NCERT Exemplar Class 9 Maths Solutions Chapter 1 Number Systems 2021

Q: 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.

Q: 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.

Q: 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.

Q: 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 .

Q: 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…..)

Q: 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.

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.

Scholarship Test: Vidyamandir Intellect Quest (VIQ)

Don't Miss: JEE Main 2027: Narayana Scholarship Test Preparation Kit for Class 9

This Story also Contains

NCERT Exemplar Class 9 Maths Solutions Chapter-1: Exercise-1.1
NCERT Exemplar Maths Solutions Class 9 Chapter 1: Exercise-1.2
Question:1
NCERT Exemplar Class 9 Maths Solutions Chapter-1: Exercise-1.3
NCERT Exemplar Class 9 Maths Solutions Chapter-1: Exercise-1.4
Question:1
NCERT Exemplar Solutions Class 9 Maths Chapter 1 Number System Important Topics:
NCERT Class 9 Maths Exemplar Solutions for Other Chapters:
Features of NCERT Exemplar Class 9 Maths Solutions Chapter 1 Number System:

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

Answer:

Answer: [C]
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

Answer:

Answer: [C]
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

Answer:

Answer: [D]
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

Answer:

Answer: [D]
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

Answer:

Answer: [D]
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) $\sqrt{81}$

Answer:

Answer: [C]
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) $0.4014001400014$

Answer:

Answer: [D]
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) $1.8$

Answer:

Answer: [C]
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) $\frac{1}{9}$

Answer:

Answer. [C]
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) $4\sqrt{6}$

Answer:

Answer. [C]
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
cannot find the exact answer by adding their decimal representations.
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:

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) $\frac{\sqrt{7}+2}{45}$

Answer:

Answer. [A]
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) $3+2\sqrt{2}$

Answer:

Answer.[D]
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

After rationalizing the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}}$ we get the denominator as:
(A)13
(B)19
(C)5
(D)35

Answer:

Answer. [B]
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

Answer:

Answer. [B]
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

If $\sqrt{2}= 1\cdot 4142$ = 1.4142 then $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ is equal to :
(A)2.4142
(B)5.8282
(C)0.4142
(D)0.1718

Answer:

Answer. [C]
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)
(D)}

Answer:

Answer. [C]
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) $12\sqrt{32}$
Answer:

Answer. [B]
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) $\frac{1}{81}$

Answer:

Answer. [A]
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

Answer:

Answer. [A]
Solution. We have,
(256)^0.16 × (256)^0.09 [ $\because$ a^m × aⁿ = a^m+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 [ $\because$ (a^m)ⁿ = a^mn]
= 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) $x^{\frac{12}{7}}\times x^{\frac{7}{12}}$

Answer:

Answer. [C]
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.

NCERT Exemplar Maths Solutions Class 9 Chapter 1: Exercise-1.2

Question:1

Let x and y be rational and irrational number respectively. Is x + y necessarily an irrational number?
Give an example in support of your answer.

Answer:

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:

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) State whether the given statement is true or false? Justify your answer by an example. $\frac{\sqrt{2}}{3}$ is a rational number.
(ii) State whether the given statement is true or false? Justify your answer by an example. There are infinitely many integers between any two integers.
(iii) State whether the given statement is true or false? Justify your answer by an example. Number of rational numbers between 15 and 18 is finite.
(iv) State whether the given statement is true or false? Justify your answer by an example. There are numbers which cannot be written in the form $\frac{p}{q}$ , $q \neq 0$ , p and q both are integers.

(v) State whether the given statement is true or false? Justify your answer by an example. The square of an irrational number is always rational.
(vi) State whether the given statement is true or false? Justify your answer by an example. $\frac{\sqrt{12}}{\sqrt{3}}$ is not a rational number as $\sqrt{12}$ and $\sqrt{3}$ are not integers.
(vii) State whether the given statement is true or false? Justify your answer by an example. $\frac{\sqrt{15}}{\sqrt{3}}$ is written in the form $\frac{p}{q}, q \neq 0$ and so it is a rational number.

Answer:

(i)
Answer: False
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)
Answer: False
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)
Answer: False
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.

(iv) Answer: True,
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)
Answer: False
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)
Answer: False
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)
Answer: False
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

Answer:

(i)Answer. [Rational]
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.

(ii)Answer. [Irrational]
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
(iii)Answer. [Irrational]
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

(iv)Answer. [Rational]
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.
(v)Answer. [Irrational]
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

(vi)Answer. [Rational]
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.
(vii)Answer. [Rational]
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.
(viii)Answer. [Rational]
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.

