R.d Sharma 2022 Mcqs Solutions for Class 9 Maths Chapter 1 - Number System

Answer:

Given that, $x=0.\overline{3}$.

Thus,
$$\begin{gathered} x=0.3333... \\ 10x=3.3333... \\ \Rightarrow 10x-x=\left(3.3333...\right)- \left(0.3333...\right) \\ \Rightarrow 10x-x=3 \\ \Rightarrow 10x=3+x \end{gathered}$$

Also, if $10x-x=3$, then
$$\begin{gathered} 9x=3 \\ \Rightarrow x=\frac{1}{3} \end{gathered}$$

Therefore, $x=0.\overline{3}$ is a rational number, because x can be expressed in the form $\frac{p}{q}$, by solving the equation 10x = 3 + x.

Hence, the correct answer is option (a).

Answer:

Given that, some of the rational numbers between 7 and 11 are written in the form $\frac{m}{6}$.

So,
$$\begin{gathered} \Rightarrow 7<\frac{m}{6}<11 \\ \Rightarrow 7\times 6<\frac{m}{6}\times 6<11\times 6 \\ \Rightarrow 42<m<66 \end{gathered}$$

Thus, the integer values of m lie between 42 and 66.

Hence, the correct answer is option (b).

Answer:

(i) Every integer is a rational number.
It is correct as every integer can be written in the form of $\frac{p}{q}$, where q = 1.

(ii) Every rational number is an integer.
It is incorrect as every rational number cannot be written in the form of $\frac{p}{q}$ with q = 1.

(iii) A real number is either rational or irrational number.
It is correct as every real number can either be expressed as $\frac{p}{q},q\ne 0$ or it cannot be.

(iv) Every whole number is a natural number.
It is incorrect as whole numbers include positive integers along with 0 while natural numbers only include positive integers except 0.

Hence, the correct answer is option (c).

Answer:

The decimal expansion of a rational number is terminating or non-terminating repeating.

Hence, the correct answer is option (b).

Answer:

A number is an irrational if and only if its decimal representation is non-terminating and non-repeating.

Hence, the correct answer is option (c).

Answer:

A number is an irrational if and only if its decimal representation is non-terminating and non-repeating.

Here, $0.301323100523...$ has non-terminating and non-repeating decimal representation.

Hence, the correct answer is option (d).
 

Answer:

Let $x=0.\overline{4}$.

Therefore,
$$\begin{gathered} x=0.444... \\ \Rightarrow 10x=4.444... \left(\mathrm{Multiplying} \mathrm{by} 10\right) \\ \Rightarrow 10x-x=4.444...-0.444... \\ \Rightarrow 9x=4 \\ \Rightarrow x=\frac{4}{9} \end{gathered}$$

Hence, the correct answer is option (a).

 

Answer:

Let $x=1.\overline{6}$.

Therefore,
$$\begin{gathered} x=1.\overline{6} \\ \Rightarrow 10x=16.\overline{6} \left(\mathrm{Multiplying} \mathrm{by} 10 \mathrm{on} \mathrm{both} \mathrm{sides}\right) \\ \Rightarrow 10x-x=16.\overline{6}-1.\overline{6} \\ \Rightarrow 9x=15 \\ \Rightarrow x=\frac{5}{3} \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

A number is an rational if and only if its decimal representation is terminating or non-terminating repeating.

Here,
(a) π = 3.1415926535897..., which is non-terminating non-repeating. Thus, it is irrational.

(b) $\sqrt{5}$ is irrational as it cannot be expressed as $\frac{p}{q},q\ne 0$.

(c) 0.101001000100001... is non-terminating non-repeating. Thus, it is irrational.

(d) 0.835835835... = $0.\overline{835}$, which is non-terminating repeating. Thus, it is rational.

Hence, the correct answer is option (d).

Answer:

We have, $\frac{17}{7}$.

Now,


So, $\frac{17}{7}=2.42857142=2.\overline{428571}$.

Thus, there are 6 digits in the repeating block of digits in the decimal expansion of $\frac{17}{7}$.

Hence, the correct answer is option (b).

 

Answer:

The decimal expansion that a rational number is terminating or non-terminating repeating.

Here, 0.5030030003... is neither terminating nor repeating.

Hence, the correct answer is option (d).
 

Answer:

$\mathrm{\pi }$ = 3.1415926535897932... which is non-terminating non-repeating. Thus, it is an irrational number. 

$\frac{22}{7}$ is rational as it is a fraction of two integers $\frac{p}{q},q\ne 0$ and has recurring decimal value, i.e., 3.142857.

Hence, the correct answer is option (d).

Answer:

An irrational number cannot be expressed in the form of $\frac{p}{q},q\ne 0$.

