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#### Question 1:

Write the correct answer in each of the following:
Every rational number is
(A) a natural number
(B) an integer
(C) a real number
(D) a whole number

Whole numbers are the numbers from 0, 1, 2, 3...∞ .
Natural numbers start from 1, 2, 3, 4, 5, 6... ∞.
Integers contains all numbers from $-$∞ to $+$∞.
Rational numbers are all number which can be written in the form of $\frac{p}{q}$ where and q are integers.
Real numbers are all number which can or cannot be written in the form of $\frac{p}{q}$.
So, we can conclude that all rational numbers are real numbers .

Hence, the correct answer is option (C).

#### Question 2:

Write the correct answer in each of the following:
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

Between 2 rational numbers, there are infinitely many rational numbers.

Hence, the correct answer is option (C).

#### Question 3:

Write the correct answer in each of the following:
Decimal representation of a rational number cannot be
(A) terminating
(B) non-terminating
(C) non-terminating repeating
(D) non-terminating non-repeating

Decimal representation of a rational number cannot be non-terminating non-repeating.

 Rational Number Irrational Number Either terminating or Non-terminating repeating decimal expansion. Non-terminating non-repeating decimal expansion.

Hence, the correct answer is option (D).

#### Question 4:

Write the correct answer in each of the following:
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

When it comes to a product of two irrational numbers, there can be two possible cases which are mentioned as follows:-
1. $\sqrt{a}×\sqrt{a}=a$ , where $a$ is a rational number
for example: $\sqrt{2}×\sqrt{2}$ $=2$
2.  where $a$ and $b$ are rational numbers
for example: $\sqrt{3}×\sqrt{2}=\sqrt{6}$ and $\sqrt{6}$ is an irrational number

Hence, the correct answer is option (D).

#### Question 5:

Write the correct answer in each of the following:
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

Since, $\sqrt{2}$ is an irrational number, therefore its decimal expansion will be non-terminating non-recurring.
Hence, the correct answer is option (D)

#### Question 6:

Write the correct answer in each of the following:
Which of the following is irrational?

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

(B) $\frac{\sqrt{12}}{\sqrt{3}}$

(C) $\sqrt{7}$

(D) $\sqrt{81}$

$\sqrt{\frac{4}{9}}=\frac{2}{3}$; which is a rational number.
$\sqrt{\frac{12}{3}}=\sqrt{4}=2$; which is a rational number.
$\sqrt{81}=9$ which is a rational number.
$\sqrt{7}$ is a non-terminating, non-repeating decimal expansion.

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

Hence, the correct answer is option (C).

#### Question 7:

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

0.4014001400014... is an irrational number because its decimal expansion is non-terminating and non-repeating.
Hence, the correct answer is option (D).

#### Question 8:

Write the correct answer in each of the following:
A rational number between $\sqrt{2}$ and $\sqrt{3}$ is

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

(B) $\frac{\sqrt{2}·\sqrt{3}}{2}$

(C) 1.5

(D) 1.8

Value of $\sqrt{2}$ = 1.414...
Value of $\sqrt{3}$ = 1.732...
$\frac{\sqrt{2}+\sqrt{3}}{2}$ is an irrational number.
$\frac{\sqrt{2}·\sqrt{3}}{2}=\frac{\sqrt{6}}{2}$ is an irrational number.
1.5 is a rational number and lies between 1.414... and 1.732..

Hence, the correct answer is option (C).

#### Question 9:

Write the correct answer in each of the following:
The value of 1.999... in the form $\frac{p}{q},$ where p and q are integers and q ≠ 0, is
(A) $\frac{19}{10}$

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

(C) 2

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

Let x = 1.999...  (1)
Then, 10x = 19.999...  (2)
Subtracting (1) from (2), we get
10xx = (19.999...) – (1.999...)
⇒ 9x = 18
⇒ $x=\frac{18}{9}=2$

Hence, the correct answer is option (C).

#### Question 10:

Write the correct answer in each of the following:

$2\sqrt{3}+\sqrt{3}$ is equal to

(A) $2\sqrt{6}$

(B) 6

(C) $3\sqrt{3}$

(D) $4\sqrt{6}$

Hence, the correct answer is option C.