(ix)Answer. [Rational]
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.
(x)Answer. [Irrational]
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 = $\frac{17}{4}$

Answer:

(i)Answer. [Irrational]
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

(ii)Answer. [Rational]
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.
(iii)Answer. [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.
(iv)Answer. [Irrational]
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 $\frac{1}{5}$

Answer:

(i)Answer.
$-\frac{11}{10}$ , $-\frac{6}{5}$ 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 ……

(iv)Answer. Rational number: 0
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

(vii) Answer. Rational number: 3
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

(ix) Answer. Rational number: 1
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, $-\frac{3}{2},-\frac{12}{5}$

Answer:

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.
FireShot%20Capture%20003%20-%20Careers360%20CMS%20-%20cms-article%20-%20learn (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

Answer:

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.
58373
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:
$\sqrt{10}= \sqrt{3^{2}+1^{2}}$
$\sqrt{17}= \sqrt{4^{2}+1^{2}}$
583731
583732

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:

Answer:

(i) Solution. AB = 4.5 units, BC = 1 unit
58395
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 = $\sqrt{5\cdot 6}$

583951
(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 - $\sqrt{8. 1}$
583952
(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 = $\sqrt{2. 3}$
583953

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……..

Answer:

(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 ……. = $\frac{1}{7}$

Answer:

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

Answer:

(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
$= \left ( \frac{9+18+7}{3} \right )\sqrt{3}$
$= \frac{34}{3}\sqrt{3}$
$= 19\cdot 63$
Hence the answer is 19.63
(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 $5-2\sqrt{6}$

(vii) Answer. 0
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
Hence the answer is 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

Answer:

(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 : $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}$

Answer:

(i) Answer. a = 11
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:

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 : $\frac{1}{\sqrt{3}+\sqrt{2}}$

Answer:

(i) Answer. 2.3093
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
Hence the answer is 2.3093

(ii) Answer. 2.449
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 :
$= \sqrt{2}\cdot \sqrt{3}$
$= 1\cdot 414\times 1\cdot 732= 2\cdot 449$
Hence the answer is 2.449

(iii) Answer. 0.462852
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
Hence the answer is 0.462852
(iv) Answer. 0.414
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
Putting the given value of $\sqrt{2}= 1\cdot 414$
We get
= 1.414 – 1
= 0.414
Hence the answer is 0.414

(v) Answer. 0.318
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
Putting the given values, $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$
We get,
= 1.732 – 1.414
= 0.318
Hence the answer is 0.318

Question:14

(i) Simplify :- (1³ + 2³ + 3³)^1/2
(ii)Simplify :- $\left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}$
(iii) Simplify :- $\left ( \frac{1}{27} \right )^{-\frac{2}{3}}$
(iv) Simplify :- $\left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}$
(v) Simplify :- $\frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}$
(vi) Simplify :- $64^{-\frac{1}{3}}\left ( 64^{\frac{1}{3}}-64^{\frac{2}{3}} \right )$
(vii) Simplify :- $\frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$
Answer:

(i) Answer. 6
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 $= \sqrt{1+8+27}$
$=\sqrt{36}= 6$
Hence the answer is 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 $\frac{2025}{64}$

(iii) Answer. 9
Solution. Given $\left ( \frac{1}{27} \right )^{-\frac{2}{3}}$
We know that
27 = 3.3.3 = 3³
= 3² = 9
Hence the answer is 9

(iv) Answer. 5
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
Hence the answer is 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
Hence the answer is

(vi) Answer. – 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
= – 3
Hence the answer is – 3

(vii) Answer. 16
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 = 2^{5

and}
Hence the answer is 16.

NCERT Exemplar Class 9 Maths Solutions Chapter-1: Exercise-1.4

Question:1

Express 0.6 + $0\cdot \bar{7}+0\cdot \overline{47}$ in the form $\frac{p}{q}$ where p and q are integers and $q\neq 0$ .
Answer:

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

Simplify :- $\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$

Answer:

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

Question:3

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

Answer:

Answer. 2.0632
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
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
Hence the answer is 2.0632.

Question:4

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

Answer:

Answer. 7
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.

Answer:

Answer. 98
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
Hence the answer is 98.

Question:6

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

Answer:

Answer. 2⁴⁸
Solution. Given, $\left ( 256 \right )^{-\left ( 4^{\frac{-3}{2}} \right )}$
We know that,
256 = 2.2.2.2.2.2.2.2 = 2⁸
$\left ( 256 \right )^{-\left ( 4^{\frac{-3}{2}} \right )}$ $= \left ( 2^{8} \right )^{\left ( -4 \right )\times \left ( -\frac{3}{2} \right )}$
$\because \left ( \left ( a \right ) ^{m}\right )^{n}= \left ( a \right )^{mn}$
= $\left ( 2 \right )^{8\times \left ( -4 \right )\times \left ( -\frac{3}{2} \right )}$
$= 2^{8\times 4\times \frac{3}{2}}= 2^{8\times 2\times 3}= 2^{48}$
Hence the answer is 2⁴⁸

Question:7

Find the value of $\frac{4}{\left ( 216 \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 256 \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 243 \right )^{\frac{-1}{5}}}$ .

