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# NCERT Exemplar for Class 9 Math Chapter 1 - Number Systems (Book Solutions)

Last updated date: 09th Sep 2024
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## NCERT Exemplar for Class 9 Math - Number Systems - Free PDF Download

NCERT Exemplar for Class 9 Math Solutions Chapter 1 'Number Systems' is an ideal resource for scoring high in the tests. It is an absolute necessity if you are experiencing issues in understanding the chapter. The strategies and steps mentioned by our experts in taking care of the Exemplar issues are extremely simple. Our specialists have a detailed knowledge of the number system and CBSE NCERT Exemplar Class 9 Math.

Competitive Exams after 12th Science

## Access NCERT Exemplar Solutions for Class 9 Mathematics Chapter 1 - Number Systems

### Sample Questions

1. Which of the following is not equal to ${\left[ {{{\left( {\dfrac{5}{6}} \right)}^{\dfrac{1}{5}}}} \right]^{ - \dfrac{1}{6}}}$ ?

(A) ${\left[ {\left( {\dfrac{5}{6}} \right)} \right]^{\left( {\dfrac{1}{5} - \dfrac{1}{6}} \right)}}$

(B) $\dfrac{1}{{{{\left[ {{{\left( {\dfrac{5}{6}} \right)}^{\left( {\dfrac{1}{5}} \right)}}} \right]}^{\left( {\dfrac{1}{6}} \right)}}}}$

(C) ${\left( {\dfrac{6}{5}} \right)^{\left( {\dfrac{1}{{30}}} \right)}}$

(D) ${\left( {\dfrac{5}{6}} \right)^{\left( {\dfrac{1}{{30}}} \right)}}$

Ans: option b is correct

If $p$ and $q$ are rational numbers and $a$ is a positive real number, then

${\left( {{a^p}} \right)^q} = {\left( a \right)^{pq}}$ . So, this law of exponents is to be used to find the value of a given expression.

$\Rightarrow {\left[ {{{\left( {\dfrac{{{5}}}{{{6}}}} \right)}^{\dfrac{{{1}}}{{{5}}}}}} \right]^{{{ - }}\dfrac{{{1}}}{{{6}}}}}{{ = }}\left[ {{{\left( {\dfrac{{{5}}}{{{6}}}} \right)}^{\dfrac{{{1}}}{{{5}}}{{ \times }}\left( {\dfrac{{{{ - 1}}}}{{{6}}}} \right)}}} \right]$

EXERCISE 1.1

1. Every rational number is

(A) a natural number

(B) an integer

(C) a real number

(D) a whole number

Ans: C

Real Number is the union of rational numbers and the irrational numbers.

Hence, every rational number is a real number.

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) There are only rational numbers and no irrational numbers

Ans: C

Rational number is just the division of any two integers.

To find the rational number between two rational number, use the following steps:

• First, check the values of Denominators.

• If denominators are the same, then check the value of the numerator.

• If the numerators differ by a large value then write the rational numbers with an increment of one (keeping the denominator part unchanged.)

• If the numerators differ by a smaller value than the number of rational numbers to be found, simply multiply the numerators and denominators by multiples of 10.

Hence, we can find infinitely many rational numbers between two rational numbers.

3. Decimal representation of a rational number cannot be

(A) terminating

(B) non-terminating

(C) non-terminating repeating

(D) non-terminating non-repeating

Ans: D

A rational number is expressed in decimal form.

A rational number can be either terminating or non-terminating recurring decimal.

Hence, Decimal expansion of any rational number cannot be non-terminating non-repeating.

4. The product of any two irrational numbers is

(A) Always an irrational number

(B) Always a rational number

(C) Always an integer

(D) Sometimes rational, sometimes irrational

Ans: D

The product of any two irrational numbers can be rational or irrational.

Taking an example to illustrate the above statement,

Let $\left( {{{4}}{{ + }}\sqrt {{5}} } \right),{{ }}\left( {{{4}} - \sqrt {{5}} } \right){\text{ and }}\left( {{{3}} - \sqrt {{5}} } \right)$ be rational numbers.

Consider,

$\left( {{{4}}{{ + }}\sqrt {{5}} } \right)\left( {{{4}}{{ - }}\sqrt {{5}} } \right)$

$= \left[ {{{\left( {{4}} \right)}^{{2}}}{{ - }}{{\left( {\sqrt {{5}} } \right)}^{{2}}}} \right]$

${{ = }}{{1}}{{6}}{{ - }}{{5}}$

${{ = }}{{11}}$

$\left( {{{4}}{{ + }}\sqrt {{5}} } \right)\left( {{{3}}{{ - }}\sqrt {{5}} } \right)$

${{ = }}{{4}}\left( {{{3}}{{ - }}\sqrt {{5}} } \right){{ + }}\sqrt {{5}} \left( {{{3}}{{ - }}\sqrt {{5}} } \right)$

${{ = 12}}{{ - }}{{4}}\sqrt {{5}} {{ + }}{{3}}\sqrt {{5}} {{ - }}{{5}}$

${{ = 7}}{{ - }}\sqrt {{5}}$

The product is rational.

Hence, the product of any two irrational numbers is sometimes rational, sometimes irrational

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

Ans: D

The given number $\sqrt {{2}}$ is an irrational number.

We know that:

When any rational number is expressed in the form of decimal expansion then it is either terminating or non-terminating recurring.

So, its expansion will be non- terminating, non- recurring.

6. Which of the following is irrational?

(A) $\sqrt {\dfrac{4}{9}}$

(B) $\dfrac{{\sqrt {12} }}{{\sqrt 3 }}$

(C) $\sqrt 7$

(D) $\sqrt {81}$

Ans: C

On solving all the options:

$\sqrt {\dfrac{{{4}}}{{{9}}}} {{ = }}\dfrac{{{{2}}}}{{{3}}}$ rational

$\dfrac{{\sqrt {{{12}}} }}{{\sqrt {{3}} }}{{ = }}\dfrac{{{{2}}\sqrt {{3}} }}{{\sqrt {{3}} }}{{ = }}{{2}}{\text{ rational}}$

$\sqrt {{{81}}} {{ = }}{{9}}{\text{ rational}}$

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

7. Which of the following is irrational?

(A) $0.14$

(B) $0.14\overline {16}$

(C) $0.\overline {1416}$

(D) $0.4014001400014...$

Ans: D

The number ${{0}}{{.4014001400014}}...$ is non- terminating and non-recurring.

It can not be represented as a simple fraction.

Hence, the number $0.4014001400014...$ is irrational

8. A rational number between $\sqrt 2$  and $\sqrt 3$  is

(A) $\dfrac{{\sqrt 2 + \sqrt 3 }}{2}$

(B) $\dfrac{{\sqrt 2 - \sqrt 3 }}{2}$

(C) $1.5$

(D) $1.8$

Ans: C

We know that:

$\sqrt {{2}} {{ = }}{{1}}{{.414}}$

$\sqrt {{{3}}} {{ = }}{{1}}{{.732}}$

The number ${{1}}{{.5}}$ lies between 1.414 and 1.732.

Hence, a rational number between $\sqrt {{2}}$  and $\sqrt {{3}}$  is 1.5.

9. The value of $1.999...$ in the form $\dfrac{p}{q}$ where $p$ and $q$ are integers and $q{\text{ }} \ne {\text{ }}0{\text{ }},$ is

(A) $\dfrac{{19}}{{10}}$

(B) $\dfrac{{1999}}{{1000}}$

(C) $2$

(D) $\dfrac{1}{9}$

Ans: C

Assume ${{x}}{{ = }}{{1}}{{.999}}......{\text{ }}(1)$

Multiply equation (1) by 10, we get:

${{10}}{{x}}{{ = }}{{19}}{{.999}}........(2)$

Subtracting $\left( {{1}} \right)$ from $\left( {{2}} \right)$

${{10}}x - x{{ = }}\left( {{{19}}{{.999}}...} \right) - \left( {{{1}}{{.999}}...} \right)$

${{9}}x{{ = }}{{18}}$

$x{{ = }}\dfrac{{{{18}}}}{{{9}}}$

$x{{ = }}{{2}}$

10. $2\sqrt 3 + \sqrt 3$ is equal to

(A) $2\sqrt 6$

(B) $6$

(C) $3\sqrt 3$

(D) $4\sqrt 6$

Ans: C

Let ${{2}}\sqrt {{3}} {{ + }}\sqrt {{3}}$

Taking $\sqrt {{3}}$ as common

${{2}}\sqrt {{3}} {{ + }}\sqrt {{3}}$

${{ = }}\sqrt {{3}} \left( {{{2 + 1}}} \right)$

${{ = 3}}\sqrt {{3}}$

Hence, ${{2}}\sqrt {{3}} {{ + }}\sqrt {{3}}$= ${{3}}\sqrt {{3}}$

11.  $\sqrt {10} \times \sqrt {15}$ is equal to

(A) $6\sqrt 5$

(B) $5\sqrt 6$

(C) $\sqrt {25}$

(D) $10\sqrt 5$

Ans: B

Let a and b be positive real numbers, then

$\sqrt a \sqrt b = \sqrt {ab}$

$\sqrt {10} {{ }} \times {{ }}\sqrt {15} = {{ }}\sqrt {10 \times 15}$

$= \sqrt {2 \times 5 \times 3 \times 5}$

$= \sqrt {25 \times 6}$

$= 5\sqrt 6$

12. The number obtained on rationalizing the denominator of $\dfrac{1}{{\sqrt 7 - 2}}$ is

(A) $\dfrac{{\sqrt 7 + 2}}{3}$

(B)  $\dfrac{{\sqrt 7 - 2}}{3}$

(C)  $\dfrac{{\sqrt 7 + 2}}{5}$

(D)  $\dfrac{{\sqrt 7 + 2}}{{45}}$

Ans: A

To find the number obtained on rationalizing the denominator of $\dfrac{1}{{\sqrt 7 - 2}}{{ }}$ , multiply both numerator and denominator by $\sqrt 7 + 2$

