1 }{ 2 } =\frac { 12 }{ 24 } ,\\$

$\frac { 9 }{ 24 } ,\frac { 10 }{ 24 } \ and\ \frac { 11 }{ 24 } \quad are\quad three\quad rational\quad numbers\quad lying\quad between\quad x\quad and\quad y.\quad \quad \\$

State True or False.

$\sqrt{4}$ is an irrational number.

0%
True
0%
False

Explanation

We know that, a rational number is a real number that can be represented in the form of

$\dfrac pq$ where both

$p$ and

$q$ are co-prime.

Given number is

$\sqrt 4$

$\Rightarrow \sqrt4=2$

We can write

$2$ as

$\dfrac21$ where both

$2$ and

$1$ are co-prime.

So,

$\sqrt4$ is a rational number.

Hence, the given statement is false.

$\displaystyle 64^{a}=\frac{1}{256^{b}}$ , then

$3a + 4b$ equals:

0%
$2$
0%
$4$
0%
$8$
0%
$0$

Explanation

$64^{a}=\dfrac{1}{256^{b}}$

$\Rightarrow (2^{6})^{a}=\dfrac{1}{(2^{8})^{b}}$

$\Rightarrow 2^{6a}\times 2^{8b}=1$

$\Rightarrow 2^{6a+8b}=2^{0}$

$\Rightarrow 6a+8b=0$

$\Rightarrow 3a+4b=0$

Give an example of two irrational numbers, whose sum is an irrational number.

0%
$2\sqrt{5},3\sqrt{5}$
0%
$2\sqrt{5},-2\sqrt{5}$
0%
$2+\sqrt{5},2-\sqrt{5}$
0%
$2+\sqrt{5},3-\sqrt{5}$

Explanation

Consider option

$A$ .

The number are

$2\sqrt{5}$ and

$3\sqrt{5}$ .
Their sum

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

Consider option

$B$ .

The number are

$2\sqrt{5}$ and

$-2\sqrt{5}$ .
Their sum

$=2\sqrt{5} +(-2\sqrt{5}) = 0$ , which is a rational number.

Consider option

$C$ .

The number are

$2+\sqrt{5}$ and

$2-\sqrt{5}$ .
Their sum

$=2+\sqrt{5} + 2-\sqrt{5} = 4$ , which is a rational number.

Consider option

$D$ .

The number are

$2+\sqrt{5}$ and

$3-\sqrt{5}$ .
Their sum

$=2+\sqrt{5} + 3-\sqrt{5} = 5$ , which is a rational number.

Therefore, option

$A$ is correct.

$x = 2$ and

$y = 3$ , then find the value of

$\left[ \displaystyle\frac { 1 }{ x^{ x } } +\displaystyle\frac { 1 }{ y^{ y } } \right]$ .

0%
$\displaystyle\frac { -31 }{108 }$
0%
$\displaystyle\frac { 31 }{108 }$
0%
$\displaystyle\frac { 125 }{171}$
0%
$\displaystyle\frac { 153 }{222}$

Explanation

$\cfrac { 1 }{ x^{ x } } +\cfrac { 1 }{ y^{ y } } =\cfrac { 1 }{ 2^{ 2 } } +\cfrac { 1 }{ 3^{ 3 } }$

$=\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 27 } \\ =\cfrac { 27+4 }{ 108 }$

$=\cfrac { 31 }{ 108 }$

By how much does

$\displaystyle \sqrt{12}+\sqrt{18}$ exceed

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

0%
$\displaystyle \sqrt{3}+2\sqrt{2}$
0%
$\displaystyle \sqrt{2}+2\sqrt{3}$
0%
$\displaystyle \sqrt{3}+\sqrt{2}$
0%
$\displaystyle \sqrt{3}-2\sqrt{2}$

Explanation

Required difference

$\displaystyle =(\sqrt{12}+\sqrt{18})-(\sqrt{3}+\sqrt{2})$

$\displaystyle =(\sqrt{4\times 3}+\sqrt{9\times 2})-(\sqrt{3}+\sqrt{2})=2\sqrt{3}+3\sqrt{2}-\sqrt{3}-\sqrt{2}$

$\displaystyle =(2\sqrt{3}-\sqrt{3})+(3\sqrt{2}-\sqrt{2})=\sqrt{3}+2\sqrt{2}$

Give an example of two irrational numbers, whose sum is a rational number.

0%
$4 +\sqrt{5},-\sqrt{5}$
0%
$4 +\sqrt{5},\sqrt{5}$
0%
$4 -\sqrt{5},-\sqrt{5}$
0%
$2+\sqrt{5},2+\sqrt{5}$

Explanation

Consider option

$A$ .

The number are

$4+\sqrt{5}$ and

$-\sqrt{5}$ .
Their sum

$\left(4+\sqrt{5}\right) + \left(-\sqrt{5}\right)$

$=4+\sqrt{5}-\sqrt{5} = 4$ , which is a rational number.

Consider option

$B$ .

The number are

$4+\sqrt{5}$ and

$\sqrt{5}$ .
Their sum

$\left(4+\sqrt{5}\right) + \left(\sqrt{5}\right)$

$=4+2\sqrt{5}$ , which is an irrational number.

Consider option

$C$ .

The number are

$4-\sqrt{5}$ and

$-\sqrt{5}$ .
Their sum

$\left(4-\sqrt{5}\right) + \left(-\sqrt{5}\right)$

$=4-2\sqrt{5}$ , which is an irrational number.