Answer:

Answer. 214
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
Hence the answer is 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.

NCERT Class 9 Exemplar Solutions for Other Subjects:

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 Solution Subject Wise

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

Frequently Asked Questions (FAQs)

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.

Also Read

NCERT Solutions for Class 9 - Subject-wise PDFs Updated for 2024-25

NCERT Solutions for Class 9 Science

NCERT Solutions for Class 9 Maths

NCERT Syllabus for Class 9 Maths 2023 - Download Syllabus Pdf

NCERT Solutions For Class 9 Science Chapter 1 Matter in Our Surroundings

NCERT Solutions for Exercise 2.5 Class 9 Maths Chapter 2 Polynomials

NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

NCERT Exemplar Class 9 Maths Solutions - Chapter Wise Problems with Solutions

NCERT Exemplar Class 9 Solutions - Subject Wise Problems with Solutions

NCERT Exemplar Class 9 Science Solutions Chapter 15 Improvement in Food Resources

NCERT Exemplar Class 9 Science Solutions Chapter 14 Natural Resources

NCERT Exemplar Class 9 Science Solutions Chapter 13 Why do we fall ill?

NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound

Articles

Silverzone ABHO Result 2024-25 - Check Akhil Bhartiya Hindi Olympiad Results Here

Nov 20, 2024

Silverzone SKGKO Result 2024-25 – Check Your Scores and Performance

Nov 20, 2024

Education Loan for School Students in India – Eligibility, Benefits & Application Process

Nov 20, 2024

Silverzone ABHO Answer Key 2024 | Download Akhil Bhartiya Hindi Olympiad Solutions

Nov 20, 2024

Silverzone SKGKO Answer Key 2024-25: Download Smart Kid GK Olympiad Answer Key @ silverzone.org

Nov 20, 2024

Sainik School Entrance Exam 2025: Application, Exam Date, How to Apply, Syllabus

Nov 20, 2024

ICICI Education Loan: Eligibility, Interest Rates & Application Process

Nov 19, 2024

MPSOS 10th Result 2024 - Check MPSOS Class 10 December 2024 Exam Result @mpsos.nic.in

Nov 16, 2024

Upcoming School Exams

National Institute of Open Schooling 12th Examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Admit Card

National Institute of Open Schooling 10th examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Mock Test

Mizoram Board of School Education 12th Examination

Application Date:07 October,2024 - 22 November,2024

Exam Pattern

Result

Mock Test

Mizoram Board 10th Examination

Application Date:07 October,2024 - 22 November,2024

Exam Pattern

Admit Card

Result

Punjab Board of Secondary Education 10th Examination

Application Correction Date:08 October,2024 - 27 November,2024

Exam Pattern

Result

Admit Card

View All School Exams

Get answers from students and experts

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 9 Maths Solutions Chapter 1 Number Systems

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

NCERT Exemplar Maths Solutions Class 9 Chapter 1: Exercise-1.2

Question:1

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

NCERT Exemplar Class 9 Maths Solutions Chapter-1: Exercise-1.4

Question:1

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

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

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

NCERT Class 9 Exemplar Solutions for Other Subjects:

Check NCERT Solutions for questions given in book

Also, check NCERT Solution Subject Wise

Also, check NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

Also Read

Articles

Upcoming School Exams

National Institute of Open Schooling 12th Examination

National Institute of Open Schooling 10th examination

Mizoram Board of School Education 12th Examination

Mizoram Board 10th Examination

Punjab Board of Secondary Education 10th Examination

Popular Questions

Colleges After 12th

Popular Course After 12th

Applications for Admissions are open.

VMC VIQ Scholarship Test

JEE Main Important Physics formulas

JEE Main Important Chemistry formulas

TOEFL ® Registrations 2024

Pearson | PTE

JEE Main high scoring chapters and topics

Explore on Careers360

NCERT Solutions

NCERT Exemplars

NCERT Books

NCERT Notes

CBSE

CBSE Class 10

CBSE Class 12th

CISCE Board 10th

IMO

IEO

NSO

NMMS

Schools By Medium

By Ownership

By City

By State

NVS

JNVST

Download Careers360 App

All this at the convenience of your phone

Scan and download the app