NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

Number systems class 9 questions and answers - Exercise 1.1

Q1 Is zero a rational number? Can you write it in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0?

Answer:

Any number that can represent in the form of $\frac{p}{q}$ $(whereq\ne 0)$ is a rational number

Now, we can write 0 in the form of $\frac{p}{q}$ for eg. $\frac{0}{1},\frac{0}{2},\frac{0}{-1}$ etc.

Therefore, 0 is a rational number.

Q2 Find six rational numbers between 3 and 4.

Answer:

There are an infinite number of rational numbers between 3 and 4. one way to take them is
$\Rightarrow 3=\frac{21}{7}and4=\frac{28}{7}$
Therefore, six rational numbers between 3 and 4 are $\frac{22}{7},\frac{23}{7},\frac{24}{7},\frac{25}{7},\frac{26}{7},\frac{27}{7}$

Q3 Find five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$ .

Answer:

We can write
$\Rightarrow \frac{3}{5}=\frac{3\times 6}{5\times 6}=\frac{18}{30}$
And
$\Rightarrow \frac{4}{5}=\frac{4\times 6}{5\times 6}=\frac{24}{30}$
Therefore, five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$ . are $\frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30},$

Q4 (i) State whether the following statements are true or false. Give reasons for your answers. (i)Every natural number is a whole number.

Answer:

(i) TRUE
The number that is starting from 1, i.e 1, 2, 3, 4, 5, 6, .................. are natural numbers
The number that is starting from 0. i.e, 0, 1, 2, 3, 4, 5.............are whole numbers
Therefore, we can clearly see that the collection of whole numbers contains all natural numbers.

Q4 (ii) State whether the following statements are true or false. Give reasons for your answers. (ii) Every integer is a whole number.

Answer:

(ii) FALSE
Because integers may be negative or positive but whole numbers are always positive. for eg. -1 is an integer but not a whole number.

Q4 (iii) State whether the following statements are true or false. Give reasons for your answers.(iii) Every rational number is a whole number.

Answer:

(iii) FALSE
Numbers that can be represented in the form of $\frac{p}{q}(whereq\ne 0)$ are a rational number.
And numbers that are starting from 0 i.e. 0,1,2,3,4,......... are whole numbers
Therefore, we can clearly see that every rational number is not a whole number for eg. $\frac{3}{4}$ is a rational number but not a whole number

Class 9 maths chapter 1 question answer - Exercise 1.2

Q1 (i) State whether the following statements are true or false. Justify your answers. (i) Every irrational number is a real number.

Answer:

(i) TRUE
Since the real numbers are the collection of all rational and irrational numbers.

Q1 (ii) State whether the following statements are true or false. Justify your answers. (ii) Every point on the number line is of the form $\sqrt{m}$ , where m is a natural number.

Answer:

(ii) FALSE
Because negative numbers are also present on the number line and no negative number can be the square root of any natural number

Q1 (iii) State whether the following statements are true or false. Justify your answers. (iii) Every real number is an irrational number.

Answer:

(iii) FALSE
As real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number.

For eg. 4 is a real number but not an irrational number

Q2 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.

Answer:

NO, Square root of all positive integers is not irrational. For the eg square of 4 is 2 which is a rational number.

Q3 Show how $\sqrt{5}$ can be represented on the number line.

Answer:

We know that
$\sqrt{5}=\sqrt{(2{)}^2+(1{)}^2}$
Now,
Let OA be a line of length 2 unit on the number line. Now, construct AB of unit length perpendicular to OA. and join OB.
Now, in right angle triangle OAB, by Pythagoras theorem
$OB=\sqrt{(2{)}^2+(1{)}^2}=\sqrt{5}$
Now, take O as centre and OB as radius, draw an arc intersecting number line at C. Point C represent $\sqrt{5}$ on a number line.

