Rs Aggarwal 2017 Solutions for Class 9 Math Chapter 1 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 9 students for Math Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 9 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Question 1:

What are rational numbers? Give ten examples of rational numbers.

The numbers that can be written in the $\frac{p}{q}$ form, where p and q are integers and q ≠ 0 are known as rational numbers.

Examples of rational numbers:
(1)  $\frac{4}{5}$      (2) $\frac{1}{5}$     (3) $\frac{5}{4}$     (4) $\frac{4}{1}$ = 4       (5) $\frac{5}{2}$     (6) $\frac{1}{7}$      (7) $\frac{0}{1}$ = 0        (8) $\frac{9}{5}$        (9) $\frac{5}{5}$ = 1       (10) $\frac{4}{9}$

#### Question 2:

Represent each of the following rational numbers on the number line:
(i) 5
(ii) −3
(iii) $\frac{5}{7}$
(iv) $\frac{8}{3}$
(v) 1.3
(vi) −2.4
(vii) $\frac{23}{6}$

(i) (ii) (iii) (iv) $\frac{8}{3}=2\frac{2}{3}$ (v) $1.3=\frac{13}{10}=1\frac{3}{10}$ (vi) $-2.4=\frac{-24}{10}=\frac{-12}{5}=-2\frac{2}{5}$ (vii) $\frac{23}{6}=3\frac{5}{6}$ #### Question 3:

Find a rational number lying between
(i)
(ii)
(iii) 1.3 and 1.4
(iv) 0.75 and 1.2
(v) $-1\mathrm{and}\frac{1}{2}$
(vi) $-\frac{3}{4}\mathrm{and}-\frac{2}{5}$

(i) Let:
x = $\frac{1}{4}$ and y = $\frac{1}{3}$
Rational number lying between x and y:

= $\frac{7}{24}$

(ii) Let:
x = $\frac{3}{8}$ and y = $\frac{2}{5}$
Rational number lying between x and y:

=

(iii) Let:
x = 1.3 and y = 1.4
Rational number lying between x and y:

=

(iv) Let:
x = 0.75 and y = 1.2
Rational number lying between x and y:

=

(v) Let:
x = $-$1 and y = $\frac{1}{2}$
Rational number lying between x and y:

= $-\frac{1}{4}$

(vi) Let:
x$-\frac{3}{4}$ and y = $-\frac{2}{5}$
Rational number lying between x and y:

=

#### Question 4:

Find three rational numbers lying between $\frac{1}{5}\mathrm{and}\frac{1}{4}$.

Let:
x
= $\frac{1}{5}$, y = $\frac{1}{4}$ and n = 3
We know:
d =

So, three rational numbers lying between x and y are:

(x + d), (x + 2d) and (x + 3d)

=

=

#### Question 5:

Find five rational numbers lying between $\frac{2}{5}\mathrm{and}\frac{3}{4}$.

Let:
x = $\frac{2}{5}$ , y = $\frac{3}{4}$ and n = 5
We know:
d =

So, five rational numbers between x and y are:
(x + d), (x + 2d), (x + 3d), (x + 4d) and (x + 5d)

=

=

#### Question 6:

Insert six rational numbers between 3 and 4.

Let:
x = 3, y = 4 and n = 6
We know:
d =
So, six rational numbers between x and y are:
(x + d), (x + 2d), (x + 3d), (x + 4d), (x + 5d) and (x + 6d)
=

=

#### Question 7:

Insert 16 rational numbers between 2.1 and 2.2.

Let:
x = 2.1, y = 2.2 and n = 16

We know:
d = $\frac{y-x}{n+1}=\frac{2.2-2.1}{16+1}=\frac{0.1}{17}=\frac{1}{170}$= 0.005 (approx.)
So, 16 rational numbers between 2.1 and 2.2 are:
(x + d), (x + 2d), ...(x + 16d)
= [2.1 + 0.005], [2.1 + 2(0.005)],...[2.1 + 16(0.005)]
= 2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17, 2.175 and 2.18

#### Question 1:

Without actual division, find which of the following rationals are terminating decimals.
(i) $\frac{13}{80}$
(ii) $\frac{7}{24}$
(iii) $\frac{5}{12}$
(iv) $\frac{8}{35}$
(v) $\frac{16}{125}$

(i) Denominator of $\frac{13}{80}$ is 80.
And,
80 = 24$×$5
Therefore, 80 has no other factors than 2 and 5.
Thus, $\frac{13}{80}$ is a terminating decimal.

(ii) Denominator of $\frac{7}{24}$ is 24.
And,
24 = 23$×$3
So, 24 has a prime factor 3, which is other than 2 and 5.
Thus, $\frac{7}{24}$ is not a terminating decimal.

(iii)  Denominator of $\frac{5}{12}$ is 12.
And,
12 = 22$×$3
So, 12 has a prime factor 3, which is other than 2 and 5.
Thus, $\frac{5}{12}$ is not a terminating decimal.

(iv) Denominator of $\frac{8}{35}$ is 35.
And,
35 = 7$×$5
So, 35 has a prime factor 7, which is other than 2 and 5.
Thus, $\frac{8}{35}$ is not a terminating decimal.

(v) Denominator of $\frac{16}{125}$ is 125.
And,
125 = 53
Therefore, 125 has no other factors than 2 and 5.
Thus, $\frac{16}{125}$ is a terminating decimal.

#### Question 2:

Convert each of the following into a decimal.
(i) $\frac{5}{8}$
(ii) $\frac{9}{16}$
(iii) $\frac{7}{25}$
(iv) $\frac{11}{24}$
(v) $2\frac{5}{12}$

(i) $\frac{5}{8}$ = 0.625
By actual division, we have: (ii) $\frac{9}{16}$ = 0.5625
By actual division, we have: (iii) $\frac{7}{25}$ = 0.28
By actual division, we have: (iv) $\frac{11}{24}$ = By actual division, we have: (v)  2$\frac{5}{12}$ = $\frac{29}{12}$ = By actual division, we have: #### Question 3:

Express each of the following as a fraction in simplest form.
(i) $0.\overline{)3}$
(ii) $1.\overline{)3}$
(iii) $0.\overline{)34}$
(iv) $3.\overline{)14}$
(v) $0.\overline{)324}$
(vi) $0.\overline{)17}$
(vii) $0.\overline{)54}$
(viii) $0.\overline{)163}$

(i) $0.\overline{)3}$
Let $x=0.\overline{3}$
∴ x = 0.3333...                           ...(i)
10x = 3.3333...                           ...(ii)
On subtracting (i) from (ii), we get:
9x = 3
= x = $\frac{1}{3}$

∴

(ii) $1.\overline{)3}$
Let $x=1.\overline{3}$
∴ x = 1.3333...                    ...(i)
10x = 13.3333...                  ...(ii)
On subtracting (i) from (ii), we get:
9x = 12
= x = $\frac{4}{3}$

$1.\overline{)3}=\frac{4}{3}$

(iii) $0.\overline{)34}$
Let x$0.\overline{)34}$
x = 0.3434...                    ...(i)
100x = 34.3434...                ...(ii)
On subtracting (i) from (ii), we get:
99x = 34
= x = $\frac{34}{99}$

