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CBSE Questions for Class 10 Maths Real Numbers Quiz 1 - MCQExams.com
CBSE
Class 10 Maths
Real Numbers
Quiz 1
The non-terminating non-recurring decimal cannot be represented as:
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irrational numbers
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rational numbers
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real numbers
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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.
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None of the common factors
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Lowest common factor
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Highest common factor
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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
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An irrational number
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A composite number
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A rational number
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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 $$(2 + \sqrt {2})$$?
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A rational number
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An irrational number
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A natural number
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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 $$22$$?
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$$13$$
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$$22$$
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$$1$$
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$$286$$
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:
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one decimal place
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two decimal places
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three decimal places
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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
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Terminating
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Non-terminating
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Non-terminating and Repeating
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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
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Sum of two irrational numbers is rational
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Sum of two irrational numbers is irrational
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Sum of a rational and an irrational number is irrational
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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 _________.
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$$12$$
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$$24$$
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$$48$$
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$$96$$
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.
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True
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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
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True
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False
Explanation
$$7+\sqrt7$$ is irrational
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True
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False
$$\sqrt{7}+7$$ is a rational number
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True
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False
State whether the given statement is True or False :
$$3+\sqrt { 2 } $$ is an irrational number.
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True
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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
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True
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False
Explanation
$$P = \sqrt {15} \,\,is\,\,a\,\,irrational\,\,number.$$
$$\dfrac {5+\sqrt {2}}{3}$$ is an irrational number.
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True
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False
Explanation
As we know that sum of a rational and an irrational number is irrational as well like the numerator of the given fraction.
And when an irrational number is divided by a rational number the result is irrational as well hence the given number is an irrational number.
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 .................
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$$Z$$
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$$N \cup$${$$0$$}
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$$N$$
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$$R$$
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
$$\displaystyle \frac{15}{1600}$$
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Terminating decimal expansion
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Non-terminating decimal expansion
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Cannot be determined
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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
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True
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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
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True
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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 _______
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$$\dfrac {7}{2^{4}\times 5}$$
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$$\dfrac {7}{2\times 5^{4}}$$
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$$\dfrac {7}{2^{4}\times 5^{4}}$$
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$$\dfrac {5^{3}\times 7}{2^{3}\times 5^{4}}$$
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 $${x}^{2}$$
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It is rational
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It can be irrational.
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It can be rational.
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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?
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$$\sqrt{41616}$$
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$$23.232323$$
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$$\displaystyle\frac{(1+\sqrt{3})^3-(1-\sqrt{3})^3}{\sqrt 3}$$
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$$23.10100100010000...$$
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.
If $$p$$ is prime, then $$\sqrt{p}$$ is irrational. So
$$\sqrt{7}$$ is:
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a rational number
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an irrational number
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not a real number
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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?
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$$\frac{2}{3}$$
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$$\frac{7}{15}$$
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$$\frac{2}{5}$$
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$$\frac{5}{13}$$
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.
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True
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False
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Neither
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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?
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$$\displaystyle \frac{77}{210}$$
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$$\displaystyle \frac{23}{30}$$
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$$\displaystyle \frac{125}{441}$$
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$$\displaystyle \frac{23}{8}$$
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.
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True
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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
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rational
number
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irrational
number
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natural number
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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.
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True
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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.
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