$\dfrac{1}{{\sqrt 7 - 2}}{{ }}$$= \dfrac{1}{{\sqrt 7 - 2}}{{ }} \times \dfrac{{\sqrt 7 + 2}}{{\sqrt 7 + 2}} {{ = }}\dfrac{{\sqrt 7 + 2}}{{\left( {\sqrt 7 - 2} \right)\left( {\sqrt 7 + 2} \right)}} = \dfrac{{\sqrt 7 + 2}}{{{{\left( {\sqrt 7 } \right)}^2}{{ }} - {{\left( 2 \right)}^2}}}....................(a + b)(a - b) = ({a^2} - {b^2}) = \dfrac{{\sqrt 7 + 2}}{{{{7}} - 4}} = \dfrac{{\sqrt 7 + 2}}{3} Hence, the number obtained on rationalizing the denominator of \dfrac{1}{{\sqrt 7 - 2}}{{ }} is \dfrac{{\sqrt 7 + 2}}{3}. 13. \dfrac{1}{{\sqrt 9 - \sqrt 8 }} is equal to (A) \dfrac{1}{2}\left( {3 - 2\sqrt 2 } \right) (B) \dfrac{1}{{3 + 2\sqrt 2 }} (C) 3 - 2\sqrt 2 (D) 3 + 2\sqrt 2 Ans: D To find the value of \dfrac{1}{{\sqrt 9 - \sqrt 8 }}, first we need to rationalize it. This can be done by multiplying both numerator and denominator by \sqrt 9 + \sqrt 8 = \dfrac{1}{{\sqrt 9 - \sqrt 8 }} = \dfrac{1}{{\sqrt 9 - \sqrt 8 }} \times \dfrac{{\sqrt 9 + \sqrt 8 }}{{\sqrt 9 + \sqrt 8 }} = \dfrac{{3 + 2\sqrt 2 }}{{9 - 8}} = \dfrac{{3 + 2\sqrt 2 }}{1} = 3 + 2\sqrt 2 The value of \dfrac{1}{{\sqrt 9 - \sqrt 8 }} is $3 + 2\sqrt 2$. 14. After rationalizing the denominator of $\dfrac{7}{{3\sqrt 3 - 2\sqrt 2 }}$, we get the denominator as (A) $13$ (B) $19$ (C) $5$ (D) $35$ Ans: B To find the number obtained on rationalizing the denominator of $\dfrac{7}{{3\sqrt 3 - 2\sqrt 2 }}$ , multiply both numerator and denominator by 3\sqrt 3 + 2\sqrt 2. $\dfrac{7}{{3\sqrt 3 - 2\sqrt 2 }}$ = \dfrac{7}{{3\sqrt 3 - 2\sqrt 2 }} \times \dfrac{{3\sqrt 3 + 2\sqrt 2 }}{{3\sqrt 3 + 2\sqrt 2 }} = \dfrac{{7\left( {3\sqrt 3 + 2\sqrt 2 } \right)}}{{{{\left( {3\sqrt 3 } \right)}^2} - {{\left( {2\sqrt 2 } \right)}^2}}} = \dfrac{{7\left( {3\sqrt 3 + 2\sqrt 2 } \right)}}{{27 - 8}} = \dfrac{{7\left( {3\sqrt 3 + 2\sqrt 2 } \right)}}{{19}} Therefore, the denominator after rationalization is 19 15. The value of $\dfrac{{\sqrt {32} + \sqrt {48} }}{{\sqrt 8 + \sqrt {12} }}$ is equal to (A) $\sqrt 2$ (B) $2$ (C) $4$ (D) $8$ Ans: B $\dfrac{{\sqrt {32} + \sqrt {48} }}{{\sqrt 8 + \sqrt {12} }}$ = \dfrac{{4\sqrt 2 + 4\sqrt 3 }}{{2\sqrt 2 + 2\sqrt 3 }} = \dfrac{{4\left( {\sqrt 2 {{ + }}\sqrt 3 } \right)}}{{2\left( {\sqrt 2 {{ + }}\sqrt 3 } \right)}} = \dfrac{4}{2} = 2 Hence, the value of $\dfrac{{\sqrt {32} + \sqrt {48} }}{{\sqrt 8 + \sqrt {12} }}$ is 2. 16. If \sqrt 2 = 1.4142, then \sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} is equal to (A) 2.4142 (B) 5.8282 (C) 0.4142 (D) 0.1718 Ans: C \sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} = \sqrt {\dfrac{{1.4142 - 1}}{{1.4142 + 1}}} = \sqrt {\dfrac{{0.4142}}{{2.4142}}} = \sqrt {0.1715} = 0.4142 17. \sqrt[4]{{\sqrt[3]{{{2^2}}}}} equals (A) {2^{\dfrac{{ - 1}}{6}}} (B) {2^{ - 6}} (C) {2^{\dfrac{1}{6}}} (D) {2^6} Ans: C Simplify it term by term. \sqrt[4]{{\sqrt[3]{{{2^2}}}}} = {\left( {{{\left( {{2^2}} \right)}^{\dfrac{1}{3}}}} \right)^{\dfrac{1}{4}}} = \left( {{{\left( {{2^2}} \right)}^{\dfrac{1}{3} \times }}^{\dfrac{1}{4}}} \right) = {\left( {{2^2}} \right)^{\dfrac{1}{{12}}}} = {\left( 2 \right)^{\dfrac{2}{{12}}}} = {\left( 2 \right)^{\dfrac{1}{6}}} 18. The product \sqrt[3]{2}.\sqrt[4]{2}.\sqrt[{12}]{{32}} equals (A) \sqrt 2 (B) 2 (C) \sqrt[{12}]{2} (D) \sqrt[{12}]{{32}} Ans: B \sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[{12}]{{32}} = {\left( 2 \right)^{\dfrac{1}{3}}} \cdot {\left( 2 \right)^{\dfrac{1}{4}}} \cdot {\left( 2 \right)^{\dfrac{5}{{12}}}} ={{\left( 2 \right)}^{\dfrac{1}{3}+\dfrac{1}{4}+ \dfrac{5}{2}}} = {\left( 2 \right)^{\dfrac{{4 + 3 + 5}}{{12}}}} = {\left( 2 \right)^{\dfrac{{12}}{{12}}}} = 2 Hence, the value of \sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[{12}]{{32}} is 2. 19. Value of \sqrt[4]{{{{\left( {81} \right)}^{ - 2}}}} is (A) \dfrac{1}{9} (B) \dfrac{1}{3} (C) 9 (D) \dfrac{1}{{81}} Ans: A Let a be any integer {(a{}^m)^n} = {a^{mn}} $\sqrt[4]{{{{\left( {81} \right)}^{ - 2}}}}$ = \sqrt[4]{{{{\left( {\dfrac{1}{{81}}} \right)}^2}}} = {\left( {{{\left( {\dfrac{1}{{81}}} \right)}^2}} \right)^{\dfrac{1}{4}}} = {\left( {\dfrac{1}{{81}}} \right)^{2 \times \dfrac{1}{4}}} = {\left( {\dfrac{1}{{81}}} \right)^{\dfrac{1}{2}}} = \dfrac{1}{9} The value of \sqrt[4]{{{{\left( {81} \right)}^{ - 2}}}} is $\dfrac{1}{9}$ . 20. Value of ${\left( {256} \right)^{0.16}} \times {\left( {256} \right)^{0.09}}$ is (A) $4$ (B) $16$ (C) $64$ (D) $256.25$ Ans: A Let a and b are integers then {a^m}.{a^n} = {a^{m + n}} {\left( {256} \right)^{0.16}} \times {\left( {256} \right)^{0.09}} = {\left( {256} \right)^{0.16 + 0.09}} = {\left( {256} \right)^{0.25}} = {\left( {256} \right)^{\dfrac{1}{4}}} = {\left( {{{\left( 4 \right)}^4}} \right)^{\dfrac{1}{4}}} = {\left( 4 \right)^{\dfrac{4}{4}}} = 4 Hence, the value of ${\left( {256} \right)^{0.16}} \times {\left( {256} \right)^{0.09}}$ is 4. 21. Which of the following is equal to$x$? (A) ${x^{\dfrac{{12}}{7}}} - {x^{\dfrac{5}{7}}}$ (B) $\sqrt[{12}]{{{{\left( {{x^4}} \right)}^{\dfrac{1}{3}}}}}$ (C) ${\left( {\sqrt {{x^3}} } \right)^{\dfrac{2}{3}}}$ (D) ${x^{\dfrac{{12}}{7}}} \times {x^{\dfrac{7}{{12}}}}$ Ans: C (A) ${x^{\dfrac{{12}}{7}}} - {x^{\dfrac{5}{7}}} \ne x$ (B) \sqrt[{12}]{{{{\left( {{x^4}} \right)}^{\dfrac{1}{3}}}}} ={{\left( x \right)}^{4\times \dfrac{1}{3}\times \dfrac{1}{12}}} = {x^{\dfrac{1}{9}}} \ne x (C) {\left( {\sqrt {{x^3}} } \right)^{\dfrac{2}{3}}} = {\left( {{x^3}} \right)^{\dfrac{1}{2} \times }}^{\dfrac{2}{3}} = {x^{\dfrac{3}{3}}} = x (D) ${x^{\dfrac{{12}}{7}}} \times {x^{\dfrac{7}{{12}}}} \ne x$ So, option (C) is correct ### Sample Questions 1. Are there two irrational numbers whose sum and product both are rational? Justify. Ans: Yes, the sum and product of two irrational numbers are rational numbers. Assume, 4 + \sqrt 7 \& 4 - \sqrt 7 {\text{are two irrational numbers}} {{(4}} + \sqrt 7 ) + (4 - \sqrt 7 ) = 8\text{ rational} (4 + \sqrt 7 ) \times (4 - \sqrt 7 ) = 16 - 7 = 9\text{ rational} Hence we can conclude that two irrational numbers whose sum and product both are rational. 2. State whether the following statement is true: There is a number ${{x}}$ such that ${{{x}}^{{2}}}$ is irrational but ${{{x}}^{{4}}}$ is rational. Justify your answer by an example. Ans: True, the Statement given is true there is a number ${{x}}$ such that ${{{x}}^{{2}}}$ is irrational but ${{{x}}^{{4}}}$ is rational. Assume, $x = \sqrt[4]{2}$ ${x^2} = {(\sqrt[4]{2})^2} = \sqrt 2 \text{ an irrational number}{\text{.}}$ ${x^4} = {(\sqrt[4]{2})^4} = 2,\text{ a rational number}{\text{.}}$ Hence, the statement is “There is a number ${{x}}$ such that ${{{x}}^{{2}}}$ is irrational but ${{{x}}^{{4}}}$ is rational.” ### EXERCISE 1.2 1. Let $x$ and $y$ be rational and irrational numbers, respectively. Is $x + y$ necessarily an irrational number? Give an example in support of your answer. Ans: Yes, if ${{x}}$ and ${{y}}$ be rational and irrational numbers, respectively. Is ${{x + y}}$ necessarily an irrational number Assume, x = 5\& y = \sqrt 7 x + y = 5 + \sqrt 7 ( = p){\text{say}} {{{{p}}^{{2}}}{{ = }}{{{{(5}}{{ + }}\sqrt {{7}} {{)}}}^{{2}}}} {{{{p}}^{{2}}}{{ = }}{{{5}}^{{2}}}{{ + }}{{{{(}}\sqrt {{7}} {{)}}}^{{2}}}{{ + }}{{2}}{{(5)}}{{(}}\sqrt {{7}} {{)}}} {{{{p}}^{{2}}}{{ = }}{{25}}{{ + }}{{7}}{{ + }}{{10}}\sqrt {{7}} \Rightarrow \dfrac{{{{{p}}^{{2}}}{{ - }}{{32}}}}{{{{10}}}}{{ = }}\sqrt {{7}} } ${{x + y}}$ is an irrational number. 2. Let $x$ be rational and $y$ be irrational. Is $xy$ necessarily irrational? Justify your answer by an example. Ans: No, if ${{x}}$ be rational and ${{y}}$ be irrational then, $x{{y}}$ is not necessarily irrational If we suppose $x = 2$ be a non-zero rational and y = \sqrt 5 be an irrational. Then, x \times y = {{2}}{{ \times }}\sqrt {{5}} {{ = }}{{2}}\sqrt {{5}} {\text{ irrational}} Hence, if ${{x}}$ be rational and ${{y}}$ be irrational then, $x{{y}}$ is not necessarily irrational 3. State whether the following statements are true or false? Justify your answer (i). \dfrac{{\sqrt 2 }}{3} is a rational number. Ans: False, the division of an irrational number by a non-zero rational number gives an irrational number and here \sqrt 2 is an irrational number and 3 is a rational number. (ii). There are infinitely many integers between any two integers. Ans: False, As among two consecutive integers there is no existence of any integer. (iii). Number of rational numbers between $15$ and $18$ is finite. Ans: False, Among two rational numbers there exists an infinite number of rational numbers between them. (iv). There are numbers which cannot be written in the form \dfrac{p}{q} , $q{{ }} \ne {{ }}0 ,$ $p, q$ both are integers. Ans: True, As there exists an infinite numbers which cannot be written in the form of \dfrac{p}{q} where p and q are integers andq \ne {{0}}. (v). The square of an irrational number is always rational. Ans: False, it cannot be always true as for an irrational number {\left( {\sqrt[{{4}}]{{{3}}}} \right)^{{2}}} = \sqrt 3 which is not a rational number. (vi). $\dfrac{{\sqrt {12} }}{{\sqrt 3 }}$ is not a rational number as \sqrt {12} and \sqrt 3 are not integers. Ans: False, on solving $\dfrac{{\sqrt {{{12}}} }}{{\sqrt {{3}} }}$ we get, $\dfrac{{\sqrt {12} }}{{\sqrt 3 }} = \dfrac{{\sqrt {4 \times 3} }}{{\sqrt 3 }} = \dfrac{{\sqrt 4 \times \sqrt 3 }}{{\sqrt 3 }} = 2 \times 1 = 2$ . Here, 2 is a rational number. (vii). \dfrac{{\sqrt {15} }}{{\sqrt 3 }} is written in the form \dfrac{p}{q}, $q{\text{ }} \ne {\text{ }}0$ and so it is a rational number. Ans: False, on solving \dfrac{{\sqrt {{{15}}} }}{{\sqrt {{3}} }} we get, $\dfrac{{\sqrt {15} }}{{\sqrt 3 }} = \dfrac{{\sqrt {5 \times 3} }}{{\sqrt 3 }} = \dfrac{{\sqrt 5 \times \sqrt 3 }}{{\sqrt 3 }} = \sqrt 5$. Here, \sqrt 5 is an irrational number. 4. Classify the following numbers as rational or irrational with justification: (i). \sqrt {196} Ans: The given number is \sqrt {{{196}}} Here, \sqrt {{{196}}} {{ = }}\sqrt {{{\left( {{{14}}} \right)}^{{2}}}} {{ = }}{{14}} , rational number. \sqrt {{{196}}} is a rational number. (ii). 