NCERT Exemplar Class 9 Maths Solutions Chapter 1 Number Systems

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NCERT Exemplar Class 9 Maths Solutions Chapter 1 Number Systems

Edited By Ravindra Pindel | Updated on Aug 31, 2022 11:38 AM IST

NCERT Exemplar Class 9 Maths solutions chapter 1 provide detailed answers for questions related to Number System. These problems and their solutions are devised for a better knowledge of the chapter 1 of NCERT Class 9 Maths Book. These solutions are extremely useful in understanding the concepts and know-how of real numbers. Experts have created these NCERT exemplar solutions for Class 9 Maths chapter 1 with a thorough analysis.

The NCERT exemplar Class 9 Maths chapter 1 solutions provided for exemplar problems are detailed and expressive. These are helpful to pinpoint the critical understanding of concepts. The NCERT exemplar Class 9 Maths solutions chapter 1 are in accord with the CBSE syllabus for Class 9.

Also, read - NCERT Solutions for Class 9 Maths

NCERT Exemplar Class 9 Maths Solutions Chapter-1: Exercise-1.1

Question:1

Every rational number is:
(A)A natural number
(B)An integer
(C)A real number
(D) A whole number

Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Examples: $4/5, 2/3, -6/7$
(A) All the positive integers from 1 to infinity are natural numbers. These cannot be fractions.
Examples: 1, 2, 3, 4 and so on
(B) An integer is a number which can be written without fractional components or decimal representation. They can be positive, negative or zero.
Examples: $\left \{ .......,-4,-3,-2,-1,0,1,2,3,4,......... \right \}$
(C) All numbers that we generally use are real numbers. They can be positive, negative, decimal, whole, natural, integer etc.
Examples: $0.123, -2.5,0,2,5,6.3456$
(D) All the positive integers from 0 to infinity are whole numbers. These cannot be fractions.
Examples: 0, 1, 2, 3, 4 and so on
From the above definitions we can easily see that every rational number is a real number. Therefore option (C) is correct.

Question:2

Between two rational numbers:

(A)There is no rational number
(B)there is exactly one rational number
(C)there are infinitely many rational numbers
(D)no irrational number

Solution.
Firstly let us define a rational number.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Examples: $4/5, 2/3, -6/7$
So we can say that it is a type of real number.
Similarly we can see that irrational numbers are real numbers which cannot be represented as simple fractions.
Examples: $\sqrt{2},\sqrt{3},\pi$
So we can say that between two rational numbers, there are infinitely many rational numbers.
For Example: between rational numbers 5 and 4, there are rational numbers like $4.1, 4.11, 4.12, 4.13$………and so on.
Therefore option (C) is correct.

Question:3

Decimal representation of a rational number cannot be :
(A)Terminating
(B)non-terminating
(C)non-terminating repeating
(D)non-terminating non-repeating

Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Examples: $4/5, 2/3, -6/7$
Terminating decimals have a finite number of digits after decimal point,
Examples: $1/2 = 0.5, 3/5 = 0.6$
Non terminating decimals are the ones which keep on continuing after decimal point.
Examples: $1/3 = 0.3333..., 5/11 = 0.454545....$
Recurring decimals are those non terminating decimals which have a particular pattern/sequence that keeps on repeating itself after the decimal point. They are also called repeating decimals.
Examples: $1/3 = 0.33333....., 4/11 = 0.363636....$
Non-Recurring decimals are those non terminating decimals which do not have a particular pattern/sequence after the decimal point. They are also called non repeating decimals.
Examples:
$\sqrt{2}=1.414213562373$
$\sqrt{3}=1.732050807568$
$\pi =3.14159265359$
So by the above definitions we can see that the decimal representation of a rational number cannot be non-terminating non repeating because decimal expansion of rational number is either terminating or non-terminating recurring (repeating).
Therefore option (D) is correct.

Question:4

The product of any two irrational number is
(A)always an irrational number
(B)always a rational number
(C)always an integer
(D)sometimes rational, sometimes irrational

Solution.
Irrational numbers are real numbers which cannot be represented as simple fractions.
Examples: $\sqrt{2},\sqrt{3},\pi$
The product of two irrational numbers can be rational or irrational depending on the two numbers.
For example, $\sqrt{2}\times \sqrt{2}$ is 2 which is a rational number
whereas $\sqrt{2}\times \sqrt{3}$ is $\sqrt{6}$ which is an irrational number.
Therefore option (D) is correct.

Question:5

The decimal expansion of the number $\sqrt{2}$ is.
(A) A finite decimal
(B) $1.41421$
(C) non-terminating recurring
(D) non-terminating non-recurring

Solution.
Terminating decimals have a finite number of digits after decimal point,
Examples: $1/2 = 0.5, 3/5 = 0.6$
Non terminating decimals are the ones which keep on continuing after decimal point.
Examples: $1/3 = 0.33333...., 5/11 = 0.454545...$
Recurring decimals are those non terminating decimals which have a particular pattern/sequence that keeps on repeating itself after the decimal point. They are also called repeating decimals.
Examples: $1/3 = 0.33333..., 4/11 = 0.363636....$
Non-Recurring decimals are those non terminating decimals which do not have a particular pattern/sequence after the decimal point. They are also called non repeating decimals.
Examples:
$\sqrt{2}=1.414213562373$
$\sqrt{3}=1.732050807568$
$\pi =3.14159265359$
So, the decimal expansion of the number $\sqrt{2}$ is non-terminating non-recurring. It is an irrational number which is a non-terminating non-recurring decimal expansion.
Therefore option (D) is correct.

Question:6

Which of the following is irrational?
(A) $\sqrt{\frac{4}{9}}$
(B) $\frac{\sqrt{12}}{\sqrt{3}}$
(C) $\sqrt{7 }$
(D) $\sqrt{81}$

Solution.

Irrational numbers are real numbers which cannot be represented as simple fractions.
Examples: $\sqrt{2},\sqrt{3},\pi$
$\sqrt{\frac{4}{9}}=\frac{2}{3}$ (rational)
$\frac{\sqrt{12}}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3}}=2$ (rational)
$\sqrt{81}=9$ (rational)
but $\sqrt{7}$ is an irrational number.
Therefore option (C) is correct.

Question:7

Which of the following is irrational?
(A) $0.14$
(B) $0.14\overline{16}$
(C) $0.\overline{1416}$
(D) $0.4014001400014$

Solution.

Irrational numbers are real numbers which cannot be represented as simple fractions.
Examples: $\sqrt{2},\sqrt{3},\pi$
Irrational cannot be expressed in form of p/q form and its decimal expansion is non- terminating non-recurring decimal expansions.
Non-Recurring non terminating decimals do not have a particular pattern/sequence after the decimal point. They are also called non repeating decimals.
Examples:
$\sqrt{2}=1.414213562373$
$\sqrt{3}=1.732050807568$
$\pi =3.14159265359$
In the given options, only option D is non-terminating non-recurring decimal as it satisfies the above definition.
Therefore option (D) is correct.

Question:8

A rational number between $\sqrt{2}$ and $\sqrt{3}$ is –
(A) $\frac{\sqrt{2}+\sqrt{3}}{2}$
(B) $\frac{\sqrt{2}.\sqrt{3}}{2}$
(C) $1.5$
(D) $1.8$

Solution.

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Irrational numbers are real numbers which cannot be represented as simple fractions.
Now, we have $\sqrt{2}=1.414$ and $\sqrt{3}= 1.732$
(A) $\frac{\sqrt{2}+\sqrt{3}}{2}$ cannot be represented as simple fraction hence it is irrational.
(B) $\frac{\sqrt{2}.\sqrt{3}}{2}$ $=\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{\frac{3}{2}}=\sqrt{1.5}$ cannot be represented as simple fraction hence it is irrational.
(C) 1.5 is a rational number between $\sqrt{2}=1.414$ and $\sqrt{3}= 1.732$
Q $1.414 < 1.5 < 1.732$
(D) 1.8 is a rational number but does not lie between
$\sqrt{2}=1.414$ and $\sqrt{3}= 1.732$.
Q $1.732 < 1.8$

Therefore option (C) is correct.

Question:9

The value of 1.999 ___in the form $\frac{p}{q}$ where p and q are integers and q $\neq$0, is
(A)$\frac{19}{10}$
(B)$\frac{1999}{1000}$
(C)2
(D)$\frac{1}{9}$

Solution.

Let x = 1.999……
Since, one digit is repeating, we multiply x by 10
we get, 10x = 19.999……..
so, 10x = 18 + 1.999………
10x = 18 + x
Therefore, 10x – x = 18, i.e., 9x = 18
i.e., $x= \frac{18}{9}= \frac{2}{1}= 2$
Hence option C is correct answer.

Question:10

$2\sqrt{3}+\sqrt{3}$ is equal to –
(A).$2\sqrt{6}$
(B) 6
(C)$3\sqrt{3}$
(D) $4\sqrt{6}$

Solution.
We can see that both the numbers $\sqrt{3}$ and $2\sqrt{3}$ are irrational numbers.
So their decimal representation will be non terminating, non repeating. So we
Now $2\sqrt{3}+\sqrt{3}$ can be written as
$= \sqrt{3}\left ( 2+1 \right )$ Taking $\sqrt{3}$ common
Hence the correct answer is $3\sqrt{3}$

Question:11

$\sqrt{10}\times \sqrt{15}$is equal to :
(A)$6\sqrt{5}$
(B)$5\sqrt{6}$
(C)$\sqrt{25}$
(D)$10\sqrt{5}$

Answer. [B] $5\sqrt{6}$
Solution.
We know that
$10= 5\times 2$
$15= 5\times 3$
So we have,
$\sqrt{10}\times \sqrt{15}= \sqrt{5\times 2}\times \sqrt{5\times 3}$
This can be written as
$= \sqrt{5\times 5\times2\times3}$
$=5 \sqrt{2\times 3}$
$=5 \sqrt{6}$
Therefore option (B) is correct.

Question:12

The number obtained on rationalizing the denominator of $\frac{1}{\sqrt{7}-2}$ is :
(A) $\frac{\sqrt{7}+2}{3}$
(B)$\frac{\sqrt{7}-2}{3}$
(C)$\frac{\sqrt{7}+2}{5}$
(D)$\frac{\sqrt{7}+2}{45}$

Solution
.
We have, $\frac{1}{\sqrt{7}-2}$
We have to rationalize it
$\frac{1}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2}$ [Multiplying numerator and denominator by $\sqrt{7}+2$]
= $\frac{\sqrt{7}+2}{\left ( \sqrt{7} \right )^{2}-\left ( 2 \right )^{2}}$ [$\because$ (a – b) (a + b) = a2 – b2]
$= \frac{\sqrt{7}+2}{7-4}$
$= \frac{\sqrt{7}+2}{3}$
Hence option A is correct.

