MCQ Questions for Class 10 Maths Real Numbers Quiz 6

Without actually performing the long division, state whether the following rational number will have terminating decimal expansion or a non-terminating repeating decimal expansion. Also, find the numbers of places of decimals after which the decimal expansion terminates.

$\dfrac { 13 }{ 3125 }$

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

The given value is

$\dfrac{13}{3125}$ the denominator is 3125 which can be written as:

$3125=2^0 \times 5^5$ it is in the form of

$2^m \times 5^n$

$max(m,n)=5$

$\therefore$ the expansion is terminating decimal it terminates after

$max(m,n)=5$ places from the decimal [since

$m=0,n=5$ ]

If HCF of numbers

$408$ and

$1032$ can be expressed in the form of

$1032x -408 \times 5$ , then find the value of

$x$ .

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

Explanation

$408=2\times2\times2\times3\times17$

$1032=2\times2\times2\times3\times43$

Hence,

$HCF=2\times2\times2\times3=24$

Now,

$1032x-408\times5=24\Rightarrow 1032x=2064\Rightarrow x=2$

Find the

$HCF$ of integer pairs

$963$ &

$654$ .

0%
$3$
0%
$5$
0%
$9$
0%
$1$

State whether the given statement is true/false:

$\sqrt{p} + \sqrt{q}$ , is irrational, where p,q are primes.

0%
True
0%
False

Explanation

Assume that $\sqrt p + \sqrt q$ is rational. So,

$\sqrt p + \sqrt q = \dfrac{a}{b}$

$p + q + 2\sqrt {pq} = \dfrac{{{a^2}}}{{{b^2}}}$

$\sqrt {pq} = \dfrac{1}{2}\left( {\dfrac{{{a^2}}}{{{b^2}}} – p - q} \right)$

Since the RHS of the above equation is rational but $\sqrt {pq}$ is an irrational number, so the assumption is wrong.

Therefore, it is true that $\sqrt p + \sqrt q$ is irrational.

GCF of

$99$ and

$100$ is __________

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

when we divide

$100$ with

$99$ we will get remainder

$1$ and hence next we divide 99 with

$1$ which gives us remainder

$0$ .

therefore

$1$ is the HCF or GCF of

$99$ and

$100$

State true or false.

$\dfrac{6}{{200}}$ has non terminating repeating decimal expansion.

0%
True
0%
False

Explanation

$\dfrac{p}{q}$ is terminating if

$q$ is of the form

${2}^{n}{5}^{m}$ in the simplest ratio possible

$\dfrac{p}{q}$ where

$p$ and

$q$ are co-primes.

(

$m$ and

$n$ are non-negative integers)

Consider

$\dfrac{6}{200}$

$(i)6$ and

$200$ are not co-prime

$\because$ HCF

$\left(6,200\right)=2$

$\dfrac{6}{200} =\dfrac{3\times 2}{100 \times 2}$

$\dfrac{3}{100}=0.03$

Hence

$\dfrac{6}{200}=0.03$ has terminating decimal expansion.

State whether given statement is true or false:

$3\sqrt{2}$ is a rational number.

0%
True
0%
False

Which of the following numbers are irrational numbers?

0%
$2- \sqrt 5$
0%
$\left( {3 + \sqrt {23} } \right) - \left( {\sqrt {23} } \right)$
0%
$\frac{1}{\sqrt 2}$
0%
$2\pi$

Explanation

$A$ is a irrational number as it cannot be expressed of the form

$\cfrac{p}{q}$

$B$ is a rational number. As it can be expressed in the form of

$\cfrac{3}{1}$

$C$ is a irrational number as it cannot be expressed of the form

$\cfrac{p}{q}$

$D$ is a irrational number as

$\pi$ is itself irrational and the product of irrational with rational is always irrational.

State whether the following statement is true or false.

The following number is irrational

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

0%
True
0%
False

$5-2\sqrt {3}$ is irrational.

0%
True
0%
False

State whether the following statement is true or false.

The following number is irrational

$6+\sqrt {2}$

0%
True
0%
False

Explanation

An Irrational Number is a real number that cannot be written as a simple fraction.

Let us assume

$6+\sqrt{2}$ is rational.

Hence,

$6+\sqrt{2}$ can be written in form

$\dfrac{a}{b}$

Where,

$a$ and

$b$

$(n\ne 0)$ are co-prime.

Hence,

$6+\sqrt{2}=\dfrac{a}{b}$

$\Rightarrow$

$\sqrt{2}=\dfrac{a}{b}-6$

$\Rightarrow$

$\sqrt{2}=\dfrac{a-6b}{b}$

Here,

$\dfrac{a-6b}{b}$ is a rational number, but

$\sqrt{2}$ is irrational.

