MCQ Questions for Class 10 Maths Real Numbers Quiz 1

The non-terminating non-recurring decimal cannot be represented as:

0%
irrational numbers
0%
rational numbers
0%
real numbers
0%
none of these

Explanation

Irrational numbers are non-terminating and non-recurring decimal, and they can not be represented as a quotient of two integers, i.e. as a rational number.
Therefore,

$B$ is the correct answer.

Euclids division lemma can be used to find the

$...........$ of any two positive integers and to show the common properties of numbers.

0%
None of the common factors
0%
Lowest common factor
0%
Highest common factor
0%
Common factor

Explanation

As seen by dividing the two positive integers we get a quotient and a remainder and using this lemma, we get the highest common factor at the end of the procedure.
Therefore

$,$

$C$ is the correct answer.

$m$ is not a perfect square, then

$\sqrt {m}$ is

0%
An irrational number
0%
A composite number
0%
A rational number
0%
None of these

Explanation

$\sqrt {m}$ will be an irrational number if

$m$ is not a perfect square number.

Therefore, option

$A$ is the correct.

What is the square of

$(2 + \sqrt {2})$ ?

0%
A rational number
0%
An irrational number
0%
A natural number
0%
A whole number

Explanation

We know that ,

$(a+b)^2=a^2+2ab+b^2$

$\therefore (2+√2)^2= (4+2\times 2\times√2+2)$

$= 6+4√2$

which is an irrational number

What is the HCF of

$13$ and

$22$ ?

0%
$13$
0%
$22$
0%
$1$
0%
$286$

Explanation

Let's first write the prime factorization of given numbers

$13=13$

$22=2\times 11$

So, we see that there is no common factor in the prime factorization of

$13$ and

$22$

$\implies 1$ is the only factor that is common to

$13$ and

$22$

$\implies$ H.C.F is

$1$

The decimal expansion of the rational number

$\dfrac {33}{2^2\cdot 5}$ will terminate after:

0%
one decimal place
0%
two decimal places
0%
three decimal places
0%
more than $3$ decimal places

Explanation

$\dfrac { 33 }{ 2^{ 2 }\cdot 5 } =\dfrac { 33\times5 }{ 4\times 5\times 5 }$

$=\dfrac { 165 }{ 100 }$

$=1.65$ .
Thus the decimal expansion will terminate after two decimal places.

Therefore, option

$B$ is correct.

The decimal representation of

$\dfrac { 93 }{ 1500 }$ will be

0%
Terminating
0%
Non-terminating
0%
Non-terminating and Repeating
0%
Non-terminating and Non-repeating

Explanation

Checking the termination of

$\cfrac{93}{1500}$ is same as checking the termination of

$\cfrac{31}{500}$ which is equal to

$\cfrac{62}{1000}$

As the value of

$\cfrac{62}{1000}$ is

$0.062$ , we can say that the fraction is terminating.

$\therefore \cfrac{93}{1500}$ is terminating.

Hence, option

$A$ is correct.

The fact that

$3+2\sqrt{5}$ is irrational is because

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

Explanation

$3$ is a rational number and

$2\sqrt5$ is an irrational number

Sum of a rational and irrational number is irrational, so

$3+2\sqrt5$ is irrational.

$HCF$ of

$(10224, 1608)$ is _________.

0%
$12$
0%
$24$
0%
$48$
0%
$96$

Explanation

First step is to write

$10224$ and

$1608$ as the product of its prime factors.

$10224 = 2^{4}\times3^{2}\times71$

$1608 \;\;= 2^{3}\times3\times67$

$\therefore HCF(10224,1608) = 2^{3}\times 3$

$=24$

State true or false:

$\dfrac {\sqrt 2}{3}$ is an irrational number.

0%
True
0%
False

Explanation

Given number is

$\dfrac {\sqrt 2}{3}$

Here,

$\sqrt{2}$ is an irrational number and

$3$ is a rational number.

We know,

$\dfrac{Irrational\ number}{Rational \ number}=Irrational\ number$

So,

$\dfrac {\sqrt 2}{3}$ is an irrartional number.

Hence, the given statement is

$True$

State whether the statement is true/false.

$\sqrt{72}$ is irrational

0%
True
0%
False

$7+\sqrt7$ is irrational

0%
True
0%
False

$\sqrt{7}+7$ is a rational number

0%
True
0%
False

State whether the given statement is True or False :

$3+\sqrt { 2 }$ is an irrational number.

0%
True
0%
False

Explanation

$3 \text{ is a rational number and }$

$\sqrt2$ is a

$irrational$ number

And

$\text{the addition of rational and irrational is always an irrational number}$

So that

$(3+\sqrt2)$ is an irrational number.

hence option A is correct

Statement

$p:\sqrt {15}$ is a rational number

0%
True
0%
False

Explanation

$P = \sqrt {15} \,\,is\,\,a\,\,irrational\,\,number.$
1204382_1315090_ans_a8cd30e8f9b746788ca1aaff933f2240.jpg

$\dfrac {5+\sqrt {2}}{3}$ is an irrational number.