#### Question 11:

Write the correct answer in each of the following:

$\sqrt{10}×\sqrt{15}$ is equal to

(A) $6\sqrt{5}$

(B) $5\sqrt{6}$

(C) $\sqrt{25}$

(D) $10\sqrt{5}$

Hence, the correct answer is option B.

#### Question 12:

Write the correct answer in each of the following:
The number obtained on rationalising the denominator of $\frac{1}{\sqrt{7}-2}$ is
(A) $\frac{\sqrt{7}+2}{3}$

(B) $\frac{\sqrt{7}-2}{3}$

(C) $\frac{\sqrt{7}+2}{5}$

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

$\frac{1}{\sqrt{7}-2}$
$=\frac{1}{\sqrt{7}-2}×\frac{\sqrt{7}+2}{\sqrt{7}+2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+2}{{\left(\sqrt{7}\right)}^{2}-{2}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+2}{7-4}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+2}{3}$

Hence, the correct answer is option A.

#### Question 13:

Write the correct answer in each of the following:

$\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to

(A) $\frac{1}{2}\left(3-2\sqrt{2}\right)$

(B) $\frac{1}{3+2\sqrt{2}}$

(C) $3-2\sqrt{2}$

(D) $3+2\sqrt{2}$

$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{3-2\sqrt{2}}$
By Rationalizing the denominator, we get

Hence, the correct answer is option D.

#### Question 14:

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

Thus, we get 19 as the denominator, after rationalization of denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}}$.

Hence, the correct answer is option B.

#### Question 15:

Write the correct answer in each of the following:

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

$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{16×2}+\sqrt{16×3}}{\sqrt{4×2}+\sqrt{3×4}}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{4\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}\phantom{\rule{0ex}{0ex}}=2$

Hence, the correct answer is option B.

#### Question 16:

Write the correct answer in each of the following:

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

$\sqrt{\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)}$
$=\sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}×\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{{\left(\sqrt{2}-1\right)}^{2}}{{\left(\sqrt{2}\right)}^{2}-{1}^{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{{\left(\sqrt{2}-1\right)}^{2}}{1}}\phantom{\rule{0ex}{0ex}}=\sqrt{2}-1$
= 1.4142... – 1
= 0.4142 ...

Hence, the correct answer is option C.

#### Question 17:

Write the correct answer in each of the following:

$\sqrt[4]{\sqrt[3]{{2}^{2}}}$equals

(A) ${2}^{-\frac{1}{6}}$

(B) 2–6

(C) ${2}^{\frac{1}{6}}$

(D) 26

Hence, the correct answer is option C.

#### Question 18:

Write the correct answer in each of the following:

The product $\sqrt[3]{2}·\sqrt[4]{2}·\sqrt[12]{32}$ equals

(A) $\sqrt{2}$

(B) 2

(C) $\sqrt[12]{2}$

(D) $\sqrt[12]{32}$

LCM of 3, 4 and 12 = 12
and $\sqrt[12]{32}=\sqrt[12]{{2}^{5}}$
Thus, the product of

Hence, the correct answer is option B.

#### Question 19:

Write the correct answer in each of the following:

Value of $\sqrt[4]{{\left(81\right)}^{-2}}$ is

(A) $\frac{1}{9}$

(B) $\frac{1}{3}$

(C) 9

(D) $\frac{1}{81}$

Hence, the correct answer is option A.

#### Question 20:

Write the correct answer in each of the following:
Value of (256)0.16 × (256)0.09 is
(A) 4
(B) 16
(C) 64
(D) 256.25

(256)0.16 × (256)0.09

#### Question 21:

Write the correct answer in each of the following:
Which of the following is equal to x?

(A) ${x}^{\frac{12}{7}}-{x}^{\frac{5}{7}}$

(B) $\sqrt[12]{{\left({x}^{4}\right)}^{\frac{1}{3}}}$

(C) ${\left(\sqrt{{x}^{3}}\right)}^{\frac{2}{3}}$

(D) ${x}^{\frac{12}{7}}×{x}^{\frac{7}{12}}$

$\begin{array}{rcl}{\left(\sqrt{{x}^{3}}\right)}^{\frac{2}{3}}& =& {\left({x}^{3}\right)}^{\frac{1}{2}×\frac{2}{3}}\\ & =& {\left({x}^{3}\right)}^{\frac{1}{3}}\\ & =& {x}^{3×\frac{1}{3}}\\ & =& {x}^{1}\\ & =& x\end{array}$

$\begin{array}{rcl}{x}^{\frac{12}{7}}×{x}^{\frac{7}{12}}& =& {x}^{\left(\frac{12}{7}+\frac{7}{12}\right)}\\ & =& {x}^{\frac{193}{84}}\\ & \ne & x\end{array}$

Hence, the correct answer is option C.