Since, Rational

$\ne$ Irrational

This is a contriadition

$\therefore$ Our assumption is incorrect.

$\therefore$

$6+\sqrt{2}$ is irrational number.

The decimal expansion of the rational number

$\dfrac{43}{2^4\times 5^3}$ will terminate after:

0%
$3$
0%
$4$
0%
$2$
0%
$1$

Each of the following numbers is irrational
i)

$(5 + 3\sqrt{2})$
ii)

$3 \sqrt{7}$
iii)

$\dfrac{3}{\sqrt{5}}$
iv)

$(2 - 3\sqrt{5})$
v)

$(\sqrt{3} + \sqrt{5})$

0%
True
0%
False

Explanation

1. Let us assume that

$5 + 3 \sqrt 2$ is irrational number.

$\Rightarrow 5 + 3 \sqrt 2 = \dfrac {a}{b}$ [ where a & b are coprime and b

$\neq$ 0 ]

$3 \sqrt2 = \dfrac{a}{b} - 5 = \dfrac{a-5b}{b} \\ \sqrt2 = \dfrac{a-5b}{3b}$

As we know that 3, 5, P & q are integers.

The result of

$\dfrac{a-5b}{3b}$ will be rational. But

$\sqrt2$ is irrational.

We can conclude that

$\sqrt2 \neq \dfrac{a-5b}{3b}$

So this contradiction has arisen because of our wrong assumption that

$5 + 3 \sqrt 2$ is rational.

2. Let us assume that

$3 \sqrt 7$ is a rational number.

$\Rightarrow 3 \sqrt 7= \dfrac {a}{b}$ [ where a & b are integers ]

$\sqrt 7 = \dfrac{a}{3b}$

Since a and b are integers

$\therefore \dfrac{a}{3b}$ is also rational but this contradicts the fact as

$\sqrt7$ is irrational.

Our assumption was wrong.

Hence

$3 \sqrt 7$ is irrational.

3. Let

$\dfrac{3}{\sqrt5}$ be a rational number.

$\dfrac{3}{\sqrt5} = \dfrac{a}{b}$ [a & b are co-prime]

$\dfrac{3b}{a} = \sqrt 5$

A & B are coprime :

$\dfrac{3b}{a}$ is rational but it contradicts are fact as

$\sqrt 5 \neq \dfrac{3b}{a}$ because

$\sqrt 5$ is irrational.

$\Rightarrow$ Our assumption was wrong

$\dfrac{3}{\sqrt5}$ is irrational number.

State whether the following statement is true or false.

The following number is irrational

$7\sqrt {5}$

0%
True
0%
False

Explanation

An Irrational Number is a real number that cannot be written as a simple fraction.

Let us assume

$7\sqrt{5}$ is rational.

Hence,

$7\sqrt{5}$ can be written in form

$\dfrac{a}{b}$

Where,

$a$ and

$b$

$(n\ne 0)$ are co-prime.

Hence,

$7\sqrt{5}=\dfrac{a}{b}$

$\Rightarrow$

$\sqrt{5}=\dfrac{1}{7}\times\dfrac{a}{b}$

Here,

$\dfrac{a}{7b}$ is a rational number, but

$\sqrt{5}$ is irrational.

Since, Rational

$\ne$ Irrational

This is a contriadition

$\therefore$ Our assumption is incorrect.

$\therefore$

$7\sqrt{5}$ is irrational number.

Given that

$\sqrt {3}$ is irrational. Then "

$2 + \sqrt {3}$ is irrational. "is true/false

0%
True
0%
False

$\sqrt{3}-\sqrt{5}$ is an rational number.

0%
True
0%
False

A rectangular veranda is of dimension

$18$ m

$72$ cm

$\times 13$ m

$20$ cm. Square tiles of the same dimensions are used to cover it. Find the least number of such tiles.

0%
$4290$
0%
$4540$
0%
$4620$
0%
$4230$

Explanation

The edge of rectangular veranda are

$18\ m\ 72\ cm=1872\ cm$ and

$13\ m\ 20\ cm=1320\ cm$ .

On taking

$HCF$ of

$1872$ and

$1320$ using prime factorisation, we get

$HCF=24$

Therefore,

No. of tiles required

$=$

$\dfrac{Area\ of\ Veranda}{Area\ of\ tiles}$

$=\dfrac{1872\times 1320}{24\times 24}$

$=4290$

Hence, this is the answer.

State true or false.

$\sqrt { 3 } + \sqrt { 4 }$ is an rational number.

