# NCERT Solutions For Class 9 Maths

In order to help you in your Class IX preparation, NTSE Guru has prepared the detailed NCERT Solutions for class 9 Maths including all the questions of the important topic of your Mathematics book. NTSEGuru experts provide you with descriptive solutions for the chapter ‘Number System’ of Class IX Maths NCERT book.

## Class 9th Maths NCERT Solutions

Get class 9th Maths NCERT solutions here:

EXERCISE – 1

1. Is zero a rational number? Can you write it in the form $\frac{p}{q},$ where p and q are integers and $q\ne$ 0?

Sol. Yes. Zero is a rational number as it can be represented as $\frac{0}{1}$ or $\frac{0}{2}$ or $\frac{0}{3}$ etc.

2. Find six rational numbers between 3 and 4.

Sol. There are infinite rational numbers in between 3 and 4.

3 and 4 can be represented as $\frac{21}{7}$ and $\frac{28}{7}$ respectively.

Therefore, rational numbers between 3 and 4 are

$\frac{22}{7},\,\,\frac{23}{7},\,\,\frac{24}{7},\,\,\frac{25}{7},\,\,\frac{26}{7},\,\,\frac{27}{7}$

3.    Find five rational numbers between $\frac{3}{5}&space;and&space;\frac{4}{5}.$

Sol.                        $\frac{3}{5}=\frac{3\times&space;6}{5\times&space;6}=\frac{18}{30}$

$\frac{4}{5}=\frac{4\times&space;6}{5\times&space;6}=\frac{24}{30}$

Therefore, rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$

$\frac{19}{30},\,\frac{20}{30},\,\frac{21}{30},\,\frac{22}{30},\,\frac{23}{30}$

4.    State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

(ii) Every integer is a whole number.

(iii) Every rational number is a whole number

Sol.    (i) True; since the collection of whole numbers contains all natural numbers.

(ii) False; as integers may be negative but whole numbers are positive. For example: −3 is an integer but not a whole number.

(iii) False; as rational numbers may be fractional but whole numbers may not be.

For example: $\frac{1}{5}$ is a rational number but not a whole number.

EXERCISE – 2

1.    State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

(ii) Every point on the number line is of the form $\sqrt{m},$ where m is a natural number.

(iii) Every real number is an irrational number.

Sol.    (i) True; since the collection of real numbers is made up of rational and irrational numbers.

(ii) False; as negative numbers cannot be expressed as the square root of any other number.

(iii) False; as real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number.

2.  Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

Sol. If numbers such as $\sqrt{4}=2,\,\,\sqrt{9}=3$ are considered,

Then here, 2 and 3 are rational numbers. Thus, the square roots of all positive integers are not irrational.

3. Show how $\sqrt{5}$ can be represented on the number line.

Sol. To represent $\sqrt{5}$ on number line, take OB = 2 units and make a perpendicular AB at B such that AB = 1 unit.

Now by Pythagoras theorem, the length of OA is $\sqrt{5}$. Taking O as centre and OA as radius, mark an arc on OB, which intersects at C. Hence, OC = $\sqrt{5}$.

EXERCISE – 3

1.  Write the following in decimal form and say what kind of decimal expansion each has :

(i) $\frac{36}{100}$                                       (ii) $\frac{1}{11}$                                      (iii) $4\frac{1}{8}$

(iv) $\frac{3}{13}$                                        (v) $\frac{2}{11}$                                      (vi) $\frac{329}{400}$

Sol.

(i) $\frac{36}{100}=0.36$ Terminating.

(ii) $\frac{1}{11}=0.\overline{09},$ Recurring & Non-terminating.

(iii) $4\frac{1}{8}=4.125,$ Terminating.

(iv) $\frac{3}{13}=0.\overline{230769},$ Recurring & Non-terminating.

(v)  $\frac{2}{11}=0.\overline{18},$ Recurring & Non-terminating.

(vi) $\frac{329}{400}\,=0.8225,$ Terminating.

2. You know that $\frac{1}{7}=0.\overline{142857}$. Can you predict what the decimal expansions of $\frac{3}{2},\,\frac{3}{7},\frac{4}{7},\,\frac{5}{7},\,\frac{6}{7}$ are, without actually doing the long division? If so, how?

[Hint: Study the remainders while finding the value of carefully.]