Consider option

$D$ .

The number are

$2+\sqrt{5}$ and

$2+\sqrt{5}$ .
Their sum

$=2+\sqrt{5} + 2+\sqrt{5} = 4+2\sqrt{}5$ , which is an irrational number.

Therefore, option

$A$ is correct.

The value of

$512^\frac {-2}{9}$ is

0%
$\displaystyle \frac {1}{2}$
0%
$2$
0%
$4$
0%
$\displaystyle \frac {1}{4}$

Explanation

$\begin{aligned}{}{\left( {512} \right)^{ - {\textstyle{2 \over 9}}}} &= {\left( {{2^9}} \right)^{ - {\textstyle{2 \over 9}}}}\\&={2^{9 \times \frac{{ - 2}}{9}}} \quad\quad\quad\because[(a^b)^c=a^{bc}]\\&= {2^{ - 2}}\\ &= \frac{1}{4}\end{aligned}$

Hence, option

$D$ is correct.

By how much does

$(\displaystyle 5\sqrt{7}-2\sqrt{5})$ exceeds

$(\displaystyle 3\sqrt{7}-4\sqrt{5})$ ?

0%
$\displaystyle 5(\sqrt{7}+\sqrt{5})$
0%
$\displaystyle\sqrt{7}+\sqrt{5}$
0%
$\displaystyle 2(\sqrt{7}+\sqrt{5})$
0%
$\displaystyle 7(\sqrt{2}+\sqrt{5})$

Explanation

$(5\sqrt{7}-2\sqrt{5})-(3\sqrt{7}-4\sqrt{5})$

$=(5\sqrt{7}-2\sqrt{5}-3\sqrt{7}+4\sqrt{5})$

$=2\sqrt{7}+2\sqrt{5}$

$=2(\sqrt{7}+\sqrt{5})$

The rational number

${\cfrac{0}{7}}$ :

0%
Has a positive numerator
0%
Has a negative numerator
0%
Has either a positive numerator or a negative numerator
0%
Has neither a positive numerator nor a negative numerator

Explanation

In the given question numerator is

$0$ and

$0$ is neither positive nor negative. So the correct answer will be option D.

If A : The quotient of two integers is always a rational number, and

R :

$\displaystyle \frac{1}{0}$ is not rational, then which of the following statements is true ?

0%
A is True and R is the correct explanation of A
0%
A is False and R is the correct explanation of A
0%
A is True and R is False
0%
Both A and R are False

Explanation

Since

$\displaystyle \frac{1}{0}$ is not rational, the quotient of two integers is not rational.

Choose the rational number which does not lie between rational numbers

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

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

0%
$\displaystyle \frac{46}{75}$
0%
$\displaystyle \frac{47}{75}$
0%
$\displaystyle \frac{49}{75}$
0%
$\displaystyle \frac{50}{75}$

Explanation

For

$\dfrac{3}{5}$ multiply numerator and denominator by

$15$ to make denominator

$75$ that comes into

$\dfrac{45}{75}.$
Similarly doing for second then we have

$\dfrac{50}{75}.$
Now question is asking about rational lying between them.

So, we need to check the numerator only that lies in between

$45$ and

$50$ or not.
Clearly

$D$ is correct.

The value of

$\displaystyle \frac{9}{\sqrt{11}+\sqrt{2}}$ is equal to

0%
$\displaystyle 9\left ( \sqrt{11}+\sqrt{2} \right )$
0%
$9 \displaystyle \left ( \sqrt{11}-\sqrt{2} \right )$
0%
$\displaystyle \sqrt{11}-\sqrt{2}$
0%
$\dfrac{9}{ \displaystyle \left ( \sqrt{11}-\sqrt{2} \right ) }$

Explanation

$\displaystyle \frac{9}{\sqrt{11}+\sqrt{2}}=\frac{9}{\sqrt{11}+\sqrt{2}}\times \frac{\sqrt{11}-\sqrt{2}}{\sqrt{11}-\sqrt{2}}$ ..... [By rationalizing]

$=\displaystyle \frac{9\left ( \sqrt{11}-\sqrt{2} \right )}{11-2}$

$=\displaystyle \frac{9\left ( \sqrt{11}-\sqrt{2} \right )}{9}=\sqrt{11}-\sqrt{2}$

Choose the rational number which does not lie between rational numbers

$-\cfrac {2}{5}$ and

$-\cfrac {1}{5}$

0%
$-\dfrac {1}{4}$
0%
$-\dfrac {3}{10}$
0%
$\dfrac {3}{10}$
0%
$-\dfrac {7}{20}$

Explanation

Since the given rational numbers

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

$-\dfrac {1}{5}$ are negative rational numbers, therefore, none of the positive rational number can lie between them.

Hence, the rational number

$\dfrac {3}{10}$ does not lie between the rational numbers

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

$-\dfrac {1}{5}$

Which of the following numbers is different from others?

0%
$\sqrt 2$
0%
$\sqrt 3$
0%
$\sqrt 4$
0%
$\sqrt 5$

Explanation

$\sqrt{2}, \sqrt{3}, \sqrt{5}$ are irrational numbers.
But

$\sqrt{4}=2$ is a rational number.
So,

$\sqrt{4}$ is different from others.

CBSE Questions for Class 9 Maths Number Systems Quiz 7 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Number Systems Quiz 7 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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