#### Question 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.

Since it is given that x and y are a rational and irrational number.
Let, ,

then $x+y=4+\sqrt{2}=4+1.4142...=4.4142...$, which is non-terminating and non-recurring.
Hence, x$+$y is an irrational number

#### Question 2:

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

No, xy can only be irrational if xa.
Let x be a non zero rational y be an irrational. Then for xy to be irrational. Assume that xy be a rational number. Now, quotient of two non-zero rational number is a rational number.
So, $\frac{xy}{x}$is a rational number y is a rational number.
But this contradicts the fact that y is an irrational number.
Thus, our assumption is wrong.
Hence, xy is an irrational number.
Exception occurs when x = 0, then xy = 0, i.e. rational number.

#### Question 3:

State whether the following statements are true or false? Justify your answer.

(i) $\frac{\sqrt{2}}{3}$ is a rational number.

(ii) There are infinitely many integers between any two integers.

(iii) Number of rational numbers between 15 and 18 is finite.

(iv) There are numbers which cannot be written in the form both are integers.

(v) The square of an irrational number is always rational.

(vi) $\frac{\sqrt{12}}{\sqrt{3}}$ is not a rational number as $\sqrt{12}$ and $\sqrt{3}$ are not integers.

(vii) $\frac{\sqrt{15}}{\sqrt{3}}$ is written in the form  and so it is a rational number.

(i) False, $\sqrt{2}$ is a an irrational number and 3 is a rational number. When we divide irrational number by non-zero rational number it will always give irrational number.
(ii) False, because between 2 consecutive integers there doesn't exists any other integer.
(iii) False, because between any 2 rational numbers there exist infinitely many rational numbers.
(iv) True, because there are infinitely many numbers which cannot be written in the form $\frac{p}{q}$, q ≠ 0, p, q both are integers and these numbers are called irrational numbers.
(v) False, say an irrational number be $\sqrt{3}$ and $\sqrt[4]{3}$
(a) ${\left(\sqrt{3}\right)}^{2}=3$, which is a rational number
(b) ${\left(\sqrt[4]{3}\right)}^{2}=\sqrt{3}$, which is not a rational number
Hence, square of an irrational number is not always a rational number
(vi) False, $\frac{\sqrt{12}}{\sqrt{3}}=\frac{\sqrt{4×3}}{\sqrt{3}}=\sqrt{4}=2$, which is rational number.
(vii) False, $\frac{\sqrt{15}}{\sqrt{3}}=\sqrt{\frac{15}{3}}=\sqrt{5}$, which is an irrational number.

#### Question 4:

Classify the following numbers as rational or irrational with justification :

(i) $\sqrt{196}$

(ii) $3\sqrt{18}$

(iii) $\sqrt{\frac{9}{27}}$

(iv) $\frac{\sqrt{28}}{\sqrt{343}}$

(v) $–\sqrt{0.4}$

(vi) $\frac{\sqrt{12}}{\sqrt{75}}$

(vii) 0.5918

(viii) $\left(1+\sqrt{5}\right)-\left(4+\sqrt{5}\right)$

(ix) 10.124124...

(x) 1.010010001...