0%
True
0%
False

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 (the number on top) and the denominator (the number on the bottom) are whole numbers.

$\Rightarrow$ All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction.

$\sqrt{3}=1.732$ is an irrational.

$\sqrt{4}=2$ is rational.

Now,

$\Rightarrow$

$\sqrt{3}+\sqrt{4}=1.732+2$

$=3.732$

$3.732$ cannot be converted into fraction.

$\therefore$

$\sqrt{3}+\sqrt{4}$ is an irrational number.

$\therefore$ Given statement is false.

$n^2 -1$ is divisible by

$8$ , if n is

0%
an integer
0%
a natural number
0%
an odd integer
0%
an even integer

Explanation

Given,

Any odd positive integer

$n$ can be written in form of

$4q + 1$ or

$4q + 3$ .

$n = 4q + 1$ , when

$n^2 - 1 = (4q + 1)^2 - 1 = 16q^2 + 8q + 1 - 1 = 8q(2q + 1)$ which is divisible by

$8.$

$n = 4q + 3$ , when

$n^2 - 1 = (4q + 3)^2 - 1 = 16q^2 + 24q + 9 - 1 = 8(2q^2 + 3q + 1)$ which is divisible by

$8$ .

So, it is clear that

$n^2 - 1$ is divisible by

$8$ , if

$n$ is an odd positive integer.

$\dfrac { 317 } { 3125 }$ represents ______.

0%
A terminating decimal
0%
A non-recurring decimal
0%
A recurring decimal
0%
An Integer

Explanation

If the denominator can be in the powers of either

$2$ or

$5$ or both, then it will be a terminating decimal, otherwise non-terminating.

Now, the denominator

$3125=5^5$ , i.e. is in the form

$2^m\times5^n$ ; where

$n,m$ are non-negative.

Hence, the rational number is terminating.

Therefore,

$\dfrac{317}{3125}$ represents a terminating decimal.

State whether the following statement is true or not:

$7-\sqrt { 2 }$ is irrational.

0%
True
0%
False

Explanation

Let us suppose

$7-\sqrt 2$ is rational.

Therefore,

$=>7-\sqrt 2$ is in the form of

$\dfrac pq$ where

$p$ and

$q$ are integers and

$q\neq0$

$=>\sqrt2=-\dfrac pq+7$

$=>\sqrt2=\dfrac{-p+7q}{q}$

$p, q$ and

$7$ are integers therefore,

$\dfrac{-p+7q}{q}$ is a rational number.

Hence,

$\sqrt 2$ is a rational number.

But we know that

$\sqrt 2$ is an irrational number.

So this is a contradiction.

This contradiction has arisen because of our wrong assumption that

$7-\sqrt 2$ is a rational number.

Hence,

$7- \sqrt2$ is an irrational number.

The HCF of

$420$ and

$130$ is

0%
$10$
0%
$15$
0%
$5$
0%
$3$

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

0%
True
0%
False

Explanation

$\textbf{Step1-Apply definition of rational function}$

$\text{Square root of a prime number is always an irrational number as it cannot be }$

$\text{represented in the form}$

$\dfrac pq$

$\text{where $$p,q$$ are integers and $$q$$ is non-zero.}$

$\text{As $$2$$ is a prime number}$

$\sqrt 2$

$\text{ is an irrational number.}$

$\textbf{Hence option B is correct}$

Find the H.C.F. of :

0%
$8$ and $40$
0%
$15$ and $25$
0%
$16$ and $60$
0%
$14$ and $35$
0%
$15,$ $25$ and $60$
0%
$16,$ $48$ and $80$

Explanation

${\textbf{Step - 1: Solving A}}$

${\text{Prime factorizing 8 and 40,}}$

$\Rightarrow {\text{ 8 = 2 }} \times {\text{ 2 }} \times {\text{ 2}}$

$\Rightarrow {\text{ 40 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 5}}$

${\text{Multiplying common prime factors to get HCF, }}$

${\text{H}}{\text{.C}}{\text{.F = 2 }} \times {\text{ 2 }} \times {\text{ 2 = 8}}$

${\textbf{Step - 2: Solving B}}$

${\text{Prime factorizing 15 and 25,}}$

$\Rightarrow {\text{ 15 = 3 }} \times {\text{ 5}}$

$\Rightarrow {\text{ 25 = 5 }} \times {\text{ 5}}$

${\text{Multiplying common prime factors to get HCF, }}$

${\text{H}}{\text{.C}}{\text{.F = 5}}$

${\textbf{Step - 3: Solving C}}$

${\text{Prime factorizing 16 and 60,}}$

$\Rightarrow {\text{ 16 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2}}$