$0.\overline{)34}=\frac{34}{99}$

(iv) $3.\overline{)14}$
Let x = $3.\overline{)14}$
x = 3.1414...                    ...(i)
100x = 314.1414...              ...(ii)
On subtracting (i) from (ii), we get:
99x = 311
= x = $\frac{311}{99}$

$3.\overline{)14}$ = $\frac{311}{99}$

(v) $0.\overline{)324}$
Let x = $0.\overline{)324}$
x = 0.324324...                    ...(i)
1000x = 324.324324...            ...(ii)
On subtracting (i) from (ii), we get:
999x = 324
= x = $\frac{324}{999}$=$\frac{12}{37}$

$0.\overline{)324}$= $\frac{12}{37}$

(vi) $0.1\overline{)7}$
Let x = $0.1\overline{)7}$
x = 0.1777...
10x = 1.777...              ...(i)
100x = 17.777....         ...(ii)
On subtracting (i) from (ii). we get:
90x = 16
= x = $\frac{8}{45}$
$0.1\overline{)7}$ = $\frac{8}{45}$

(vii) $0.5\overline{)4}$
Let x = $0.5\overline{)4}$
x = 0.5444...
10x = 5.4444...               ...(i)
100x = 54.4444...           ...(ii)
On subtracting (i) from (ii), we get:
90x = 49
= x = $\frac{49}{90}$
$0.5\overline{)4}$= $\frac{49}{90}$

(viii) $0.1\overline{)63}$
Let x$0.1\overline{)63}$
x = 0.16363...
10x = 1.6363...               ...(i)
1000x = 163.6363...       ...(ii)
On subtracting (i) from (ii), we get:
990x = 162
= x = $\frac{162}{990}=\frac{9}{55}$

$0.1\overline{)63}$ = $\frac{9}{55}$

#### Question 4:

Write, whether the given statement is true or false. Give reasons.
(i) Every natural number is a whole number.
(ii) Every whole number is a natural number.
(iii) Every integer is a rational number.
(iv) Every rational number is a whole number.
(v) Every terminating decimal is a rational number.
(vi) 0 is a rational number.

(i) True
Natural numbers start from 1 to infinity and whole numbers start from 0 to infinity; hence, every natural number is a whole number.

(ii) False
0 is a whole number but not a natural number, so every whole number is not a natural number.

(iii) True
Every integer can be expressed in the $\frac{p}{q}$ form.

(iv) False
Because whole numbers consist only of numbers of the form $\frac{p}{1}$, where p is a positive number. On the other hand, rational numbers are the numbers whose denominator can be anything except 0.

(v) True
Every terminating decimal can be easily expressed in the $\frac{p}{q}$ form.

(vi) True
Every terminating decimal can be easily expressed in the $\frac{p}{q}$ form.

(vii) True
0 can be expressed in the form $\frac{p}{q}$, so it is a rational number.

#### Question 1:

What are irrationl numbers? How do they differ from rational numbers? Give examples.

A number that can neither be expressed as a terminating decimal nor be expressed as a repeating decimal is called an irrational number. A rational number, on the other hand, is always a terminating decimal, and if not, it is a repeating decimal.
Examples of irrational numbers:
0.101001000...
0.232332333...

#### Question 2:

Classify the following numbers as rational or irrational. Give reasons to support your answer:
(i) $\sqrt{4}$
(ii) $\sqrt{196}$
(iii) $\sqrt{21}$
(iv) $\sqrt{43}$
(v) $3+\sqrt{3}$
(vi) $\sqrt{7}-2$
(vii) $\frac{2}{3}\sqrt{6}$
(viii) $0.\overline{)6}$
(ix) 1.232332333...
(x) 3.040040004....
(xi) 3.2576
(xii) 2.356565656...
(xiii) $\mathrm{\pi }$
(xiv) $\frac{22}{7}$

#### Question 3:

Represent on the real line. Let X'OX be a horizontal line taken as the x-axis and O be the origin representing 0.
Take OA = 1 unit and AB ⊥ OA such that AB = 1 unit.
Join OB.
Now,
OB = units
Taking O as the centre and OB as the radius, draw an arc, meeting OX at P.
We have:
OP = OB = $\sqrt{2}$ units
Thus, point P represents $\sqrt{2}$ on the number line.
Now, draw BC ⊥ OB such that BC = 1 unit.
Join OC.
We have:
OC = units
Taking O as the centre and OC as the radius, draw an arc, meeting OX at Q.
We have:
OQ = OC = $\sqrt{3}$units
Thus, point Q represents $\sqrt{3}$ on the number line.
Now, draw CD ⊥OC such that CD = 1 unit.
Join OD.
We have:
OD = units
Now, draw DE ⊥OD such that DE = 1 unit.
Join OE.
We have:
OE = units
Taking O as the centre and OE as the radius, draw an arc, meeting OX at R.
We have:
OR = OE = $\sqrt{5}$ units
Thus, point R represents $\sqrt{5}$ on the number line.

#### Question 4:

Represent $\sqrt{6}\mathrm{and}\sqrt{7}$on the real line. Let X'OX be a horizontal line taken as the x-axis and O be the origin representing 0.
Take OA = 2 units and AB ⊥ OA such that AB = 1 unit.
Now, join OB.
We have: OB = units
Taking O as the centre and OB as the radius, draw an arc, meeting OX at P.
Thus, we have:
OP = OB = $\sqrt{5}$ units
Here, point P represents $\sqrt{5}$ on the number line.
Now, draw BC ⊥ OB such that BC = 1 unit.
Join OC.
We have: OC = units
Taking O as the centre and OC as the radius, draw an arc, meeting OX at Q.
Thus, we have:
OQ = OC = $\sqrt{6}$units
Here, point Q represents $\sqrt{6}$ on the number line.
Now, draw CD ⊥OC such that CD =1 unit.
Join OD.
We have: OD = units
Taking O as the centre and OE as the radius, draw an arc, meeting OX at R.
Now,
OR = OD = $\sqrt{7}$ units
Thus, point R represents $\sqrt{7}$ on the number line.

#### Question 5:

Giving reason in each case, show that each of the following numbers is irrational.
(i) $4+\sqrt{5}$
(ii) $\left(-3+\sqrt{6}\right)$
(iii) $5\sqrt{7}$
(iv) $-3\sqrt{8}$
(v) $\frac{2}{\sqrt{5}}$
(vi) $\frac{4}{\sqrt{3}}$

#### Question 6:

State in each case, whether the given statement is true of false.
(i) The sum of two rational numbers is rational.
(ii) The sum of two irrational numbers is irrational.
(iii) The product of two rational numbers is rational.
(iv) The product of two irrational number is irrational.
(v) The sum of a rational number and an irrational number is irrational.
(vi) The product of a nonzero rational number and an irrational number is a rational number.
(vii) Every real number is rational.
(viii) Every real number is either rational or irrational.
(ix) $\mathrm{\pi }$is irrational and$\frac{22}{7}$is rational.

(i) True

(ii) False
Example:

(iii) True

(iv) False

(v) True

(vi) False
Example:

(vii) False
Real numbers can be divided into rational and irrational numbers.

