NCERT Solutions for Class 9 Mathematics Chapter 1- Number Systems

$\mathrm{Q.1\; Is}\mathrm{zero}\mathrm{a}\mathrm{rational}\mathrm{number}?\mathrm{Can}\mathrm{you}\mathrm{write}\mathrm{it}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{}\frac{\mathrm{p}}{\mathrm{q}},\mathrm{where}\mathrm{p}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\mathrm{q}\hspace{0.17em}\mathrm{are}\mathrm{integers}\mathrm{and}\hspace{0.17em}\mathrm{q}\ne 0?$

Ans- 

Yes, 0 is a rational number. It can be represented as (0/1), (0/2), (0/3) etc.

Q.2

Find six rational numbers between 3 and 4.

Ans-

There are infinite rational numbers between 3 and 4. 3 and 4 can be represented as 24/8 and 32/8 respectively. The rational numbers between 3 and 4 are 25/8, 26/8, 27/8, 28/8, 29/8, 30/8.

Q.3 Find five rational numbers between 35 and 45.

Ans-

$\begin{array}{l}\text{There\hspace{0.17em}}\text{are infinite natural numbers between}\frac{3}{5}\text{and}\frac{4}{5}.\\ \frac{3}{5}=\frac{3\times 6}{5\times 6}=\frac{18}{30}\\ \frac{4}{5}=\frac{4\times 6}{5\times 6}=\frac{24}{30}\\ \text{Therefore, rational numbers between}\frac{3}{5}\text{and}\frac{4}{5}\text{are}\\ \frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30}\end{array}$

Q.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.

Ans-

(i) True; since the collection of whole numbers contains all natural numbers.
(ii) False; since integers may be negative but whole numbers are positive. For example: – 5 is an integer it is not a whole number.
(iii) False; as rational number may be a fraction but whole number may not be a fraction.
For example: 4/5 is a rational number and it is not a whole number.

$\mathrm{Q.5}$

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

$\sqrt{\text{m}}$

where m is a natural number.
(iii) Every real number is an irrational number.

Ans-

(i) True; because real number is a collection of rational and irrational number.
(ii) False; as negative numbers cannot be represented as the square root of any other number.
(iii) False; as real numbers include both rational and irrational numbers i.e., irrational number is a part of real number. Therefore, every real number cannot be an irrational number.

Q.6 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.

Ans-

No, the square roots of all positive integers are not irrational. For example: The square roots of 4 and 9 are 2 and 3 respectively.

$\mathrm{Q.7}$

Show

how 5 can be represented on the number line.

Ans-

$\begin{array}{l}\sqrt{5}\text{can be written as}\\ \sqrt{5}=\sqrt{4+1}\\ =\sqrt{{\left(2\right)}^2+1^2}\\ \text{Mark a point}‘\text{A}’\text{representing 2 on number line}.\\ \text{Now},\text{construct AB of unit length perpendicular to OA}.\\ \text{Then},\text{taking O as centre and OB as radius},\text{draw an arc which intersect the number line at C}.\\ \text{OC represents the length of}\sqrt{5}\text{The point C represents}\sqrt{5}\end{array}$

$\mathrm{Q.8}$

Write the following in decimal form and say what kind of

decimal expansion each has:i 36100  ii 111  iii 418  iv 313  v 211  vi 329400

Ans-

$\begin{array}{l}\left(\mathrm{i}\right)\frac{36}{100}=0.36\\ \text{It is a terminating decimal because remainder is zero.}\\ \left(\mathrm{ii}\right)\frac{1}{11}=0.0909090909\dots \\ \text{It is a non-terminating decimal because remainder is not zero.}\\ \left(\mathrm{iii}\right)4\frac{1}{8}=4.125\\ \text{It is a terminating decimal because remainder is zero.}\\ \left(\mathrm{iv}\right)\frac{3}{13}=0.230769230769\dots \\ =0.\overline{230769}\\ \text{It is a non terminating decimal because it is a repeating}\\ \\ \left(\mathrm{v}\right)\frac{2}{11}=0.181818181818\dots \\ =0.\overline{18}\\ \text{It is a terminating decimal because it is a repeating}\\ \\ \left(\mathrm{vi}\right)\frac{329}{400}=0.8225\\ \text{It is a terminating decimal because remainder is zero.}\end{array}$ $\begin{array}{l}\end{array}$ $\mathrm{Q.9}$

You know that

17 = 0.142857¯. Can you predict what thedecimal expansions of 27 ,37 ,47, 57, 67 are, without actuallydoing the long division? If so, how?