(i) $\sqrt{196}=\sqrt{{14}^{2}}=14$
Hence, it is a rational number.
(ii) $3\sqrt{18}=3\sqrt{{3}^{2}×2}=3×3\sqrt{2}=9\sqrt{2}$
Hence, it is an irrational number.
(iii) $\sqrt{\frac{9}{27}}=\sqrt{\frac{3×3}{3×3×3}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}$
Hence, it is an irrational number.
(iv) $\frac{\sqrt{28}}{\sqrt{343}}=\sqrt{\frac{2×2×7}{7×7×7}}=\sqrt{\frac{{2}^{2}}{{7}^{2}}}=\frac{2}{7}$
Hence, it is a rational number.
(v) $-\sqrt{0.4}=-\sqrt{\frac{4}{10}}=\frac{-2}{\sqrt{10}}$
Hence, it is an irrational number.
(vi) $\frac{\sqrt{12}}{\sqrt{75}}=\frac{\sqrt{4×3}}{\sqrt{25×3}}=\sqrt{\frac{4}{25}}=\frac{2}{5}$
Hence, it is a rational number.
(vii) 0.5918, it is a number with terminating decimal. Thus, it can be represented in $\frac{p}{q}$ form
$0.5918=\frac{5918}{10000}$
Hence, it is a rational number.
(viii) $\left(1+\sqrt{5}\right)-\left(4+\sqrt{5}\right)=1-4+\sqrt{5}-\sqrt{5}$ = −3 which is a rational number.
(ix) 10.124124... is a number with non-terminating recurring decimal expansion.
Hence, it is a rational number.
(x) 1.010010001..., is a number with non-terminating non-recurring decimal expansion.
Hence, it is an irrational number.

#### Question 1:

Find which of the variables x, y, z and u represent rational numbers and which irrational numbers :

(i) x2 = 5

(ii) y2 = 9

(iii) z2 = .04

(iv) ${u}^{2}=\frac{17}{4}$

(i) x2 = 5
Square root both sides, we get
$x=±\sqrt{5}$
Hence, the variable represents irrational number.
(ii) y2 = 9
Square root both sides, we get
$y=±\sqrt{9}=±3$
Hence, rational number
(iii) z2 = 0.04
$⇒{z}^{2}=\frac{4}{100}$
Square root both sides, we get
$z=±\sqrt{\frac{4}{100}}=±\frac{2}{10}\phantom{\rule{0ex}{0ex}}z=±\frac{1}{5}$
Hence, rational number

(iv) ${u}^{2}=\frac{17}{4}$
Square root both sides, we get
$u=±\sqrt{\frac{17}{4}}=\frac{\sqrt{17}}{2}$
Hence, irrational number.

#### Question 2:

Find three rational numbers between

(i) –1 and –2

(ii) 0.1 and 0.11

(iii)

(iv)

(i) Let x = –1 and y = –2

(ii) x = 0.1 and y = 0.11

.

(iii)
Multiplying the numerator and denominator by 4.
$\frac{5×4}{7×4}=\frac{20}{28}$

$\frac{6×4}{7×4}=\frac{24}{28}$
Thus, the rational numbers between are $\frac{21}{28},\frac{22}{28},\frac{23}{28}$.

Hence, the required numbers are $\frac{3}{4},\frac{11}{14},\frac{23}{28}$.

(iv)
Multiply x by 20 in numerator & denominator
$x=\frac{1}{5}×\frac{20}{20}=\frac{20}{100}$

Multiply y by 25 in numerator & denominator

$y=\frac{1}{4}×\frac{25}{25}=\frac{25}{100}$
Thus, 3 rational numbers between

Hence, the required numbers are .

#### Question 3:

Insert a rational number and an irrational number between the following :

(i) 2 and 3

(ii) 0 and 0.1

(iii) $\frac{1}{3}$ and $\frac{1}{2}$

(iv) $\frac{-2}{5}$ and $\frac{1}{2}$

(v) 0.15 and 0.16

(vi) $\sqrt{2}$ and $\sqrt{3}$

(vii) 2.357 and 3.121

(viii) .0001 and .001

(ix) 3.623623 and 0.484848

(x) 6.375289 and 6.375738

(i) A rational number between 2 and 3 is 2.5. Irrational number is 2.050050005...
(ii) A rational number between 0 and 0.1 is 0.05. Irrational number is 0.056005600056...
(iii) A rational number between $\frac{1}{3}$ and $\frac{1}{2}$ is $\frac{5}{12}$ Irrational number is 0.4014001400014...
(iv) A rational number between $\frac{-2}{5}$ and $\frac{1}{2}$ is 0. Irrational number is 0.130130013...
(v) A rational number between 0.15 and 0.16 is 0.155. Irrational number is 0.150150015...
(vi) A rational number between $\sqrt{2}$ and $\sqrt{3}$ is 1.5. Irrational number is 1.00150015...
(vii) A rational number between 2.357 and 3.121 is 3. Irrational number is 2.707007...
(viii) A rational number between 0.0001 and 0.001 is 0.00014. Irrational number is 0.00010130013...
(ix) A rational number between 3.623673 and 0.484848 is 2. Irrational number is 1.90900...
(x) A rational number between 6.375289 and 6.375738 is 6.3755. Irrational number is 6.385321002100021...