Question:13

$\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to :
(A) $\frac{1}{2}\left ( 3-2\sqrt{2} \right )$
(B) $\frac{1}{3+2\sqrt{2}}$
(C)$3-2\sqrt{2}$
(D) $3+2\sqrt{2}$

Solution.
We have, $\frac{1}{\sqrt{9}-\sqrt{8}}$
We have to rationalize it
$\frac{1}{\sqrt{9}-\sqrt{8}}\times \frac{\sqrt{9}+\sqrt{8}}{\sqrt{9}+\sqrt{8}}$ [Multiplying and dividing by $\sqrt{9}+\sqrt{8}$]
= $\frac{\sqrt{9}+\sqrt{8}}{\left ( \sqrt{9} \right )^{2}-\left ( \sqrt{8} \right )^{2}}$ [$\because$ (a – b) (a + b) = a2 – b2]
=$\frac{3+2\sqrt{2}}{9-8}$
$= 3+2\sqrt{2}$
Hence option D is correct.

Question:14

Solution.

We have,$\frac{7}{3\sqrt{3}-2\sqrt{2}}$
We have to rationalize it
$\frac{7}{3\sqrt{3}-2\sqrt{2}}\times \frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}$
[Multiplying numerator and denominator by $3\sqrt{3}+2\sqrt{2}$]
= $\frac{7\left ( 3\sqrt{3}+2\sqrt{2} \right )}{\left ( 3\sqrt{3} \right )^{2}-\left ( 2\sqrt{2} \right )^{2}}$ [$\because$ (a – b) (a + b) = a2 – b2]
$\frac{7\left ( 3\sqrt{3} \right )+2\sqrt{2}}{27-8}$
= $\frac{7\left ( 3\sqrt{3} \right )+2\sqrt{2}}{19}$
Therefore we get the denominator as 19.
Hence (B) is the correct option.

Question:15

The value of $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is equal to :
(A) $\sqrt{2}$
(B) 2
(C)4
(D) 8

Solution.
$\sqrt{32}+\sqrt{48}= \sqrt{16\times 2}+\sqrt{16\times3}= 4\left ( \sqrt{2}+\sqrt{3} \right )$
$\sqrt{8}+\sqrt{12}= \sqrt{4\times 2}+\sqrt{4\times3}= 2\left ( \sqrt{2}+\sqrt{3} \right )$
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}= \frac{4\left ( \sqrt{2}+\sqrt{3} \right )}{2\left ( \sqrt{2} \right )+\sqrt{3}}= 2$

Hence (B) is the correrct option.

Question:16

Solution

We have, $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$
We have to rationalize it
$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}}$ [Multiplying numerator and denominator by $\sqrt{2}-1$]
= $\frac{\sqrt{\left ( \sqrt{2}-1 \right )\times\left ( \sqrt{2} -1\right ) }}{\sqrt{\left ( \sqrt{2} \right )^{2}-\left ( 1 \right )^{2}}}$ [$\because$ (a – b) (a + b) = a2 – b2]
= $\frac{\sqrt{\left ( \sqrt{2}-1 \right )^{2}}}{1}$
$\sqrt{\left ( \sqrt{2}-1 \right )^{2}}$
=$\sqrt{2}-1$
=$1\cdot 4142-1$
=0.4142
Hence option C is correct.

Question:17

$\sqrt[4]{\sqrt[3]{2^{2}}}$ equals :-
(A) $2^{-\frac{1}{6}}$
(B)2-6
(C) $2^{\frac{1}{6}}$
(D) $2^{6}$

Solution.

We have, $\sqrt[4]{\sqrt[3]{2^{2}}}$ =$\left ( \left ( 2^{2} \right )^{}\frac{1}{3} \right )^{\frac{1}{4}}$
$\because \sqrt[n]{a}= \left ( a \right )^{\frac{1}{n}}$
$\because \left ( a^{m} \right )^{n}= a^{m\times n}$
So we get,
$\left ( 2 \right )^{2\times \frac{1}{3}\times \frac{1}{4}}$
$= \left ( 2 \right )^{\frac{1}{3}\times \frac{1}{2}}$
$= \left ( 2 \right )^{\frac{1}{6}}$
Hence option C is correct.

Question:18

The product $\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}$ equals:
(A)$\sqrt{2}$
(B)2
(C)$12\sqrt{2}$
(D)$12\sqrt{32}$

Solution.
We have, $\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}$
We know that $\sqrt[n]{a}= \left ( a \right )^{\frac{1}{n}}$
$\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}$=$\left ( 2 \right )^{\frac{1}{3}}\left ( 2 \right )^{\frac{1}{4}}\left ( 2\cdot 2\cdot 2\cdot 2\cdot 2 \right )^{\frac{1}{12}}$
$= \left ( 2 \right )^{\frac{1}{3}}\left ( 2 \right )^{\frac{1}{4}}\left ( 2^{5} \right )^{\frac{1}{12}}$
$= \left ( 2 \right )^{\frac{1}{3}}\left ( 2 \right )^{\frac{1}{4}}\left ( 2 \right )^{\frac{5}{12}}$ $\because \left ( a^{m} \right )^{n}= a^{m\times n}$
$= \left ( 2 \right )^{\frac{1}{3}+ \frac{1}{4}+\frac{5}{12}}$
$\because a^{m}\times a^{n}= a^{m+n}$
Now, $\frac{1}{3}+\frac{1}{4}+\frac{5}{12}= \frac{4+3+5}{12}= \frac{12}{12}= 1$
So, $\left ( 2 \right )^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}=2^{1}= 2$
Hence option B is correct.

Question:19

The value of $\sqrt[4]{\left ( 81 \right )^{-2}}$is :
(A)$\frac{1}{9}$
(B)$\frac{1}{3}$
(C)9
(D)$\frac{1}{81}$

Solution.
We have, $\sqrt[4]{\left ( 81 \right )^{-2}}$
We know that $\sqrt[n]{a}= \left ( a \right )^{\frac{1}{n}}$
So,
$\sqrt[4]{\left ( 81 \right )^{-2}}= \left ( \left ( 81 \right )^{-2} \right )^{\frac{1}{4}}$
$=\left ( 81 \right )^{-2\times \frac{1}{4}}$ $\because \left ( a^{m} \right )^{n}= a^{m\times n}$
$= 81^{-\frac{1}{2}}= \left ( \frac{1}{81} \right )^{\frac{1}{2}}$ $\because \left ( a^{-m}= \left ( \frac{1}{a} \right )^{m} \right )$
$= \sqrt{\frac{1}{81}}$
$=\frac{1}{9}$
Hence option A is correct

Question:20

Value of (256)0.16 × (256)0.09 is :
(A) 4
(B) 16
(C) 64
(D) 256.25

Solution.
We have,
(256)0.16 × (256)0.09 [$\because$ am × an = am+n]
= (256)0.16 + 0.09
= (256)0.25
= $\left ( 256 \right )^{\frac{25}{100}}$
= $\left ( 256 \right )^{\frac{1}{4}}$
Now 256 = 28 = (22)4 = 44
= (44)1/4 [$\because$ (am)n = amn]
= 4
Hence option A is correct.

Question:21

Which of the following is equal to x?
(A)$x^{\frac{12}{7}}+x^{\frac{5}{7}}$
(B)$\sqrt[12]{\left ( x^{4} \right )^{\frac{1}{3}}}$
(C)$\left ( \sqrt{x^{3}} \right )^{\frac{2}{3}}$
(D)$x^{\frac{12}{7}}\times x^{\frac{7}{12}}$

Solution.
(A) We have,
$x^{\frac{12}{7}}+x^{\frac{5}{7}}= x^{\frac{1}{7}\left ( 12 \right )}+x^{\frac{1}{7}\left ( 5 \right )}$
$= x^{\frac{1}{7}}\left ( x^{12} +x^{5}\right )\neq x$
(B) We have,
$\sqrt[12]{\left ( x^{4} \right )^{\frac{1}{3}}}=\left ( \left ( x^{4} \right ) ^{\frac{1}{3}}\right )^{\frac{1}{12}}$ $\left ( \sqrt[n]{a} = \left ( a \right )^{\frac{1}{n}}\right )$
$= x^{4\times ^{\frac{1}{3}\times \frac{1}{12}}}= x^{\frac{1}{9}}$ $\because \left ( a^{m} \right )^{n}= a^{m\times n}$
$\neq x$
(C) We have,
$\left ( \sqrt{x^{3}} \right )^{\frac{2}{3}}= \left ( \left ( x^{3} \right ) ^{\frac{2}{3}}\right )^{\frac{1}{2}}$ $\left ( \sqrt[n]{a} = \left ( a \right )^{\frac{1}{n}}\right )$
$= x^{3\times \frac{2}{3}\times \frac{1}{2}}$ $\because \left ( a^{m} \right )^{n}= a^{m\times n}$
= x
(D) We have,
$x^{\frac{12}{7}}\times x^{\frac{7}{12}}= x^{\frac{12}{7}+\frac{7}{12}}= x^{\frac{144+49}{84}}\neq x$
$\because a^{m}\times a^{n}= a^{m+n}$
Hence option C is correct.

Question:1

Let x and y be rational and irrational number respectively. Is x + y necessarily an irrational number?

Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Irrational numbers are real numbers which cannot be represented as simple fractions.
Given that x and y be rational and irrational number respectively.
So, $(x + y)$ is necessarily an irrational number.
For example, let $x=2,y=\sqrt{3}$
Then, $x + y = 2+\sqrt{3}$
If possible, let us assume $x + y = 2+\sqrt{3}$ be a rational number.

Consider $a = 2 + \sqrt{3}$
On squaring both sides, we get
$a^{2}=(2+\sqrt{3})^{2}$
(using identity $(a + b)^{2} = a^{2} + b^{2} + 2ab$)
$\Rightarrow a^{2}=2^{2}+(\sqrt{3})^{2}+2(2)(\sqrt{3})$
$\Rightarrow a^{2}=4+3+4\sqrt{3}$
$\Rightarrow \frac{a^{2}-7}{4}=\sqrt{3}$
But we have assumed a is rational
$\Rightarrow \frac{a^{2}-7}{4}$is rational
$\Rightarrow \sqrt{3}$ is rational which is not true.
Hence our assumption was incorrect, so $2+ \sqrt{3}$is irrational.
Hence proved

Question:2

Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example.

Answer: [xy is not necessarily an irrational number.]
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Irrational numbers are real numbers which cannot be represented as simple fractions.
Given that x and y be rational and irrational number respectively.
Let x = 0 (a rational number) and $y=\sqrt{3}$be an irrational number. then,
$xy=0(\sqrt{3})=0$, which is not an irrational number.
Hence, xy is not necessarily an irrational number.

Question:3

(i)
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Examples: $4/5, 2/3, -6/7$
Irrational numbers are real numbers which cannot be represented as simple fractions.
Examples: $\sqrt{2},\sqrt{3},\pi$
The given number is $\frac{\sqrt{2}}{3}$.
Here $\sqrt{2}$ is an irrational number and 3 is a rational number, we know that when we divide irrational number by non-zero rational it always gives an irrational number.
Hence the given number is irrational.
Therefore the given statement is False.