3\sqrt {18} Ans: The given number is {{3}}\sqrt {{{18}}} ${{3}}\sqrt {{{18}}} {{ = }}{{3}}\sqrt {{{{{(3)}}}^{{2}}}{{ \times }}{{2}}} {{ = }}{{3}}{{ \times }}{{3}}\sqrt {{2}} {{ = }}{{9}}\sqrt {{2}}$ $\sqrt {{2}}$ is a non-repeating and non-terminating number. $\Rightarrow {{9}}\sqrt {{2}}$ is a non-repeating and non-terminating number. $\Rightarrow {{3}}\sqrt {{{18}}}$ is a non-repeating and non-terminating number. Hence, the number ${{3}}\sqrt {{{18}}}$cannot be represented as a ratio of two integers. Therefore, the number ${{3}}\sqrt {{{18}}}$ is an irrational number. (iii). \sqrt {\dfrac{9}{{27}}} Ans: The given number is\sqrt {\dfrac{{{9}}}{{{{27}}}}}. $\sqrt {\dfrac{{{9}}}{{{{27}}}}} {{ = }}\sqrt {\dfrac{{{9}}}{{{{9}}{{ \times }}{{3}}}}} {{ = }}\dfrac{{{1}}}{{\sqrt {{3}} }}$ $\sqrt {{3}}$ is an irrational number. $\Rightarrow \dfrac{{{1}}}{{\sqrt {{3}} }}$ is an irrational number. \Rightarrow \sqrt {\dfrac{{{9}}}{{{{27}}}}}is an irrational number. (iv). \sqrt {\dfrac{{28}}{{343}}} Ans: The given number is \sqrt {\dfrac{{{{28}}}}{{{{343}}}}} $\dfrac{{\sqrt {{{28}}} }}{{\sqrt {{{343}}} }}{{ = }}\dfrac{{\sqrt {{{2}}{{ \times }}{{2}}{{ \times }}{{7}}} }}{{\sqrt {{{7}}{{ \times }}{{7}}{{ \times }}{{7}}} }}{{ = }}\dfrac{{{{2}}\sqrt {{7}} }}{{{{7}}\sqrt {{7}} }}{{ = }}\dfrac{{{2}}}{{{7}}}$, simple fraction And we know that rational numbers are the real numbers that can be written in simple fractions. The number \sqrt {\dfrac{{{{28}}}}{{{{343}}}}} is a rational number. (v). - \sqrt {0.4} Ans: The given number is - \sqrt {{{0}}{{.4}}} $- \sqrt {{{0}}{{.4}}} {{ = }} - \sqrt {\dfrac{{{4}}}{{{{10}}}}} {{ = }} - \dfrac{{{2}}}{{\sqrt {{{10}}} }}$ Here, $\sqrt {{{10}}}$ is an irrational number. $\Rightarrow - \dfrac{2}{{\sqrt {{{10}}} }}$ is an irrational number. \Rightarrow - \sqrt {{{0}}{{.4}}}is an irrational number. (vi). \dfrac{{\sqrt {12} }}{{\sqrt {75} }} Ans: The given number is \dfrac{{\sqrt {{{12}}} }}{{\sqrt {{{75}}} }}. $\dfrac{{\sqrt {{{12}}} }}{{\sqrt {{{75}}} }}{{ = }}\dfrac{{\sqrt {{{4 \times 3}}} }}{{\sqrt {{{25 \times 3}}} }}{{ = }}\dfrac{{\sqrt {{4}} \sqrt {{3}} }}{{\sqrt {{{25}}} \sqrt {{3}} }}{{ = }}\dfrac{{{2}}}{{{5}}}$, simple fraction And we know that rational numbers are the real numbers that can be written in simple fractions. The number \dfrac{{\sqrt {{{12}}} }}{{\sqrt {{{75}}} }} is rational number. (vii). $0.5918$ Ans: The given number is ${{0}}{{.5918}}$. ${{0}}{{.5918}}$ is a terminating decimal so it can be written in the form of \dfrac{p}{q} where p and q are integers andq \ne {{0}} Hence, the number 0.5918 is a rational number. (viii). $\left( {1 + \sqrt 5 } \right){{ - }}\left( {4 + \sqrt 5 } \right)$ Ans: The given number is $\left( {{{1}}{{ + }}\sqrt {{5}} } \right) - \left( {{{4}}{{ + }}\sqrt {{5}} } \right)$ $(1 + \sqrt 5 ) - (4 + \sqrt 5 ) = 1 - 4 + \sqrt 5 - \sqrt 5 = - 3$, rational number. Hence, $\left( {{{1}}{{ + }}\sqrt {{5}} } \right) - \left( {{{4}}{{ + }}\sqrt {{5}} } \right)$ is rational number. (ix). $10.124124...$ Ans: The given number is ${{10}}{{.124124}}...$ When any rational number is expressed in the form of decimal expansion then it is either terminating or non-terminating recurring. ${{10}}{{.124124}}...$ is a non-terminating recurring decimal expansion so is a rational number. (x). $1.010010001...$ Ans: The given number is ${{1}}{{.010010001}}...$ ${{1}}{{.010010001}}...$ is a non-terminating non-recurring decimal expansion so is an irrational number. ${{1}}{{.010010001}}...$ is an irrational number. ### Sample Questions 1. Locate \sqrt {13} on the number line. Ans: (i). Write 13 as a square of two natural numbers ${{13}}{{ = }}{{9}}{{ + }}{{4 = }}{{{3}}^{{2}}}{{ + }}{{{2}}^{{2}}}$ (ii). Draw OA of measurement {{3}}{\text{units}} on the number line and then BA of {{2}}{\text{units}} in perpendicular to OA. (iii). Taking O as a centre and radius equal to OB, draw an arc which will intersect the number line at point P. (iv). A right triangle with known leg lengths determines a hypotenuse that is of equal length to the number you want to place on the number line. (v). Point P is refer to \sqrt {{{13}}} 2. Express $0.12\overline 3$ in the form \dfrac{p}{q} where $p$ and $q$ are integers and $q{\text{ }} \ne 0.$ Ans: Assume, ${{x}}{{ = 0}}{{.12\bar 3}}$ 10x = 1.2\overline 3 {{10x - x = 1}}{{.2}}\overline {{3}} {{ - 0}}{{.12}}\overline {{3}} {{ = 1}}{{.233}}\overline {{3}} {{ }}...{{ - 0}}{{.1233}}\overline {{3}} {{ }}... {{9x = 1}}{{.11}} {{x = }}\dfrac{{{{1}}{{.11}}}}{{{9}}} {{x}}{{ = }}\dfrac{{{{111}}}}{{{{900}}}} {{x}}{{ = }}\dfrac{{{{37}}}}{{{{300}}}} Hence, the number ${{0}}{{.12}}\overline 3$ in the form \dfrac{p}{q}is $\dfrac{{{{37}}}}{{{{300}}}}$ . 3. Simplify: (3\sqrt 5 - 5\sqrt 2 )(4\sqrt 5 + 3\sqrt 2 ) Ans: The given expression is {{(3}}\sqrt {{5}} {{ - }}{{5}}\sqrt {{2}} {{)}}{{(4}}\sqrt {{5}} {{ + }}{{3}}\sqrt {{2}} {{)}} \Rightarrow {{12}}{{ \times }}{{5}}{{ - }}{{20}}\sqrt {{2}} {{ \times }}\sqrt {{5}} {{ + }}{{9}}\sqrt {{5}} {{ \times }}\sqrt {{2}} {{ - }}{{15 \times 2}} {\text{Add the like terms, we get:}} \Rightarrow {{60}}{{ - }}{{20}}\sqrt {{{10}}} {{ + }}{{9}}\sqrt {{{10}}} {{ - }}{{30}} \Rightarrow {{30}}{{ - }}{{11}}\sqrt {{{10}}} Hence, {{(3}}\sqrt {{5}} {{ - 5}}\sqrt {{2}} {{)}}{{(4}}\sqrt {{5}} {{ + 3}}\sqrt {{2}} {{) = 30}}{{ - }}{{11}}\sqrt {{{10}}} 4. Find the value of $a$ in the following: \dfrac{6}{{3\sqrt {2} - 2\sqrt 3 }} = 3\sqrt 2 - a\sqrt 3 Ans: Given expression is \dfrac{{{6}}}{{{{3}}\sqrt {{{2}}} - {{2}}\sqrt {{3}} }}{{ = }}{{3}}\sqrt {{2}} - {{a}}\sqrt {{3}} Consider the LHS. \dfrac{{{{6}}}}{{{{3}}\sqrt {{2}} {{ - }}{{2}}\sqrt {{3}} }} Multiply the numerator and denominator by {{3}}\sqrt {{2}} {{ + }}{{2}}\sqrt {{3}} , we get: \Rightarrow \dfrac{{{6}}}{{{{3}}\sqrt {{{2}}} {{ - }}{{2}}\sqrt {{3}} }}{{ \times }}\dfrac{{{{3}}\sqrt {{2}} {{ + }}{{2}}\sqrt {{3}} }}{{{{3}}\sqrt {{2}} {{ + }}{{2}}\sqrt {{3}} }} On solving, \Rightarrow \dfrac{{{{6}}{{(3}}\sqrt {{2}} {{ + }}{{2}}\sqrt {{3}} {{)}}}}{{{{{{(3}}\sqrt {{2}} {{)}}}^{{2}}}{{ - }}{{{{(2}}\sqrt {{3}} {{)}}}^{{2}}}}} \Rightarrow \dfrac{{{{6}}{{(3}}\sqrt {{{2}}} {{ + }}{{2}}\sqrt {{3}} {{)}}}}{{{{18}}{{ - }}{{12}}}} \Rightarrow \dfrac{{{{6}}{{(3}}\sqrt {{{2}}} {{ + }}{{2}}\sqrt {{3}} {{)}}}}{{{6}}} = {{3}}\sqrt {{2}} {{ + }}{{2}}\sqrt {{3}} On equating L.H.S and R.H.S {{3}}\sqrt {{2}} {{ + }}{{2}}\sqrt {{3}} {{ = }}{{3}}\sqrt {{2}} - {{a}}\sqrt {{3}} {\text{Comparing the like terms, we get:}} a = - 2 5. Simplify: {\left[ {5{{\left( {{8^{\dfrac{1}{3}}} + 2{7^{\dfrac{1}{3}}}} \right)}^3}} \right]^{\dfrac{1}{4}}} Ans: {\left( {{a^p}} \right)^q} = {\left( a \right)^{pq}} {\left[ {{{5}}{{\left( {{{{8}}^{\dfrac{{{1}}}{{{3}}}}}{{ + }}{{2}}{{{7}}^{\dfrac{{{1}}}{{{3}}}}}} \right)}^{{3}}}} \right]^{\dfrac{{{1}}}{{{4}}}}} {{ = }}{\left[ {{{5}}{{\left( {{{\left( {{{{2}}^{{3}}}} \right)}^{\dfrac{{{1}}}{{{3}}}}}{{ + }}{{\left( {{{{3}}^{{3}}}} \right)}^{\dfrac{{{1}}}{{{3}}}}}} \right)}^{{3}}}} \right]^{\dfrac{{{1}}}{{{4}}}}} {{ = }}{\left[ {{{5}}{{{{(2}}{{ + }}{{3)}}}^{{3}}}} \right]^{\dfrac{{{1}}}{{{4}}}}} {{ = }}{\left[ {{{5}}{{{{(5)}}}^{{3}}}} \right]^{\dfrac{{{1}}}{{{4}}}}} {{ = }}{\left[ {{{{5}}^{{4}}}} \right]^{\dfrac{{{1}}}{{{4}}}}}{{ = }}{{5}} ### EXERCISE 1.3 1. Find which of the variables $x$,$y$,$z$ and $u$ represent rational numbers and which irrational numbers: (i) {x^2} = 5 Ans: {x^2} = 5 \therefore x = \pm \sqrt 5 ,{\text{ irrational number}} So, $x$ represents irrational number (ii){y^2} = 9 Ans: {y^2} = 9 \therefore y = \pm 3,{\text{ rational number}} So,$y$ represents rational number (iii) ${z^2} = .04$ Ans: {z^2} = .04 z = \pm 0.2,{\text{ rational number}} So, $z$ represents rational number (iv) ${u^2} = \dfrac{{17}}{4}$ Ans: {u^2} = \dfrac{{17}}{4} \therefore u = \pm \sqrt {\dfrac{{17}}{4}} \therefore u = \pm \dfrac{{\sqrt {17} }}{2},{\text{ irrational number}} So, $u$ represents irrational number 2. Find three rational numbers between (i) $- 1$ and $2$ Ans: Rational numbers can be easily found between $- 1$ and ${\text{2}}$ and those are $\dfrac{1}{2}$,$\dfrac{2}{4}$ and $\dfrac{4}{8}$ (ii) $0.1$ and $0.11$ Ans: $0.1 = \dfrac{1}{{10}}$ and $0.11 = \dfrac{{11}}{{100}}$ So, $\dfrac{1}{{10}} \times \dfrac{{100}}{{100}} = \dfrac{{100}}{{1000}}$ and $\dfrac{{11}}{{100}} \times \dfrac{{10}}{{10}} = \dfrac{{110}}{{1000}}$ Hence, three rational numbers between $\dfrac{{100}}{{1000}}$ and $\dfrac{{110}}{{1000}}$ are $\dfrac{{102}}{{1000}}$,$\dfrac{{104}}{{1000}}$ and $\dfrac{{106}}{{1000}}$. (iii) $\dfrac{5}{7}$ and $\dfrac{6}{7}$ Ans: $\dfrac{5}{7} \times \dfrac{{10}}{{10}} = \dfrac{{50}}{{70}}$ and $\dfrac{6}{7} \times \dfrac{{10}}{{10}} = \dfrac{{60}}{{70}}$ Therefore, three rational numbers between $\dfrac{{50}}{{70}}$ and $\dfrac{{60}}{{70}}$ are $\dfrac{{51}}{{70}}$, $\dfrac{{52}}{{70}}$ and $\dfrac{{53}}{{70}}$ (iv) $\dfrac{1}{4}$ and $\dfrac{1}{5}$ Ans: $\dfrac{1}{4} \times \dfrac{{50}}{{50}} = \dfrac{{50}}{{200}}$ and $\dfrac{1}{5} \times \dfrac{{40}}{{40}} = \dfrac{{40}}{{200}}$ So, three rational numbers between $\dfrac{{50}}{{200}}$ and $\dfrac{{40}}{{200}}$ are $\dfrac{{41}}{{200}}$, $\dfrac{{42}}{{200}}$ and $\dfrac{{43}}{{200}}$ 3. Insert a rational number and an irrational number between the following: (i). $2$ and $3$ Ans: The given numbers are 2 and 3. Rational number between $2$ and $3$ is 2.1 Irrational number between $2$ and $3$ is 2.050050005…… (ii) $0$ and $0.1$ Ans: The given numbers are 0 and 0.1. Rational number between 0 and 1is 0.05 Irrational number between 0 and 1 is 0.003000300003…… (iii) $\dfrac{1}{3}$ and $\dfrac{1}{2}$ Ans: The given numbers are $\dfrac{1}{3}$ and $\dfrac{1}{2}$. Rational number between $\dfrac{1}{3}$ and $\dfrac{1}{2}$ is $\dfrac{5}{{12}}$. Irrational number between $\dfrac{1}{3}$ and $\dfrac{1}{2}$ is 0.4242242224…….. (iv) $\dfrac{{ - 2}}{5}$ and $\dfrac{1}{2}$ Ans: The given numbers are $\dfrac{{ - 2}}{5}$ and $\dfrac{1}{2}$ Rational numbers between $\dfrac{{ - 2}}{5}$ and $\dfrac{1}{2}$ is 0. Irrational number between $\dfrac{{ - 2}}{5}$ and $\dfrac{1}{2}$ is 0.272272227…. (v) $0.15$ and $0.16$ Ans: The given numbers are $0.15$ and $0.16$. Rational number between $0.15$ and $0.16$ is 0.153 Irrational number between $0.15$ and $0.16$ is 0.1535535553. (vi) $\sqrt 2$ and $\sqrt 3$ Ans: The given numbers are $\sqrt 2$ and $\sqrt 3$. i.e. The given numbers are 1.4142…. and 1.7320 …. Rational number between 1.4142…. and 1.7320 ….is 1.5656 Irrational number between 1.4142…. and 1.7320 …. Is 1.565565556….. (vii) $2.357$ and $3.121$ Ans: The given numbers are $2.357$ and $3.121$. Rational number between $2.357$ and $3.121$ is 2.8 Irrational number between $2.357$ and $3.121$ is 2.858858885…. (viii) $.0001$ and $.001$ Ans: The given numbers are $0.0001$ and $0.001$. Rational number between $0.0001$ and $0.001$ is 0.00022 Irrational number between $0.0001$ and $0.001$ is 0.0002232223…… (ix) $3.623623$ and $0.484848$ Ans: The given numbers are $3.623623$ and $0.484848$ Rational number between $3.623623$ and $0.484848$ is 2. Irrational number between $3.623623$ and $0.484848$ is 2.101101110…… (x) $6.375289$ and $6.375738$ Ans: The given numbers are $6.375289$ and $6.375738$. Rational number between $6.375289$ and $6.375738$ is 6.3753.. Irrational number between $6.375289$ and $6.375738$ is 6.375414114111……… 4. Represent the following numbers on the number line: 7,7.2,\dfrac{{ - 3}}{2},\dfrac{{ - 12}}{5} Ans: Firstly we draw a number line whose midpoint is O. Mark positive numbers on the right hand side of O and negative numbers on the 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 from 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 - \dfrac{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 - \dfrac{{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. 5. Locate \sqrt 5 ,\sqrt {10} and \sqrt {17} on the number line. Ans: Presentation of \sqrt 5 on number line: We write 5 as the sum of the square of two natural numbers: 5 = 1 + 4 = {1^2} + {2^2} On the number line, take {\text{OA}} = 2 units. Draw {\text{BA}} = 1 unit, perpendicular to {\text{OA}}. Join {\text{OB}}. By Pythagoras theorem, {\text{OB}} = \sqrt 5 Using a compass with centre {\text{O}} and radius {\text{OB}}, draw an arc which intersects the number line at the point C. Then, C corresponds to \sqrt 5. Presentation of \sqrt {10} on the number line: We write 10 as the sum of the square of two natural numbers: 10 = 1 + 9 = {1^2} + {3^2} On the number line, take {\text{OA}} = 3 units. Draw {\text{BA}} = 1 unit, perpendicular to {\text{OA}}, Join {\text{OB}}. By Pythagoras theorem, {\text{OB}} = \sqrt {10} Using a compass with centre {\text{O}} and radius {\text{OB}}, draw an arc which intersects the number line at the point C. Then, C corresponds to \sqrt {10}. Presentation of \sqrt {17} on the number line: We write 17 as the sum of the square of two natural numbers. 17 = 1 + 16 = {1^2} + {4^2} On the number line, take {\text{OA}} = 4 units. Draw {\text{BA}} = 1 units, perpendicular to {\text{OA}}. Join {\text{OB}}. By Pythagoras theorem, {\text{OB}} = \sqrt {17} Using a compass with centre {\text{O}} and radius {\text{OB}}, draw an arc which intersects the number line at the point C. Then, C corresponds to \sqrt {17}. 6. Represent geometrically the following numbers on the number line: (i) \sqrt {4.5} Ans: Mark the distance 4.5 units from a fixed point A on a given line to obtain a point B such that AB = 4.5 units. {\text{ From }}B,{\text{ mark a distance of }}1{\text{ unit and mark the new point as }}C{\text{. }} Find the midpoint of AC and mark that point as O. Draw a semicircle with centre O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D. Then, BD = \sqrt 4 .5. Draw an arc with centre B and radius B D, meeting AC produced at E, then BE = BD = \sqrt 4 .5 units. (ii) \sqrt {5.6} Ans: 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. {\text{ From B, mark a distance of }}1{\text{ unit and mark the new point as C}}{\text{. }} Find the midpoint of A C and mark that point as O. Draw a semicircle with centre O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D. Then, BD = \sqrt 5 .6. Draw an arc with centre B and radius B D, meeting A C produced at E, then BE = BD = \sqrt 5 \cdot 6 units. (iii) \sqrt {8.1} Ans: 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. {\text{ From B, mark a distance of }}1{\text{ unit and mark the new point as C}}{\text{. }} Find the midpoint of A C and mark that point as O. Draw a semicircle with centre O and radius OC. Draw a line perpendicular to A C passing through B and intersecting the semicircle at D. Then, BD = \sqrt {8.1}. Draw an arc with centre B and radius B D, meeting A C produced at E, then BE = BD = \sqrt {8.1} units. (iv) \sqrt {2.3} Ans: Mark the distance 2.3 units from a fixed point A on a given line to obtain a point B such that AB = 2.3 units. {\text{ From B, mark a distance of }}1{\text{ unit and mark the new point as C}}{\text{. }} Find the midpoint of A C and mark that point as O. Draw a semicircle with centre O and radius OC. Draw a line perpendicular to A C passing through B and intersecting the semicircle at D. Then, BD = \sqrt {2.3}. Draw an arc with centre B and radius B D, meeting A C produced at E, then BE = BD = \sqrt {2.3} units. 7. Express the following in the form \dfrac{p}{q}, where p and q are integers and q \ne 0 (i) 0.2 Ans: 0.2 0.2 = \dfrac{2}{{10}} \Rightarrow 0.2 = \dfrac{1}{5} (ii) 0.888… Ans: 0.888… Let x = 0.888 \ldots \Rightarrow x = 0.8 Multiplying both sides by 10, we get 10x = 8.8 \ldots \ldots \ldots \ldots \ldots (2) Subtracting equation (1) from equation (2), we get 10x - x = 8.8 - 0.8 \Rightarrow 9x = 8.0 \Rightarrow {\text{x}} = \dfrac{8}{9} (iii) 5.\bar 2 Ans: 5.\bar 2 Let x = 5.2 \ldots \ldots \ldots Multiplying both sides by 10 , we get 10x = 52.2 \ldots \ldots \ldots \ldots \ldots ..(2) Subtracting equation (1) from equation (2), we get 10x - x = 52.2 - 5.2 \Rightarrow 9x = 47 \Rightarrow x = \dfrac{{47}}{9} (iv) 0.\overline {001} Ans: 0.\overline {001} Let x = 0.001. Multiplying both sides by 1000 , we get 1000x = 1.001 Subtracting equation (1) from equation (2), we get 1000x - x = 1.001 - 0.001 \Rightarrow 999x = 1 \Rightarrow {{x}} = \dfrac{1}{{999}} (v) 0.2555… Ans: 0.2555… Let x = 0.2555 \ldots \Rightarrow x = 0.25. Multiplying both sides by 10 , we get 10x = 2.5 \ldots \ldots \ldots \ldots \ldots .(2) Multiplying both sides by 100 , we get 100x = 25.5 \ldots \ldots \ldots \ldots .(3) Subtracting equation (2) from equation (3), we get 100x - 10x = 25.5 - 2.5 \Rightarrow 90x = 23 \Rightarrow {{x}} = \dfrac{{23}}{{90}} (vi) 0.1\overline {34} Ans: 0.1\overline {34} Let x = 0.134 \ldots \ldots \ldots \ldots Multiplying both sides by 10 , we get 10x = 1.34.. Multiplying both sides by 1000 , we get 1000x = 134.34 \ldots \ldots \ldots \ldots (3) Subtracting equation (2) from equation (3), we get 1000x - 10x = 134.34 - 1.34 \Rightarrow 990x = 133 \Rightarrow {{x}} = \dfrac{{133}}{{990}} (vii) .00323232 \ldots Ans: .00323232 \ldots Let x = 0.00323232 \ldots \Rightarrow x = 0.0032 Multiplying both sides by 100 , we get 100 100x = 0.32 \ldots \ldots \ldots \ldots \ldots (2) Multiplying both sides by 10000 , we get 10000x = 32.32 \ldots \ldots \ldots \ldots ..(3) Subtracting equation (2) from equation (3), we get 10000x - 100x = 32.32 - 0.32 \Rightarrow 9900x = 32 \Rightarrow {{x}} = \dfrac{{32}}{{9900}} = \dfrac{8}{{2475}} (viii) .404040 \ldots Ans: .404040 \ldots Let x = 0.404040 \ldots \Rightarrow x = 0.40 \ldots \ldots \ldots \ldots \ldots \ldots .(1) Multiplying both sides by 100 , we get 100x = 40.40 \ldots \ldots \ldots \ldots (2) Subtracting equation (1) from equation (2), we get 100x - x = 40.40 - 0.40 \Rightarrow 99x = 40 \Rightarrow {{x}} = \dfrac{{40}}{{99}} 8. Show that 0.142857142857 \ldots = \dfrac{1}{7} Ans: Let x = 0.142857142857 \ldots \Rightarrow x = 0.142857,m - x Multiplying both sides by 1000000 , we get 1000000x = 142857.142857 \ldots \ldots \ldots \ldots ..(2) Subtracting equation (1) from equation (2), we get 1000000x - x = 142857.142857 - 0.142857 \Rightarrow 999999x = 142857 \Rightarrow {{x}} = \dfrac{{142857}}{{999999}} = \dfrac{1}{7} Hence, proved. 9. Simplify the following: (i) \sqrt {45} - 3\sqrt {20} + 4\sqrt 5 Ans: \sqrt {45} - 3\sqrt {20} + 4\sqrt 5 {\sqrt {45} - 3\sqrt 2 0 + 4\sqrt 5 } { = \sqrt 3 \times \sqrt 3 \times \sqrt 5 - 3\sqrt 2 \times \sqrt 2 \times \sqrt 5 + 4\sqrt 5 } { = 3\sqrt 5 - 6\sqrt 5 + 4\sqrt 5 } { = \sqrt 5 } (ii) \dfrac{{\sqrt {24} }}{8} + \dfrac{{\sqrt {54} }}{9} Ans: \dfrac{{\sqrt {24} }}{8} + \dfrac{{\sqrt {54} }}{9} \dfrac{{\sqrt 2 4}}{8} + \dfrac{{\sqrt {54} }}{9} = \dfrac{{\sqrt {2 \times 2 \times 2 \times 3} }}{8} + \dfrac{{\sqrt {2 \times 3 \times 3 \times 3} }}{9} = \dfrac{{2\sqrt 6 }}{8} + \dfrac{{3\sqrt 6 }}{9} = \dfrac{{\sqrt 6 }}{4} + \dfrac{{\sqrt 6 }}{3} = \dfrac{{3\sqrt 6 + 4\sqrt 6 }}{{12}} = \dfrac{{7\sqrt 6 }}{{12}} (iii) \sqrt[4]{{12}} \times \sqrt[7]{6} Ans: \sqrt[4]{{12}} \times \sqrt[7]{6} {4\sqrt {12} \times 7\sqrt 6 } { = (4 \times \sqrt 6 \times \sqrt 2 ) \times (7 \times \sqrt 6 )} { = (4 \times 6 \times 7 \times \sqrt 2 ) = 168\sqrt 2 } (iv) 4\sqrt {28} \div 3\sqrt 7 \div \sqrt[3]{7} Ans: 4\sqrt {28} \div 3\sqrt 7 \div \sqrt[3]{7} = 4\sqrt {4 \times 7} \times \dfrac{1}{{3\sqrt 7 \times \sqrt[3]{7}}} = 8\sqrt 7 \times \dfrac{1}{{3\sqrt 7 \times \sqrt[3]{7}}} = \dfrac{8}{{3\sqrt[3]{7}}} (v) 3\sqrt 3 + 2\sqrt {27} + \dfrac{7}{{\sqrt 3 }} Ans: 3\sqrt 3 + 2\sqrt {27} + \dfrac{7}{{\sqrt 3 }} {3\sqrt 3 + 2\sqrt {27} + \dfrac{7}{{\sqrt 3 }}} { = 3\sqrt 3 + 2\sqrt {9 \times 3} + \dfrac{7}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }}} { = 3\sqrt 3 + 6\sqrt 3 + \dfrac{{7\sqrt 3 }}{3}} { = \dfrac{{3 \times 9\sqrt 3 + 7\sqrt 3 }}{3}} { = \dfrac{{34\sqrt 3 }}{3}} (vi) {(\sqrt 3 - \sqrt 2 )^2} Ans: {(\sqrt 3 - \sqrt 2 )^2} = {(\sqrt 3 )^2} + {(\sqrt 2 )^2} - 2(\sqrt 3 )(\sqrt 2 )$$\left\{ {} \right.$ Using $\left. {{{(a - b)}^2} = {a^2} + {b^2} - 2ab} \right\}$