Here,
(a) $\sqrt{\frac{4}{9}}=\frac{2}{3}$, which is a rational number.

(b) $\frac{\sqrt{1250}}{\sqrt{8}}=\frac{25\sqrt{2}}{2\sqrt{2}}=\frac{25}{2}$, which is a rational number.

(c) $\sqrt{8}=2\sqrt{2}$, which is an irrational number.

(d) $\frac{\sqrt{24}}{\sqrt{6}}=\sqrt{\frac{24}{6}}=\sqrt{4}=2$, which is a rational number.

Hence, the correct answer is option (c).

Answer:

A rational number can be expressed in the form of $\frac{p}{q},q\ne 0$.

Here,
(a) $\sqrt{3}+1$
The sum of an irrational and a rational number is always an irrational number. Thus, it is an irrational number.

(b) π = 3.1415926535897932...
The decimal expansion of π is non-terminating non-repeating. Thus, it is an irrational number.

(c) $2\sqrt{3}$ is an irrational number as it cannot be expressed as $\frac{p}{q},q\ne 0$.

(d) 0 = $\frac{0}{1}$, which is of the form $\frac{p}{q},q\ne 0$. Thus, it is a rational number. 

Hence, the correct answer is option (d).

Answer:

Given that, the two rational numbers are –3 and 3 and another rational number between them is to be found.

Here,
(a) 0 lies between –3 and 3 and is a rational number as 0 = $\frac{0}{1}$, i.e., it is of the form $\frac{p}{q},q\ne 0$.

(b) –4.3 does not lie between –3 and 3.

(c) –3.4 does not lie between –3 and 3.

(d) 1.101100110001... is an irrational number as it has non-terminating non-repeating decimal expansion.

Hence, the correct answer is option (a).

Answer:

A rational number has terminating or non-terminating repeating decimal expansion.

Here,
(a) 3.14
It is a rational number as it has terminating decimal expansion.

(b) 3.141414... = $3.\overline{14}$
It is a rational number as it has terminating or non-terminating repeating decimal expansion.

(c) 3.1444444... = $3.1\overline{4}$
It is a rational number as it has terminating or non-terminating repeating decimal expansion.

(d) 3.14114111411114...
It is an irrational number as it has non-terminating non-repeating decimal expansion.

Hence, the correct answer is options (d).
 

Answer:

The product of a non-zero rational number with an irrational number is always an irrational number.

Example: Let a = 2 and b = $\sqrt{2}$. Then,
ab = $2\sqrt{2}$

And, $2\sqrt{2}$ is an irrational number.

Hence, the correct answer is option (a).

Answer:

Let the two numbers be $x=0.\overline{2} \mathrm{and} y=0.\overline{5}$.

Now,
$$\begin{gathered} x=0.\overline{2} \\ \Rightarrow 10x=2.\overline{2} \left(\mathrm{Multiplying} \mathrm{by} 10 \mathrm{on} \mathrm{both} \mathrm{sides}\right) \\ \Rightarrow 10x-x=2.\overline{2}-0.\overline{2} \\ \Rightarrow 9x=2 \\ \Rightarrow x=\frac{2}{9} \end{gathered}$$

Similarly,
$$\begin{gathered} y=0.\overline{5} \\ \Rightarrow 10y=5.\overline{5} \left(\mathrm{Multiplying} \mathrm{by} 10 \mathrm{on} \mathrm{both} \mathrm{sides}\right) \\ \Rightarrow 10y-y=5.\overline{5}-0.\overline{5} \\ \Rightarrow 9y=5 \\ \Rightarrow y=\frac{5}{9} \end{gathered}$$

Thus,
$\begin{array}{rcl}x+y& =& \frac{2}{9}+\frac{5}{9}\\ & =& \frac{7}{9}\end{array}$

Hence, the correct answer is option (b).

Answer:

Let the two numbers be $x=0.\overline{2} \mathrm{and} y=0.\overline{5}$.

Now,
$$\begin{gathered} x=0.\overline{2} \\ \Rightarrow 10x=2.\overline{2} \left(\mathrm{Multiplying} \mathrm{by} 10 \mathrm{on} \mathrm{both} \mathrm{sides}\right) \\ \Rightarrow 10x-x=2.\overline{2}-0.\overline{2} \\ \Rightarrow 9x=2 \\ \Rightarrow x=\frac{2}{9} \end{gathered}$$

Similarly,
$$\begin{gathered} y=0.\overline{5} \\ \Rightarrow 10y=5.\overline{5} \left(\mathrm{Multiplying} \mathrm{by} 10 \mathrm{on} \mathrm{both} \mathrm{sides}\right) \\ \Rightarrow 10y-y=5.\overline{5}-0.\overline{5} \\ \Rightarrow 9y=5 \\ \Rightarrow y=\frac{5}{9} \end{gathered}$$

Thus,
$\begin{array}{rcl}xy& =& \frac{2}{9}\times \frac{5}{9}\\ & =& \frac{10}{81}\end{array}$

Hence, the correct answer is option (c).