(viii) True

(ix) True

#### Question 1:

(i) $\left(2\sqrt{3}-5\sqrt{2}\right)\mathrm{and}\left(\sqrt{3}+2\sqrt{2}\right)$
(ii) $\left(2\sqrt{2}+5\sqrt{3}-7\sqrt{5}\right)\mathrm{and}\left(3\sqrt{3}-\sqrt{2}+\sqrt{5}\right)$
(iii) $\left(\frac{2}{3}\sqrt{7}-\frac{1}{2}+6\sqrt{11}\right)\mathrm{and}\left(\frac{1}{3}\sqrt{7}+\frac{3}{2}\sqrt{2}-\sqrt{11}\right)$

Multiply:
(i)
(ii)
(iii)
(iv)
(v)
(vi)

Divide:
(i)
(ii)
(iii)

#### Question 4:

Simplify:
(i) $\left(4+\sqrt{2}\right)\left(4-\sqrt{2}\right)$
(ii) $\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)$
(iii) $\left(6-\sqrt{6}\right)\left(6+\sqrt{6}\right)$
(iv) $\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$
(v) ${\left(\sqrt{5}-\sqrt{3}\right)}^{2}$
(vi) ${\left(3-\sqrt{3}\right)}^{2}$

#### Question 5:

Represent $\sqrt{3.2}$geometrically on the number line.

Draw a line segment AB = 3.2 units and extend it to C such that BC = 1 unit.
Now, find the midpoint O of AC.
Taking O as the centre and OA as the radius, draw a semicircle.
Now, draw BD ⊥ AC, intersecting the semicircle at D.
Here,
BD = $\sqrt{3.2}$ units
Taking B as the centre and BD as the radius, draw an arc meeting AC produced at E.
Thus, we have:
BE = BD = $\sqrt{3.2}$ units #### Question 6:

Represent $\sqrt{7.28}$ geometrically on the number line.

Draw a line segment AB = 7.28 units and extend it to C such that BC = 1 unit.
Now, find the midpoint O of AC.
Taking O as the centre and OA as the radius, draw a semicircle.
Now, draw BD ⊥ AC, intersecting the semicircle at D.
We have:
BD = $\sqrt{7.28}$ units
Now, taking B as the centre and BD as the radius, draw an arc meeting AC produced at E.
Thus, we have:
BE = BD =$\sqrt{7.28}$ units #### Question 7:

Mention the closure property, associative law, commutative law, existance of identity, existance of inverse of each real number for each of the operations (i) addition (ii) multiplication on real numbers.

(i) Closure property: The sum of two real numbers is always a real number.
(ii) Associative law: (a + b) + c = a + (b + c) for all real numbers a, b and c.
(iii) Commutative law: a + b = b + a for all real numbers a and b.
(iv) Existence of additive identity: 0 is called the additive identity for real numbers.
As, for every real number a ,  0 + a = a + 0 = a
(v) Existence of additive inverse: For each real number a, there exists  a real number ($-$a) such that  a + ($-$a) = 0 = ($-$a) + a. Here, a and ($-$a) are the additive inverse of each other.

MULTIPLICATION PROPERTIES OF REAL NUMBERS

(i) Closure property: The product of two real numbers is always a real number.
(ii) Associative law: (ab)c = a(bc)  for all real numbers a, b and c.
(iii) Commutative law: a $×$ b = b $×$ a for all real numbers a and b.
(iv) Existence of multiplicative identity: 1 is called the multiplicative identity for real numbers.
As, for every real number a ,  1 $×$ a = a $×$ 1 = a

(v) Existence of multiplicative inverse: For each real number a, there exists  a real number $\left(\frac{1}{a}\right)$ such that  a $\left(\frac{1}{a}\right)$ = 1 = $\left(\frac{1}{a}\right)$a. Here, a and $\left(\frac{1}{a}\right)$  are the multiplicative inverse of each other

#### Question 1:

Rationalise the denominator of each of the following:
$\frac{1}{\sqrt{7}}$

On multiplying the numerator and denominator of the given number by $\sqrt{7}$, we get:

#### Question 2:

Rationalise the denominator of each of the following:
$\frac{\sqrt{5}}{2\sqrt{3}}$

On multiplying the numerator and denominator of the given number by $\sqrt{3}$, we get:

#### Question 3:

Rationalise the denominator of each of the following:
$\frac{1}{\left(2+\sqrt{3}\right)}$

On multiplying the numerator and denominator of the given number by $2-\sqrt{3}$, we get:

#### Question 4:

Rationalise the denominator of each of the following:
$\frac{1}{\left(\sqrt{5}-2\right)}$

On multiplying the numerator and denominator of the given number by $\sqrt{5}+2$, we get:

#### Question 5:

Rationalise the denominator of each of the following:
$\frac{1}{\left(5+3\sqrt{2}\right)}$

On multiplying the numerator and denominator of the given number by $5-3\sqrt{2}$, we get:

#### Question 6:

Rationalise the denominator of each of the following:
$\frac{1}{\left(\sqrt{6}-\sqrt{5}\right)}$

On multiplying the numerator and denominator of the given number by $\sqrt{6}+\sqrt{5}$, we get:

#### Question 7:

Rationalise the denominator of each of the following:
$\frac{4}{\left(\sqrt{7}+\sqrt{3}\right)}$

On multiplying the numerator and denominator of the given number by $\sqrt{7}-\sqrt{3}$, we get:

#### Question 8:

Rationalise the denominator of each of the following:
$\frac{\sqrt{3}-1}{\sqrt{3}+1}$

On multiplying the numerator and denominator of the given number by $\sqrt{3}-1$, we get:

#### Question 9:

Rationalise the denominator of each of the following:
$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$

On multiplying the numerator and denominator of the given number by $3-2\sqrt{2}$, we get:

#### Question 10:

Find the values of a and b in each of the following.
$\frac{\sqrt{3}+1}{\sqrt{3}-1}=a+b\sqrt{3}$

We have:
$\frac{\sqrt{3}+1}{\sqrt{3}-1}$

Now, rationalising the denominator of the given number by multiplying both the numerator and denominator with $\sqrt{3}+1$, we get:

∴ $2+\sqrt{3}$= a + b$\sqrt{3}$
So, on comparing the LHS and the RHS, we get:
a = 2 and b = 1

#### Question 11:

Find the values of a and b in each of the following.
$\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}$

We have:
$\frac{3+\sqrt{2}}{3-\sqrt{2}}$

Now, rationalising the denominator of the given number by multiplying both the numerator and denominator with $3+\sqrt{2}$, we get:
$\frac{3+\sqrt{2}}{3-\sqrt{2}}=\frac{3+\sqrt{2}}{3-\sqrt{2}}×\frac{3+\sqrt{2}}{3+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(3+\sqrt{2}\right)}^{2}}{{\left(3\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{9+2+6\sqrt{2}}{9-2}\phantom{\rule{0ex}{0ex}}=\frac{11+6\sqrt{2}}{7}\phantom{\rule{0ex}{0ex}}=\frac{11}{7}+\frac{6}{7}\sqrt{2}$

= a + b$\sqrt{2}$
So, on comparing the LHS and the RHS, we get:
a$\frac{11}{7}$ and b = $\frac{6}{7}$