$= 3 + 2 - 2\sqrt 6$$= 5 - 2\sqrt 6 (vii) +\sqrt[4]{{81}} - 8\sqrt[3]{{216}} + 15\sqrt[5]{{32}} + \sqrt {225} Ans: +\sqrt[4]{{81}} - 8\sqrt[3]{{216}} + 15\sqrt[5]{{32}} + \sqrt {225} {\sqrt[4]{{81}} - 8\sqrt[3]{{216}} + 15\sqrt[5]{{32}} + \sqrt {225} } { = \sqrt[4]{{3 \times 3 \times 3 \times 3}} - 8\sqrt[3]{{6 \times 6 \times 6}} + 15\sqrt[5]{{2 \times 2 \times 2 \times 2 \times 2}} + \sqrt {15 \times 15} } { = 3 - 8 \times 6 + 15 \times 2 + 15} { = 3 - 48 + 30 + 15} { = - 45 + 45} { = 0} (viii) \dfrac{3}{{\sqrt 8 }} + \dfrac{1}{{\sqrt 2 }} Ans: \dfrac{3}{{\sqrt 8 }} + \dfrac{1}{{\sqrt 2 }} {\dfrac{3}{{\sqrt 8 }} + \dfrac{1}{{\sqrt 2 }}} { = \dfrac{3}{{\sqrt {4 \times 2} }} + \dfrac{1}{{\sqrt 2 }}} { = \dfrac{3}{{2\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}} { = \dfrac{{3 + 2}}{{2\sqrt 2 }}} { = \dfrac{5}{{2\sqrt 2 }}} (ix) \dfrac{{2\sqrt 3 }}{3} - \dfrac{{\sqrt 3 }}{6} Ans: \dfrac{{2\sqrt 3 }}{3} - \dfrac{{\sqrt 3 }}{6} {\dfrac{{2\sqrt 3 }}{3} - \dfrac{{\sqrt 3 }}{6}} { = \dfrac{{4\sqrt 3 - \sqrt 3 }}{6}} { = \dfrac{{3\sqrt 3 }}{6} = \dfrac{{\sqrt 3 }}{2}} 10. Rationalise the denominator of the following: (i) \dfrac{2}{{3\sqrt 3 }} Ans: {\dfrac{2}{{3\sqrt 3 }}} { = \dfrac{2}{{3\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} = \dfrac{{2\sqrt 3 }}{9}} (ii) \dfrac{{\sqrt {40} }}{{\sqrt 3 }} Ans: {\dfrac{{\sqrt {40} }}{{\sqrt 3 }}} { = \dfrac{{2\sqrt {10} }}{{\sqrt 3 }} = \dfrac{{2\sqrt {10} }}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} = \dfrac{{2\sqrt {30} }}{3}} (iii) \dfrac{{3 + \sqrt 2 }}{{4\sqrt 2 }} Ans: {\dfrac{{3 + \sqrt 2 }}{{4\sqrt 2 }}} { = \dfrac{{(3 + \sqrt 2 )}}{{4\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }} = \dfrac{{3\sqrt 2 + 2}}{8}} (iv) \dfrac{{16}}{{\sqrt {41} - 5}} Ans: {\dfrac{{16}}{{\sqrt {41} - 5}}} { = \dfrac{{16}}{{\sqrt {41} - 5}} \times \dfrac{{\sqrt {41} + 5}}{{\sqrt {41} + 5}}} {\left\{ {{\text{ Using }}({\text{a}} - {\text{b}})({\text{a}} + {\text{b}}) = {{\text{a}}^2} - {{\text{b}}^2}} \right\}} { = \dfrac{{16(\sqrt {41} + 5)}}{{41 - 25}}} { = \dfrac{{16\sqrt {41} + 80}}{{16}}} { = \sqrt {41} + 5} (v) \dfrac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} Ans: \dfrac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} \times \dfrac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }}$$\left\{ {} \right.$ Using $(a - b)(a + b) = {a^2} - {b^2}$ and $\left. {{{(a + b)}^2} = {a^2} + {b^2} + 2ab} \right\}$

