MCQ Questions for Class 9 Maths Number Systems Quiz 5

Which of the following is an irrational number?

0%
$\dfrac{11}{2}$
0%
$\sqrt{16}$
0%
$\sqrt{9}$
0%
$\sqrt{11}$

Explanation

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

Option

$A$ is a rational number.

Option

$B$ and

$C$ are

$\sqrt{16}$ and

$\sqrt{9}$ , i.e.

$4$ and

$3$ respectively.

$D$ cannot be expressed as a ratio of integers.

$D$ is the correct answer.

Which of the following rational numbers lies between

$\dfrac {3}{2}$ and

$4$ ?

0%
$\dfrac {1}{2}$
0%
$3$
0%
$\dfrac {8}{2}$
0%
$\dfrac {9}{2}$

Explanation

We have to find a rational number between

$\dfrac {3}{2}$ and

$4$ . The L.C.M. of the denominators of both numbers is

$2$ .

$\therefore \dfrac {3}{2} = \dfrac {3}{2}$ and

$4\times \dfrac {2}{2} = \dfrac {8}{2}$ .

$\therefore$ From the above given options, only

$\dfrac {6}{2}$
i.e.

$=3$ lies between

$\dfrac {3}{2}$ and

$4$ .

$5$ is a rational number. It can be written as ..........

0%
$\dfrac {5}{1}$
0%
$\dfrac {1}{5}$
0%
$\dfrac {5}{5}$
0%
$\dfrac {5}{25}$

Explanation

Rational numbers are defined as all the numbers of the form

$\dfrac {p}{q}$ where

$q\neq 0$ .

$5$ being a natural number can be written as

$\dfrac {5}{1}$ . A natural number is a rational number.
Therefore,

$A$ is the correct answer.

$\pi = 3.14159265358979........$ is an

0%
rational number
0%
whole number
0%
irrational number
0%
all of the above

Explanation

$\pi = 3.14159265358979........$ is a non-terminating and non-repeating irrational number.

Hence, option

$C$ is correct.

A terminating decimal has a ............ number of terms after the decimal point.

0%
zero
0%
infinite
0%
finite
0%
none of the above

Explanation

A terminating decimal is a decimal that ends after a finite number of steps. Hence, is a decimal with a finite number of digits.
Therefore,

$C$ is the correct answer.

To simplify the following expression correctly, what must be done with the exponents?

${ { 5 }^{ a }\times { 5 }^{ b }\times }{ 5 }^{ c }$

0%
Add the exponents.
0%
Multiply the exponents.
0%
Divide the exponents
0%
Subtract the exponents.

Explanation

Since the base is same, the exponents can be added (Product rule of exponents).

$\therefore 5^a \times 5^b \times 5^c = 5^{(a+b+c)}$

Hence, option

$A$ is correct.

Solve the following using Product Law of Exponents.

$a\times { a }^{ 2 }\times { a }^{ \tfrac { 1 }{ 2 } }$

0%
${ a }^{ 7 }$
0%
${ a }^{ \tfrac { 2 }{ 7 } }$
0%
${ a }^{ 3 }$
0%
${ a }^{ \tfrac { 7 }{ 2 } }$

Explanation

By product law of exponents

$a\times { a }^{ 2 }\times { a }^{ \tfrac { 1 }{ 2 } } = { a }^{ \left( 1+2+\tfrac { 1 }{ 2 } \right) }$

$={ a }^{ \tfrac { 7 }{ 2 } }$
Therefore option

$D$ is the correct answer.

Simplify the following:

${ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }\times { \left( \cfrac { 1 }{ 2 } \right) }^{ 5 }\times { \left( \cfrac { 1 }{ 2 } \right) }^{ 6 }$

0%
${ \left( \cfrac { 1 }{ 2 } \right) }^{ 15 }$
0%
${ \left( \cfrac { 1 }{ 2 } \right) }^{ 5 }$
0%
${ \left( \cfrac { 1 }{ 2 } \right) }^{ 10 }$
0%
None of these

Explanation

we know,

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

so,

${ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }\times { \left( \cfrac { 1 }{ 2 } \right) }^{ 5 }$

$=\left(\cfrac12\right)^6$

so,

${ \left( \cfrac { 1 }{ 2 } \right) }^{ 4+5+6 }$

$=\left(\cfrac12\right)^{15}$

Simplify and give reasons:

$\left[ { \left( \cfrac { 3 }{ 2 } \right) }^{ -2 } \right] ^{ 2 }$

0%
$\cfrac{16}{81}$
0%
$\cfrac{6}{81}$
0%
$\cfrac{16}{1}$
0%
None of these

Explanation

we know that,

$\because (a^m)^n=a^{mn}$

so,

$((\dfrac32)^{-2})^2=(\dfrac32)^{-4}$

we also know

$\because (\dfrac{a}{b})^{-m}=(\dfrac{b}{a})^m$

$=(\dfrac23)^4$

$=\dfrac{16}{81}$

Simplify and give reasons:

${(-3)}^{-4}$

0%
$\cfrac{1}{81}$
0%
$\cfrac{-1}{81}$
0%
$\cfrac{3}{81}$
0%
None of these

Explanation

$(-3)^{-4}$

$=(-1*3)^{-4}$

$=(-1)^{-4}(3)^{-4}$

$\because (ab)^m=a^m*b^m$

$=1*3^{-4}$

$=3^{-4}$

$\because a^{-m}=\dfrac1{a^m}$

$=(\dfrac13)^4$

$=\dfrac1{81}$

Which irrational number is plotted on the number line in figure?

0%
$\pi$
0%
$\sqrt{2}$
0%
$\sqrt{3}$
0%
$\sqrt{5}$

Explanation

According to the given figure,

the red dot is closer to

$1.75$ and out of the options only

$\sqrt3$ is only one near to

$1.75$ .

The irrational number is plotted on the number line

$\sqrt {3}=1.732$ .

Hence, option

$C$ is correct.

State whether the following statements are true or false. Justify your answers.
Every irrational number is a real number

0%
True
0%
False

State whether the following statement is true or false:
All real numbers are irrational

0%
True
0%
False

Explanation

The real numbers include natural numbers or counting numbers, whole numbers, integers, rational numbers (fractions and repeating or terminating decimals), and irrational numbers.

So, Real number includes number like

$\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{7}....$ and numbers like

$1,2,3,4,...$ etc which are not irrational numbers.

So the statement, all real numbers are irrational is false.

Classify the following numbers as rational or irrational:

$\displaystyle \frac{\sqrt{12}}{\sqrt{75}}$ .

0%
Rational
0%
Irrational
0%
Can't be determined
0%
None of these

Explanation

$\dfrac { \sqrt { 12 } }{ \sqrt { 75 } } =\dfrac { 2\sqrt { 3 } }{ 5\sqrt { 3 } } =\dfrac { 2 }{ 5 }$ , which is a rational number.

Hence, the correct answer will be option

$A$ .

State true or false.
Every integer 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

$q \neq 0$

for example

$2$

$2$ can also be written as above form

i.e

$\dfrac{2}{1}$

$2$ is a Rational number and is Integer also.

The given statement is True .

Is the following statement True or False
Every rational number is an integer.

0%
True
0%
False

Explanation

False .

Rational numbers are special because they can be written as a fraction. More specifically, the definition of rational numbers says that any rational number can be written as the ratio of p to q, where p and q are integers and q is not zero

An integer is a whole number (not a fractional number) that can be positive, negative, or zero.

The number

$1.43,3.14,0.09 ,\dfrac{13}{4}$ are not integer but rational number

This statement is false that every rational number is an integer

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

0%
$2\sqrt{2}$
0%
$\sqrt{2}$
0%
$\displaystyle\frac{\sqrt{2}}{2}$
0%
$2$

Explanation

To find

$\dfrac{2}{\sqrt2}$ ,

Multiply numerator and denominator by

$\sqrt2$ , we get,

$\dfrac{2}{\sqrt2}$

$=\dfrac{2 \times \sqrt2 }{\sqrt2 \times \sqrt2}$

$= \dfrac{2\sqrt2 }{2}$

$= \sqrt2$

Hence, option

$B$ is correct.

Find whether the following statement are true or false.
Every integer is a rational number and vice versa.

0%
True
0%
False

Are the following statements true or false? Given reasons for your answer.
Every rational number is a whole number.

0%
True
0%
False

Explanation

By the definition of rational number,

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

$q\neq 0$ .
We know that every whole number can be represented as

$\dfrac {a}{1}$ .

$\therefore$ From the above two statements we can conclude that every whole number is a rational number.

${ \left( \dfrac { 2 }{ 3 } \right) }^{ 0 }=?$

0%
$\dfrac{3}{2}$
0%
$\dfrac{2}{3}$
0%
$1$
0%
$0$

Explanation

Given

${ \left( \dfrac { 2 }{ 3 } \right) }^{ 0 }$

By using tha law of exponents i.e.,

$\left(\dfrac{a}{b}\right)^0=1$

$\therefore { \left( \dfrac { 2 }{ 3 } \right) }^{ 0 }=1$

Which one of the following is an irrational number?