Answer:

Given that, there is a number x such that x² is irrational but x⁴ is rational.

This is possible if the square of a number has square root of any number and then, its 4^th power will be a rational number.

Here,
(a) $\sqrt{5}$
Its square will be a rational number.

(b) $\sqrt{2}$
Its square will be a rational number.

(c) $\sqrt[3]{2}$
Now,
$$\begin{gathered} {\left(\sqrt[3]{2}\right)}^2=2^{\frac{2}{3}} \\ {\left(2^{\frac{2}{3}}\right)}^2=2^{\frac{4}{3}}, \mathrm{which} \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{rational} \mathrm{number} \end{gathered}$$

(d) $\sqrt[4]{5}$
Now,
$$\begin{gathered} {\left(\sqrt[4]{5}\right)}^2=\sqrt{5}, \mathrm{which} \mathrm{is} \mathrm{irrational} \\ {\left(\sqrt{5}\right)}^2=5, \mathrm{which} \mathrm{is} \mathrm{rational} \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

The decimal expansion of an irrational number is non-terminating and non- repeating. Thus, we can say that a number, whose decimal expansion is non-terminating and non- repeating, called irrational number. And the decimal expansion of rational number is either terminating or repeating. Thus, we can say that a number, whose decimal expansion is either terminating or repeating, is called a rational number.

Hence the correct option is .

Answer:

Since, and are two irrational number and

Therefore, sum of two irrational numbers may be rational

Now, letandbe two irrational numbers and

Therefore, sum of two irrational number may be irrational

Hence the correct option is .

Answer:

The sum of irrational number and rational number is always irrational number.

Let a be a rational number and b be an irrational number.

Then,

As 2ab is irrational therefore is irrational.

Hence is irrational.

Therefore answer is .

Answer:

Since we know that the product of rational and irrational numbers is always irrational.
For example: Let $\frac{1}{2} \mathrm{and} \sqrt{3}$ are rational and irrational numbers respectively and their product is $\frac{\sqrt{3}}{2}$.

Hence the correct is option (b).

Answer:

Given that

And 7 is not a perfect square.

Hence the correct option is.

Answer:

Given that

Here is non-terminating or non-repeating. So it is an irrational number.

Hence the correct option is.

Answer:

Given that

Here,, this is the form of . So this is a rational number

Hence the correct option is.

Answer:

Since the given number is repeating, so it is rational number because rational number is always either terminating or repeating

Hence the correct option is.

Answer:

The term “natural number” refers either to a member of the set of positive integer.

And natural number starts from one of counting digit .Thus, if n is a natural number then sometimes n is a perfect square and sometimes it is not.

Therefore, sometimesis a natural number and sometimes it is an irrational number

Hence the correct option is.

Answer:

Given that

Here is repeating but non-terminating. 

Hence the correct option is.

Answer:

In basic mathematics, number line is a picture of straight line on which every point is assumed to correspond to real number.

Hence the correct option is.

Answer:

Given decimal numbers are

Here the number is non terminating or non-repeating.

Hence the correct option is.

Answer:

Given that

The correct option is.

Answer:

Given number is

The correct option is

Answer:

Given that

Now we have to express this number into form

Let X =$0.3\overline{2}$

The correct option is

Answer:

Given that

Now we have to express this number into the form of
Let

$$\begin{gathered} x = 23.43 \\ x = 23+0.4343... \\ x = 23+\frac{43}{99} \\ x=\frac{2277+43}{99}=\frac{2320}{99} \end{gathered}$$

The correct option is

Answer:

Given that

Now we have to express this number into form

The correct option is

Answer:

Given that

Let

Now we have to find the value of

The correct option is  

Answer:

Let 

Here a and b are rational numbers. So we observe that in first decimal place a and b have distinct. According to question a < b.so an irrational number between 2 and 2.5 is OR

Hence the correct answer is.

Answer:

We are given the following expression and asked to find out the number of consecutive zeros

We basically, will focus on the powers of 2 and 5 because the multiplication of these two number gives one zero. So

$$\begin{gathered} 2^3\times 3^4\times 5^4\times 7=2^3\times 5^4\times 3^4\times 7 \\ =2^3\times 5^3\times 5\times 3^4\times 7 \\ ={\left(2\times 5\right)}^3\times 5\times 3^4\times 7 \\ ={10}^3\times 5\times 3^4\times 7 \\ =5\times 81\times 7\times 1000 \\ =2835000 \end{gathered}$$

Therefore the consecutive zeros at the last is 3

So the option (a) is correct

Answer:

Give number is. Now multiplying by in the given number, we have

Hence the correct option is

Answer:

Statement-2 (Reason): The decimal expansion of $\sqrt{2}$ is non-terminating non-recurring.
Now, $\sqrt{2}$ is given as:


Therefore, $\sqrt{2}=1.4142...$, i.e., its decimal expansion is non-terminating non-recurring. 