#### Question 4:

Represent the following numbers on the number line :

#### Question 5:

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

(i) 5 = 22 +12
In ΔABC
AB = 2 units, BC = 1 units
∠ABC = 90º

By Pythagoras theorem,
$\begin{array}{rcl}\mathrm{AC}& =& \sqrt{{\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}}\\ & =& \sqrt{{2}^{2}+{1}^{2}}\\ \mathrm{AC}& =& \sqrt{5}\end{array}$
Taking AC as radius we represent $\sqrt{5}$ on the number line with point P.
(ii) 10 = 32 + 12
In ΔABC
AB = 3 units, BC = 1 unit
∠ABC = 90º

By Pythagoras theorem,
$\begin{array}{rcl}\mathrm{AC}& =& \sqrt{{\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}}\\ & =& \sqrt{{3}^{2}+{1}^{2}}\\ \mathrm{AC}& =& \sqrt{10}\end{array}$
Taking AC as radius we represent $\sqrt{10}$ on the number line with point P.
(iii) 17 = 42 + 1
In ΔABC
AB = 4 units, BC = 1 unit
∠ABC = 90º

By Pythagoras theorem,
$\begin{array}{rcl}\mathrm{AC}& =& \sqrt{{\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}}\\ & =& \sqrt{{4}^{2}+{1}^{2}}\\ \mathrm{AC}& =& \sqrt{17}\end{array}$
Taking AC as radius, we represent $\sqrt{17}$ on the number line with point P.

#### Question 6:

Represent geometrically the following numbers on the number line :

(i) $\sqrt{4.5}$

(ii) $\sqrt{5.6}$

(iii) $\sqrt{8.1}$

(iv) $\sqrt{2.3}$

Step 1: Draw line segment AB of length equal to the number inside the root and extend it to C such that BC = 1.
Step 2: Draw a semi circle with centre O. O being mid point of AC and radius is OA.
Step 3: Draw a perpendicular line from B to cut the semi circle at D.
Step 4: Draw an arc with centre B and radius BD meeting AC produced at E.

(i)

Point E represent $\sqrt{4.5}$ on number line.
(ii)
Point E represent $\sqrt{5.6}$ on number line.
(iii)
Point E represent $\sqrt{8.1}$ on number line.
(iv)
Point E represent $\sqrt{2.3}$ on number line.

#### Question 7:

Express the following in the form $\frac{p}{q},$ where p and q are integers and q ≠ 0 :
(i) 0.2

(ii) 0.888...

(iii) $5.\overline{)2}$

(iv) $0.\overline{001}$

(v) 0.2555...

(vi) $0.1\overline{34}$

(vii) .00323232...

(viii) .404040...

(i) $0.2=\frac{2}{10}=\frac{1}{5}$

(ii) x = 0.888...      (1)
multiply by 10 on both sides
10x = 8.888...        (2)
Subtract (1) from (2)
9x = 8.888... – 0.888...
9x = 8

(iii) x = 5.222...     (1)
multiply by 10 on both sides
10x = 52.222...     (2)
Subtract (1) from (2)
9x = 52.222... – 5.222...
9x = 47
$x=\frac{47}{9}$

(iv) x = 0.001001...     (1)
multiply both sides by 1000
1000x = 1.001001...    (2)
Subtract (1) from (2)
$999x=1⇒x=\frac{1}{999}$

(v) x = 0.25555...    (1)
multiply both sides by 10
10x = 2.555...         (2)
Again multiply both sides by 10
100x = 25.555...     (3)
Subtract (2) from (3)
100x – 10x = (25.555...) – (2.555...)
90x = 23

(vi) x = 0.1343434...     (1)
10x = 1.343434...            (2)
Again multiply both sides by 1000.
1000x = 134.3434...     (3)
Subtract (2) from (3)
1000x – 10x = (134.3434...) – (1.3434...)
990x = 133