Ans-

$\begin{array}{l}\mathrm{Given}:\frac{1}{7}=0.\overline{142857}\\ \therefore \frac{2}{7}=2\times 0.\overline{142857}\\ =0.\overline{285714}\\ \frac{3}{7}=3\times 0.\overline{142857}\\ =0.\overline{428571}\\ \frac{4}{7}=4\times 0.\overline{142857}\\ =0.\overline{571428}\\ \frac{5}{7}=5\times 0.\overline{142857}\\ =0.\overline{714285}\\ \frac{6}{7}=6\times 0.\overline{142857}\\ =0.\overline{857142}\end{array}$ $\begin{array}{l}\end{array}$ $\mathrm{Q.10}$

Express the following in the form

pq, where p and q areintegers and q¹0.i 0  ii 0.47 ¯  iii 0.001¯

An-

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Let}\text{x}=0.\overline{6}...\left(\mathrm{i}\right)\\ \text{Multiplying both sides by 10, we get}\\ 10\mathrm{x}=10\times 0.\overline{6}\\ 10\mathrm{x}=6.\overline{6}...\left(\mathrm{ii}\right)\\ \text{Subtracting equation}\left(\mathrm{i}\right)\text{from equation}\left(\mathrm{ii}\right),\text{we get}\\ 10\mathrm{x}-\mathrm{x}=6.\overline{6}-0.\overline{6}\\ 9\mathrm{x}=6\\ \mathrm{x}=\frac{6}{9}\\ \mathrm{x}=\frac{2}{3}\\ \therefore 0.\overline{6}=\frac{2}{3}\\ \left(\mathrm{ii}\right)\\ \mathrm{Let}\text{x}=0.4\overline{7}...\left(\mathrm{i}\right)\\ \text{Multiplying both sides of equation}\left(\mathrm{i}\right)\text{by 10, we get}\\ 10\mathrm{x}=10\times 0.4\overline{7}\\ 10\mathrm{x}=4.\overline{7}...\left(\mathrm{ii}\right)\\ \mathrm{Multiplying}\text{both sides of equation}\left(\mathrm{ii}\right)\text{by 10, we get}\\ 100\mathrm{x}=10\times 4.\overline{7}\\ 100\mathrm{x}=47.\overline{7}...\left(\mathrm{iii}\right)\\ \text{Subtracting equation}\left(\mathrm{ii}\right)\text{from equation}\left(\mathrm{iii}\right),\text{we get}\\ 100\mathrm{x}-10\mathrm{x}=47.\overline{7}-4.\overline{7}\\ 90\mathrm{x}=43\\ \mathrm{x}=\frac{43}{90}\\ \therefore 0.4\overline{7}=\frac{43}{90}\\ \left(\mathrm{iii}\right)\mathrm{Let}\text{x}=0.\overline{001}...\left(\mathrm{i}\right)\\ \text{Multiplying both sides of equation}\left(\mathrm{i}\right)\text{by 1000, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1000x}=1000\times 0.\overline{001}\\ =1.\overline{001}...\left(\mathrm{ii}\right)\\ \text{Subtracting equation}\left(\mathrm{i}\right)\text{from equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{1000x\hspace{0.17em}}-\mathrm{x}=1.\overline{001}-0.\overline{001}\\ 999\mathrm{x}=1\\ \mathrm{x}=\frac{1}{999}\\ \therefore 0.\overline{001}=\frac{1}{999}\end{array}$ $\begin{array}{l}\end{array}$ $\mathrm{Q.11}$

Express 0.99999, in the form

pq. Are you surprised by your answer?With your teacher and classmates discussway the answer makes sense.