Sol.  Without actual long division, the decimal expansions of $\frac{2}{7},\,\frac{3}{7},\,\frac{4}{7},\,\frac{5}{7},\&space;\frac{6}{7}$ are as follows:

$\frac{2}{7}=2\times&space;\frac{1}{7}=2\times&space;0.\overline{142857}=0.\overline{285714}$

$\frac{3}{7}=3\times&space;\frac{1}{7}=3\times&space;0.\overline{142857}=0.\overline{428571}$

$\frac{4}{7}=4\times&space;\frac{1}{7}=4\times&space;0\overline{.142857}=0.\overline{571428}$

$\frac{5}{7}=5\times&space;\frac{1}{7}=5\times&space;0.\overline{142857}=0.\overline{714285}$

$\frac{6}{7}=6\times&space;\frac{1}{7}=6\times&space;0.\overline{142857}=0.\overline{857142}$

3.    Express the following in the form $\frac{p}{q},$ where p and q are integers and $q\ne&space;0$.

(i) $0.\overline{6}$                                   (ii) $0.\overline{47}$                              (iii) $0.\overline{001}$

Sol.   (i) $0.\overline{6}$

Let, $x=0.\overline{6}$

$\Rightarrow \,\,x=0.6666...$                               …(i)

Multiplying equation (i) by 10 both sides

10x = 6.6666…

$\Rightarrow$ 10x = 6 + 0.6666…

$\Rightarrow$ 10x = 6 + x                                       [From equation (i)]

$\Rightarrow$ 10x – x = 6

$\Rightarrow$ 9x = 6

$\Rightarrow x=\frac{6}{9}=\frac{2}{3}$

(ii) $0.\overline{47}$

Let, $x=0.\overline{47}$

x=0.47777…                              …(i)

Multiplying equation (i) by 10 both sides

$\Rightarrow$ 10x = 4.7777…                     …(ii)

Multiplying equation (ii) by 10 both sides

$\Rightarrow$ 100x = 47.7777…

$\Rightarrow$ 100x = 43 + 10x                   [From equation (ii)]

$\Rightarrow$ 100x-10x=43

$\Rightarrow$ 90x=43

$\Rightarrow$ $x=\frac{43}{90}$

(iii) $0.\overline{001}$

Let $x=0.\overline{001}$

$\Rightarrow \,\,x=0.001001001...$            …(i)

Multiplying equation (i) by 1000 both sides

1000x=1.001001001…

$\Rightarrow 1000x=1+0.001001001...$

$\Rightarrow 1000x=1+x$               [From equation (i)]

$\Rightarrow 1000x-x=1$

$\Rightarrow 999x=1$

$\Rightarrow \,x=\frac{1}{999}$

4.    Express 0.99999…. in the form $\frac{p}{q}$ . Are you surprised by your answer? With yourteacher and classmates discuss why the answer makes sense.

Sol.    0.99999…

Let x=0.99999…             …(1)

Multiplying equation (i) by 10 both sides

10x=9.99999…

$\Rightarrow \,10x=9+0.99999...$

$\Rightarrow \,10x=9+x$                        [From equation (i)]

$\Rightarrow \,10x-x=9$

$\Rightarrow \,\,9x=9$

$\Rightarrow \,\,x=\frac{9}{9}=1$

The answer makes sense as 0.99999… is very close to 1, that is why we can say that 0.99999 = 1.

5.  What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$  ?

Sol. The maximum number of digits that can be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$  is 16 (less than 17).

$\frac{1}{17}=0.\overline{0588235294117647}$

By performing the actual division, we get

So, the maximum number of digits that can be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$  is 16.

6.    Look at several examples of rational numbers in the form $\frac{p}{q}(q\ne&space;0),$ where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Sol.                             $\frac{2}{5}=0.4,$  $\frac{1}{10}\,=0.1,$  $\frac{3}{2}=1.5,$  $\frac{7}{8}=0.875$

The denominator of all the rational numbers is in the form of  ${{2}^{m}}\times {{5}^{n}},$ where m and n are integers.

7.    Write three numbers whose decimal expansions are non-terminating non-recurring.

Sol.    Three non-terminating non-recurring decimals:

(1) 0.414114111411114…

(2) 2.01001000100001…

(3) $\pi =3.1416...$

8.    Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$.

Sol.  $\frac{5}{7}=0.\overline{714285}$ and $\frac{9}{11}=0.\overline{81}$

We know that three are infinite many irrational numbers between two rational numbers.

So, the three irrational numbers are:

(i) 0.72722722272222

(ii) 0.73733733373333

(iii) 0.74744744474444

9.    Classify the following numbers as rational or irrational :

(i) $\sqrt{23}$                                           (ii) $\sqrt{225}$                                    (iii) 0.3796

(iv) 7.478478…                              (v) 1.101001000100001…

Sol. (i) $\sqrt{23},$ Irrational number

(ii) $\sqrt{225}=15,$ Rational number

(iii) 0.3796, Rational number

(iv) $7.478478...=7.\overline{478},$ Rational number

(v) 1.101001000100001 …, Irrational number

EXERCISE – 4

1.    Visualise 3.765 on the number line, using successive magnification.

Sol.