(ii)
Solution.
An integer is a number which can be written without fractional components or decimal representation. They can be positive, negative or zero.
Examples: $\left \{ .....-4,-3,-2,-1,0,1,2,3,4,..... \right \}$
The given statement is “There are infinitely many integers between any two integers.”
This is false, because between two integers (like 1 and 9), there does not exist infinite integers.
Also, if we consider two consecutive integers (like 8 and 9), there does not exist any integer between them.
Therefore the given statement is False.

(iii)
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Examples: $4/5, 2/3, -6/7$
So, it is a type of real number.
The given statement is: Number of rational numbers between 15 and 18 is finite.
If we see the definition of rational numbers as mentioned above, the given statement is false, because between any two rational numbers there exist infinitely many rational numbers.
Here we have rational numbers between 15 and 18 as:
$16,16.1\left ( =\frac{161}{10} \right ),16.2\left ( =\frac{162}{10} \right ),16.12\left ( =\frac{1612}{100} \right ),.....$ and infinitely more.
Therefore the given statement is False.

Solution:
There are infinitely many numbers which cannot be written in the form $\frac{p}{q}$ ,$q \neq 0$, p and q both are integers. These numbers are called irrational numbers
(v)
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number.
Irrational numbers are real numbers which cannot be represented as simple fractions.
The given statement is: The square of an irrational number is always rational.
This is False, e.g., let us consider irrational numbers $\sqrt{2}$ and $\sqrt[4]{2}$
$(a)(\sqrt{2})^{2}=2,$which is a rational number.
$(b)(\sqrt[4]{2})^{2}=\sqrt{2},$ which is an irrational number.
Hence, square of an irrational number is not always a rational number.
Therefore the given statement is False.
(vi)
Solution:
$\frac{\sqrt{12}}{\sqrt{3}}=\frac{\sqrt{4 \times 3}}{\sqrt{3}}=\frac{\sqrt{4} \times \sqrt{3}}{\sqrt{3}}=2 \times 1=2$which is a rational number.
(vii)
Solution:
$\frac{\sqrt{15}}{\sqrt{3}}=\frac{\sqrt{5 \times 3}}{\sqrt{3}}=\frac{\sqrt{5} \times \sqrt{3}}{\sqrt{3}}=\sqrt{5}$which is an irrational number.

Question:4

(i) Classify the given number as rational or irrational with justification.$\sqrt{196}$.
(ii) Classify the given number as rational or irrational with justification $3\sqrt{18}$
(iii) Classify the given number as rational or irrational with justification.$\sqrt{\frac{9}{27}}$
(iv) Classify the given number as rational or irrational with justification.$\frac{\sqrt{28}}{\sqrt{343}}$
(v) Classify the given number as rational or irrational with justification.$-\sqrt{0\cdot 4}$
(vi) Classify the given number as rational or irrational with justification.$\frac{\sqrt{12}}{\sqrt{75}}$
(vii) Classify the given number as rational or irrational with justification.0.5918
(viii) Classify the given number as rational or irrational with justification.$\left ( 1+\sqrt{5} \right )-\left ( 4+\sqrt{5} \right )$
(ix) Classify the given number as rational or irrational with justification.10.124124 ………..
(x) Classify the given number as rational or irrational with justification.1.010010001 ………….

Solution.

We have,
$\sqrt{196}$ = 14 = $\frac{14}{1}$ which follows rule of rational number.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.

So, $\sqrt{196}$ is a rational number.

Solution.

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
We have,
$3\sqrt{18}= 3\sqrt{9\times 2}$
$= 3\sqrt{9}\sqrt{2}$
$= 3\times 3\sqrt{2}= 9\sqrt{2}$
So, it can be written in the form of $\frac{p}{q}$ as $\frac{9\sqrt{2}}{1}$
But we know that $9\sqrt{2}$ is irrational
(Irrational numbers are real numbers which cannot be represented as simple fractions.)
Hence, $3\sqrt{18}$ is an irrational number
Solution.

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
We have,
$\sqrt{\frac{9}{27}}= \sqrt{\frac{3\times 3}{3\times 3\times 3}}$
$= \sqrt{\frac{1}{3}}= \frac{\sqrt{1}}{\sqrt{3}}= \frac{1}{\sqrt{3}}$
So this can be written in the form of $\frac{p}{q}$ as $\frac{1}{\sqrt{3}}$ but we can see that $\sqrt{3}$ (denominator) is irrational.
(Irrational numbers are real numbers which cannot be represented as simple fractions.)
Hence $\sqrt{\frac{9}{27}}$ is irrational

Solution.

We have,
$\frac{\sqrt{28}}{\sqrt{343}}= \frac{\sqrt{4\times 7}}{\sqrt{49\times 7}}$
$= \frac{2\times \sqrt{7}}{7\times \sqrt{7}}= \frac{2}{7}$
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Hence $\frac{\sqrt{28}}{\sqrt{343}}$ is a rational number.
Solution.

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
We have,
$-\sqrt{0\cdot 4}= -\sqrt{\frac{4}{10}}$
$= -\sqrt{\frac{2}{5}}= -\frac{\sqrt{2}}{\sqrt{5}}$
So, it can be written in the form of $\frac{p}{q}$ as $\frac{-\sqrt{2}}{\sqrt{5}}$
But we know that both $\sqrt{2},\sqrt{5}$ are irrational
(Irrational numbers are real numbers which cannot be represented as simple fractions.)
Hence, $-\sqrt{0\cdot 4}$ is an irrational number

Solution.
We have,
$\frac{\sqrt{12}}{\sqrt{75}}= \frac{\sqrt{4\times 3}}{\sqrt{25\times 3}}$
$= \frac{\sqrt{4}\sqrt{3}}{\sqrt{25}\sqrt{3}}= \frac{\sqrt{4}}{\sqrt{25}}$
$= \frac{2}{5}$
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q
should be rational when the fraction is expressed in the simplest form.
So, $\frac{\sqrt{12}}{\sqrt{75}}$ is a rational number.
Solution.

We have,
$0\cdot 5918= \frac{0\cdot 5918\times 10000}{1\times 10000}$
$= \frac{5918}{10000}= \frac{2959}{5000}$
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Also we can see that 0.5918 is a terminating decimal number hence it must be rational.
So, 0.5918 is a rational number.
Solution.

We have,
$\left ( 1+\sqrt{5} \right )-\left ( 4+\sqrt{5} \right )$
$= 1+\sqrt{5}-4-\sqrt{5}$
$= -3= \frac{-3}{1}$
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
So, $\left ( 1+\sqrt{5} \right )-\left ( 4+\sqrt{5} \right )$ is a rational number.

Solution.

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
All non-terminating recurring decimal numbers are rational numbers.
Non terminating Recurring decimals are those decimals which have a particular pattern/sequence that keeps on repeating
itself after the decimal point. They are also called repeating decimals.
Examples: 1/3 = 0.33333…, 4/11 = 0.363636…
Now, 10.124124 ………. is a decimal expansion which is a non-terminating recurring.
So, it is a rational number.
Solution
.
Non terminating Recurring decimals are those decimals which have a particular pattern/sequence that keeps on repeating itself after the decimal point.
All non-terminating recurring decimal numbers are rational numbers.
Non terminating Non Recurring decimals are those decimals which do not have a particular pattern/sequence after the decimal point and it does not end.
All non-terminating non-recurring decimal numbers are irrational numbers.
1.010010001 ………. is non-terminating non-recurring decimal number, therefore it cannot be written in the form $\frac{p}{q};q\neq 0$,with p,q both as integers.
Thus, 1.010010001 ……….. is an irrational number.

NCERT Exemplar Class 9 Maths Solutions Chapter-1: Exercise-1.3

Question:1

(i)Find whether variable x represent rational number or an irrational number.x2 = 5
(ii)Find whether variable x represent rational number or an irrational number.y2 = 9
(iii)Find whether variable x represent rational number or an irrational number.z2 = 0.04
(iv)Find whether variable x represent rational number or an irrational number.u2 = $\frac{17}{4}$

Solution.

Given that
x2 = 5
On taking square root on both sides, we get
$\Rightarrow x= \pm \sqrt{5}$
Irrational numbers are real numbers which cannot be represented as simple fractions.
So, x is an irrational number

Solution.

We have
y2 = 9
On taking square root on both sides, we get
$\Rightarrow x= \pm 3$
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q
should be rational when the fraction is expressed in the simplest form.
So, y can be written as $\frac{3}{1},\frac{-3}{1}$ where 1, 3, -3 are rational numbers.
Hence y is rational.
Solution.

We have
z2 = 0.04 =$\frac{4}{100}$
On taking square root on both sides, we get
$\Rightarrow z= \pm \frac{2}{10}= \pm \frac{1}{5}$
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q
should be rational when the fraction is expressed in the simplest form.
So, z can be written as $\pm \frac{1}{5}$ where 1, 5, -5 are rational numbers.
Hence z is rational.
Solution.

Given that
u2 = $\frac{17}{4}$
On taking square root on both sides, we get
$\Rightarrow u= \sqrt{\frac{17}{4}}= \pm \frac{\sqrt{17}}{2}$
Any number which can be represented in the form of p/q where q is not equal to zero
is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
So, u can be written as $\pm \frac{\sqrt{17}}{2}$ where $\sqrt{17}$ is irrational.
Hence u is irrational.

Question:2

(i) Find three rational numbers between – 1 and – 2
(ii) Find three rational numbers between 0.1 and 0.11
(iii)Find three rational numbers between $\frac{5}{7}$ and $\frac{6}{7}$
(iv)Find three rational numbers between $\frac{1}{4}$ and $\frac{1}{5}$

$-\frac{11}{10}$, $-\frac{6}{5}$
and$-\frac{5}{4}$
Solution.

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.

Between -1 and -2, many rational number can be written as:
$-1\cdot 1=-\frac{11}{10}$
$-1\cdot 2=-\frac{12}{10}= -\frac{6}{5}$
$-1\cdot 25=-\frac{125}{100}= -\frac{5}{4}$
$-1\cdot 3=-\frac{13}{10}$
$-1\cdot 4=-\frac{14}{10}= -\frac{7}{5}$
(ii) Answer: $\frac{103}{1000}$,$\frac{104}{1000}$,$\frac{105}{1000}$
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.

Between 0.1 and 0.11, many rational number can be written as:

0.103 = $\frac{103}{1000}$

0.104 = $\frac{104}{1000}$

0.105 = $\frac{105}{1000}$

(iii)Answer. $\frac{51}{70},\frac{52}{70}$and$\frac{53}{70}$
Solution.