(vii) x = 0.003232...     (1)
Multiply both sides by 100.
100x = 0.323232...       (2)
Again multiply both sides by 10000.
10000x = 32.3232..      (3)
Subtract (2) from (3)
10000x – 100x = (32.3232...) – (0.3232...)
9900x = 32

(viii) x = 0.404040...    (1)
Multiply both sides by 100.
100x = 40.404040...     (2)
Subtract (1) from (2)
100xx = (40.4040...) – (0.4040...)
99x = 40

#### Question 8:

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

Let x = 0.142857142857...

x = $0.\overline{)142857}$ ......(1)

Multiplying equation (1) by 1000000

1000000 x = $142857.\overline{)142857}$  .... (2)

Subtracting equation (1)  from equation (2)

10,00,000$-$ x =

999999x = 142857

x = $\frac{142857}{999999}$

x = $\frac{1}{7}$

Hence proved.

#### Question 9:

Simplify the following:

(i) $\sqrt{45}-3\sqrt{20}+4\sqrt{5}$

(ii) $\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}$

(iii) $\sqrt[4]{12}×\sqrt[7]{6}$

(iv) $4\sqrt{28}÷3\sqrt{7}÷\sqrt[3]{7}$

(v) $3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}$

(vi) ${\left(\sqrt{3}-\sqrt{2}\right)}^{2}$

(vii)

(viii) $\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$

(ix) $\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}$

#### Question 10:

Rationalise the denominator of the following:

(i) $\frac{2}{3\sqrt{3}}$

(ii) $\frac{\sqrt{40}}{\sqrt{3}}$

(iii) $\frac{3+\sqrt{2}}{4\sqrt{2}}$

(iv) $\frac{16}{\sqrt{41}-5}$

(v) $\frac{2+\sqrt{3}}{2-\sqrt{3}}$

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

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

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

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

(i) For rationalization multiply numerator and denominator by $\sqrt{3}.$
$\begin{array}{rcl}& ⇒& \frac{2}{3\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3×3}\\ & =& \frac{2\sqrt{3}}{9}\end{array}$
(ii) Multiply numerator and denominator by $\sqrt{3}.$
$\begin{array}{rcl}& ⇒& \frac{\sqrt{40}}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{40×3}}{{\left(\sqrt{3}\right)}^{2}}\\ & =& \frac{\sqrt{120}}{3}=\frac{2}{3}\sqrt{30}\end{array}$
(iii) Multiply numerator by $\sqrt{2}.$
$\begin{array}{rcl}& ⇒& \frac{3+\sqrt{2}}{4\sqrt{2}}×\frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}+2}{4×2}\\ & =& \frac{3\sqrt{2}+2}{8}\end{array}$
(iv) Multiply numerator and denominator by $\sqrt{41}+5$
$\begin{array}{rcl}& ⇒& \frac{16}{\sqrt{41}-5}×\frac{\sqrt{41}+5}{\sqrt{41}+5}=\frac{16\left(\sqrt{41}+5\right)}{{\left(\sqrt{41}\right)}^{2}-{5}^{2}}\\ & =& \frac{16\left(\sqrt{41}+5\right)}{41-25}=\frac{16\left(\sqrt{41}+5\right)}{16}\\ & =& \sqrt{41}+5\end{array}$
(v) Multiply numerator denominator by $2+\sqrt{3}$

(vi) Multiply numerator and denominator by $\sqrt{2}-\sqrt{3}$
$⇒\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}×\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{{\left(\sqrt{2}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{-1}=\sqrt{18}-\sqrt{12}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}-2\sqrt{3}$

(vii) Multiply numerator and denominator by $\sqrt{3}+\sqrt{2}$

(viii) Multiply numerator and denominator by $\sqrt{5}+\sqrt{3}$
$⇒\frac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}×\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{\left(3\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{15+3\sqrt{15}+\sqrt{15}+3}{5-3}=\frac{18+4\sqrt{15}}{2}\phantom{\rule{0ex}{0ex}}=9+2\sqrt{15}$

(ix) Multiply numerator and denominator by $4\sqrt{3}-3\sqrt{2}$

#### Question 11:

Find the values of a and b in each of the following:

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

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

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

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

(i) Given,
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=a-6\sqrt{3}$
Rationalizing the denominator on LHS, by multiplying numerator & denominator by $\left(7-4\sqrt{3}\right)$ we have,
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}×\frac{7-4\sqrt{3}}{7-4\sqrt{3}}=\frac{5\left(7-4\sqrt{3}\right)+2\sqrt{3}\left(7-4\sqrt{3}\right)}{{7}^{2}-{\left(4\sqrt{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{35-20\sqrt{3}+14\sqrt{3}-24}{49-48}\phantom{\rule{0ex}{0ex}}=\frac{35-24-20\sqrt{3}+14\sqrt{3}}{1}=11-6\sqrt{3}$
Now as LHS = RHS
$11-6\sqrt{3}=a-6\sqrt{6}\phantom{\rule{0ex}{0ex}}⇒a=11$
(ii) Given, $\frac{3-\sqrt{5}}{3+2\sqrt{5}}=a\sqrt{5}-\frac{19}{11}$
Rationalizing the denominator on LHS by multiplying numerator and denominator by $\left(3-2\sqrt{5}\right)$

(iii) Given, $\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=2-b\sqrt{6}$
Rationalizing the denominator on LHS by multiplying the numerator and denominator by $\left(3\sqrt{2}+2\sqrt{3}\right)$

(iv) Given,

#### Question 12:

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

Rationalizing the denominator by $2-\sqrt{3}$
$\frac{1}{2+\sqrt{3}}×\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{{2}^{2}-{\left(\sqrt{3}\right)}^{2}}=\frac{2-\sqrt{3}}{4-3}\phantom{\rule{0ex}{0ex}}\frac{1}{a}=2-\sqrt{3}$

#### Question 13:

Rationalise the denominator in each of the following and hence evaluate by taking and $\sqrt{5}=2.236,$ upto three places of decimal.

(i) $\frac{4}{\sqrt{3}}$

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

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

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

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

(i) Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{3}$

(ii) Multiplying the numerator and denominator by $\sqrt{6}.$
$\begin{array}{rcl}& ⇒& \frac{6}{\sqrt{6}}×\frac{\sqrt{6}}{\sqrt{6}}=\frac{6\sqrt{6}}{{\left(\sqrt{6}\right)}^{2}}\\ & =& \frac{6\sqrt{6}}{6}\\ & =& \sqrt{6}=\sqrt{3}×\sqrt{2}\end{array}$

(iv)  multiply numerator and denominator by $2-\sqrt{2}$

(v) Multiply numerator and denominator by $\sqrt{3}-\sqrt{2}.$

#### Question 14:

Simplify :
(i) ${\left({1}^{3}+{2}^{3}+{3}^{3}\right)}^{\frac{1}{2}}$

#### Question 1:

Express $0.6+0.\overline{)7}+0.4\overline{)7}$ in the form $\frac{p}{q},$ where p and q are integers and q ≠ 0.

$x=0.\overline{)7}=0.77777...$         (1)
Multiplying both sides by 10
10x = 7.777...                 (2)
Subtract (1) from (2)
10xx = 7.777... – 0.777...
9x = 7
$x=\frac{7}{9}$
Now, $y=0.4\overline{)7}=0.4777...$   (3)
Multiply both sides by 10
10y = 4.777...                    (4)
Again multiply both sides by 100
100y = 47.777...               (5)
Subtract (4) from (5)
100y – 10y = 47.777... – 4.777...
90y = 43 $⇒y=\frac{43}{90}$

#### Question 2:

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

Given,
$\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{6+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$
Rationalizing the denominators of the terms
$\left[\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}×\frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}}\right]-\left[\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}×\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}\right]-\left[\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}×\frac{\sqrt{15}-3\sqrt{2}}{\sqrt{15}-3\sqrt{2}}\right]$
[Using identity (ab) (a + b) = a2b2]
$=\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{{\left(\sqrt{10}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}-\frac{2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{{\left(\sqrt{15}\right)}^{2}-{\left(3\sqrt{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{7\left(\sqrt{30}-3\right)}{10-3}-\frac{\left(2\sqrt{30}-10\right)}{6-5}-\frac{\left(3\sqrt{30}-18\right)}{15-18}\phantom{\rule{0ex}{0ex}}=\frac{\left(7\sqrt{30}-21\right)}{7}-\left(2\sqrt{30}-10\right)-\frac{3\left(\sqrt{30}-6\right)}{\left(-3\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{30}-3-2\sqrt{30}+10+\sqrt{30}-6\phantom{\rule{0ex}{0ex}}=1$