Ans-

$\begin{array}{l}\text{\hspace{0.17em}Let x}=0.99999\mathrm{\dots }\\ =0.\overline{9}\mathrm{\dots }\left(i\right)\\ \text{Multiplying both sides by 10, we get}\\ 10x=10\times 0.\overline{9}\\ 10x=9.\overline{9}\mathrm{\dots }\left(ii\right)\\ \text{Subtracting equation}\left(i\right)\text{from equation}\left(ii\right),\text{we get}\\ 10x-x=9.\overline{9}-0.\overline{9}\\ 9x=9\\ x=\frac{9}{9}\\ x=1\\ \therefore 0.99999\mathrm{\dots }=1\end{array}$ $\begin{array}{l}\end{array}$ $\mathrm{Q.12}$

What

can the maximum number of digits be in the repeatingblock of digits in the decimal expansion of 117? Perform thedivision to check your answer.

Ans-

\begin{array}{l}0.0588235294117647\\ 17\overline{)1.00}\\ \underset{¯}{085}\\ 150\\ \underset{¯}{136}\\ 140\\ \underset{¯}{136}\\ 40\\ \underset{¯}{34}\\ 60\\ \underset{¯}{51}\\ 90\\ \underset{¯}{85}\\ 50\\ \underset{¯}{34}\\ 160\\ \underset{¯}{153}\\ 70\\ \underset{¯}{68}\\ 20\\ \underset{¯}{17}\\ 30\\ \underset{¯}{17}\\ 130\\ \underset{¯}{119}\\ 110\\ \underset{¯}{102}\\ 80\\ \underset{¯}{68}\\ 120\\ \underset{¯}{119}\\ 1\\ Thus,\frac{1}{17}=0.\overline{0588235294117647}\end{array} $\begin{array}{l}\end{array}$ $\mathrm{Q.13}$

Look

at several examples of rational numbers in the formpq q≠0 , where p and q are integers with no commonfactors other than 1 and having terminating decimalrepresentations (expansions).Can you guess what propertyq must satisfy?

Ans-

$\begin{array}{l}\text{Since},\text{the rational number}\frac{p}{q}\text{will be terminating decimal if denominator q}\\ \text{is either 2, 4, 5, 8, 10 and so on}\mathrm{\dots }\\ \frac{9}{2}=4.5\\ \frac{11}{4}=2.75\\ \frac{17}{8}=2.125\\ \frac{13}{5}=2.6\\ \frac{32}{10}=3.2\\ \text{We see that rational number}\frac{\text{p}}{q}\text{will be terminating if prime factors of q}\\ \text{are either 2 only or multiple of 2 and 5 only or both}\text{.}\end{array}$ $\mathrm{Q.14}$

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

Ans-

Three numbers whose decimal expansions are non- terminating non-recurring are as follows:

0.030030012003000050004123000…

0.01200012500003500050010008879000102003…

1.5200050040060080010030010040038001…

$\mathrm{Q.15}$

Find

three different irrational numbers between therational numbers 57 and 911.

Ans-

$\begin{array}{l}\text{Since},\frac{5}{7}=0.\overline{714285}\text{and}\frac{9}{11}=0.\overline{81}\\ \text{Three irrational numbers between}\frac{5}{7}\text{and}\frac{9}{11}\text{are:}\\ \mathrm{0.72005006004000202005004\dots }\\ \mathrm{0.75005006004000202005004\dots }\\ \mathrm{0.80005006004000202005004\dots }\end{array}$ $\begin{array}{l}\end{array}$ $\mathrm{Q.16}$

Classify the following numbers as rational or irrational:

i 23  ii 225 iii 0.3796 iv 7.478478...   v 1.101001000100001...

Ans-

$\begin{array}{l}\left(\mathrm{i}\right)\sqrt{23}=4.7983152331272\dots \\ \text{Since},\text{this number is non-terminating and non-repeating, therefore}\\ \text{it is irrational number.}\\ \left(\mathrm{ii}\right)\sqrt{225}=15,\text{It is rational number.}\\ \left(\mathrm{iii}\right)0.3796,\text{It is rational number because this number is terminating.}\\ \left(\mathrm{iv}\right)7.478478\dots =7.\overline{478},\text{this is rational number as it is non-terminating}\\ \text{and repeating.}\\ \left(\mathrm{v}\right)1.101001000100001\dots ,\text{}\\ \text{It is irrational number because it is non-terminating non repeating.}\end{array}$ $\mathrm{Q.17}$

Visualise 3.765 on the number line, using successive magnification.