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
We can write $\frac{5}{7}$as $\frac{5\times 10}{7\times 10}= \frac{50}{70}$ and $\frac{6}{7}$ as $\frac{6\times 10}{7\times 10}= \frac{60}{70}$
So, three rational number between $\frac{5}{7}and\frac{6}{7}$ are $\frac{51}{70},\frac{52}{70}$and$\frac{53}{70}$

(iv)Answer. $\frac{41}{200},\frac{42}{200},\frac{43}{200}$
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
L.C.M. of 4 and 5 is 20.
We can write $\frac{1}{4}$ as $\frac{1\times 40}{4\times 50}= \frac{50}{200}$
and $\frac{1}{5}\, as\, \frac{1\times 40}{5\times 40}= \frac{40}{200}$
So, three rational number between $\frac{1}{4}\, and\,\frac{1}{5}$ are
$\frac{41}{200},\frac{42}{200},\frac{43}{200}$

Question:3

(i) Answer. Rational number: $\frac{5}{2}$
Irrational number: 2.040040004 ……….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 2 and 3
Rational number: 2.5 = $\frac{25}{10}$$= \frac{5}{2}$
and irrational number : 2.040040004

(ii) Answer. Rational number: $\frac{19}{1000}$
Irrational number 0.0105000500005 ……..
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 0 and 0.1:
0.1 can be written as 0.10
Rational number: 0.019 = $\frac{19}{1000}$
and irrational number 0.0105000500005

(iii)Answer. Rational number $\frac{21}{60}$
Irrational number : 0.414114111 ……
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions. They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
LCM of 3 and 2 is 6.
We can write $\frac{1}{3}$ as $\frac{1\times 20}{3\times 20}= \frac{20}{60}$
and $\frac{1}{2}$ as $\frac{1\times 30}{3\times 30}= \frac{30}{60}$
Also, $\frac{1}{3}$ = 0.333333….
And $\frac{1}{2}= 0\cdot 5$
So, rational number between $\frac{1}{3}$ and $\frac{1}{2}$ is $\frac{21}{60}$
and irrational number : 0.414114111 ……

Irrational number: 0.151551555 …….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
$\frac{-2}{5}= -0\cdot 4$ and $\frac{1}{2}= -0\cdot 5$
Rational number between -0.4 and 0.5 is 0
And irrational number: 0.151551555 …….
(v) Answer. Rational number: $\frac{151}{1000}$
Irrational number: 0.151551555 ……….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 0.15 and 0.16
Rational number : 0.151 = $\frac{151}{1000}$
and irrational number 0.151551555

(vi) Answer. Rational number: $\frac{3}{2}$
Irrational number: 1.585585558 ………
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions. They are non-terminating non-recurring in nature.
Between $\sqrt{2}\, and\, \sqrt{3}$
$\sqrt{2}= 1\cdot 414213562373$
$\sqrt{3}= 1\cdot 732050807568$
Rational number: 1.5 = $\frac{3}{2}$
and irrational number: 1.585585558

Irrational number: 3.101101110………
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 2.357 and 3.121
Rational number: 3
Irrational number: 3.101101110………
(viii) Answer. Rational number: $\frac{2}{10000}$
Irrational number: 0.000113133133 ……….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 0.0001 and 0.001
Rational number: 0.0002 = $\frac{2}{10000}$
Irrational number: 0.000113133133

Irrational number: 1.909009000 ……
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 3.623623 and 0.484848
A rational number between 3.623623 and 0.484848 is 1.
An irrational number between 3.623623 and 0.484848 is 1.909009000 ……

(x) Answer. A rational number is $\frac{63753}{10000}$
An irrational number is 6.375414114111……..
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 6.375289 and 6.375738:
A rational number is 6.3753 = $\frac{63753}{10000}$
An irrational number is 6.375414114111……..

Question:4

Represent the following numbers on the number line. 7, 7.2, $-\frac{3}{2},-\frac{12}{5}$

Solution.
Firstly we draw a number line whose mid-point is O. Mark positive numbers on right hand side of O and negative numbers on left hand side of O.
(i) Number 7 is a positive number. So we mark a number 7 on the right
hand side of O, which is at a 7 units distance form zero.
(ii) Number 7.2 is a positive number. So, we mark a number 7.2 on the right hand side of O, which is 7.2 units distance from zero.
(iii) Number $-\frac{3}{2}$ or – 1.5 is a negative number so, we mark a number 1.5 on the left hand side of zero, which is at 1.5 units distance from zero.
(iv) Number $-\frac{12}{5}$ or –2.4 is a negative number. So, we mark a number 2.4 on the left hand side of zero, which is at 2.4 units distance from zero.

Question:5

Locate $\sqrt{5},\sqrt{10}$ and $\sqrt{17}$ on the number line

Hint.
Solution.

Step I- Draw number line shown in the figure.
Let the point O represent 0 (zero) and point A represent 2 units from O.

Step II- Draw perpendicular AX from A on the number line and cut off arc AB = 1 unit
We have OA = 2 units and AB = 1 unit
Using Pythagoras theorem, we have.
OB2 = OA2 + AB2
OB2 = (2)2 + (1)2 = 5
OB = $\sqrt{5}$
Taking O as the centre and OB = $\sqrt{5}$ as radius draw an arc cutting the line at C.
Clearly, OC = OB = $\sqrt{5}$.
Hence, C represents $\sqrt{5}$ on the number line.
Similarly for $\sqrt{10}$ and $\sqrt{17}$we can plot the points as follows:
$\sqrt{10}= \sqrt{3^{2}+1^{2}}$
$\sqrt{17}= \sqrt{4^{2}+1^{2}}$

Question:6

(i) Represent geometrically the following numbers on the number line : $\sqrt{4\cdot 5}$
(ii) Represent geometrically the following numbers on the number line : $\sqrt{5\cdot 6}$
(iii) Presentation of $\sqrt{8\cdot 1}$ on number line :
(iv) Presentation of $\sqrt{2\cdot 3}$ on number line:

(i) Solution. AB = 4.5 units, BC = 1 unit

OC = OD = $\frac{5\cdot 5}{2}$ = 2.75 units
OD2 = OB2 + BD2
$\left ( \frac{4\cdot 5}{2} \right )^{2}= \left ( \frac{4\cdot 5}{2} -1\right )^{2}+\left ( BD \right )^{2}$
$\Rightarrow BD^{2}= \left ( \frac{4\cdot 5+1}{2} \right )^{2}- \left ( \frac{4\cdot 5-1}{2} \right )^{2}$
$\Rightarrow BD^{2}= 4. 5$
$\Rightarrow BD= \sqrt{4. 5}$
So the length of BD will be the required one so mark an arc of length BD on number line, this will result in the required length.

(ii) Solution. Presentation of $\sqrt{5. 6}$on number line.
Mark the distance 5.6 units from a fixed point A on a given line to obtain a point B such that AB = 5.6 units. From B mark a distance of 1 unit and mark a new point C. Find the mid point of AC and mark that point as O. Draw a semicircle with center O and radius OC. Draw a line
perpendicular to AC passing through B and intersecting the semicircle at O. Then BD = $\sqrt{5\cdot 6}$

(iii) Solution
Mark the distance 8.1 units from a fixed point A on a given line to obtain a point B such that AB = 8.1 units. From B mark a distance of 1 unit and mark the new point AB. Find the mid point of AC and mark a point as O. Draw a semi circle with point O and radius OC. Draw a line perpendicular to AC passing through B and intersecting
the semicircle at D. Then BD -
$\sqrt{8. 1}$

(iv) Solution
Mark the distance 2.3 unit from a fixed point A on a given line. To obtain a point B such that AB = 2.3 units. From B mark a distance of 1 unit and mark a new point as C. Find the mid point of AC and mark the point asO. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D. Then BD = $\sqrt{2. 3}$

Question:7

(i) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$. 0.2
(ii) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$.
0.888…….
(iii) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$.$5.\bar{2}$
(iv) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$.$0\cdot \overline{001}$
(v) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$.0.2555……
(vii) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$..00323232…..
(viii) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$..404040……..

(i) Answer.$\frac{1}{5}$
Solution. We know that
0.2 can be written as $\frac{2}{10}$
Now,
$\frac{2}{10}= \frac{1}{5}$
Hence the answer is $\frac{1}{5}$

(ii) Answer.$\frac{8}{9}$
Solution. Let x = 0.888….. .…(i)
Multiply RHS and LHS by 10
10 x = 8.88……. …(ii)
Subtracting equation (i) from (ii)
We get
10x – x = 8.8 – 0.8
$\Rightarrow 9x= 8$
$\Rightarrow x= \frac{8}{9}$
Hence answer is $\frac{8}{9}$

(iii) Answer.$\frac{47}{9}$
Solution. Let x = $5\cdot \bar{2}$ …eq. (1)
Multiply by 10 on both sides
10x = $52\cdot \bar{2}$ …eq (2)
Subtracting equation (1) from (2)
We get
10x – x = $52\cdot \bar{2}$$5\cdot \bar{2}$
$\Rightarrow$ 9x = 47
$\Rightarrow$ x = $\frac{47}{9}$
Hence the answer is $\frac{47}{9}$

(iv) Answer.$\frac{1}{999}$
Solution. Let x = $0\cdot \overline{001}$ …. Eq. (1)
Multiply by 1000 on both sides
1000 x = $1\cdot \overline{001}$ …eq.(2)
Subtracting eq. (1) from (2)
We get
1000 x – x = $1\cdot \overline{001}$$1\cdot \overline{001}$
$\Rightarrow$ 999x = 1
$\Rightarrow$ x = $\frac{1}{999}$
Hence the answer is $\frac{1}{999}$
(v) Answer.$\frac{23}{90}$
Solution. Let x = 0.2555 ….. …eq.(1)
Multiply by 10 on both sides
10x = 2.555… …eq.(2)
Multiply by 100 on both sides
100x = 25.55… …eq. (3)
Subtracting eq. (2) from (3),
We get
100x – 10x = 25.555… – 2.555…
$\Rightarrow$ 90x = 23
$\Rightarrow$ x = $\frac{23}{90}$
Hence the answer is $\frac{23}{90}$

(vii) Answer.$\frac{8}{2475}$
Solution. Let x = 0.00323232….. …eq.(1)
Multiply with 100 on both sides.
We get
100x = 0.3232… …eq(2)
Multiply again with 100 on both sides,
10000x = 32.3232… …eq(3)
Equation (3) – (2)
we get,
10000 x – 100x = 32.3232… – 0.3232…
$\Rightarrow$ 9900x = 32
$\Rightarrow$ x = $\frac{32}{9900}$
x = $\frac{8}{2475}$
Hence the answer is $\frac{8}{2475}$

(viii) Answer.$\frac{40}{99}$
Solution. Let x = 0.404040……. …(1)
Multiplying by 100 on both sides
we get
100x = 40.40… …(2)
Subtracting equation (1) from (2)
we get
100x – x = 40.40 – 0.40
$\Rightarrow$99x = 40
$\Rightarrow$ x = $\frac{40}{99}$
Hence the answer is $\frac{40}{99}$