$= \dfrac{{4 + 3 + 4\sqrt 3 }}{{4 - 3}}= 7 + 4\sqrt 3$

(vi) $\dfrac{{\sqrt 6 }}{{\sqrt 2 + \sqrt 3 }}$

Ans: ${\dfrac{{\sqrt 6 }}{{\sqrt 2 + \sqrt 3 }}}$

${ = \dfrac{{\sqrt 6 }}{{\sqrt 2 + \sqrt 3 }} \times \dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}}$

${\left\{ {{\text{ Using }}({\text{a}} - {\text{b}})({\text{a}} + {\text{b}}) = {{\text{a}}^2} - {{\text{b}}^2}} \right\}}$

${ = \dfrac{{\sqrt {18} - \sqrt {12} }}{{3 - 2}}}$

${ = \dfrac{{3\sqrt 2 - 2\sqrt 3 }}{1}}$

${ = 3\sqrt 2 - 2\sqrt 3 }$

(vii) $\dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$

Ans: $\dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$

$= \dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times \dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$$\left\{ {} \right. Using (a - b)(a + b) = {a^2} - {b^2} and \left. {{{(a + b)}^2} = {a^2} + {b^2} + 2ab} \right\}$$= \dfrac{{3 + 2 + 2\sqrt 6 }}{{3 - 2}}$

$= 5 + 2\sqrt 6$

(viii) $\dfrac{{3\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}$

Ans: ${\dfrac{{3\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}}$

${ = \left( {\dfrac{{3\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}} \right) \times \dfrac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}}$

${\left\{ {{\text{ Using }}({\text{a}} - {\text{b}})({\text{a}} + {\text{b}}) = {{\text{a}}^2} - {{\text{b}}^2}} \right\}}$

${ = \dfrac{{15 + \sqrt {15} + 3\sqrt {15} + 3}}{{5 - 3}}}$

${ = \dfrac{{18 + 4\sqrt {15} }}{2}}$

${ = 9 + 2\sqrt {15} }$

(ix) $\dfrac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {48} + \sqrt {18} }}$

Ans: ${\dfrac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {48} + \sqrt {18} }}}$

${ = \dfrac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {4 \times 4 \times 3} + \sqrt {9 \times 2} }}}$

${ = \dfrac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }}}$

${ = \dfrac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} \times \dfrac{{4\sqrt 3 - 3\sqrt 2 }}{{4\sqrt 3 - 3\sqrt 2 }}}$

${\left\{ {{\text{ Using }}(a - b)(a + b) = {a^2} - {b^2}} \right\}}$

${ = \dfrac{{16 \times 3 + 20\sqrt 6 - 12\sqrt 6 - 30}}{{48 - 18}}}$

${ = \dfrac{{18 + 8\sqrt 6 }}{{30}}}$

${ = \dfrac{{9 + 4\sqrt 6 }}{{15}}}$

11. Find the values of $a$ and $b$ in each of the following:

(i) $\dfrac{{5 + 2\sqrt 3 }}{{7 + 4\sqrt 3 }} = a - 6\sqrt 3$

Ans: ${\dfrac{{5 + 2\sqrt 3 }}{{7 + 4\sqrt 3 }} = a - 6\sqrt 3 }$

${ \Rightarrow \dfrac{{5 + 2\sqrt 3 }}{{7 + 4\sqrt 3 }} \times \dfrac{{7 - 4\sqrt 3 }}{{7 - 4\sqrt 3 }} = a - 6\sqrt 3 }$

${\left\{ {{\text{ Using }}(a - b)(a + b) = {a^2} - {b^2}} \right\}}$

${ \Rightarrow \dfrac{{35 + 14\sqrt 3 - 20\sqrt 3 - 24}}{{49 - {{(4\sqrt 3 )}^2}}} = a - 6\sqrt 3 }$

${ \Rightarrow \dfrac{{11 - 6\sqrt 3 }}{{49 - 48}} = a - 6\sqrt 3 }$

${ \Rightarrow 11 - 6\sqrt 3 = a - 6\sqrt 3 }$

${ \Rightarrow a = 11}$

(ii) $\dfrac{{3 - \sqrt 5 }}{{3 + 2\sqrt 5 }} = a\sqrt 5 - \dfrac{{19}}{{11}}$

Ans:  ${\dfrac{{3 - \sqrt 5 }}{{3 + 2\sqrt 5 }} = a\sqrt 5 - \dfrac{{19}}{{11}}}$

${ \Rightarrow \dfrac{{3 - \sqrt 5 }}{{3 + 2\sqrt 5 }} \times \dfrac{{3 - 2\sqrt 5 }}{{3 - 2\sqrt 5 }} = a\sqrt 5 - \dfrac{{19}}{{11}}}$

${\left\{ {{\text{ Using }}(a - b)(a + b) = {a^2} - {b^2}} \right\}}$

${ \Rightarrow \dfrac{{9 - 3\sqrt 5 - 6\sqrt 5 + 10}}{{9 - {{(2\sqrt 5 )}^2}}} = a\sqrt 5 - \dfrac{{19}}{{11}}}$

${ \Rightarrow \dfrac{{19 - 9\sqrt 5 }}{{9 - 20}} = a\sqrt 5 - \dfrac{{19}}{{11}}}$

${ \Rightarrow \dfrac{{19 - 9\sqrt 5 }}{{ - 11}} = a\sqrt 5 - \dfrac{{19}}{{11}}}$

$\Rightarrow \dfrac{{9\sqrt 5 }}{{11}} - \dfrac{{19}}{{11}} = a\sqrt 5 - \dfrac{{19}}{{11}}$

$\Rightarrow a = \dfrac{9}{{11}}$

(iii).  $\dfrac{{\sqrt 2 + \sqrt 3 }}{{3\sqrt 2 - 2\sqrt 3 }} = 2 - b\sqrt 6$

Ans:  ${\dfrac{{\sqrt 2 + \sqrt 3 }}{{3\sqrt 2 - 2\sqrt 3 }} = 2 - {{b}}\sqrt 6 }$

${ \Rightarrow \dfrac{{\sqrt 2 + \sqrt 3 }}{{3\sqrt 2 - 2\sqrt 3 }} \times \dfrac{{3\sqrt 2 + 2\sqrt 3 }}{{3\sqrt 2 + 2\sqrt 3 }} = 2 - {{b}}\sqrt 6 }$

${\left\{ {{\text{ Using }}({\text{a}} - {\text{b}})({\text{a}} + {\text{b}}) = {{\text{a}}^2} - {{\text{b}}^2}} \right\}}$

${ \Rightarrow \dfrac{{6 + 2\sqrt 6 + 3\sqrt 6 + 6}}{{{{(3\sqrt 2 )}^2} - {{(2\sqrt 3 )}^2}}} = 2 - {\text{b}}\sqrt 6 }$

${ \Rightarrow \dfrac{{12 + 5\sqrt 6 }}{{18 - 12}} = 2 - {\text{b}}\sqrt 6 }$

${ \Rightarrow \dfrac{{12 + 5\sqrt 6 }}{6} = 2 - {\text{b}}\sqrt 6 }$

$\Rightarrow 2 + \dfrac{{5\sqrt 6 }}{6} = 2 - {\text{b}}\sqrt 6$

$\Rightarrow b = \dfrac{{ - 5}}{6}$

(iv) $\dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }} - \dfrac{{7 - \sqrt 5 }}{{7 + \sqrt 5 }} = a + \dfrac{7}{{11}}\sqrt 5 b$

Ans:

${\dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }} - \dfrac{{7 - \sqrt 5 }}{{7 + \sqrt 5 }} = a + \dfrac{7}{{11}}\sqrt 5 b}$

${ \Rightarrow \dfrac{{7 + \sqrt 5 \times 7 + \sqrt 5 - (7 - \sqrt {5)} \times (7 - \sqrt {5)} }}{{(7 - \sqrt 5 )(7 + \sqrt 5 )}} = a + \dfrac{7}{{11}}\sqrt 5 b}$

${\left\{ {{\text{ Using }}(a - b)(a + b) = {a^2} - {b^2}} \right\}}$

${ \Rightarrow \dfrac{{(49 + 5 + 14\sqrt 5 ) - (49 + 5 - 14\sqrt 5 )}}{{49 - 5}} = a + \dfrac{7}{{11}}\sqrt 5 b}$

${ \Rightarrow \dfrac{{14\sqrt 5 }}{{44}} = a + \dfrac{7}{{11}}\sqrt 5 b}$

${ \Rightarrow \dfrac{1}{2} \times \dfrac{{7\sqrt 5 }}{{11}} = a + \dfrac{7}{{11}}\sqrt 5 b}$

${ \Rightarrow {\text{a}} = 0,{{b}} = \dfrac{1}{2}}$

12. If  $a = 2 + \sqrt 3$, then find the value of  $a - \dfrac{1}{a}$.

Ans: Let $a = 2 + \sqrt 3$,

$\therefore$ We have $\dfrac{1}{a} = \dfrac{1}{{2 + \sqrt 3 }}$

$\Rightarrow \dfrac{1}{{\text{a}}} = \dfrac{1}{{2 + \sqrt 3 }} \times \dfrac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }}$

Using $(a - b)(a + b) = {a^2} - {b^2}$, we get

$\Rightarrow \dfrac{1}{a} = \dfrac{{2 - \sqrt 3 }}{{4 - 3}}$

$\Rightarrow \dfrac{1}{a} = 2 - \sqrt 3$

Now ${\text{a}} - \dfrac{1}{{\text{a}}} = 2 + \sqrt 3 - (2 - \sqrt 3 )$

$\Rightarrow a - \dfrac{1}{a} = 2\sqrt 3$

13. Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt 2 = 1.414,\sqrt 3 = 1.732$ and $\sqrt 5 = 2.236$, upto three places of decimal.

(i)  $\dfrac{4}{{\sqrt 3 }}$

Ans: ${ = \dfrac{4}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }}}$

${ = \dfrac{{4\sqrt 3 }}{3}}$

${ = \dfrac{{4 \times 1.732}}{3}}$

${ = 2.3093}$

(ii) $\dfrac{6}{{\sqrt 6 }}$

Ans: ${ = \dfrac{{\sqrt 6 \times \sqrt 6 }}{{\sqrt 6 }}}$

${ = \sqrt 6 }$

${ = \sqrt 2 \times \sqrt 3 }$

${ = 1.414 \times 1.732}$

${ = 2.449}$

(iii) $\dfrac{{\sqrt {10} - \sqrt 5 }}{2}$

Ans: ${ = \dfrac{{\sqrt 2 \times \sqrt 5 - \sqrt 5 }}{2}}$

${ = \dfrac{{1.414 \times 2.236 - 2.236}}{2}}$

${ = 0.462852}$

(iv) $\dfrac{{\sqrt 2 }}{{2 + \sqrt 2 }}$

Ans: $= \dfrac{{\sqrt 2 }}{{2 + \sqrt 2 }} \times \dfrac{{2 - \sqrt 2 }}{{2 - \sqrt 2 }}$

$\left\{ {} \right.$ Using $\left. {(a - b)(a + b) = {a^2} - {b^2}} \right\}$

$= \dfrac{{2\sqrt 2 - 2}}{{4 - 2}}$$= \dfrac{{2(\sqrt 2 - 1)}}{2} = \sqrt 2 - 1$$= 1.414 - 1 = 0.414$

(v) $\dfrac{1}{{\sqrt 3 + \sqrt 2 }}$

Ans: $\dfrac{1}{{\sqrt 3 + \sqrt 2 }}$

${ = \dfrac{1}{{\sqrt 3 + \sqrt 2 }} \times \dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}\left\{ {{\text{ Using }}({\text{a}} - {\text{b}})({\text{a}} + {\text{b}}) = {{\text{a}}^2} - {{\text{b}}^2}} \right\}}$