#### Question 3:

If  then find the value of $\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}.$

$\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\left(3\sqrt{3}+2\sqrt{2}\right)+3\left(3\sqrt{3}-2\sqrt{2}\right)}{\left(3\sqrt{3}-2\sqrt{2}\right)\left(3\sqrt{3}+2\sqrt{2}\right)}\phantom{\rule{0ex}{0ex}}=\frac{12\sqrt{3}+8\sqrt{2}+9\sqrt{3}-6\sqrt{2}}{{\left(3\sqrt{3}\right)}^{2}-{\left(2\sqrt{2}\right)}^{2}}$
[Using identity, (+ b) (ab) = a2b2]
$\frac{21\sqrt{3}+2\sqrt{2}}{27-8}=\frac{21\sqrt{3}+2\sqrt{2}}{19}\phantom{\rule{0ex}{0ex}}=\frac{21×1.732+2×1.414}{19}=\frac{36.372+2.828}{19}\phantom{\rule{0ex}{0ex}}=\frac{39.2}{19}=2.063$

#### Question 4:

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

Given, $a=\frac{3+\sqrt{5}}{2}$
then $\frac{1}{a}=\frac{2}{3+\sqrt{5}}$
Rationalizing the denominator, we have
$\begin{array}{rcl}\frac{2}{3+\sqrt{5}}×\frac{3-\sqrt{5}}{3-\sqrt{5}}& =& \frac{6-2\sqrt{5}}{{3}^{2}-{\left(\sqrt{5}\right)}^{2}}=\frac{6-2\sqrt{5}}{9-5}=\frac{6-2\sqrt{5}}{4}\\ & =& \frac{3-\sqrt{5}}{2}\end{array}$

$\begin{array}{rcl}& =& {\left(\frac{6}{2}\right)}^{2}-2={3}^{2}-2\\ & =& 9-2\\ & =& 7\end{array}$

#### Question 5:

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

Given,

Now, multiplying y, numerator and denominator by $\sqrt{3}-\sqrt{2}$

Now, squaring both equation (1) and (2)
$\begin{array}{rcl}{x}^{2}& =& {\left(5+2\sqrt{6}\right)}^{2}={5}^{2}+20\sqrt{6}+24\\ & =& 25+24+20\sqrt{6}\\ & =& 49+20\sqrt{6}\end{array}$
$\begin{array}{rcl}{y}^{2}& =& {\left(5-2\sqrt{6}\right)}^{2}={5}^{2}-20\sqrt{6}+24\\ & =& 49-20\sqrt{6}\end{array}$

#### Question 6:

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

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

Using

#### Question 7:

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

Using $\frac{\mathit{1}}{{\mathit{a}}^{\mathit{-}\mathit{m}}}$ = ${a}^{\mathit{m}}$

= 4 $×$ ${\left(216\right)}^{\frac{2}{3}}$ $+$ 1 $×$ ${\left(256\right)}^{\frac{3}{4}}$ $+$$×$ ${\left(243\right)}^{\frac{1}{5}}$

Now, 216 = ${\left(6\right)}^{3}$, 256 = ${\left(4\right)}^{4}$, 243 = ${\left(3\right)}^{5}$

= 4 $×$ ${\left\{{\left(6\right)}^{3}\right\}}^{\frac{2}{3}}$ $+$ 1 $×$ ${\left\{{\left(4\right)}^{4}\right\}}^{\frac{3}{4}}$ $+$$×$ ${\left\{{\left(3\right)}^{5}\right\}}^{\frac{1}{5}}$

Using
= 4 $×$ ${\left(6\right)}^{3×\frac{2}{3}}$ $+$ 1 $×$ ${\left(4\right)}^{4×\frac{3}{4}}$ $+$$×$ ${\left(3\right)}^{5×\frac{1}{5}}$

= 4

=

=144 + 64 + 6

= 214

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