Question:8

Show that 0.142857142857 ……. = $\frac{1}{7}$

Solution. Let, x = 0.142857142857… …(i)
Multiply (i) by 1000000, we get
1000000 x = 142857.142857… …(ii)
Subtracting equation (i) from (ii), we get
1000000 x – x = 142857.142857… - 0.142857…
999999 x = 142857
$x= \frac{142857}{999999}= \frac{47619}{333333}$$= \frac{15873}{111111}= \frac{5291}{37037}$
$= \frac{481}{3367}= \frac{1}{7}$
Hence, 0.142857 ……. $\frac{1}{7}$
Hence proved

Question:9

(i) Simplify the following : $\sqrt{45}-3\sqrt{20}+4\sqrt{5}$
(ii) Simplify the following : $\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}$
(iii) Simplify the following : $4\sqrt{12}\times 7\sqrt{6}$
(iv) Simplify the following : $4\sqrt{28}\div 3\sqrt{7}\div 3\sqrt{7}$
(v) Simplify the following : $3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}$
(vi) Simplify the following : $\left ( \sqrt{3}-\sqrt{2} \right )^{2}$
(vii) Simplify the following : $\sqrt[4]{81}- 8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{255}$
(viii) Simplify the following : $\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$
(ix) Simplify the following : $\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}$

(i) Answer.$\sqrt{5}$
Solution. $\sqrt{45}-3\sqrt{20}+4\sqrt{5}$
We know that,
45 = $3\times 3\times 5$
20 = $2\times 2\times 5$
So we get
$\sqrt{3\times3\times5 }-3\sqrt{2\times2\times5}+4\sqrt{5}$
$= 3\sqrt{5}-3\left ( 2\sqrt{5} \right )+4\sqrt{5}$
$= 3\sqrt{5}-6\sqrt{5}+4\sqrt{5}$
$= 7\sqrt{5}-6\sqrt{5}$
$= \sqrt{5}$
Hence the answer is $\sqrt{5}$

(ii) Answer. $\frac{7\sqrt{6}}{12}$
Solution. We have, $\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}$
We know that,
$24= 6\times 4= 3\times 2\times 2\times 2$
$54= 9\times 6= 3\times 3\times 3\times 2$
So we get
$\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}= \frac{2\sqrt{6}}{8}+\frac{3\sqrt{6}}{9}$
$= \frac{\sqrt{6}}{4}+\frac{\sqrt{6}}{3}$
Taking LCM (3,4) = 12
$= \frac{3\sqrt{6}+4\sqrt{6}}{12}$
$= \frac{7\sqrt{6}}{12}$
(iii) Answer. $\sqrt[28]{2^{18} \times 3^{11}}$
Solution. We have
$\sqrt[4]{12}\times \sqrt[7]{6}$
We know that
12 = $2\times 2\times 3$
6 = $2\times 3$
So we get,
=$\sqrt[4]{2\times 2\times 3}\times \sqrt[7]{2\times 3}$
$=2^{1 / 4} \cdot 2^{1 / 4} \cdot 3^{1 / 4} \cdot 2^{1 / 7} \cdot 3^{1 / 7}$
$=2^{\frac{1}{4}+\frac{1}{4}+\frac{1}{7}} \times 3^{\frac{1}{4}+\frac{1}{7}}$
$=2^{9 / 14} \times 3^{11 / 28}$
=$\sqrt[28]{2^{18} \times 3^{11}}$
Hence the number is $\sqrt[28]{2^{18} \times 3^{11}}$.

(iv)Answer. $\frac{8}{\left ( 3\times \sqrt[3]{7} \right )}$
Solution. We have, $4\sqrt{28}\div 3\sqrt{7}\div \sqrt[3]{7}$
We know that
28 = $4\times 7$
So we can write,
$4\sqrt{28}\div 3\sqrt{7}\div \sqrt[3]{7}= \left [ \frac{4\sqrt{28}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}$
= $\left [ \frac{4\sqrt{4\times 7}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}$
= $\left [ \frac{4\times 2\sqrt{ 7}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}$
$= \frac{8}{3}\div 7^{\frac{1}{3}}$
$= \frac{8}{\left ( 3\times 7^{\frac{1}{3}} \right )}$
$= \frac{8}{\left ( 3\times \sqrt[3]{7} \right )}$
Hence the answer is $\frac{8}{\left ( 3\times \sqrt[3]{7} \right )}$

(v) Answer.$3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}$
We know that
27 = $3\times 3\times 3$
So, $3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}$ = $3\sqrt{3}+2\sqrt{3\times 3\times 3}+\frac{7}{\sqrt{3}}$

$= 3\sqrt{3}+2\left ( 3\sqrt{3} \right )+\frac{7}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
(Rationalising the denominator)
$= 3\sqrt{3}+6\left ( \sqrt{3} \right )+\frac{7\sqrt{3}}{3}$

$= \left ( 3+6+\frac{7}{3} \right )\sqrt{3}$ (Taking $\sqrt{3}$ common)
Now LCM (1,1,3) = 3
$= \left ( \frac{9+18+7}{3} \right )\sqrt{3}$
$= \frac{34}{3}\sqrt{3}$
$= 19\cdot 63$
(vi) Answer.$5-2\sqrt{6}$
Solution. Given, $\left ( \sqrt{3}-\sqrt{2} \right )^{2}$
We know that (a + b)2 = a2 – 2ab + b2
Comparing the given equation with the identity, we get:
$\left ( \sqrt{3}-\sqrt{2} \right )^{2}= \left ( \sqrt{3} \right )^{2}-2\left ( \sqrt{3} \right )\left ( \sqrt{2} \right )+\left ( \sqrt{2} \right )^2$
= 3 + 2 – $2\sqrt{3\times 2}$
$= 5-2\sqrt{6}$
Hence the answer is $5-2\sqrt{6}$

Solution. We have, $\sqrt[4]{81}-8 \sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}$
We know that
81 = $3\times 3\times3\times3$
216 = $6\times 6\times6$
32 = $2\times 2\times2\times2\times2$
225 = $15\times 15$
So,$\sqrt[4]{81}-8 \sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}$
$= \sqrt[4]{3\times3\times3\times3 }-8\sqrt[3]{6\times6\times6}+15\sqrt[5]{2\times2\times2\times2\times2}+\sqrt{15\times15}$
= 3 – 8 × 6 + 15 × 2 + 15
= 3 – 48 + 30 + 15
= – 45 + 45
= 0

(viii) Answer. $\frac{5}{2\sqrt{2}}$
Solution. We have, $\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$
We know that, 8 =$2\times 2\times 2$
So,
$\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$ $= \frac{3}{\sqrt{2\times 2\times 2}}+\frac{1}{\sqrt{2}}$
$= \frac{3}{2\sqrt{2}}+\frac{1}{\sqrt{2}}$
$= \frac{3}{2\sqrt{2}}+\frac{2}{2\sqrt{2}}$
$= \frac{5}{2\sqrt{2}}$
Hence the answer is $\frac{5}{2\sqrt{2}}$

(ix) Answer. $\frac{\sqrt{3}}{2}$
Solution. We have, $\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}$
LCM (3,6) = 6
$\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}= \frac{4\sqrt{3}}{6}-\frac{\sqrt{3}}{6}$
$= \frac{4\sqrt{3}-\sqrt{3}}{6}$
$= \frac{3\sqrt{3}}{6}$
$= \frac{\sqrt{3}}{2}$
Hence the answer is $\frac{\sqrt{3}}{2}$.

Question:10

(i) Rationalise the denominator of the following : $\frac{2}{3\sqrt{3}}$
(ii)Rationalise the denominator of the following : $\frac{\sqrt{40}}{\sqrt{3}}$
(iii) Rationalise the denominator of the following : $\frac{3+\sqrt{2}}{4\sqrt{2}}$
(iv)Rationalise the denominator of the following :$\frac{16}{\sqrt{41}-5}$
(v) Rationalise the denominator of the following : $\frac{2+\sqrt{3}}{2-\sqrt{3}}$
(vi) Rationalise the denominator of the following : $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$
(vii) Rationalise the denominator of the following : $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
(viii) Rationalise the denominator of the following : $\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$
(ix) Rationalise the denominator of the following : $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$

(i) Answer.$\frac{2\sqrt{3}}{9}$
Solution. We have, $\frac{2}{3\sqrt{3}}$
Rationalising the denominator, we get:
$\frac{2}{3\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{2\sqrt{3}}{3\sqrt{3}\sqrt{3}}$
$= \frac{2\sqrt{3}}{9}$
Hence the answer is $\frac{2\sqrt{3}}{9}$

(ii) Answer.$\frac{2\sqrt{30}}{3}$
Solution. We have ,$\frac{\sqrt{40}}{\sqrt{3}}$
We know that, 40 = (2) (2) (10)
$\frac{\sqrt{40}}{\sqrt{3}}= \frac{\sqrt{2\cdot 2\cdot 10}}{\sqrt{3}}= \frac{2\sqrt{10}}{\sqrt{3}}$
Rationalising the denominator, we get:
$= \frac{2\sqrt{10}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{2\sqrt{10}\sqrt{3}}{\sqrt{3}\sqrt{3}}$
$= \frac{2\sqrt{30}}{3}$
Hence the answer is: $\frac{2\sqrt{30}}{3}$

(iii) Answer. $\frac{3\sqrt{2}+2}{8}$
Solution. We have $\frac{3+\sqrt{2}}{4\sqrt{2}}$
Rationalising the denominator, we get:
$\frac{3+\sqrt{2}}{4\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$
$= \frac{\left ( 3+\sqrt{2} \right )\sqrt{2}}{4\sqrt{2}\sqrt{2}}$
$= \frac{3\sqrt{2}+2}{8}$
Hence the answer is $\frac{3\sqrt{2}+2}{8}$

(iv) Answer. $\sqrt{41}+5$
Solution. We have $\frac{16}{\sqrt{41}-5}$
Rationalising the denominator, we get:
$\frac{16}{\sqrt{41}-5}\times \frac{\sqrt{41}+5}{\sqrt{41}+5}$
$= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41}-5 \right )\sqrt{41}+5}$
Using the identity (a – b) (a + b) = a2 – b2
We get:
$= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41} \right )^{2}-\left ( 5 \right )^{2}}$
$= \frac{16\left ( \sqrt{41}+5 \right )}{41-25}$
$= \frac{16\left ( \sqrt{41}+5 \right )}{16}$
$= \sqrt{41}+5$
Hence the answer is $\sqrt{41}+5$