${ = \dfrac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}}}$

${ = \sqrt 3 - \sqrt 2 }$

${ = 1.732 - 1.414}$

${ = 0.318}$

14. Simplify:

(i) ${\left( {{1^3} + {2^3} + {3^3}} \right)^{\dfrac{1}{2}}}$

Ans:  ${{{\left( {{1^3} + {2^3} + {3^3}} \right)}^{\dfrac{1}{2}}}}$

${ = \sqrt {1 + 8 + 27} = \sqrt {36} = 6}$

(ii) ${\dfrac{3}{5}^4}{\dfrac{8}{5}^{ - 12}}{\dfrac{{32}}{5}^6}$

Ans: ${{{\left( {\dfrac{3}{5}} \right)}^4}{{\left( {\dfrac{8}{5}} \right)}^{ - 12}}{{\left( {\dfrac{{32}}{5}} \right)}^6}}$

${ = {{\left( {\dfrac{3}{5}} \right)}^4} \times {{\left( {\dfrac{5}{8}} \right)}^{12}} \times {{\left( {\dfrac{{32}}{5}} \right)}^6}}$

${ = {3^4} \times {5^{12 - 6 - 4}} \times {{32}^6} \times {8^{ - 12}}}$

${ = 81 \times {5^2} \times {{(2)}^{5 \times 6}} \times {2^{ - 36}}}$

${ = 81 \times 25 \times {2^{30 - 36}}}$

${\left\{ {{\text{ Using }}{a^x}.{a^y} = {a^{x + y}}} \right\}}$

${ = 2025 \times {2^{ - 6}} = \dfrac{{2025}}{{64}}}$

(iii) ${\dfrac{1}{{27}}^{\dfrac{{ - 2}}{3}}}$

Ans: ${{{\dfrac{1}{{27}}}^{\dfrac{{ - 2}}{3}}}}$

${ = {{\dfrac{1}{{{3^3}}}}^{\dfrac{{ - 2}}{3}}}}$

${ = {3^{3 \times \dfrac{2}{3}}}}$

${ = {3^2} = 9}$

(iv) ${{(625)}^{-\dfrac{1}{2}-\dfrac{1}{4}}}$

Ans:  ${{{(625)}^{\dfrac{{ - 1}}{2}}} \times \dfrac{{ - 1}}{4} \times 2}$

${ = {{(625)}^{\dfrac{1}{4}}}}$

${ = {{\left( {{5^4}} \right)}^{\dfrac{1}{4}}}}$

${ = 5}$

(v) $\dfrac{{{9^{\dfrac{1}{3}}} \times 2{7^{ - \dfrac{1}{2}}}}}{{{3^{\dfrac{1}{6}}} \times {3^{ - \dfrac{2}{3}}}}}$

Ans: $\dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{1}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}}$

$= \dfrac{{{{(3)}^{2 \times \dfrac{1}{3}}} \times {{(3)}^{3 \times \dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{1}{6} \times 3}}\dfrac{{ - 2}}{2}}}\left\{ {} \right.$ Using $\left. {{a^x}.{a^y} = {a^{x + y}}} \right\}$

$= \dfrac{{{3^{\dfrac{2}{3} - \dfrac{3}{2}}}}}{{{3^{\dfrac{1}{6} - \dfrac{2}{3}}}}} = \dfrac{{{3^{\dfrac{{4 - 9}}{6}}}}}{{{3^{\dfrac{{1 - 4}}{6}}}}}$$= \dfrac{{{3^{\dfrac{{ - 5}}{6}}}}}{{{3^{\dfrac{{ - 3}}{6}}}}} = {3^{\dfrac{{ - 5}}{6} + \dfrac{3}{6}}}$$= {3^{\dfrac{{ - 2}}{6}}}$

$= \dfrac{{{1^{\dfrac{1}{3}}}}}{3}$$= \sqrt[2]{{\dfrac{1}{3}}} (vi){{{64}^{\dfrac{{ - 1}}{3}}}\left( {{{64}^{\dfrac{1}{3}}} - {{64}^{\dfrac{2}{3}}}} \right)} Ans: {{{64}^{\dfrac{{ - 1}}{3}}}\left( {{{64}^{\dfrac{1}{3}}} - {{64}^{\dfrac{2}{3}}}} \right)} { = {4^{3 \times \dfrac{{ - 1}}{3}}}\left( {{4^{3 \times \dfrac{1}{3}}} - {4^{3 \times \dfrac{2}{3}}}} \right)} { = {4^{ - 1}}\left( {4 - {4^2}} \right)} { = \dfrac{1}{4}( - 12)} { = - 3} (vii) \dfrac{{{8^{\dfrac{1}{3}}} \times 1{6^{\dfrac{1}{3}}}}}{{3{2^{ - \dfrac{1}{3}}}}} Ans: {\dfrac{{{8^{\dfrac{1}{3}}} \times {{16}^{\dfrac{1}{3}}}}}{{{{32}^{\dfrac{{ - 1}}{3}}}}}} { = \dfrac{{{2^{3 \times \dfrac{1}{3}}} \times {2^{4 \times \dfrac{1}{3}}}}}{{{2^{\dfrac{{5 \times - 1}}{3}}}}}} = {2^{1 + \dfrac{4}{3} + \dfrac{5}{3}}}$$\left\{ {} \right.$ Using $\left. {{a^x} \cdot {a^y} = {a^{x + y}}} \right\}$

$= {2^{\dfrac{{3 + 4 + 5}}{3}}}$$= {2^{\dfrac{{12}}{3}}} = {2^4}$$= 16$

Sample Questions

1. If $a = 5 + 2\sqrt 6$ and $b = \dfrac{1}{a}$, then what will be the value of ${a^2} + {b^2}?$

Ans:  $a = 5 + 2\sqrt 6$

$b = \dfrac{1}{a} = \dfrac{1}{{5 + 2\sqrt 6 }} = \dfrac{1}{{5 + 2\sqrt 6 }} \times \dfrac{{5 - 2\sqrt 6 }}{{5 - 2\sqrt 6 }} = \dfrac{{5 - 2\sqrt 6 }}{{{5^2} - {{(2\sqrt 6 )}^2}}} = \dfrac{{5 - 2\sqrt 6 }}{{25 - 24}} = 5 - 2\sqrt 6$

Here,

${{a^2} + {b^2} = {{(a + b)}^2} - 2ab}$

${a + b = (5 + 2\sqrt 6 ) + (5 - 2\sqrt 6 ) = 10}$

${ab = (5 + 2\sqrt 6 )(5 - 2\sqrt 6 ) = {5^2} - {{(2\sqrt 6 )}^2} = 25 - 24 = 1}$

${{a^2} + {b^2} = {{10}^2} - 2 \times 1 = 100 - 2 = 98}$

### EXERCISE 1.4

1. Express $0.6 + 0.\overline 7 + 0.4\overline 7$ in the form $\dfrac{p}{q}$ where $p$ and $q$ are integers and $q \ne 0.$

Ans: Assume,

$x{\text{ = }}{{0}}{\text{.}}\overline {{7}} {\text{ = }}{{0}}{{.777}}........................{{(1)}}$

Multiply the equation (1) by 10

${{10}}x{{ = }}{{7}}{{.77}}........................{{(2)}}$

Subtract equation (1) from equation (2)

$10x - x = (7.77....) - (0.77..)$

$\Rightarrow 9x = 7$

$\Rightarrow x = \dfrac{7}{9}$

Now, let

$y = {{0}}{{.4}}\overline {{7}} {{ = }}{{0}}{{.4777}}..........................(3)$

Multiply equation (3) by 10

${{10y }}{{ = }}{{4}}{{.777}}.................{{(4)}}$

Multiply equation (4) by 10

$100y = \,47.777..................(5)$

Subtract equation (4) from (5)

$(100y - 10y) = (47.777...) - (4.777...)$

$\Rightarrow 90y = 43$

$\Rightarrow y = \dfrac{{43}}{9}$

Now, substituting the values in expression ${{0}}{{.6}}{{ + }}{{ 0}}{{.}}\overline {{7}} {{ + }}{{0}}{{.4}}\overline {{7}}$

$\Rightarrow \dfrac{{{6}}}{{{{10}}}}{{ + }}\dfrac{{{7}}}{{{9}}}{{ + }}\dfrac{{{{43}}}}{{{{90}}}}$

${{ = }}\dfrac{{{{54}}{{ + }}{{70}}{{ + }}{{43}}}}{{{{90}}}}$

${{ = }}\dfrac{{{{167}}}}{{{{90}}}}$

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

Ans: $\dfrac{{7\sqrt 3 }}{{\sqrt {10} + \sqrt 3 }} - \dfrac{{2\sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} - \dfrac{{3\sqrt 2 }}{{\sqrt {15} + 3\sqrt 2 }}$

Rationalize the terms of given expression separately,

$\Rightarrow \dfrac{{7\sqrt 3 }}{{\sqrt {10} + \sqrt 3 }} \times \dfrac{{\sqrt {10} - \sqrt 3 }}{{\sqrt {10} - \sqrt 3 }} = \dfrac{{7\sqrt {30} - 21}}{{{{(\sqrt {10} )}^2} - {{(\sqrt 3 )}^2}}}$

$\Rightarrow \dfrac{{7\sqrt {30} - 21}}{{10 - 3}} = \dfrac{{7(\sqrt {30} - 3)}}{7} = \sqrt {30} - 3$

Now, consider

$\dfrac{{2\sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} = \dfrac{{2\sqrt 5 }}{{(\sqrt 6 + \sqrt 5 )}} \times \dfrac{{(\sqrt 6 - \sqrt 5 )}}{{(\sqrt 6 - \sqrt 5 )}}$

$\Rightarrow \dfrac{{2\sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} = \dfrac{{2\sqrt {30} - 10}}{{{{(\sqrt 6 )}^2} - {{(\sqrt 5 )}^2}}}$

$\Rightarrow \dfrac{{2\sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} = \dfrac{{2\sqrt {30} - 10}}{{6 - 5}}$

$\Rightarrow \dfrac{{2\sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} = 2\sqrt {30} - 10$

Consider,

$\dfrac{{3\sqrt 2 }}{{\sqrt {15} + 3\sqrt 2 }} = \dfrac{{3\sqrt 2 }}{{\sqrt {15} + 3\sqrt 2 }} \times \dfrac{{\sqrt {15} - 3\sqrt 2 }}{{\sqrt {15} - 3\sqrt 2 }}$

$\Rightarrow \dfrac{{3\sqrt {30} - 18}}{{{{(\sqrt {15} )}^2} - {{(3\sqrt 2 )}^2}}}$

$\Rightarrow \dfrac{{3(\sqrt {30} - 6)}}{{15 - 18}} = \dfrac{{3(\sqrt {30} - 6)}}{{ - 3}} = ( - \sqrt {30} + 6) = 6 - \sqrt {30}$

Now, substituting all the values of terms in the expression,

$\dfrac{{7\sqrt 3 }}{{\sqrt {10} + \sqrt 3 }} - \dfrac{{2\sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} - \dfrac{{3\sqrt 2 }}{{\sqrt {15} + 3\sqrt 2 }}$

$= \left( {\sqrt {30} - 3} \right) - \left( {2\sqrt {30} - 10} \right) - \left( {6 - \sqrt {30} } \right)$