Thus, Statement-2 is true.

Statement-1 (Assertion): $\sqrt{2}$ is an irrational number.
This statement is true as the decimal expansion of $\sqrt{2}$ is non-terminating non-recurring.

Thus, Statement-1 is true and Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-1 (Assertion): The sum of any two irrational numbers is an irrational number.
Consider two irrational numbers $\sqrt{2}$ and $\sqrt{3}$.

Their sum = $\sqrt{2}$ + $\sqrt{3}$
Their sum is also an irrational number.

Now, consider two irrational numbers $\sqrt{2}$ and $-\sqrt{2}$.

Their sum = $\sqrt{2}+\left(-\sqrt{2}\right)=\sqrt{2}-\sqrt{2}=0$
Their sum is a rational number. Thus, Statement-1 is false.

Statement-2 (Reason): There are two irrational numbers whose sum is a rational number.
From the above example, we get that Statement-2 is true.

Hence, the correct answer is option (d).

Answer:

Statement-1 (Assertion): The product of any two irrational numbers is an irrational number.
Consider two irrational numbers $\sqrt{2}$ and $\sqrt{3}$.

$\begin{array}{rcl}\mathrm{Their} \mathrm{product}& =& \sqrt{2}\times \sqrt{3}\\ & =& \sqrt{6}\end{array}$
Their product is also an irrational number.

Now, consider two irrational numbers $\sqrt{2}$ and $\sqrt{2}$.

$\begin{array}{rcl}\mathrm{Their} \mathrm{product}& =& \sqrt{2}\times \sqrt{2}\\ & =& 2\end{array}$
Their product is a rational number. So, the product of any two irrational numbers can be a rational or an irrational number.
Thus, Statement-1 is false.

Statement-2 (Reason): There are two irrational numbers whose product is not an irrational number.
From the above example, we get that Statement-2 is true.

Hence, the correct answer is option (d).

Answer:

Statement-1 (Assertion): $\sqrt{3}$ is an irrational number.

Now, $\sqrt{3}$ is given as:


Therefore, $\sqrt{3}=1.732...$, i.e., it has non-terminating non-repeating decimal expansion. So, $\sqrt{3}$ is an irrational number. 
Thus, Statement-1 is true.

Statement-2 (Reason): The square root of a positive integer which is not a perfect square is an irrational number.
From the above example, we get that the square root of a positive integer which is not a perfect square is an irrational number.
Thus, Statement-2 is true and is the correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-1 (Assertion): $\sqrt{2}$ is an irrational number.

Now, $\sqrt{2}$ is given as:


Since its decimal expansion is non-terminating non-recurring, $\sqrt{2}$ is an irrational number.
Thus, Statement-1 is true.

Statement-2 (Reason): The sum of a rational number and an irrational number is an irrational number.
Consider two numbers $\sqrt{2}$ and 3.

Their sum = $\sqrt{2}$ + 3, which is also an irrational number.
Thus, Statement-2 is true but is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

Answer:

Statement-1 (Assertion): There are two rational numbers whose sum and product both are rationals.
Consider two rational numbers $\frac{1}{2}$ and 2.
Their sum = 2 + $\frac{1}{2}$
                  = $\frac{5}{2}$, which is a rational number

Their product = $2\times \frac{1}{2}$
                       = 1, which is a rational number
Therefore, the sum and product of two rational number is a rational number. Thus, Statement-1 is true. 

Statement-2 (Reason): There are numbers which cannot be written in the form $\frac{p}{q}, q\ne 0, p, q$ both are integers.
The numbers which cannot be written in the form $\frac{p}{q}, q\ne 0, p, q$ both are integers are called irrational numbers. Thus, Statement-2 is true but is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

Answer:

Statement-1 (Assertion): The decimal representation of $\frac{3}{8}$ is terminating.
Now, 
$\frac{3}{8}=\frac{3}{2^3\times 5^0}$

Therefore, the decimal representation of $\frac{3}{8}$ is terminating after 3 decimal places. Thus, Statement-1 is true.

Statement-2 (Reason): If the denominator of a rational number is of the form 2^m × 5ⁿ, where m, n are not-negative integers, then its decimal representation is terminating.
This statement is true as the denominator of an irrational number cannot be expressed as 2^m × 5ⁿ, where m, n are not-negative integers.

Thus, Statement-2 is true and is the correct explanation for Statement-1.

Hence, the correct answer is option (a).