(v) Answer.$7+4\sqrt{3}$
Solution. We have, $\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Rationalising the denominator, we get:
$\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2-\sqrt{3}}$
$= \frac{\left ( 2+\sqrt{3} \right )^{2}}{\left ( 2-\sqrt{3} \right )\left ( 2+\sqrt{3} \right )}$
Using (a – b) (a + b) = a2 – b2
and (a + b)2 = a2 + b2 + 2ab
$= \frac{2^{2}+\left ( \sqrt{3} \right )^{2}+2\cdot 2\cdot \sqrt{3}}{2^{2}-\left ( \sqrt{3} \right )^{2}}$
$= \frac{4+3+4\sqrt{3}}{4-3}$
$= 7+4\sqrt{3}$
Hence the answer is $7+4\sqrt{3}$
(vi)Answer. $3\sqrt{2}-2\sqrt{3}$
Solution. We have, $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$
Rationalising the denominator, we get:
$= \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$= \frac{\sqrt{6}\left ( \sqrt{3}-\sqrt{2} \right )}{\left ( \sqrt{2}+\sqrt{3} \right )\left ( \sqrt{3}-\sqrt{2} \right )}$
Using the identity (a – b) (a + b) = a2 – b2
We get:
$= \frac{\sqrt{18}-\sqrt{12}}{\left ( \sqrt{3} \right )^{2}\left ( \sqrt{2} \right )^{2}}$
$= \frac{\sqrt{3\cdot 3\cdot 2}-\sqrt{2\cdot 2\cdot 3}}{3-2}$
$= \frac{3\sqrt{2}-2\sqrt{3}}{1}$
$= 3\sqrt{2}-2\sqrt{3}$
Hence the answer $3\sqrt{2}-2\sqrt{3}$
(vii) Answer. $5+2\sqrt{6}$
Solution. We have, $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
Rationalising the denominator, we get:
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$= \frac{\left ( \sqrt{3} +\sqrt{2}\right )^{2}}{\left ( \sqrt{3} -\sqrt{2} \right )\left ( \sqrt{3} +\sqrt{2} \right )}$
Using (a – b) (a + b) = a2 – b2
and (a + b)2 = a2 + b2 + 2ab
$= \frac{\left ( \sqrt{3} \right )^{2}+\left ( \sqrt{2} \right )^{2}+2\sqrt{3}\sqrt{2}}{\left ( \sqrt{3} \right )^{2}-\left ( \sqrt{2} \right )^{2}}$
$= \frac{3+2+2\sqrt{6}}{3-2}$
$= 5+2\sqrt{6}$
Hence the answer is $5+2\sqrt{6}$
(viii)
Answer:$9+2 \sqrt{15}$
Solution:
We have $\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$
Rationalize
$=\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{3 \sqrt{5}(\sqrt{5}+\sqrt{3})+\sqrt{3}(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}$
$=\frac{15+3 \sqrt{15}+\sqrt{15}+3}{5-3}=\frac{18+4 \sqrt{15}}{2}$
$=9+2 \sqrt{15}$
(ix) Answer:$\frac{9+4 \sqrt{6}}{15}$
Solution:
We have $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$
$=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{16 \times 3}+\sqrt{9 \times 2}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}}$
Rationalize
$=\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}} \times \frac{(4 \sqrt{3}-3 \sqrt{2})}{(4 \sqrt{3}-3 \sqrt{2})}$
$=\frac{4 \sqrt{3}(4 \sqrt{3}-3 \sqrt{2})+5 \sqrt{2}(4 \sqrt{3}-3 \sqrt{2})}{(4 \sqrt{3})^{2}-(3 \sqrt{2})^{2}}$
$=\frac{48-12 \sqrt{6}+20 \sqrt{6}-30}{30}$
$=\frac{18+8 \sqrt{6}}{30}$
$=\frac{9+4 \sqrt{6}}{15}$

Question:11

(i) Find the values of a in each of the following : $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}= a-6\sqrt{3}$
(ii) Find the values of a in the following : $\frac{3-\sqrt{5}}{3+2\sqrt{5}}= a\sqrt{5}-\frac{19}{11}$
(iii) Find the values of b in the following : $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}$
(iv) Find the values of a and b in the following : $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}$

Solution. We have, $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}= a-6\sqrt{3}$
LHS = $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$
Rationalising the denominator, we get:
$= \frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}$
$= \frac{\left ( 5+2\sqrt{3} \right )\left ( 7-4\sqrt{3} \right )}{\left ( 7+4\sqrt{3} \right )\left ( 7-4\sqrt{3} \right )}$
{Using (a – b) (a + b) = a2 – b2}

$= \frac{35+14\sqrt{3}-20\sqrt{3}-24}{7^{2}-\left ( 4\sqrt{3} \right )^{2}}$
$= \frac{11-6\sqrt{3}}{49-48}$
$= 11-6\sqrt{3}$
Now RHS $= a-6\sqrt{3}$
$\Rightarrow 11-6\sqrt{3}= a-6\sqrt{3}$
$\Rightarrow 11-6\sqrt{3}= a-6\sqrt{3}$
$\Rightarrow a= 11$
Hence a = 11 is the required answer

(ii)Answer.$a= \frac{9}{11}$
Solution. Given that, $\frac{3-\sqrt{5}}{3+2\sqrt{5}}= a\sqrt{5}-\frac{19}{11}$
LHS = $\frac{3-\sqrt{5}}{3+2\sqrt{5}}$
Rationalising the denominator, we get:
LHS $= \frac{3-\sqrt{5}}{3+2\sqrt{5}}\times \frac{3-2\sqrt{5}}{3-2\sqrt{5}}$
{Using (a – b) (a + b) = a2 – b2}
$= \frac{9-3\sqrt{5}-6\sqrt{5}+10}{3^{2}-\left ( 2\sqrt{5} \right )^{2}}$
$= \frac{19-9\sqrt{5}}{9-20}$
Now RHS $= a\sqrt{5}-\frac{19}{11}$
$\Rightarrow \frac{9\sqrt{5}}{11}-\frac{19}{11}= a\sqrt{5}-\frac{19}{11}$
Comparing both , we get
$\Rightarrow a= \frac{9}{11}$
Hence $a= \frac{9}{11}$is the correct answer
(iii) Answer:$b = -\frac{5 }{6}$
Solution:
Given:
$\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}$
LHS = $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}$
Rationalize
= $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}} \times \frac{3 \sqrt{2}+2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}$
=$\frac{\sqrt{2}(3 \sqrt{2}+2 \sqrt{3})+\sqrt{3}(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2})^{2}-(2 \sqrt{3})^{2}}$
= $\frac{6+2 \sqrt{6}+3 \sqrt{6}+6}{18-12}$
= $2+\frac{5 \sqrt{6}}{6}$
$2+\frac{5 \sqrt{6}}{6}=2-b \sqrt{6}$
$b = -\frac{5 }{6}$

(iv) Answer. a = 0, b = 1
Solution. Given,
$\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}$
LHS $= \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}$
$=\frac{\left ( 7+\sqrt{5} \right )\times\left ( 7+\sqrt{5} \right )-\left ( 7-\sqrt{5} \right ) \times \left ( 7-\sqrt{5} \right )}{\left (7 -\sqrt{5} \right )\left (7 +\sqrt{5} \right )}$
Using (a – b) (a + b) = a2 – b2
(a + b)2 = a2 + b2 + 2ab
(a - b)2 = a2 + b2 - 2ab
$= \frac{\left ( 7^{2} +\sqrt{5}^{2}+2\cdot 7\cdot \sqrt{5}\right )-\left ( 7^{2} +\sqrt{5}^{2}-2\cdot 7\cdot \sqrt{5} \right )}{7^{2}-\sqrt{5}^{2}}$
$= \frac{\left ( 49+5+14\sqrt{5} \right )-\left ( 49+5-14\sqrt{5} \right )}{49-5}$
$= \frac{54+14\sqrt{5}-54+14\sqrt{5}}{44}$
$= \frac{28\sqrt{5}}{44}$
RHS $= a+\frac{7}{11}\sqrt{5b}$
Now LHS = RHS
$\Rightarrow \frac{28\sqrt{5}}{44}= a+\frac{7}{11}\sqrt{5b}$
$\Rightarrow 0+\left ( \frac{4}{4} \right )\frac{7}{11}\sqrt{5}= a+\frac{7}{11}\sqrt{5b}$
$\Rightarrow$ a = 0, b = 1
Hence the answer is a = 0, b = 1

Question:12

If a = 2 + $\sqrt{3}$ then find the value of $a-\frac{1}{a}$

Answer. $2\sqrt{3}$
Solution. Given that a = $2\sqrt{3}$
$\therefore$ We have $\frac{1}{a}= \frac{1}{2+\sqrt{3}}$
Rationalising,
$\Rightarrow \frac{1}{a}= \frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
Using (a – b) (a + b) = a2 – b2
$\Rightarrow \frac{1}{a}= \frac{2-\sqrt{3}}{2^{2}-\sqrt{3}^{2}}= \frac{2-\sqrt{3}}{4-3}$
$\Rightarrow \frac{1}{a}=2-\sqrt{3}$
Now, $a- \frac{1}{a}=2+\sqrt{3}-\left ( 2-\sqrt{3} \right )$
$\Rightarrow a- \frac{1}{a}=2\sqrt{3}$
Hence the answer is $2\sqrt{3}$

Question:13

(i) Rationalize the denominator in each of the following and hence evaluate by taking $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$ and $\sqrt{5}= 2\cdot 236$, upto three places of decimal : $\frac{4}{\sqrt{3}}$
(ii) Rationalize the denominator in each of the following and hence evaluate by taking $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$ and $\sqrt{5}= 2\cdot 236$, upto three places of decimal : $\frac{6}{\sqrt{6}}$
(iii) Rationalize the denominator in each of the following and hence evaluate by taking $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$ and $\sqrt{5}= 2\cdot 236$, upto three places of decimal : $\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}$
(iv) Rationalize the denominator in each of the following and hence evaluate by taking $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$ and $\sqrt{5}= 2\cdot 236$, upto three places of decimal : $\frac{2}{2+\sqrt{2}}$
(v) Rationalize the denominator in each of the following and hence evaluate by taking $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$ and $\sqrt{5}= 2\cdot 236$, upto three places of decimal : $\frac{1}{\sqrt{3}+\sqrt{2}}$

Solution. Given: $\frac{4}{\sqrt{3}}$
Rationalising,
$\frac{4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}= \frac{4\sqrt{3}}{3}$
(Given that $\sqrt{3}= 1.732$)
$= \frac{4\times 1\cdot 732}{3}$
= 2.3093

Solution. Given: $\frac{6}{\sqrt{6}}$
Rationalising,
$\frac{6}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}= \frac{6\sqrt{6}}{6}$
$= \frac{6\times \sqrt{2}\sqrt{3}}{6}$
Putting the given values,
$\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$
We get :
$= \sqrt{2}\cdot \sqrt{3}$
$= 1\cdot 414\times 1\cdot 732= 2\cdot 449$