0%
True
0%
False

To get the terminating decimal expansion of a rational number

$\dfrac{p}{q}$ . if

$q = 2^m 5^n$ then

$m$ and

$n$ must belong to .................

0%
$Z$
0%
$N \cup$ { $0$ }
0%
$N$
0%
$R$

Explanation

To get the terminating decimal expansion of a rational number

$\dfrac{p}{q}$ ,

$q$ have to be an integer or natural number.

Now,

$q = 2^m 5^n$ will be an integer or natural if

$m$ and

$n$ both are naturals. So, to get

$q = 2^m 5^n$ as integer

$m$ and

$n$ must belong to

$N$ .

Hence, option C is correct.

Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non -terminating decimal expansion

$\displaystyle \frac{15}{1600}$

0%
Terminating decimal expansion
0%
Non-terminating decimal expansion
0%
Cannot be determined
0%
None

Explanation

Factorize the denominator we get

$1600=2 \times 2 \times 2 \times2 \times 2 \times 2 \times 5 \times 5 = 2^{6} \times 5^{2}$
so denominator is in form of

$2^n \times 5^m$
Hence

$\frac{15}{1600}$ is terminating.

State whether the following statements are true or false. Justify your answers.
Every real number need not be a rational number

0%
True
0%
False

Explanation

Real number are all numbers on number line

And a rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. Since q may be equal to 1, every integer is a rational number.

And other numbers are not rational number are called irrational number.

Then every real number need not be a rational is true

Eg:

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

State the following statement is True or False
35.251252253...is an irrational number

0%
True
0%
False

Explanation

$35.251522253...\Rightarrow$ It is a non-terminating decimal. This number cannot be written as a simple fraction.

Therefore, it's an irrational number.

$\dfrac {p}{q}$ form of

$0.0875$ is _______

0%
$\dfrac {7}{2^{4}\times 5}$
0%
$\dfrac {7}{2\times 5^{4}}$
0%
$\dfrac {7}{2^{4}\times 5^{4}}$
0%
$\dfrac {5^{3}\times 7}{2^{3}\times 5^{4}}$

Explanation

Since,

$0.0875=\displaystyle \frac {875}{10000}=\frac {175}{2000}=\frac{35}{400} =\frac {7}{80}=\frac 7{16\times 5}=\frac 7{2^4\times 5}$

Option

$A$ is correct.

Let

$x$ be an irrational number then what can be said about

${x}^{2}$

0%
It is rational
0%
It can be irrational.
0%
It can be rational.
0%
Both $B$ and $C$

Explanation

$x$ is any irrational number
Let

$x=\sqrt [ 4 ]{ 3 }$

$\Rightarrow { x }^{ 2 }=\sqrt { 3 }$
which is irrational so option

$B$ is correct.
Now let

$x=\sqrt 3$

$\Rightarrow {x}^{2}=3$
which is rational so option

$C$ is correct.
So correct answer is

$D$

Which of the following is an irrational number?

0%
$\sqrt{41616}$
0%
$23.232323$
0%
$\displaystyle\frac{(1+\sqrt{3})^3-(1-\sqrt{3})^3}{\sqrt 3}$
0%
$23.10100100010000...$

Explanation

Op-(A)

$\sqrt{41616}=204$

$\sqrt{41616}$ is a perfect square, so it is a rational number.

Op-(B)

$23.232323....$

It is recurring decimal, so it is a rational number.

Op-(C)

$\dfrac{(1+\sqrt{3})^3-(1-\sqrt{3})^3}{\sqrt{3}}=\dfrac{2\sqrt{3}(1+2\sqrt{3}+3-2+1-2\sqrt{3}+3)}{\sqrt{3}}=\dfrac{2\sqrt{3}(6)}{\sqrt{3}}=\dfrac{12\sqrt{3}}{\sqrt{3}}=12$ .

Only, the number

$23.10100100010000.......$ is not terminating.

So, D is the irrational number.

$p$ is prime, then

$\sqrt{p}$ is irrational. So

$\sqrt{7}$ is:

0%
a rational number
0%
an irrational number
0%
not a real number
0%
terminating decimal

Explanation

A rational number can be represented in the form of

$\dfrac pq,$ where

$p$ and

$q$ are integers and

$q$ is a non-zero integer.

Here,

$\sqrt 7$ is not a perfect square and thus cannot be expressed in the form of

$\dfrac pq,$ thus it is an irrational number.

Which of the following are non-terminating numbers?