#### Question 12:

Find the values of a and b in each of the following.
$\frac{5-\sqrt{6}}{5+\sqrt{6}}=a-b\sqrt{6}$

We have:
$\frac{5-\sqrt{6}}{5+\sqrt{6}}$

Now, rationalising the denominator of the given number by multiplying both the numerator and denominator with $5-\sqrt{6}$, we get:
$\frac{5-\sqrt{6}}{5+\sqrt{6}}=\frac{5-\sqrt{6}}{5+\sqrt{6}}×\frac{5-\sqrt{6}}{5-\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(5-\sqrt{6}\right)}^{2}}{{\left(5\right)}^{2}-{\left(\sqrt{6}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{25+6-10\sqrt{6}}{25-6}\phantom{\rule{0ex}{0ex}}=\frac{31-10\sqrt{6}}{19}\phantom{\rule{0ex}{0ex}}=\frac{31}{19}-\frac{10}{19}\sqrt{6}$

= a $-$ b$\sqrt{6}$
So, on comparing the LHS and the RHS, we get:
a$\frac{31}{19}$ and b = $\frac{10}{19}$

#### Question 13:

Find the values of a and b in each of the following.
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=a-b\sqrt{3}$

We have :
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$

Now, rationalising the denominator of the given number by multiplying both the numerator and denominator with $7-4\sqrt{3}$, we get:
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=\frac{5+2\sqrt{3}}{7+4\sqrt{3}}×\frac{7-4\sqrt{3}}{7-4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{35+14\sqrt{3}-20\sqrt{3}-24}{{\left(7\right)}^{2}-{\left(4\sqrt{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{11-6\sqrt{3}}{49-48}\phantom{\rule{0ex}{0ex}}=11-6\sqrt{3}$

$11-6\sqrt{3}$ = a $-$ b$\sqrt{3}$
So, on comparing the LHS and the RHS, we get:
a = 11 and b = 6

#### Question 14:

Simplify:$\left(\frac{\sqrt{5}-1}{\sqrt{5}+1}+\frac{\sqrt{5}+1}{\sqrt{5}-1}\right)$

#### Question 15:

Simplify: $\left(\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}\right)$.

$\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(4+\sqrt{5}\right)}^{2}+{\left(4-\sqrt{5}\right)}^{2}}{\left(4-\sqrt{5}\right)\left(4+\sqrt{5}\right)}\phantom{\rule{0ex}{0ex}}=\frac{{4}^{2}+{\left(\sqrt{5}\right)}^{2}+2×4×\sqrt{5}+{4}^{2}+{\left(\sqrt{5}\right)}^{2}-2×4×\sqrt{5}}{16-5}\phantom{\rule{0ex}{0ex}}=\frac{16+5+8\sqrt{5}+16+5-8\sqrt{5}}{16-5}\phantom{\rule{0ex}{0ex}}=\frac{42}{11}$

#### Question 16:

If $x=\left(4-\sqrt{15}\right)$, find the value of $\left(x+\frac{1}{x}\right)$.

#### Question 17:

If $x=\left(2+\sqrt{3}\right)$, find the value of $\left({x}^{2}+\frac{1}{{x}^{2}}\right)$.

#### Question 18:

Show that $\frac{1}{\left(3-\sqrt{8}\right)}-\frac{1}{\left(\sqrt{8}-\sqrt{7}\right)}+\frac{1}{\left(\sqrt{7}-\sqrt{6}\right)}-\frac{1}{\left(\sqrt{6}-\sqrt{5}\right)}+\frac{1}{\left(\sqrt{5}-2\right)}=5$.

$=\frac{1}{3-\sqrt{8}}×\frac{3+\sqrt{8}}{3+\sqrt{8}}-\left(\frac{1}{\sqrt{8}-\sqrt{7}}×\frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}\right)+\frac{1}{\sqrt{7}-\sqrt{6}}×\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}-\left(\frac{1}{\sqrt{6}-\sqrt{5}}×\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}\right)+\frac{1}{\sqrt{5}-2}×\frac{\sqrt{5}+2}{\sqrt{5}+2}\phantom{\rule{0ex}{0ex}}=\frac{3+\sqrt{8}}{9-8}-\left(\frac{\sqrt{8}+\sqrt{7}}{8-7}\right)+\frac{\sqrt{7}+\sqrt{6}}{7-6}-\left(\frac{\sqrt{6}+\sqrt{5}}{6-5}\right)+\frac{\sqrt{5}+2}{5-4}\phantom{\rule{0ex}{0ex}}=3+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+2\phantom{\rule{0ex}{0ex}}=3+2\phantom{\rule{0ex}{0ex}}=5$

So, LHS = RHS

#### Question 1:

Simplify:
(i) $\left({6}^{2/5}×{6}^{3/5}\right)$
(ii) $\left({3}^{1/2}×{3}^{1/3}\right)$
(iii) $\left({7}^{5/6}×{7}^{2/3}\right)$

#### Question 2:

Simplify:
(i) $\frac{{6}^{1/4}}{{6}^{1/5}}$
(ii) $\frac{{8}^{1/2}}{{8}^{2/3}}$
(iii) $\frac{{5}^{6/7}}{{5}^{2/3}}$

#### Question 3:

Simplify:
(i) ${3}^{1/4}×{5}^{1/4}$
(ii) ${2}^{5/8}×{3}^{5/8}$
(iii) ${6}^{1/2}×{7}^{1/2}$

#### Question 4:

Simplify:
(i) (34)1/4
(ii) (31/3)4
(iii) $\left(\frac{1}{{3}^{4}}\right)$

Evaluate:
(i) (49)1/2
(ii) (125)1/3
(iii) (64)1/6

Evaluate:
(i) (25)3/2
(ii) (32)2/5
(iii) (81)3/4

Evaluate:
(i) (64)−1/2
(ii) (8)−1/3
(iii) (81)−1/4

#### Question 1:

Which of the following is an irrational number?
(a) 3.14
(b) $3.\overline{)14}$
(c) $3.1\overline{)4}$
(d) 3.141141114...

(d) 3.141141114...

Because 3.141141114... is neither a repeating decimal nor a terminating decimal, it is an irrational number.

#### Question 2:

Which of the following is an irrational number?
(a) $\sqrt{49}$
(b) $\sqrt{\frac{9}{16}}$
(c) $\sqrt{5}$
(d) $\frac{\sqrt{20}}{\sqrt{5}}$

(c) $\sqrt{5}$

Because $\sqrt{5}$ cannot be expressed in the $\frac{p}{q}$ form where p and q are integers (q$\ne$0), it is an irrational number. The remaining options have a perfect root; they can be expressed in the $\frac{p}{q}$ form, so they are rational numbers.

#### Question 3:

Which of the following is an irrational number?
(a) $0.\overline{)32}$
(b) $0.3\overline{)21}$
(c) $0.\overline{)321}$
(d) 0.3232232223...

(d) 0.3232232223...

Because 0.3232232223... is neither a repeating decimal nor a terminating decimal, it is an irrational number.

#### Question 4:

Which of the following is a rational number?
(a) $\sqrt{2}$
(b) $\sqrt{23}$
(c) $\sqrt{225}$
(d) 0.1010010001....