Solution. Given that $\frac{\sqrt{10}-\sqrt{5}}{2}$
This can be written as
$\frac{\sqrt{2}\times \sqrt{5}-\sqrt{5}}{2}$
Now putting the given values,
$\sqrt{2}= 1\cdot 414,\sqrt{5}= 2\cdot 236$
We get :
$\Rightarrow \frac{1\cdot 414\times 2\cdot 236-2\cdot 236}{2}$
= 0.462852
Solution. Given: $\frac{\sqrt{2}}{2+\sqrt{2}}$
Rationalising,
$\frac{\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}$
Using (a – b) (a + b) = a2 – b2
$= \frac{\sqrt{2}\left ( 2-\sqrt{2} \right )}{2^{2}-\sqrt{2}^{2}}$
$= \frac{2\sqrt{2}-2}{4-2}$
$= \frac{2\left ( \sqrt{2}-1 \right )}{2}$
$= \sqrt{2}-1$

Putting the given value of $\sqrt{2}= 1\cdot 414$
We get
= 1.414 – 1
= 0.414

Solution. Given that $\frac{1}{\sqrt{3}+\sqrt{2}}$
Rationalising,
$\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
Using (a – b) (a + b) = a2 – b2
$= \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}^{2}-\sqrt{2}^{2}}$
$= \frac{\sqrt{3}-\sqrt{2}}{3-2}$
$= \sqrt{3}-\sqrt{2}$

Putting the given values, $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$
We get,
= 1.732 – 1.414
= 0.318

Question:14

Solution. (13 + 23 + 33)1/2
We know that
13 = 1.1.1 = 1
23 = 2.2.2 = 8
33 = 3.3.3 = 27
Putting these values we get
(13 + 23 + 33)1/2$= \sqrt{1+8+27}$
$=\sqrt{36}= 6$

(ii) Answer.$\frac{2025}{64}$
Solution.$\left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}$
We know that
8 = 2.2.2 = 23
32 = 2.2.2.2.2 = 25
$\left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}= \left ( \frac{3}{5} \right )^{4}\left ( \frac{2^{3}}{5} \right )^{-12}\left ( \frac{2^{5}}{5} \right )^{6}$
$= \frac{3^{4}\left ( 2^{3} \right )^{-12}\left ( 2^{5} \right )^{6}}{5^{4}5^{-12}5^{6}}$ $\because \left ( \frac{a}{b} \right )^{m}= \frac{a^{m}}{b^{m}}$
$= \frac{3^{4}2^{-36}2^{30}}{5^{4}5^{-12}5^{6}}$ $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$
$= \frac{3^{4}\times 2^{-36+30}}{5^{4-12+6}}$ $\because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}$
$= \frac{3^{4}\times 2^{-6}}{5^{-2}}$
$= \frac{3^{4}\times 5^{2}}{2^{6}}$ $\because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}$
$= \frac{81\times 25}{64}$
$= \frac{2025}{64}$
Hence the answer is $\frac{2025}{64}$

Solution. Given $\left ( \frac{1}{27} \right )^{-\frac{2}{3}}$
We know that
27 = 3.3.3 = 33
$\left ( \frac{1}{27} \right )^{-\frac{2}{3}}= \left ( \frac{1}{3^{3}} \right )^{-\frac{2}{3}}$
$= \left ( 3^{3} \right )^{\frac{2}{3}}$ $\because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}$
$= \left ( 3 \right )^{3\times \frac{2}{3}}$ $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$

= 32 = 9

Solution. Given $\left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}$
We know that
$625= \left ( 25 \right )\left ( 25 \right )= 5\cdot 5\cdot 5\cdot 5= 5^{4}$
$\left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}= \left [ \left \{ \left ( \left ( 5 \right )^{4} \right )^{-\frac{1}{2}} \right \} ^{-\frac{1}{4}}\right ]^{2}$
$= 5^{4\times \frac{-1}{2}\times \frac{-1}{4}\times 2}$ $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$
= 51 = 5

(v) Answer.$\sqrt[3]{\frac{1}{3}}$
Solution. We have $\frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}$
Now we know that
9 = 3.3 = 32
27 = 3.3.3 = 33
$\frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}= \frac{\left ( 3^{2} \right )^{\frac{1}{3}}\times \left ( 3^{3} \right )^{\frac{1}{2}}}{\left ( 3 \right )^{\frac{1}{6}}\times \left ( 3^{2} \right )^{\tfrac{-2}{3}}}$
$= \frac{\left ( 3\right )^{2\times \frac{1}{3}}\times \left ( 3 \right )^{3\times \frac{-1}{2}}}{\left ( 3 \right )^{\frac{1}{6}}\times \left ( 3 \right )^{\tfrac{-2}{3}}}$ $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$
$= \frac{\left ( 3 \right )^{\frac{2}{3}-\frac{3}{2}}}{\left ( 3 \right )^{\frac{1}{6}-\frac{2}{3}}}$ $\because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}$
$= \frac{3^{\frac{4-9}{6}}}{3^{\frac{1-4}{6}}}$
$=\frac{3^{-\frac{5}{6}}}{3^{-\frac{3}{6}}}$
$= 3^{\frac{-5}{6}-\left ( \frac{-3}{6} \right )}$ $\because \frac{\left ( a \right )^{m}}{\left ( a \right )^{n}}= \left ( a \right )^{m-n}$
$= 3^{-\frac{2}{6}}$
$= \left ( \frac{1}{3} \right )^{\frac{1}{3}}$ $\because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}$
$= \sqrt[3]{\frac{1}{3}}$

Hence the answer is $\sqrt[3]{\frac{1}{3}}$

Solution. We have ,$64^{-\frac{1}{3}}\left ( 64^{\frac{1}{3}}-64^{\frac{2}{3}} \right )$
We know that 64 =4.4.4=43
$= \left ( 4^{3} \right )^{\frac{-1}{3}}\left \{ \left ( \left ( 4^{3} \right ) ^{\frac{1}{3}}-\left ( 4^{3} \right ) ^{\frac{2}{3}}\right )\right \}$
$= \left ( 4 \right )^{3\times \frac{-1}{3}}\left \{ \left ( \left ( 4 \right )^{3\times \frac{1}{3}}-\left ( 4 \right )^{3\times \frac{2}{3}} \right ) \right \}$ $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$
$= 4^{-1}\left ( 4-4^{2} \right )$
$= \frac{1}{4}\left ( 4-16 \right )$ $\because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}$
$= \frac{1}{4}\left ( -12 \right )$

= – 3
Hence the answer is – 3

Solution.
Given ,$\frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$
We know that
8 = 2.2.2 = 23
16 = 2.2.2.2 = 24
32 = 2.2.2.2.2 = 25
$\frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}= \frac{\left ( 2^{3} \right )^{\frac{1}{3}}\times\left ( 2^{4} \right )^{\frac{1}{3}}}{\left ( 2^{5} \right )^{-\frac{1}{3}}}$
$= \frac{2^{3\times\frac{1}{3}}\times2^{4\times\frac{1}{3}}}{2^{5\times\frac{-1}{3}}}$ $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$
$= 2^{1+\frac{4}{3}+\frac{5}{3}}$ $\because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}$ and
$\because \frac{\left ( a \right )^{m}}{\left ( a \right )^{n}}= \left ( a \right )^{m-n}$
$= 2^{\frac{3+4+5}{3}}= 2^{\frac{12}{3}}$
$= 2^{4}= 16$

Question:1

Answer. $\frac{167}{90}$
Solution. Let x = 0.6
Multiply by 10 on LHS and RHS
10x = 6
$x= \frac{6}{10}$
$x= \frac{3}{5}$
So, the $\frac{p}{q}$ from of 0.6 = $\frac{3}{5}$
Let y = $0\cdot \bar{7}$
Multiply by 10 on LHS and RHS
10y = 7.7777 …….
10y – y = $7\cdot \bar{7}-0\cdot \bar{7}$
= 7.77777 ….. – 0.77777 ……
9y = 7
$y= \frac{7}{9}$
So the $\frac{p}{q}$ from of 0.7777 = $\frac{7}{9}$
Let z = 0.47777…
Multiply by 10 on both side
10z = 4.7777 ….
10z – z = $4\cdot \bar{7}-0\cdot4\bar{7}$
9z = 4.3
$z\approx \frac{4\cdot 3}{9}$
$z= \frac{43}{90}$
So the $\frac{p}{q}$from of 0.4777 …. = $\frac{43}{90}$
Therefore, $\frac{p}{q}$ form of 0.6 + $0\cdot \bar{7}+0\cdot4\bar{7}$ is,
$x+y+z= \frac{3}{5}+\frac{7}{9}+\frac{43}{90}$
$= \frac{\left ( 54+70+43 \right )}{90}$
$= \frac{167}{90}$
Hence the answer is $\frac{167}{90}$

Question:2

Solution.
$\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$
Rationalise the denominators:
$\Rightarrow \left ( \frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}\times \frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}} \right )-\left ( \frac{2\sqrt{5}}{\sqrt{6}+\sqrt{3}} \times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}\right )-\left ( \frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2} } \times\frac{\sqrt{15}-3\sqrt{2}}{\sqrt{15}-3\sqrt{2}}\right )$
$\Rightarrow \frac{7\sqrt{3}\left ( \sqrt{10} -\sqrt{3}\right )}{10-3}-\frac{2\sqrt{5}\left ( \sqrt{6}-\sqrt{5} \right )}{6-5}-\frac{3\sqrt{2}\left ( \sqrt{15}-3\sqrt{2} \right )}{15-8}$
$\left [ \because a^{2}-b^{2} = \left ( a+b \right )\left ( a-b \right )\right ]$
$\Rightarrow \frac{7\sqrt{3}\left ( \sqrt{10} -\sqrt{3}\right )}{7}-\frac{2\sqrt{5}\left ( \sqrt{6}-\sqrt{5} \right )}{1}-\frac{3\sqrt{2}\left ( \sqrt{15}-3\sqrt{2} \right )}{3}$
$\Rightarrow \frac{7\sqrt{30}-21}{7}-\frac{2\sqrt{30}-10}{1}+\frac{3\sqrt{30}-18}{3}$
$\Rightarrow \frac{21\sqrt{30}-63-42\sqrt{30}+210+21\sqrt{30}-126}{21}$
$\Rightarrow \frac{21}{21}= 1$

Question:3

If $\sqrt{2}= 1\cdot 414$ and $\sqrt{3}= 1\cdot 732$ then find the value of $\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}$

Solution. Given that :
$\sqrt{2}= 1\cdot 414$,$\sqrt{3}= 1\cdot 732$
$\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}$
$= \frac{4\left ( 3\sqrt{3}+2\sqrt{2} \right )}{\left ( 3\sqrt{3}-2\sqrt{2} \right )\left ( 3\sqrt{3}+2\sqrt{2} \right )}+\frac{3\left ( 3\sqrt{3}-2\sqrt{2} \right )}{\left ( 3\sqrt{3}+2\sqrt{2} \right )\left (3\sqrt{3}-2\sqrt{2} \right )}$
$= \frac{4\left ( 3\sqrt{3}+2\sqrt{2} \right )+3\left ( 3\sqrt{3}-2\sqrt{2} \right )}{\left ( 3\sqrt{3}-2\sqrt{2} \right )\left ( 3\sqrt{3}+2\sqrt{2} \right )}$
Using (a – b) (a + b) = a2 – b2
$= \frac{12\sqrt{3}+8\sqrt{2}+9\sqrt{3}-6\sqrt{2}}{\left ( 3\sqrt{3} \right )^{2}-\left ( 2\sqrt{2} \right )^{2}}$
$= \frac{21\sqrt{3}+2\sqrt{2}}{27-8}$
$= \frac{21\sqrt{3}+2\sqrt{2}}{19}$