• First of all, we observe that 3.765 lie between 3 and 4. Divide this portion into 10 equal parts.
• In the next step, we locate 3.765 between 3.7 and 3.8.
• To get a more accurate visualization of representation, we divide this portion of number line into 10 equal parts and use a magnifying glass to visualize that 3.765 lies between 3.76 and 3.77.
• Now to visualize 3.765 still more accurately, we divide the portion between 3.76 and 3.77 into 10 equal parts and locate 3.765.

2.    Visualise $4.\overline{26}$ on the number line, up to 4 decimal places.

Sol.

• First of all, we observe that $4.2626\,\,(4.\overline{26})$ lies between 4 and 5. Divide this portion into 10 equal parts.
• In the next step, we locate 4.2626 between 4.2 and 4.3.’
• To get a more accurate visualization of representation, we divide this portion of number line into 10 equal parts and use a magnifying glass to visualize that 4.2626 lies between 4.262 and 4.263.
• Now to visualize 4.2626 still more accurately, we divide the portion between 4.262 and 4.263 into 10 equal parts and locate 4.2626.

EXERCISE – 5

1.    Classify the following numbers as rational or irrational:

(i) $2-\sqrt{5}$                             (ii) $\left(&space;3+\sqrt{23}&space;\right)\,-\sqrt{23}$        (iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$

(iv) $\frac{1}{\sqrt{2}}$                                   (v) $2\pi$

Sol.    (i) $2-\sqrt{5}$ Irrational number.

(ii) $(3-\sqrt{23})-\sqrt{23}=3$ Rational number.

(iii) $\frac{2\sqrt{7}}{7\sqrt{7}}\,=\frac{2}{7}$ Rational number.

(iv) $\frac{1}{\sqrt{2}}$ Irrational number.

(v) $2\pi$ Irrational number.

2.    Simplify of the following expressions:

(i) $(3+\sqrt{3})(2+\sqrt{2})$                    (ii) $(3+\sqrt{3})\,\,(3-\sqrt{3})$

(iii) ${{(\sqrt{5}+\sqrt{2})}^{2}}$                              (iv) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$

Sol. (i)$(3+\sqrt{3})(2+\sqrt{2})=6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}$

(ii) $(3+\sqrt{3})(3-\sqrt{3})\,={{3}^{2}}-{{(\sqrt{3})}^{2}} \left[ \because \,\,(a+b)(a-b)\,={{a}^{2}}-{{b}^{2}} \right] =9-3=6$

(iii) ${{(\sqrt{5}+\sqrt{2})}^{2}}={{(\sqrt{5})}^{2}}+{{(\sqrt{2})}^{2}}\,+2\times \sqrt{5}\,\times \sqrt{2}\,=7+2\sqrt{10} \left[ \because \,\,{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \right]$

(iv) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})\,={{(\sqrt{5})}^{2}}\,={{(\sqrt{2})}^{2}}=5-2=3$ $[\because \,\,(a-b)\,(a+b)\,={{a}^{2}}-{{b}^{2}}]$

3.    Recall, $\pi$ is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, $\pi&space;=\frac{c}{d}$ this seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?

Sol.      With a scale or tape we get only an approximate rational number as the result of our measurement. That is why $\pi$ can be approximately represented as a quotient of two rational numbers. As a matter of mathematical truth it is irrational.

4.    Represent $\sqrt{9.3}$ on the number line.

Sol.      To represent $\sqrt{9.3}$ on the number line, draw AB = 9.3 units. Now produce AB to C, such that BC = 1 unit. Draw the perpendicular bisector of AC which intersects AC at O. Taking O as centre and OA as radius; draw a semi-circle which intersects D to the perpendicular at B. Now taking 0 as centre and OD as radius, draw an arc, which intersects AC produced at E. Hence, $OE=\sqrt{9.3}$.

5.    Rationalise the denominators of the following:

(i) $\frac{1}{\sqrt{7}}$                                                              (ii) $\frac{1}{\sqrt{7}-\sqrt{6}}$

(iii) $\frac{1}{\sqrt{5}+\sqrt{2}}$                                                (iv) $\frac{1}{\sqrt{7}-2}$

Sol.