(c) $\sqrt{225}$

Because 225 is a square of 15, i.e., $\sqrt{225}$ = 15, and it can be expressed in the $\frac{p}{q}$ form, it is a rational number.

#### Question 5:

Every rational number is
(a) a natural number
(b) a whole number
(c) an integer
(d) a real number

(d) a real number

Every rational number is a real number, as every rational number can be easily expressed on the real number line.

#### Question 6:

Between any two rational numbers there
(a) is no rational number
(b) is exactly one rational numbers
(c) are infinitely many rational numbers
(d) is no irrational number

(c) are infinitely many rational numbers

Because the range between any two rational numbers can be easily divided into any number of divisions, there can be an infinite number of rational numbers between any two rational numbers.

#### Question 7:

The decimal representation of a rational number is
(a) always terminating
(b) either terminating or repeating
(c) either terminating or non-repeating
(d) neither terminating nor repeating

(b) either terminating or repeating

As per the definition of rational numbers, they are either repeating or terminating decimals.

#### Question 8:

The decimal representation of an irrational number is
(a) always terminating
(b) either terminating or repeating
(c) either terminating or non-repeating
(d) neither terminating nor repeating

(d) neither terminating nor repeating

As per the definition of irrational numbers, these are neither terminating nor repeating decimals.

#### Question 9:

Decimal expansion of $\sqrt{2}$is
(a) a finite decimal
(b) a terminating or repeating decimal
(c) a non-terminating and non-repeating decimal
(d) none of these

(c) a non-terminating and non-repeating decimal

Because $\sqrt{2}$ is an irrational number, its decimal expansion is non-terminating and non-repeating.

#### Question 10:

The product of two irrational number is
(a) always irrational
(b) always rational
(c) always an integer
(d) sometimes rational and sometimes irrational

(d) sometimes rational and sometimes irrational

For example:
$\sqrt{2}$ is an irrational number, when it is multiplied with itself  it results into 2, which is a rational number.
$\sqrt{2}$  when multiplied with $\sqrt{3}$, which is also an irrational number, results into $\sqrt{6}$, which is an irrational number.

#### Question 11:

Which of the following is a true statment?
(a) The sum of two irrational numbers is an irrational number
(b) The product of two irrational numbers is an irrational number
(c) Every real number is always rational
(d) Every real number is either rational or irrational

(d) Every real number is either rational or irrational.

Because a real number can be further categorised into either a rational number or an irrational number, every real number is either rational or irrational.

#### Question 12:

Which of the following is a true statment?
(a) are both rationals
(b) are both irrationals
(c) $\mathrm{\pi }$ is rational and $\frac{22}{7}$is irrational
(d) $\mathrm{\pi }$ is irrational and $\frac{22}{7}$is rational

(d) $\mathrm{\pi }$ is irrational and $\frac{22}{7}$ is rational.
Because the value of $\mathrm{\pi }$ is neither repeating nor terminating, it is an irrational number. $\frac{22}{7}$, on the other hand, is of the form $\frac{p}{q}$, so it is a rational number.

#### Question 13:

A rational number between is
(a) $\frac{1}{2}\left(\sqrt{2}+\sqrt{3}\right)$
(b) $\frac{1}{2}\left(\sqrt{3}-\sqrt{2}\right)$
(c) 2.5
(d) 1.5

(d) 1.5
Because $\sqrt{2}$ = 1.414... and $\sqrt{3}$ = 1.732..., a rational number between these two values is 1.5

#### Question 14:

(125)−1/3 = ?
(a) 5
(b) −5
(c) $\frac{1}{5}$
(d) $\frac{-1}{5}$

(c) $\frac{1}{5}$
We have:

#### Question 15:

$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=?$
(a) $\sqrt{2}$
(b) 2
(c) 4
(d 8

(b) 2

We have:

#### Question 16:

$\sqrt{2}×\sqrt{2}×\sqrt{32}=?$
(a) 2
(b) $\sqrt{2}$
(c) $2\sqrt{2}$
(d) $4\sqrt{2}$

(a) 2

We have:

#### Question 17:

${\left(\frac{81}{16}\right)}^{-3/4}=?$
(a) $\frac{4}{9}$
(b) $\frac{9}{4}$
(c) $\frac{27}{8}$
(d) $\frac{8}{27}$

(d) $\frac{8}{27}$

$\because {\left(\frac{81}{16}\right)}^{-\frac{3}{4}}\phantom{\rule{0ex}{0ex}}\therefore {\left(\frac{16}{81}\right)}^{\frac{3}{4}}\phantom{\rule{0ex}{0ex}}={\left\{{\left(\frac{16}{81}\right)}^{\frac{1}{4}}\right\}}^{3}\phantom{\rule{0ex}{0ex}}={\left[{\left\{{\left(\frac{2}{3}\right)}^{4}\right\}}^{\frac{1}{4}}\right]}^{3}\phantom{\rule{0ex}{0ex}}={\left(\frac{2}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\frac{8}{27}$

#### Question 18:

$\sqrt{{\left(64\right)}^{-2}}=?$
(a) 4
(b) $\frac{1}{4}$
(c) 8
(d) $\frac{1}{8}$

(d) $\frac{1}{8}$

We have:

#### Question 19:

$\frac{1}{\left(\sqrt{4}-\sqrt{3}\right)}=?$
(a) $\left(2+\sqrt{3}\right)$
(b) $\left(2-\sqrt{3}\right)$
(c) 1
(d) none of these

(a) (2 + $\sqrt{3}$)
On rationalising the denominator, we get:

#### Question 20:

$\frac{1}{\left(3+2\sqrt{2}\right)}=?$
(a) $\frac{3-2\sqrt{2}}{17}$
(b) $\frac{\left(3-2\sqrt{2}\right)}{13}$
(c) $\left(3-2\sqrt{2}\right)$
(d) none of these

(c) $\left(3-2\sqrt{2}\right)$
On rationalising, we get:
$\frac{1}{3+2\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3+2\sqrt{2}}×\frac{3-2\sqrt{2}}{3-2\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{3-2\sqrt{2}}{\left(3{\right)}^{2}-{\left(2\sqrt{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3-2\sqrt{2}}{9-8}\phantom{\rule{0ex}{0ex}}=3-2\sqrt{2}$

#### Question 21:

If =?
(a) $8\sqrt{3}$
(b) 14
(c) 49
(d) 48

(b) 14
x = $7+4\sqrt{3}$
Thus, we have:
$\frac{1}{x}=\frac{1}{7+4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{1}{7+4\sqrt{3}}×\frac{1}{7+4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{7-4\sqrt{3}}{{\left(7\right)}^{2}-{\left(4\sqrt{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=7-4\sqrt{3}$
∴

#### Question 22:

If $\sqrt{2}=1.41$, then $\frac{1}{\sqrt{2}}$=?
(a) 0.075
(b) 0.75
(c) 0.705
(d) 7.05

(c) 0.705
$\sqrt{2}$ = 1.41
We have:

#### Question 23:

If $\sqrt{7}=2.646$, then$\frac{1}{\sqrt{7}}$=?
(a) 0.375
(b) 0.378
(c) 0.441
(d) none of these

(b) 0.378
$\sqrt{7}$ = 2.646
We have:

#### Question 24:

$\sqrt{10}×\sqrt{15}=?$
(a) $\sqrt{25}$
(b) $5\sqrt{6}$
(c) $6\sqrt{5}$
(d) none of these

(b) $5\sqrt{6}$

#### Question 25:

(625)0.16 × (625)0.09 =?
(a) 5
(b) 25
(c) 125
(d) 625.25

(a) 5

We have:
${\left(625\right)}^{0.16}×{\left(625\right)}^{0.09}\phantom{\rule{0ex}{0ex}}={\left(625\right)}^{0.16+0.09}\phantom{\rule{0ex}{0ex}}={\left(625\right)}^{0.25}\phantom{\rule{0ex}{0ex}}={\left(625\right)}^{\frac{25}{100}}\phantom{\rule{0ex}{0ex}}={\left(625\right)}^{\frac{1}{4}}\phantom{\rule{0ex}{0ex}}={\left\{\left(5{\right)}^{4}\right\}}^{\frac{1}{4}}\phantom{\rule{0ex}{0ex}}=5$

#### Question 26:

If $\sqrt{2}=1.414$, then $\sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}}$=?
(a) 0.207
(b) 2.414
(c) 0.414
(d) 0.621

(c) 0.414

On rationalising  $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ by multiplying it by $\left(\sqrt{2}-1\right)$ both in the numerator and denominator, we get:

#### Question 27:

The simplest for of $1.\overline{)6}$ is
(a) $\frac{833}{500}$
(b) $\frac{8}{5}$
(c) $\frac{5}{3}$
(d) none of these

(c) $\frac{5}{3}$
Let x = 1.6666666...       ...(i)
Multiplying by 10 on both sides, we get:
10x = 16.6666666...       ...(ii)
Subtracting (i) from (ii), we get:
9x = 15
$⇒$x =

#### Question 28:

The simplest form of $0.\overline{)54}$ is
(a) $\frac{27}{50}$
(b) $\frac{6}{11}$
(c) $\frac{4}{7}$
(d) none of these

(b) $\frac{6}{11}$
Let x = 0.545454...               ...(i)
Multiplying both sides by 100, we get:
100x = 54.5454545...           ...(ii)
Subtracting (i) from (ii), we get:
99x = 540
$⇒$x$\frac{54}{99}$ = $\frac{6}{11}$

#### Question 29:

The simplest form of $0.3\overline{)2}$is
(a) $\frac{16}{45}$
(b) $\frac{32}{99}$
(c) $\frac{41}{333}$
(d) none of these

(c) $\frac{29}{90}$
Let x = 0.3222222222...          ...(i)
Multiplying by 10 on both sides, we get:
10x = 3.222222222...              ...(ii)
Again, multiplying by 10 on both sides, we get:
100x = 32.222222222...          ...(iii)
On subtracting (ii) from (iii), we get:
90x = 29
x = $\frac{29}{90}$

#### Question 30:

The simplest form of $0.12\overline{)3}$ is
(a) $\frac{41}{330}$
(b) $\frac{37}{330}$
(c) $\frac{41}{333}$
(d) none of these

(d) none of these
Let x = 0.12333333333...         ...(i)
Multiplying by 100 on both sides, we get:
100x = 12.33333333...             ...(ii)
Multiplying by 10 on both sides, we get:
1000x = 123.33333333...         ...(iii)
Subtracting (ii) from (iii), we get:
900x = 111

$⇒$x = $\frac{111}{900}$

#### Question 31:

An irrational number between 5 and 6 is
(a) $\frac{1}{2}\left(5+6\right)$
(b) $\sqrt{5+6}$
(c) $\sqrt{5×6}$
(d) none of these

(c) $\sqrt{5×6}$

An irrational number between a and b is given as $\sqrt{ab}$.

#### Question 32:

An irrational number between is
(a) $\left(\sqrt{2}+\sqrt{3}\right)$
(b) $\sqrt{2}×\sqrt{3}$
(c) 51/4
(d) 61/4

(d) 61/4
An irrational number between

#### Question 33:

An irrational number between $\frac{1}{7}\mathrm{and}\frac{2}{7}$is
(a) $\frac{1}{2}\left(\frac{1}{7}+\frac{2}{7}\right)$
(b) $\left(\frac{1}{7}×\frac{2}{7}\right)$
(c) $\sqrt{\frac{1}{7}×\frac{2}{7}}$
(d) none of these

(c) $\sqrt{\frac{1}{7}×\frac{2}{7}}$

An irrational number between a and b is given as $\sqrt{ab}$.

#### Question 34:

Assertion: Three rational numbers between .
Reason: A rational number between two rational numbers p and q is $\frac{1}{2}\left(p+q\right)$.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

So, Assertion and Reason are correct (property of rational numbers). Also, Reason is the correct explanation of Assertion.

#### Question 35:

Assertion: $\sqrt{3}$is an irrational number.
Reason: Square root a positive integer which is not a perfect square is an irrational number.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

3 3

#### Question 36:

Assertion: e is irrational number.
Reason: $\mathrm{\pi }$ is an irrational number.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

It is known that e and $\mathrm{\pi }$ are irrational numbers, but Reason is not the correct explanation.

#### Question 37:

Assertion: $\sqrt{3}$is an irrational number.
Reason: The sum of rational number and an irrational number is an irrational number.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

#### Question 38:

Match the following columns:

 Column I Column II (a) $6.\overline{)54}$ is ....... . (p) 14 (b) $\mathrm{\pi }$ is ...... . (q) 6 (c) The length of period of $\frac{1}{7}$=...... . (r) a rational number (d) If $x=\left(2-\sqrt{3}\right)$, then $\left({x}^{2}+\frac{1}{{x}^{2}}\right)$....... . (s) an irrational number
(a) ........
(b) ........
(c) ........
(d) ........

(a) Because it is a non-terminating and repeating decimal, it is a rational number.

(b) $\mathrm{\pi }$ is an irrational number.

(c) $\frac{1}{7}=.142857142857$...
Hence, its period is 6.

(d)

#### Question 39:

Match the following columns:

 Column I Column II (a) (p) 4 (b) If ${\left(\frac{a}{b}\right)}^{x-2}={\left(\frac{b}{a}\right)}^{x-4}$, then x = ........ . (q) $\frac{2}{9}$ (c) If $x=\left(9+4\sqrt{5}\right)$, then $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)$= ...... . (r) $\frac{1}{9}$ (d) ${\left(\frac{81}{16}\right)}^{-3/4}×{\left(\frac{64}{27}\right)}^{-1/3}=?$ (s) 3

(a) ......
(b) ......
(c) ......
(d) ......

(a)
${\left({\left(81\right)}^{-2}\right)}^{\frac{1}{4}}\phantom{\rule{0ex}{0ex}}={\left({\left(9\right)}^{-4}\right)}^{\frac{1}{4}}={\left(9\right)}^{-4×\frac{1}{4}}={\left(9\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}$

(b)

(c)

(d)

${\left(\frac{3}{2}\right)}^{4×\frac{-3}{4}}×{\left(\frac{4}{3}\right)}^{3×\frac{-1}{3}}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{2}\right)}^{-3}×{\left(\frac{4}{3}\right)}^{-1}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{2}\right)}^{-3}×\frac{3}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\left(\frac{{3}^{-3}}{{2}^{-3}}\right)×\frac{3}{{2}^{2}}\phantom{\rule{0ex}{0ex}}=\left(\frac{{3}^{-3}×3}{{2}^{-3}×{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{3}^{-2}}{{2}^{-1}}\phantom{\rule{0ex}{0ex}}=\frac{2}{9}$

#### Question 40:

Give an example of two irrational numbers whose sum as well as the product is rational.