Ans-

3.765 can be visualised as in the following steps.

$\mathrm{Q.18}$

Visualise

4.26 ¯on the number line, upto 4 decimal places.

Ans-

$\begin{array}{l}4.\overline{26}\text{can be written as:}\\ 4.\overline{26}=4.2626\end{array}$

$\mathrm{Q.19}$

Classify the following numbers as rational or irrational:

i 2−5  ii 3+23−23  iii 2777iv 12 v 2π

Ans-

$\begin{array}{l}\left(\mathrm{i}\right)2-\sqrt{5}=\mathrm{Irrational}\text{ number because}\sqrt{5}\text{is irrational number.}\\ \left(\mathrm{ii}\right)(3+\sqrt{23})-\sqrt{23}=3,\text{which is a rational number.}\\ \left(\mathrm{iii}\right)\frac{2\sqrt{7}}{7\sqrt{7}}=\frac{2}{7},\text{which is in the form of}\frac{\mathrm{p}}{\mathrm{q}}\text{form and it is a rational}\\ \text{number.}\\ \left(\mathrm{iv}\right)\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ =\frac{\sqrt{2}}{2},\text{which is an irrational number becuase}\sqrt{2}\text{is}\\ \text{irrational number.}\\ \left(\mathrm{v}\right)2\mathrm{\pi }=2(3.1415\dots )\\ =6.2830\dots ,\text{which is non-terminating and}\\ \text{non-repeating. So it is irrational number.}\end{array}$ $\begin{array}{l}\end{array}$ $\mathrm{Q.20}$

Simplify

each of the following expressions:(i)(3+3)(2+2) (ii) (3+3)(3−3)(iii) (5+2)2(iv) (5−2)(5+2)

Ans-

$\begin{array}{l}\left(\mathrm{i}\right)(3+\sqrt{3})(2+\sqrt{2})=6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}\\ \\ \left(\mathrm{ii}\right)(3+\sqrt{3})(3-\sqrt{3})={\left(3\right)}^2-{\left(\sqrt{3}\right)}^2\\ =9-3=6\\ \left(\mathrm{iii}\right){(\sqrt{5}+\sqrt{2})}^2=5+2\sqrt{5}\sqrt{2}+2\\ =7+2\sqrt{10}\\ \left(\mathrm{iv}\right)(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})={\left(\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2\\ =5-2\\ =3\end{array}$ $\begin{array}{l}\end{array}$ $\mathrm{Q.21}$

Recall

, π is defined as the ratio of the circumference (say c)of a circle to its diameter say d.That is, π=cd.This seemsto contradict the fact that π is irrational.How will youresolve this contradiction?

Ans-

There is no contradiction. Remember that when you measure a length with a scale or any other device, you only get an approximate rational value. So, you may not realise that either c or d is irrational.

$\mathrm{Q.22}$

Represent

9.3 on the number line.

Ans-

Mark a line segment AB = 9.3 on number line. Further, take BC of 1 unit. Draw a semi-circle on AC as diameter. Draw a perpendicular to line AC passing through point B. Let it intersect the semi circle at D. Taking B as centre and BD as radius, draw an arc intersecting number line at E. BE =

$\sqrt{9.3}.$ $\begin{array}{l}\end{array}$ $\mathrm{Q.23}$

Rationalise

the denominators of the following:(i)17(ii)17−6(iii)15+2(iv)17−2

Ans-

$\begin{array}{l}\left(i\right)\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}\\ =\frac{\sqrt{7}}{7}\\ \left(ii\right)\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}[\because (a-b)(a+b)=a^2-b^2]\\ =\frac{\sqrt{7}+\sqrt{6}}{7-6}\\ =\sqrt{7}+\sqrt{6}\\ \left(iii\right)\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}\\ =\frac{\sqrt{5}-\sqrt{2}}{5-2}[\because (a-b)(a+b)=a^2-b^2]\\ =\frac{\sqrt{5}-\sqrt{2}}{3}\\ \left(iv\right)\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2}\\ =\frac{\sqrt{7}+2}{7-4}[\because (a-b)(a+b)=a^2-b^2]\\ =\frac{\sqrt{7}+2}{3}\end{array}$ $$ $\mathrm{Q.24}$