(i) $\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\,\times&space;\frac{\sqrt{7}}{\sqrt{7}}\,=\frac{\sqrt{7}}{7}$

(ii) $\frac{1}{\sqrt{6}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\,\times&space;\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}\,=\frac{\sqrt{7}+\sqrt{6}}{{{(\sqrt{7})}^{2}}-{{(\sqrt{6})}^{2}}}$

$=\frac{\sqrt{7}+\sqrt{6}}{7-6}\,=\sqrt{7}+\sqrt{6}$

(iii) $\frac{1}{\sqrt{5}+\sqrt{2}}\,=\frac{1}{\sqrt{5}+\sqrt{2}}\times&space;\,\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}\,=\frac{\sqrt{5}-\sqrt{2}}{{{(\sqrt{5})}^{2}}-{{(\sqrt{2})}^{2}}}\,=\frac{\sqrt{5}-\sqrt{2}}{5-2}$

$=\frac{\sqrt{5}-\sqrt{2}}{3}$

(iv) $\frac{1}{\sqrt{7}-2}\,=\frac{1}{\sqrt{7}-2}\,\times&space;\frac{\sqrt{7}+2}{\sqrt{7}+2}\,=\frac{\sqrt{7}+2}{(\sqrt{7})-{{(2)}^{2}}}\,=\frac{\sqrt{7}+2}{7-4}\,$ $=\frac{\sqrt{7}+2}{3}$

EXERCISE – 6

1.    Find:

(i) ${{64}^{\frac{1}{2}}}$                                (ii) ${{32}^{\frac{1}{5}}}$                                   (iii) ${{125}^{\frac{1}{3}}}$

Sol.    (i) ${{64}^{\frac{1}{2}}}\,={{({{8}^{2}})}^{\frac{1}{2}}}\,={{8}^{2\times \frac{1}{2}}}=8$

(ii) ${{32}^{\frac{1}{5}}}\,={{({{2}^{5}})}^{\frac{1}{5}}}\,={{2}^{5\times \frac{1}{5}}}=2$

(iii) ${{125}^{\frac{1}{3}}}={{({{5}^{3}})}^{\frac{1}{3}}}={{5}^{3\times \frac{1}{3}}}=5$

2.    Find:

(i) ${{9}^{\frac{3}{2}}}$                        (ii) ${{32}^{\frac{2}{5}}}$                     (iii) ${{16}^{\frac{3}{4}}}$                   (iv) ${{125}^{\frac{-1}{3}}}$

Sol.    (i) ${{9}^{\frac{3}{2}}}\,={{({{3}^{2}})}^{\frac{3}{2}}}\,={{3}^{2\times \frac{3}{2}}}\,={{3}^{3}}=27$

(ii) ${{32}^{\frac{2}{5}}}\,={{({{2}^{5}})}^{\frac{2}{5}}}\,={{2}^{5\times \frac{2}{5}}}\,={{2}^{2}}=4$

(iii) ${{16}^{\frac{3}{4}}}\,={{({{2}^{4}})}^{\frac{3}{2}}}={{2}^{4\times \frac{3}{4}}}\,={{2}^{3}}=8$

(iv) ${{125}^{-\frac{1}{3}}}={{({{5}^{3}})}^{-\frac{1}{3}}}={{5}^{3\times -\frac{1}{3}}}={{5}^{-1}}=\frac{1}{5}$

3.    Simplify:

(i) ${{2}^{\frac{2}{3}}}.\,{{2}^{\frac{1}{5}}}$                                (ii) ${{\left(&space;\frac{1}{{{3}^{3}}}&space;\right)}^{7}}$

(iii) $\frac{{{11}^{\frac{1}{2}}}}{{{11}^{\frac{1}{4}}}}$                                  (iv) ${{7}^{\frac{1}{2}}}.\,{{8}^{\frac{1}{2}}}$

Sol. (i) ${{2}^{\frac{2}{3}}}.\,{{2}^{\frac{1}{5}}}={{2}^{\frac{2}{3}+\frac{1}{5}}}\,={{2}^{\frac{10+3}{15}}}={{2}^{\frac{13}{15}}}$

(ii) ${{\left(&space;\frac{1}{{{3}^{3}}}&space;\right)}^{7}}\,={{({{3}^{-3}})}^{7}}={{3}^{-21}}$

(iii) $\frac{{{11}^{\frac{1}{2}}}}{{{11}^{\frac{1}{4}}}}\,={{11}^{\frac{1}{2}}}\,\times&space;{{11}^{-\frac{1}{4}}}\,={{11}^{\frac{1}{2}-\frac{1}{4}}}\,={{11}^{\frac{2-1}{4}}}={{11}^{\frac{1}{4}}}$

(iv) ${{7}^{\frac{1}{2}}}.\,{{8}^{\frac{1}{2}}}={{(7\times&space;8)}^{\frac{1}{2}}}\,={{56}^{\frac{1}{2}}}$

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