#### Question 41:

If x is rational and y is irrational, then show that (x + y) is always irrational.

#### Question 42:

Is the product of a rational and irrational number always irrational? Give an example.

#### Question 43:

Give an example of a number x such that x2 is an irrational number and x4 is a rational number.

#### Question 44:

The number $4.\overline{)17}$expressed as a vulgar fraction is
(a) $\frac{417}{100}$
(b) $\frac{417}{99}$
(c) $\frac{413}{99}$
(d) $\frac{413}{90}$

(c)$\frac{413}{99}$
Let x = 4.171717171...  ...(i)
Multiplying by 100 on both sides, we get:
100x = 417.171717...    ...(ii)
Subtracting (i) from (ii), we get:
99x = 413
$⇒$x = $\frac{413}{99}$

#### Question 45:

If $x=\left(2+\sqrt{3}\right)$, find the value of ${\left({x}^{2}+\frac{1}{{x}^{2}}\right)}^{2}$.

#### Question 46:

If $\frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)}=\left(a-b\sqrt{3}\right)$, find the values of a and b.

#### Question 47:

If $\frac{\left(4+\sqrt{5}\right)}{\left(4-\sqrt{5}\right)}=\left(a+b\sqrt{5}\right)$, find the values of a and b.

#### Question 48:

If$\frac{\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)}+\frac{\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)}=\left(a+b\sqrt{5}\right)$, find the values of a and b.

#### Question 49:

If $\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}-2\sqrt{3}\right)}=\left(a+b\sqrt{6}\right)$, find the values of a and b.

$\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}-2\sqrt{3}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}-2\sqrt{3}\right)}×\frac{\left(3\sqrt{2}+2\sqrt{3}\right)}{\left(3\sqrt{2}+2\sqrt{3}\right)}\phantom{\rule{0ex}{0ex}}=\frac{6+2\sqrt{6}+3\sqrt{6}+6}{18-12}\phantom{\rule{0ex}{0ex}}=\frac{12+5\sqrt{6}}{6}$

#### Question 50:

If , find (x2 + y2).

#### Question 51:

If , show that the value of (x3 − 2x2 − 7x + 5) is 3.

#### Question 52:

If $x=\left(3+\sqrt{8}\right)$, show that $\left({x}^{2}+\frac{1}{{x}^{2}}\right)=34$.

$x=3+\sqrt{8}\phantom{\rule{0ex}{0ex}}\frac{1}{x}=\frac{1}{3+\sqrt{8}}×\frac{3-\sqrt{8}}{3-\sqrt{8}}\phantom{\rule{0ex}{0ex}}=\frac{3-\sqrt{8}}{9-8}\phantom{\rule{0ex}{0ex}}=3-\sqrt{8}$

#### Question 53:

If $x=\left(2+\sqrt{3}\right)$, show that $\left({x}^{3}+\frac{1}{{x}^{3}}\right)=52$.

#### Question 54:

If $x=\left(3-2\sqrt{2}\right)$, show that $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)=±2$.

#### Question 55:

If $x=\left(5+2\sqrt{6}\right)$, show that $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)=±2\sqrt{3}$.

#### Question 1:

Find two rational numbers lying between $\frac{1}{3}\mathrm{and}\frac{1}{2}$.

#### Question 2:

Find four rational numbers between $\frac{3}{5}\mathrm{and}\frac{4}{5}$.

#### Question 3:

Write four irrational numbers between 0.1 and 0.2.

Non-repeating and non-terminating decimals are irrational numbers; thus, any such numbers lying between 0.1 and 0.2 will be irrational.
These can be 0.125986634275... ,0.165323296581422..., 0.121221222212226... and 0.13133133331333311333...

#### Question 4:

Express $\sqrt{1250}$in its simplest form.

#### Question 5:

Express $\frac{2}{3}$. $\sqrt{18}$as a pure surd.

$\frac{2}{3}×\sqrt{18}=\frac{2}{3}\sqrt{2×3×3}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}×\sqrt{2}×\sqrt{9}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}×\sqrt{2}×3\phantom{\rule{0ex}{0ex}}=2\sqrt{2}\phantom{\rule{0ex}{0ex}}=\sqrt{2×2×2}\phantom{\rule{0ex}{0ex}}=\sqrt{8}$

#### Question 6:

Divide $16\sqrt{75}\mathrm{by}5\sqrt{12}$.

$\frac{16\sqrt{75}}{5\sqrt{12}}=\frac{16×\sqrt{5×5×3}}{5\sqrt{2×2×3}}\phantom{\rule{0ex}{0ex}}=\frac{16×\sqrt{25}×\sqrt{3}}{5×\sqrt{4}×\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{16×5×\sqrt{3}}{5×2×\sqrt{3}}\phantom{\rule{0ex}{0ex}}=8$

#### Question 7:

Express $0.1\overline{)23}$ as a rational number in the form $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Thus, we have:
$0.1\overline{)23}=\frac{61}{495}$

#### Question 8:

If $\frac{6}{\left(3\sqrt{2}-2\sqrt{3}\right)}=\left(a\sqrt{2}+b\sqrt{3}\right)$, find the values of a and b.

#### Question 9:

The simplest form of ${\left(\frac{64}{729}\right)}^{-1/6}$is
(a) $\frac{2}{3}$
(b) $\frac{3}{2}$
(c) $\frac{4}{3}$
(d) $\frac{3}{4}$

(b) $\frac{3}{2}$

#### Question 10:

Which of the following is irrational?
(a) $0.\overline{)14}$
(b) $0.14\overline{)16}$
(c) $0.\overline{)1416}$
(d) 0.1401401400014.....

(d) 0.1401401400014...

Since, it is non terminating and non repeating
Therefore,  number is irrational.

#### Question 11:

Between two rational numbers
(a) there is no rational number
(b) there is exactly one rational number
(c) there are infinitely many irrational numbers
(d) there is no irrational number

(c) there are infinitely many irrational numbers

#### Question 12:

Decimals representation of an irrational number is
(a) always a terminating decimal
(b) either a terminating or a repeating decimal
(c) either a terminating or a non-repeating decimal
(d) always non-terminating and non-repeating decimal

(d) always a non-terminating and non-repeating decimal

#### Question 13:

If $x=\left(7+5\sqrt{2}\right)$, then $\left({x}^{2}+\frac{1}{{x}^{2}}\right)$=?
(a) 160
(b) 198
(c) 189
(d) 156

(b) 198

#### Question 14:

Rationalise the denominator of $\left(\frac{5\sqrt{3}-4\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\right)$.