Find

: i 6412     ii 3215     iii 12513

Ans-

$\begin{array}{l}\left(\mathrm{i}\right){64}^{\frac{1}{2}}={\left(2^6\right)}^{\frac{1}{2}}\\ =2^{6\times \frac{1}{2}}\\ =2^3\\ =8\\ \left(\mathrm{ii}\right){32}^{\frac{1}{5}}={\left(2^5\right)}^{\frac{1}{5}}\\ =2^{5\times \frac{1}{5}}\\ =2^1=2\\ \left(\mathrm{iii}\right){125}^{\frac{1}{3}}={\left(5^3\right)}^{\frac{1}{3}}\\ =5^{3\times \frac{1}{3}}\\ =5^1=5\end{array}$ $$ $\mathrm{Q.25}$

Find:

i932  ii3225  iii1634   iv125−13

Ans-

$\begin{array}{l}\left(\mathrm{i}\right)9^{\frac{3}{2}}={\left(3^2\right)}^{\frac{3}{2}}\\ =3^{2\times \frac{3}{2}}\\ =3^3\\ =27\\ \left(\mathrm{ii}\right){32}^{\frac{2}{5}}={\left(2^5\right)}^{\frac{2}{5}}\\ ={2^{5\times }}^{\frac{2}{5}}\\ =2^2\\ =4\\ \left(\mathrm{iii}\right){16}^{\frac{3}{4}}={\left(2^4\right)}^{\frac{3}{4}}\\ =2^{4\times \frac{3}{4}}\\ =2^3\\ =8\\ \left(\mathrm{iv}\right){125}^{\frac{-1}{3}}={\left(5^3\right)}^{\frac{-1}{3}}\\ ={5^{3\times }}^{\frac{-1}{3}}\\ =5^{-1}\\ =\frac{1}{5}\end{array}$ $$ $\mathrm{Q.26}$

Simplify

:    (i) 223.215     (ii) (133)7     (iii) 11121114       (iv)712.812

Ans-

$\begin{array}{l}\left(\mathrm{i}\right)2^{\frac{2}{3}}.2^{\frac{1}{5}}=2^{\frac{2}{3}+\frac{1}{5}}[\because {\mathrm{x}}^{\mathrm{a}}.{\mathrm{x}}^{\mathrm{b}}={\mathrm{x}}^{\mathrm{a}+\mathrm{b}}]\\ =2^{\frac{10+3}{15}}\\ =2^{\frac{13}{15}}\\ \\ \left(\mathrm{ii}\right){\left(\frac{1}{3^3}\right)}^7=\frac{1}{3^{3\times 7}}\\ =\frac{1}{3^{21}}\\ =3^{-21}\\ \left(\mathrm{iii}\right)\frac{{11}^{\frac{1}{2}}}{{11}^{\frac{1}{4}}}={11}^{\frac{1}{2}-\frac{1}{4}}[\because \frac{{\mathrm{x}}^{\mathrm{a}}}{{\mathrm{x}}^{\mathrm{b}}}={\mathrm{x}}^{\mathrm{a}-\mathrm{b}}]\\ ={11}^{\frac{2-1}{4}}\\ ={11}^{\frac{1}{4}}\\ \\ \left(\mathrm{iv}\right)7^{\frac{1}{2}}.8^{\frac{1}{2}}={(7\times 8)}^{\frac{1}{2}}[\because {\mathrm{x}}^{\mathrm{m}}.{\mathrm{y}}^{\mathrm{m}}={\left(\mathrm{xy}\right)}^{\mathrm{m}}]\\ ={56}^{\frac{1}{2}}\end{array}$