$= \sqrt {30} - 3 - 2\sqrt {30} + 10 - 6 + \sqrt {30}$

$= 2\sqrt {30} - 2\sqrt {30} + 10 - 9 = 1$

3. If $\sqrt 2 = 1.414,\sqrt 3 = 1.732$, then find the value of $\dfrac{4}{{3\sqrt 3 - 2\sqrt 2 }} + \dfrac{3}{{3\sqrt 3 + 2\sqrt 2 }}$

Ans: $\dfrac{4}{{3\sqrt 3 - 2\sqrt 2 }} + \dfrac{3}{{3\sqrt 3 + 2\sqrt 2 }}$

$\Rightarrow {\dfrac{{4(3\sqrt 3 + 2\sqrt 2 ) + 3(3\sqrt 3 - 2\sqrt 2 )}}{{(3\sqrt 3 - 2\sqrt 2 )(3\sqrt 3 + 2\sqrt 2 )}}}$

$\Rightarrow {\dfrac{{12\sqrt 3 + 8\sqrt 2 + 9\sqrt 3 - 6\sqrt 2 }}{{{{(3\sqrt 3 )}^2} - {{(2\sqrt 2 )}^2}}}}$

$\Rightarrow \dfrac{{21\sqrt 3 + 2\sqrt 2 }}{{27 - 8}}$

$\Rightarrow \dfrac{{21\sqrt 3 + 2\sqrt 2 }}{{19}}$

Substituting $\sqrt {{2}} {{ = }}{{1}}{{.414,}}\sqrt {{3}} {{ = }}{{1}}{{.732}}$ in above expression

${{ = }}{\dfrac{{{{21}}{{\times }}{{1}}{{.732}}{{ + }}{{2}}{{\times }}{{1}}{{.414}}}}{{{{19}}}}}$

${{ = }}{\dfrac{{{{36}}{{.372}}{{ + }}{{2}}{{.828}}}}{{{{19}}}}}$

${{ = }}{\dfrac{{{{39}}{{.2}}}}{{{{19}}}} = 2.063}$

4. If $a = \dfrac{{3 + \sqrt 5 }}{2}$, then find the value of ${a^2} + \dfrac{1}{{{a^2}}}.$

Ans: Since, $a{{ = }}\dfrac{{{{3}}{{ + }}\sqrt {{5}} }}{{{2}}}$

$\dfrac{1}{a} = \dfrac{2}{{3 + \sqrt 5 }} = \dfrac{2}{{3 + \sqrt 5 }} \times \dfrac{{3 - \sqrt 5 }}{{3 - \sqrt 5 }}$

$={\dfrac{{6 - 2\sqrt 5 }}{{{3^2} - {{(\sqrt 5 )}^2}}}}$

$={\dfrac{{6 - 2\sqrt 5 }}{{9 - 5}} = \dfrac{{6 - 2\sqrt 5 }}{4}}$

$\Rightarrow \dfrac{1}{a} = \dfrac{{2(3 - \sqrt 5 )}}{4} = \dfrac{{3 - \sqrt 5 }}{2}$

Now,

${a^2} + \dfrac{1}{{{a^2}}} = {a^2} + \dfrac{1}{{{a^2}}} + 2 - 2$

$= {\left( {a + \dfrac{1}{a}} \right)^2} - 2$

Here, 2 is added and subtracted.

Now, substituting the values of $a$ and $\dfrac{1}{a}$ in above expression we get,

$={{{\left( {\dfrac{{3 + \sqrt 5 }}{2} + \dfrac{{3 - \sqrt 5 }}{2}} \right)}^2} - 2}$

$={{{\left( {\dfrac{6}{2}} \right)}^2} - 2 = {{(3)}^2} - 2 = 9 - 2 = 7}$

5. If $x = \dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}\text{ and }y = \dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }},$ then find the value of ${x^2} + {y^2}$ .

Ans: Rationalize $x$,

$x = \dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$

$= {\dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times \dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}}$

$= {\dfrac{{{{(\sqrt 3 + \sqrt 2 )}^2}}}{{{{(\sqrt 3 )}^2} - {{(\sqrt 2 )}^2}}}}$

$= {\dfrac{{{{(\sqrt 3 )}^2} + {{(\sqrt 2 )}^2} + 2 \times \sqrt 3 \times \sqrt 2 }}{{3 - 2}}}$

$= {\dfrac{{3 + 2 + 2\sqrt 6 }}{1}}$

$= {5 + 2\sqrt 6 }$

Since, $x = \,{{5}}{{ + }}{{2}}\sqrt {{6}}$

${x^2} = {\left( {{{5}}{{ + }}{{2}}\sqrt {{6}} } \right)^2}$

$= {{{(5)}}^{{2}}}{{ + }}{{{(2}}\sqrt {{6}} {{)}}^{{2}}}{{ + }}{{2}}{{ \times }}{{5}}{{ \times }}{{2}}\sqrt {{6}}$

${{ = 25}}{{ + }}{{24}}{{ + }}{{20}}\sqrt {{6}}$

${{ = 49}}{{ + }}{{20}}\sqrt {{6}}$

Similarly rationalize $y$, $y = \dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$

$\Rightarrow \dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} = \dfrac{1}{x} = \dfrac{1}{{5 + 2\sqrt 6 }}$

$= \dfrac{1}{{5 + 2\sqrt 6 }} \times \dfrac{{5 - 2\sqrt 6 }}{{5 - 2\sqrt 6 }}$

$= \dfrac{{5 - 2\sqrt 6 }}{{{{(5)}^2} - {{(2\sqrt 6 )}^2}}} = \dfrac{{5 - 2\sqrt 6 }}{{25 - 24}} = \dfrac{{5 - 2\sqrt 6 }}{1}$

${y^2} = {(5 - 2\sqrt 6 )^2}$

${y^2} = {(5)^2} + {(2\sqrt 6 )^2} - 2 \times 5 \times 2\sqrt 6$

${y^2} = 25 + 24 - 20\sqrt 6$

${y^2} = 49 - 20\sqrt 6$

So, the value of ${x^2} + {y^2}$ is calculated as,

${x^2} + {y^2} = 49 + 20\sqrt 6 + 49 - 20\sqrt 6 = 98$

6. Simplify: ${(256)^{ - \left( {{4^{\dfrac{{ - 1}}{2}}}} \right)}}$

Ans:

$\Rightarrow (256){^{ - (4){^{ - \dfrac{3}{2}}}}}$

$= {({2^8})^{ - {{(4)}^{ - \dfrac{3}{2}}}}}$

$= {\left( {{2^8}} \right)^{ - \left( {{2^{2 \times - \dfrac{3}{2}}}} \right)}}$

$= \left( {{2^8}} \right){^{ - \left( {2{^{ - 3}}} \right)}}$

$= {\left( {{2^8}} \right)^{ - \left( { - \dfrac{1}{8}} \right)}}$

$= {2^{ - 1}}$

$= \dfrac{1}{2}$

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

Ans:

$\Rightarrow \dfrac{4}{{{{(216)}^{ - \dfrac{2}{3}}}}} + \dfrac{1}{{{{(256)}^{ - \dfrac{3}{4}}}}} + \dfrac{2}{{{{(243)}^{ - \dfrac{1}{5}}}}}$

$= \dfrac{4}{{{{\left( {{6^3}} \right)}^{ - \dfrac{2}{3}}}}} + \dfrac{1}{{\left( {{{16}^{2}}} \right){^{ - \dfrac{3}{4}}}}} + \dfrac{2}{{{{\left( {{3^5}} \right)}^{ - \dfrac{1}{5}}}}}$

$= \dfrac{4}{{{6^{3 \times \left( { - \dfrac{2}{3}} \right)}}}} + \dfrac{1}{{{{16}^{2 \times \left( { - \dfrac{3}{4}} \right)}}}} + \dfrac{2}{{{3^{5 \times \left( { - \dfrac{1}{5}} \right)}}}}$

$= \dfrac{4}{{{6^{ - 2}}}} + \dfrac{1}{{{{16}^{ - \dfrac{3}{2}}}}} + \dfrac{2}{{{3^{ - 1}}}}$

$= \,4 \times {6^2} + {16^{\dfrac{3}{2}}} + 2 \times {3^1}$

$= 4 \times 36 + {\left( {{{(4)}^2}} \right)^{\dfrac{3}{2}}} + 2 \times {3^1}$

$= 4 \times 36 + {4^3} + 6$

$= 144 + 64 + 6$

$= {{ }}214$

### Activities for Chapter 1

There are 4 activities in the chapter with an aggregate of 96 model questions and a ton of solved models, which explains every one of the fundamental concepts in the section. These questions cover every one of the significant subjects of the section like kinds of numbers, their properties, number line, whole numbers, and so on. The NCERT Exemplar issues with answers for Class 9 Math chapter 1 are meant to give abundant information to get every one of the ideas explained in detail from the chapter.

### Some Important topics from NCERT Exemplar for Class 9 Math Solutions Chapter 1 - 'Number Systems'

NCERT Exemplar issues for CBSE Class 9 Math Chapter 1 solutions principally manage numbers, their types, and their capacities. It is the first and the principal chapter of Class 9 Math, which makes the base for different sections. In the previous classes, you learned with regards to number lines and this chapter is the lengthy model of number lines. Every type of number i.e regular numbers, whole numbers, irrational numbers, rational numbers, and natural numbers are clarified profoundly in the section.

The number line and its ideas and functions are clarified minutely. The properties of numbers are depicted and clarified with the assistance of questions. With an aggregate of 4 activities and many solved models, this section contains properties of rational and irrational numbers, surd, radicals, laws of revolutionaries, properties of surds, and objective examples.

## FAQs on NCERT Exemplar for Class 9 Math Chapter 1 - Number Systems (Book Solutions)

1. What are the Important points to be noted in NCERT Math Exemplar Problems and Solutions for Class 9 Chapter 1?

• If a number ‘r’ can be written as p/q, then, at that point, ‘r’ is known as a rational number, where p and q ought to be whole numbers and q isn't equivalent to 0.

• Assuming a number s can't be written as p/q, where p and q ought to be integers and q isn't equivalent to 0.

• When we gather every objective and irrational number, then, it makes up the group of real numbers.

These are some of the important points of NCERT Math Exemplar Problems and Solutions for Class 9.

2. How many sets of questions are there in each exercise of NCERT Math Exemplar Problems and Solutions for Class 9 Chapter 1?

The questions of this chapter comprise all significant subjects of the section like kinds of numbers, number line, ration, and irrational numbers, their properties, and so forth.

• Practice 1 of the section comprises 21 problems which are for the multiple-choice questions and cover every one of the subjects of the section.

• Exercise 2 of chapter 1 consists of 4 questions.

• Exercise 3 has around 14 questions

• Exercise 4 of the chapter consists of 7 questions that are mostly long-form answers.

3. What are the Benefits of NCERT Exemplar Class 9 Math Solutions Chapter 1 by Vedantu?

The benefits are as follows:

• At Vedantu our essential target is your assistance and development in education. The NCERT Exemplar answers for Class 9 Math section 1 given by us are refreshed now and again to give you the best and reliable content.

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5. What topics are covered in NCERT Exemplar Class 9 Math Chapter 1?

The topics are as follows:

• Rational Numbers and Irrational numbers

• Finding rational numbers

• Locating irrational numbers

• Real Numbers and Their Decimal Expansions

• Finding irrational numbers

• Real numbers

• Rationalizing the denominator

• Laws of Exponents for Real Numbers

To work with simple learning and assist students with understanding the ideas examined in Chapter 1, free NCERT Exemplars are given here which can be additionally downloaded as a PDF. Students can utilize these materials as a kind of perspective device for concentrating just as rehearsing sums.