$\frac{5\sqrt{3}-4\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}=\frac{5\sqrt{3}-4\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}×\frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(5\sqrt{3}-4\sqrt{2}\right)×\left(4\sqrt{3}-3\sqrt{2}\right)}{{\left(4\sqrt{3}\right)}^{2}-{\left(3\sqrt{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(5\sqrt{3}×4\sqrt{3}\right)-\left(5\sqrt{3}×3\sqrt{2}\right)-\left(4\sqrt{2}×4\sqrt{3}\right)+\left(4\sqrt{2}×3\sqrt{2}\right)}{{\left(4\sqrt{3}\right)}^{2}-{\left(3\sqrt{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{20×3-15\sqrt{6}-16\sqrt{6}+12×2}{16×3-9×2}\phantom{\rule{0ex}{0ex}}=\frac{60-31\sqrt{6}+24}{48-18}\phantom{\rule{0ex}{0ex}}=\frac{84-31\sqrt{6}}{30}$

#### Question 15:

Simplify: $\frac{1}{{\left(27\right)}^{-1/3}}+\frac{1}{{\left(625\right)}^{-1/4}}$.

$\frac{1}{{\left(27\right)}^{\frac{-1}{3}}}+\frac{1}{{\left(625\right)}^{\frac{-1}{4}}}\phantom{\rule{0ex}{0ex}}={\left(27\right)}^{\frac{1}{3}}+{\left(625\right)}^{\frac{1}{4}}={\left({3}^{3}\right)}^{\frac{1}{3}}+{\left({5}^{4}\right)}^{\frac{1}{4}}=3+5\phantom{\rule{0ex}{0ex}}=8$

#### Question 16:

Find the smallest of the number .

We can write the given numbers in the following ways:

$\sqrt{6}={6}^{\frac{1}{3}}\phantom{\rule{0ex}{0ex}}\sqrt{24}={24}^{\frac{1}{6}}\phantom{\rule{0ex}{0ex}}\sqrt{8}={8}^{\frac{1}{4}}$

The powers of the given numbers are .
The LCM of the denominators of these powers is 12.

Because on raising all three numbers by the same power, the smallest number will still remain the smallest, raising each number to the power of 12, we get:
${6}^{\frac{1}{3}×12}={6}^{4}=1296\phantom{\rule{0ex}{0ex}}{24}^{\frac{1}{6}×12}={24}^{2}=576\phantom{\rule{0ex}{0ex}}{8}^{\frac{1}{4}×12}={8}^{3}=512$

Of these three, 512 is the smallest number, which has been derived from $\sqrt{8}$.

Hence, $\sqrt{8}$ is the smallest of all three numbers.

#### Question 17:

Match the following columns:

 Column I Column II (a) $\mathrm{\pi }$ is ...... . (p) a rational number (b) $3.\overline{)1416}$ is ...... . (q) an irrational number (c) $0.\overline{)23}$ = ....... . (r) $\frac{7}{30}$ (d) $0.2\overline{)3}$ = ...... . (s) $\frac{23}{99}$

(a) ......
(b) ......
(c) ......
(d) ......

(a) $\mathrm{\pi }$ is an irrational number.

(b) 3.141614161416... is a rational number because it is a non-terminating but repeating decimal.

(c) Let x = 0.23232323...    ...(i)
$\therefore$ 100x = 23.232323...       ...(ii)
Subtracting (i) from (ii), we get:
99x = 23
Or,
x = $\frac{23}{99}$

(d) Let x = 0.233333333..     ...(i)
$\therefore$ 10x = 2.33333333..          ...(ii)
We have:
100x = 23.333333333          ...(iii)
Subtracting (ii) from (iii), we get:
90x = 21
Or,
x = $\frac{21}{90}=\frac{7}{30}$

#### Question 18:

If , find the value of (x2 + y2).

#### Question 19:

If , find the value of $\frac{3}{\left(8\sqrt{2}+5\sqrt{5}\right)}+\frac{2}{\left(8\sqrt{2}-5\sqrt{5}\right)}$.

$\frac{3}{\left(8\sqrt{2}+5\sqrt{5}\right)}+\frac{2}{\left(8\sqrt{2}-5\sqrt{5}\right)}=\frac{3}{\left(8\sqrt{2}+5\sqrt{5}\right)}×\frac{\left(8\sqrt{2}-5\sqrt{5}\right)}{\left(8\sqrt{2}-5\sqrt{5}\right)}+\frac{2}{\left(8\sqrt{2}-5\sqrt{5}\right)}×\frac{\left(8\sqrt{2}+5\sqrt{5}\right)}{\left(8\sqrt{2}+5\sqrt{5}\right)}\phantom{\rule{0ex}{0ex}}=\frac{3×\left(8\sqrt{2}-5\sqrt{5}\right)}{64×2-25×5}+\frac{2×\left(8\sqrt{2}+5\sqrt{5}\right)}{64×2-25×5}\phantom{\rule{0ex}{0ex}}=\frac{3×\left(8\sqrt{2}-5\sqrt{5}\right)}{3}+\frac{2×\left(8\sqrt{2}+5\sqrt{5}\right)}{3}\phantom{\rule{0ex}{0ex}}=\left(8\sqrt{2}-5\sqrt{5}\right)+\frac{2×\left(8\sqrt{2}+5\sqrt{5}\right)}{3}\phantom{\rule{0ex}{0ex}}=8\sqrt{2}+\frac{2}{3}8\sqrt{2}+\frac{2}{3}5\sqrt{5}-5\sqrt{5}\phantom{\rule{0ex}{0ex}}=8\sqrt{2}\left(1+\frac{2}{3}\right)+5\sqrt{5}\left(\frac{2}{3}-1\right)\phantom{\rule{0ex}{0ex}}=8\sqrt{2}\left(\frac{5}{3}\right)+5\sqrt{5}\left(\frac{-1}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{40}{3}\left(1.41\right)-\frac{5}{3}\left(2.24\right)\phantom{\rule{0ex}{0ex}}=18.8-3.73=15.07$

#### Question 20:

Prove that ${\left(\frac{81}{16}\right)}^{-3/4}×\left\{{\left(\frac{25}{9}\right)}^{-3/2}÷{\left(\frac{5}{2}\right)}^{-3}\right\}=1$.

${\left(\frac{81}{16}\right)}^{\frac{-3}{4}}×\left\{{\left(\frac{25}{9}\right)}^{\frac{-3}{2}}÷{\left(\frac{5}{2}\right)}^{-3}\right\}\phantom{\rule{0ex}{0ex}}={\left(\frac{16}{81}\right)}^{\frac{3}{4}}×\left\{{\left(\frac{9}{25}\right)}^{\frac{3}{2}}×{\left(\frac{5}{2}\right)}^{-\left(-3\right)}\right\}\phantom{\rule{0ex}{0ex}}={\left(\frac{2}{3}\right)}^{4×\frac{3}{4}}×{\left(\frac{3}{5}\right)}^{2×\frac{3}{2}}×{\left(\frac{5}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}={\left(\frac{2}{3}\right)}^{3}×{\left(\frac{3}{5}\right)}^{3}×{\left(\frac{5}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\frac{{2}^{3}×{3}^{3}×{5}^{3}}{{3}^{3}×{5}^{3}×{2}^{3}}\phantom{\rule{0ex}{0ex}}=1$