0%
$\frac{2}{3}$
0%
$\frac{7}{15}$
0%
$\frac{2}{5}$
0%
$\frac{5}{13}$

Explanation

$We\quad know\quad the\quad factors\quad of\quad the\quad denominator\quad of\quad a\quad terminating\quad number\quad are\quad \\ only\quad 2\quad and\quad 5.\quad If\quad it\quad has\quad any\quad other\quad number\quad then\quad it\quad will\quad be\quad non\quad terminating.\\ Option\quad A\longrightarrow \quad denominator\quad is\quad 3,\quad therefore\quad it\quad is\quad non\quad terminating.\\ Option\quad B\longrightarrow \quad denominator\quad is\quad 3\times 5\quad therefore\quad it\quad is\quad non\quad terminating.\\ Option\quad C\longrightarrow \quad denominator\quad is\quad 5,\quad therefore\quad it\quad is\quad terminating.\\ Option\quad D\longrightarrow denominator\quad is\quad 13,\quad therefore\quad it\quad is\quad non\quad terminating.\\ Answer-\quad Option\quad A,B,D\\$

Let

$x=\dfrac { p }{ q }$ be a rational number, such that the prime factorization of

$q$ is of the form

$2^n 5^m$ , where

$n, m$ are non-negative integers. Then

$x$ has a decimal expansion which terminates.

0%
True
0%
False
0%
Neither
0%
Either

Explanation

The form of q is

$2^n*5^m$
q can be

$1,2,5,10,20,40....$
Any integer divided by these numbers will always give a terminating decimal number.

Which of the following will have a terminating decimal expansion?

0%
$\displaystyle \frac{77}{210}$
0%
$\displaystyle \frac{23}{30}$
0%
$\displaystyle \frac{125}{441}$
0%
$\displaystyle \frac{23}{8}$

Explanation

We know that the divisor of the forms

$2^n5^m$ always form a terminating decimal number. Let us simplify the expressions and find if the expansion have a terminating decimal expansion.

Option A :

$\dfrac{77}{210}$

$=\dfrac{7\times11}{2\times 3\times 5 \times 7 }$

$=\dfrac{11}{2\times 3 \times 5}$
Since there is a factor of 3 in the denominator, the decimal expansion will not be terminating.

Option B :

$\dfrac{23}{30}$

$=\dfrac{23}{2\times 3\times 5}$
Since, the denominator contains a power of 3 , it is non-terminating.

Option C :

$\dfrac{125}{441}$

$=\dfrac{5\times5\times5}{3\times3\times7\times7}$
This is also non-terminating.

Option D :

$\dfrac{23}{8}$

$=\dfrac{23}{2\times2\times2}$
This contains power of 2 in the denominator. Hence, the decimal expansion is terminating.

$3+2\sqrt{5}$ a rational number.

0%
True
0%
False

Explanation

Let's assume that

$3+2\sqrt5$ is rational.....

then

$3+2\sqrt5 = p/q$

$\sqrt5 =( p-3q)/(2q)$

now take

$p-3q$ to be P and

$2q$ to be Q........where P and Q are integers

which means,

$\sqrt5= P/Q$ ......

But this contradicts the fact that

$\sqrt5$ is rational

So our assumption is wrong and

$3+2\sqrt5$ is irrational.

$\sqrt3$ is

0%
rational number
0%
irrational number
0%
natural number
0%
None

Explanation

Let

$\sqrt3$ is a rational number

$\therefore \sqrt3 = \displaystyle \frac{a}{b}$ [Where a & b are co-primes]

$a^2=3b^2$ .......(i)

$\Rightarrow$

$3$ divides

$a^2$

$\Rightarrow$

$3$ also divides a

$\Rightarrow$

$a=3c$
[Where c is any non-zero positive integer]

$\Rightarrow a^2 = 9c^2$
From equation (i)

$3b^2=9c^2$

$\Rightarrow b^2 = 3c^2 \Rightarrow$

$3$ divides

$b^2$

$\Rightarrow$

$3$ also divides b
So,

$3$ is a common factor of a and b.
Our assumption is wrong, because a and b are not co - primes.
It means

$\sqrt3$ is an irrational number.

State true or false:

$\dfrac{1}{\sqrt{2}}$ is an irrational number.

0%
True
0%
False

Explanation

Let us assume that

$\sqrt 2$ is irrational

$\dfrac{1}{\sqrt 2} = \dfrac pq$ (where

$p$ and

$q$ are co prime)

$\dfrac qp = \sqrt 2$

$q = \sqrt 2p$

squaring both sides

$q^2 = 2p^2$

$...(1)$

By theorem,

$q$ is divisible by

$2$

$\therefore\ q = 2c$ (where

$c$ is an integer)

putting the value of

$q$ in equitation

$(1)$

$2p^2= q^2 =4c^2$

$p^2=\dfrac{4c^2}{2} = 2c^2$

$\dfrac{p^2}{2}= c^2$

By theorem,

$p$ is also divisible by

$2$

But

$p$ and

$q$ are coprime

This is a contradiction which has arisen due to our wrong assumption.

Hence

$\dfrac{ 1}{\sqrt2}$ is irrational.

CBSE Questions for Class 10 Maths Real Numbers Quiz 1 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Real Numbers Quiz 1 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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