$\Rightarrow {\text{ 60 = 2 }} \times {\text{ 2 }} \times {\text{ 3 }} \times {\text{ 5}}$

${\text{Multiplying common prime factors to get HCF, }}$

${\text{H}}{\text{.C}}{\text{.F = 2 }} \times {\text{ 2 = 4}}$

${\textbf{Step - 4: Solving D}}$

${\text{Prime factorizing 14 and 35,}}$

$\Rightarrow {\text{ 14 = 2 }} \times {\text{ 7}}$

$\Rightarrow {\text{ 35 = 5 }} \times {\text{ 7}}$

${\text{Multiplying common prime factors to get HCF, }}$

${\text{H}}{\text{.C}}{\text{.F = 7}}$

${\textbf{Step - 5: Solving E}}$

${\text{Prime factorizing 15, 25 and 60,}}$

$\Rightarrow {\text{ 15 = 3 }} \times {\text{ 5}}$

$\Rightarrow {\text{ 25 = }}5{\text{ }} \times {\text{ 5}}$

$\Rightarrow {\text{ 60 = 2 }} \times {\text{ 2 }} \times {\text{ 3 }} \times {\text{ 5}}$

${\text{Multiplying common prime factors to get HCF, }}$

${\text{H}}{\text{.C}}{\text{.F = }}5$

${\textbf{Step - 6: Solving F}}$

${\text{Prime factorizing 16, 48 and 80,}}$

$\Rightarrow {\text{ 16 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2}}$

$\Rightarrow {\text{ 48 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 3}}$

$\Rightarrow {\text{ 80 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 5}}$

${\text{Multiplying common prime factors to get HCF, }}$

${\text{H}}{\text{.C}}{\text{.F = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2 = 16}}$

${\textbf{Hence, we have found H.C.F of different pair of numbers using prime factorization method}}$

Mark the correct alternative of the following.
The HCF of

$100$ and

$101$ is _________.

0%
$1$
0%
$7$
0%
$37$
0%
None of these

Explanation

The given numbers

$100$ and

$101$ are consecutive numbers and respectively even and odd numbers.

$\implies$ Numbers are co-prime

So the HCF is

$1$ .

State whether the given statement is true or false.

$\pi$ is irrational and

$\dfrac{22}{7}$ is rational.

0%
True
0%
False

Explanation

True.

$\pi$ is irrational where as

$\dfrac{22}{7}$ is rational where both the numerator and denominator are integers.

The HCF of 144 and 198 is

0%
$9$
0%
$12$
0%
$6$
0%
$18$

Explanation

$\textbf{Step-1: Find the prime factors of given numbers & apply the concept of H.C.F.}$

$\text{Given numbers are}$

$144$

$\text{and}$

$198$

$\text{Representing the given numbers as the product of their prime factors:}$

$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3$

$= 2^4 \times 3^2$

$198 = 2 \times 3 \times 3 \times 11$

$= 2 \times 3^2 \times 11$

$\text{We know that}$

$HCF$

$\text{of two or more numbers is the product of all}$

$\text{ prime factors common between them.}$

$\text{So,}$

$HCF(144, 198)=2\times 3^2$

$=2\times 9$

$=18$

$\text{So, the}$

$HCF$

$\text{of}$

$144$

$\text{and}$

$198$

$\text{is}$

$18$ .

$\textbf{Hence, the correct option is D}$

The HCF of

$144, 180$ and

$192$ is

0%
$12$
0%
$16$
0%
$18$
0%
$8$

Explanation

Hence

$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$

$180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5$

$192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3$

Hence

$2^2 \times 3=12$ is the highest common factor of all the three numbers

Option (a) is the correct answer

1635275_1777832_ans_1b56fb84d8eb4880ba0fb0fd6e5e1d7a.png

H.C.F of two or more number may be one of the numbers.

0%
True
0%
False

Explanation

Consider two numbers

$17, 34$

Clearly the HCF of

$17, 34$ is

$17$ which is one of the given numbers.

$\textbf{Therefore the given statement is True}.$

State whether the following statement is true of false? Justify your answer.

$\dfrac {\sqrt {2}}{3}$ is a rational number.

0%
True
0%
False

Explanation

Rational number is a number of form

$\cfrac pq$ , where

$p$ and

$q$ are both integers and

$q\ne 0$ .

The given statement is false.

Because

$\dfrac {\sqrt {2}}{3}$ is of the form

$\dfrac {p}{q}$ but

$p = \sqrt {2}$ is not an integer but an irrational number.

CBSE Questions for Class 10 Maths Real Numbers Quiz 6 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Real Numbers Quiz 6 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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