0%
$\pi$
0%
$\sqrt{9}$
0%
$\displaystyle\frac{1}{4}$
0%
$\displaystyle\frac{1}{5}$

Explanation

$\textbf{Step 1: Simplifying the following options.}$

$\text{The numbers which are non-repeating and non-terminating are irrational numbers.}$

$\text{Option A : }\pi = 3.141... \text{ is an irrational number.}$

$\text{Option B : }\sqrt{9} = 3 \text{ is a rational number.}$

$\text{Option C : }\dfrac{1}{4} \text{ is a rational number.}$

$\text{Option D : }\dfrac{1}{5} \text{ is a rational number.}$

$\therefore \text{Option A : }\pi \text{ is an irrational number.}$

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

Find whether the following statement are true or false.
In a rational number of the form

$\cfrac { p }{ q } ,q$ must be a non zero integer.

0%
True
0%
False

Explanation

Any number which is in the form of

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

$p,q$ are integers , and

$q ≠ 0$ is called a Rational number .

Evaluate

$2^0+3^0$

0%
$2$
0%
$3$
0%
$5$
0%
$1$

State which of the following statements is/are true?
I. Numerator and denominator of a positive rational number need not to have like signs.
II. Numerator and denominator of a negative rational number should have like signs.

0%
Only I
0%
Only II
0%
Both I and II
0%
Neither I nor II

The fact that

$7\sqrt{5}$ is irrational is because

0%
Product of two irrational numbers in rational
0%
Product of a rational and an irrational number is rational
0%
Product of two rational numbers is rational
0%
Product of a rational and an irrational number is irrational

Explanation

$7$ is a rational number and

$\sqrt5$ is an irrational number

Product of a rational and irrational number is irrational, so

$7\sqrt5$ is irrational.

Which of the following rational numbers is positive?

0%
$\dfrac{-7}{9}$
0%
$\dfrac{-8}{2}$
0%
$\dfrac{6}{5}$
0%
$\dfrac{3}{-2}$

Explanation

A rational number is positive only when both its numerator and denominator are positive.

We can see that only

$\dfrac{6}{5}$ has positive numerator and denominator. Rest all the given options have either negative numerator or negative denominator.

Therefore, only

$\dfrac{6}{5}$ is a positive rational number.

Hence, option C is correct.

The expression

$2 + \sqrt{2} + \dfrac{1}{2 + \sqrt{2}} + \dfrac{1}{\sqrt{2} - 2}$ equals to

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

Explanation

$2+\sqrt { 2 } +\dfrac { 1 }{ 2+\sqrt { 2 } }+\dfrac { 1 }{ \sqrt { 2 } -2 }$

$=2+\sqrt { 2 } +\dfrac { 1 }{ 2+\sqrt { 2 } }\times \dfrac{2-\sqrt2}{2-\sqrt2}+\dfrac { 1 }{ \sqrt { 2 } -2 }.\times \dfrac { \sqrt { 2 }+2 }{\sqrt { 2 } +2 }$ .......... rationalization

$=2+\sqrt { 2 } +\dfrac { 2-\sqrt { 2 } }{ (4-2) } +\dfrac { \sqrt { 2 } +2 }{ (2-4) }$

$=2+\sqrt { 2 } +1-\dfrac { 1 }{ \sqrt { 2 } } -\dfrac { 1 }{ \sqrt { 2 } } -1$

$=2+\sqrt { 2 } -\sqrt { 2 } \\=2$

Subtraction of two irrational numbers is:

0%
a rational
0%
an irrational an integer
0%
either rational or irrational
0%
Can't determine

Explanation

The subtraction of two irrational number can be rational or irrational depending on if the irrational part is same or different in both the irrational numbers respectively.

For instance,

$(2+\sqrt3) - (1+\sqrt3) = 1$ , which is rational.

Whereas,

$\sqrt{3} - \sqrt{2}$ is irrational.

Any operation between one non-zero rational and one irrational number always gives:

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

State true or false:

An irrational number between

$3.1$ and

$3.2$ is

$\pi$ .

0%
True
0%
False

Explanation

We know,

$\pi=3.14159...$

It is also known that

$\pi$ is an irrational number.

Also,

$3.10000...<3.14159...<3.20000...$

So,

$3.10000...<\pi<3.20000...$

Hence,

$\pi$ lies between

$3.1$ and

$3.2$ .

Therefore, the statement is true.

CBSE Questions for Class 9 Maths Number Systems Quiz 5 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Number Systems Quiz 5 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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