Putting the given values,$= \frac{21\left ( 1\cdot 732 \right )+2\left ( 1\cdot 414 \right )}{19}$
$=\frac{39\cdot 2014}{19}$
= 2.0632

Question:4

If $a= \frac{3+\sqrt{5}}{2}$ then find the value of $a^{2}= \frac{1}{a^{2}}$

Solution.
Given that :- $a= \frac{3+\sqrt{5}}{2}$
$\therefore \frac{1}{a}= \frac{2}{3+\sqrt{5}}$
On rationalizing the denominator, we get
$\frac{1}{a}= \frac{2\left ( 3-\sqrt{5} \right )}{\left ( 3+\sqrt{5} \right )\left ( 3-\sqrt{5} \right )}$
Using (a – b) (a + b) = a2 – b2
$= \frac{6-2\sqrt{5}}{3^{2}-\sqrt{5}^{2}}$
$= \frac{6-2\sqrt{5}}{9-5}$
$= \frac{6-2\sqrt{5}}{4}$
$= \frac{3-\sqrt{5}}{2}$
Also, $\left ( a+\frac{1}{a} \right )^{2}= a^{2}+\frac{1}{a^{2}}+2$
Substituting the values of a and $\frac{1}{a}$
We get, $\left ( \frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2} \right )^{2}= \left ( a^{2}+\frac{1}{a^{2}}+2 \right )$
$\therefore a^{2}+\frac{1}{a^{2}}+2= \left ( \frac{3+\sqrt{5}+3-\sqrt{5}}{2} \right )^{2}$
= (3)2 = 9
$\therefore a^{2}+\frac{1}{a^{2}}= 9-2= 7$
Hence the correct answer is 7.

Question:5

If $x= \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y= \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ then find the value of x2 + y2.

Solution.We use the identity $\left ( a+b \right )^{2}= a^{2}+b^{2}+2ab$
So, $\left ( \sqrt{a}+\sqrt{b} \right )^{2}= a+2\sqrt{ab}+b$
$x^{2}+y^{2}= \left ( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \right )^{2}+\left ( \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \right )^{2}$
$= \frac{3+2\sqrt{6}+2}{3-2\sqrt{6}+2}+\frac{3-2\sqrt{6}+2}{3+2\sqrt{6}+2}$
$= \frac{5+2\sqrt{6}}{5-2\sqrt{6}}+\frac{5-2\sqrt{6}}{5+2\sqrt{6}}$
$= \frac{\left ( 5+2\sqrt{6} \right )^{2}+\left ( 5-2\sqrt{6} \right )^{2}}{\left ( 5-2\sqrt{6} \right ){\left ( 5+2\sqrt{6} \right )}}$
$= \frac{\left ( 25+20\sqrt{6}+24 \right )+\left ( 25-20\sqrt{6} +24\right )}{25-24}$

Using (a – b) (a + b) = a2– b2= 98

Question:6

Simplify : $\left ( 256 \right )^{-\left ( 4^{\frac{-3}{2}} \right )}$

Solution. Given, $\left ( 256 \right )^{-\left ( 4^{\frac{-3}{2}} \right )}$
We know that,
256 = 2.2.2.2.2.2.2.2 = 28
$\left ( 256 \right )^{-\left ( 4^{\frac{-3}{2}} \right )}$ $= \left ( 2^{8} \right )^{\left ( -4 \right )\times \left ( -\frac{3}{2} \right )}$
$\because \left ( \left ( a \right ) ^{m}\right )^{n}= \left ( a \right )^{mn}$
=$\left ( 2 \right )^{8\times \left ( -4 \right )\times \left ( -\frac{3}{2} \right )}$
$= 2^{8\times 4\times \frac{3}{2}}= 2^{8\times 2\times 3}= 2^{48}$

Question:7

Solution. We have, $\frac{4}{\left ( 216 \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 256 \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 243 \right )^{\frac{-1}{5}}}$
We know that
216 = 6.6.6 = 63
256 = 4.4.4.4 = 44
243 = 3.3.3.3.3 = 35
So, $\frac{4}{\left ( 216 \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 256 \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 243 \right )^{\frac{-1}{5}}}$
$= \frac{4}{\left ( 6^{3} \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 4^{4} \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 3 ^{5}\right )^{-\frac{1}{5}}}$
$= \frac{4}{\left ( 6 \right )^{3\times \frac{-2}{3}}}+\frac{1}{\left ( 4 \right )^{4\times \frac{-3}{4}}}+\frac{2}{\left ( 3 \right )^{5\times \frac{-1}{5}}}$
$\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$
$= \frac{4}{6^{-2}}+\frac{1}{4^{-3}}+\frac{2}{3^{-1}}$
= 4 × 62 + 43 + 2 × 3
$\because \frac{1}{\left ( a \right )^{-n}}= \left ( a \right )^{n}$
= 4 × 36 + 64 + 6
= 144 + 70
= 214

NCERT Exemplar Solutions Class 9 Maths Chapter 1 Number System Important Topics:

Topics covered in NCERT exemplar Class 9 Maths solutions chapter 1 deals with the understanding of:

◊ Real Numbers as Rational numbers and Irrational numbers.

◊ Problems to find a rational number between two given numbers or to find an irrational number between two given numbers are dealt with here.

◊ The concepts of the number line and locating rational number and irrational numbers on the number line are explained.

◊ The decimal expressions of real numbers and fractions are explained in categories like terminating or non-terminating as well as recurring or non-recurring.

◊ Questions based on operations performed on real numbers and exponents for real numbers are explained appropriately.

◊ Laws of exponents for real numbers such as dealing with multiplications of powers and addition of powers of real numbers are explained through NCERT exemplar Class 9 Maths solutions chapter 1.

◊ The concept of rationalising the denominator by multiplication of conjugate of an irrational number is explained through NCERT exemplar solutions for Class 9 Maths chapter 1.

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

 Chapter 2 Polynomials Chapter 3 Coordinate geometry Chapter 4 Linear equations in Two Variable Chapter 5 Introduction to Euclid’s Geometry Chapter 6 Lines and Angles Chapter 7 Triangles Chapter 8 Quadrilaterals Chapter 9 Area of Parallelograms and Triangles Chapter 10 Circles Chapter 11 Constructions Chapter 12 Heron’s Formula Chapter 13 Surface Areas and Volumes Chapter 14 Statistics and Probability

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 1 Number System:

These Class 9 Maths NCERT exemplar solutions chapter 1 will help the students to understand the concept and knowledge of the real numbers. Students of Class 9 can use these solutions as reference material for better study and practice sums of real numbers.

The NCERT exemplar Class 9 Maths solutions chapter 1 Number System will be adequate to solve problems of other reference books such as NCERT Class 9 Maths, RD Sharma Class 9 Maths or RS Aggarwal Class 9 Maths etcetera.

NCERT exemplar Class 9 Maths solutions chapter 1 pdf download will be made available to resolve the doubts encountered while attempting the exemplar of chapter 1.

Check NCERT Solutions for questions given in book

 Chapter No. Chapter Name Chapter 1 Number Systems Chapter 2 Polynomials Chapter 3 Coordinate Geometry Chapter 4 Linear Equations In Two Variables Chapter 5 Introduction to Euclid's Geometry Chapter 6 Lines And Angles Chapter 7 Triangles Chapter 8 Quadrilaterals Chapter 9 Areas of Parallelograms and Triangles Chapter 10 Circles Chapter 11 Constructions Chapter 12 Heron’s Formula Chapter 13 Surface Area and Volumes Chapter 14 Statistics Chapter 15 Probability

Also, check NCERT Notes Subject Wise

 NCERT Notes for Class 9 Science NCERT Notes for Class 9 Maths

Also Check NCERT Books and NCERT Syllabus here

1. What is the weightage of Number System in Class 9 Maths?

Number system weighs approximately 8-10% (discretional to the paper setters; varies from school to school for Class 9 Maths)  of the total marks of the paper, however, being one on the basic building blocks for Maths of higher Classes (Class 10, Class 11 and Class 12), the student should understand and practice NCERT exemplar Class 9 Maths chapter 1.

2. Will these problems of exemplar require additional knowledge of real number?

The NCERT exemplar Class 9 Maths solutions chapter 1 equip the student with a multidimensional approach to the problems and understanding the concept of Number System.

3. What is the weightage of Real Numbers in JEE Main and JEE Advanced?

A clear understanding of Real Number can prepare a student in solving the problems based on fundamentals of maths which ranges from 2-5% marks of the whole paper.

4. Will it be beneficial to refer NCERT exemplar Class 9 Maths solutions for competitive exams?

NCERT exemplar Class 9 Maths solutions chapter 1 will provide the student multifaceted problems to hone the skills and prepare well for the competitive exams such as JEE Main.

5. Why π is an irrational number even its value is 22/7?

Pi is an irrational number because it is a non-terminating non-recurring decimal number. Recently, a tech company employee has broken the Guinness World Record by calculating Pi till 31.4 trillion (1012) decimal places (See it is never-ending…..)

6. Is the number of rational numbers more than the number of irrational numbers?

NCERT exemplar Class 9 Maths solutions chapter 1 explains that both the sets of rational and irrational numbers are infinite; hence, it cannot be deduced.

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Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

3 Jobs Available
Critical Care Specialist

A career as Critical Care Specialist is responsible for providing the best possible prompt medical care to patients. He or she writes progress notes of patients in records. A Critical Care Specialist also liaises with admitting consultants and proceeds with the follow-up treatments.

2 Jobs Available
Ophthalmologist

Individuals in the ophthalmologist career in India are trained medically to care for all eye problems and conditions. Some optometric physicians further specialize in a particular area of the eye and are known as sub-specialists who are responsible for taking care of each and every aspect of a patient's eye problem. An ophthalmologist's job description includes performing a variety of tasks such as diagnosing the problem, prescribing medicines, performing eye surgery, recommending eyeglasses, or looking after post-surgery treatment.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Talent Director

Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Videographer

Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production.

2 Jobs Available
SEO Analyst

An SEO Analyst is a web professional who is proficient in the implementation of SEO strategies to target more keywords to improve the reach of the content on search engines. He or she provides support to acquire the goals and success of the client’s campaigns.

2 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
Quality Systems Manager

A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.

2 Jobs Available
Merchandiser

A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.

2 Jobs Available
Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

2 Jobs Available
Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

2 Jobs Available
ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available
Data Architect

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available
Security Engineer

The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

2 Jobs Available
UX Architect

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available