Class 9 maths chapter 1 question answer - Exercise: 1.3

Q1 (i) Write the following in decimal form and say what kind of decimal expansion each has : (i) $\frac{36}{100}$

Answer:

We can write $\frac{36}{100}$ as
$\Rightarrow \frac{36}{100}=0.36$
Since the decimal expansion ends after a finite number of steps. Hence, it is terminating

Q1 (ii) Write the following in decimal form and say what kind of decimal expansion each has : (ii) $\frac{1}{11}$

Answer:

We can rewrite $\frac{1}{11}$ as

$\Rightarrow \frac{1}{11}=0.09090909.....=0.\overline{09}$
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (iii) Write the following in decimal form and say what kind of decimal expansion each has : (iii) $4\frac{1}{8}$

Answer:

We can rewrite $4\frac{1}{8}$ as

$\Rightarrow 4\frac{1}{8}=\frac{33}{8}=4.125$
Since the decimal expansion ends after a finite number. Therefore, it is terminating

Q1 (iv) Write the following in decimal form and say what kind of decimal expansion each has : (iv) $\frac{3}{13}$

Answer:

We can rewrite $\frac{3}{13}$ as

$\Rightarrow \frac{3}{13}=0.230769230769=0.\overline{230769}$
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (v) Write the following in decimal form and say what kind of decimal expansion each has: (v) $\frac{2}{11}$

Answer:

We can rewrite $\frac{2}{11}$ as

$\Rightarrow \frac{2}{11}=0.181818......=0.\overline{18}$
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (vi) Write the following in decimal form and say what kind of decimal expansion each has : (vi) $\frac{329}{400}$

Answer:

We can rewrite $\frac{329}{400}$ as

$\Rightarrow \frac{329}{400}=0.8225$
Since decimal expansion ends after finite no. of figures. Hence, it is terminating.

Q2 You know that $\frac{1}{7}=0.\overline{142857}$ Can you predict what the decimal expansions of $\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}$ are, without actually doing the long division? If so, how?

Answer:

It is given that $\frac{1}{7}=0.\overline{142857}$

Therefore,

$\Rightarrow \frac{2}{7}=2\times \frac{1}{7}=2\times 0.\overline{142857}=0.\overline{285714}$

Similarly,

$\Rightarrow \frac{3}{7}=3\times \frac{1}{7}=3\times 0.\overline{142857}=0.\overline{428571}$

$\Rightarrow \frac{4}{7}=4\times \frac{1}{7}=4\times 0.\overline{142857}=0.\overline{571428}$

$\Rightarrow \frac{5}{7}=5\times \frac{1}{7}=5\times 0.\overline{142857}=0.\overline{714285}$

$\Rightarrow \frac{6}{7}=6\times \frac{1}{7}=6\times 0.\overline{142857}=0.\overline{857142}$

Q3 (i) Express the following in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0. (i) $0.\overline{6}$

Answer:

Let $x=0.\bar{6}=0.6666....$ -(i)

Now, multiply by 10 on both sides

$10x=6.6666...$

$\Rightarrow 10x=6+x\text{(}using(i))$

$\Rightarrow 9x=6$

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

Therefore, $\frac{p}{q}$ form of $0.\overline{6}$ is $\frac{2}{3}$

Q3 (ii) Express the following in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0. (ii) $0.4\overline{7}$

Answer:

We can write $0.4\bar{7}$ as

$\Rightarrow 0.4\bar{7}=\frac{4}{10}+\frac{0.777..}{10}$ -(i)

Now,

Let $x=0.\bar{7}=0.7777....$ -(ii)

Now, multiply by 10 on both sides

$10x=7.7777...$

$\Rightarrow 10x=7+x\text{(}using(ii))$

$\Rightarrow 9x=7$

$\Rightarrow x=\frac{7}{9}$
Now, put the value of x in equation (i). we will get

$\Rightarrow 0.4\bar{7}=\frac{4}{10}+\frac{7}{10\times 9}=\frac{4}{10}+\frac{7}{90}=\frac{36+7}{90}=\frac{43}{90}$

Therefore, $\frac{p}{q}$ form of $0.4\bar{7}$ is $\frac{43}{90}$

Q3 (iii) Express the following in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0. (iii) $0.\overline{001}$

Answer:

Let $x=0.\overline{001}=0.001001....$ -(i)

Now, multiply by 1000 on both sides

$1000x=1.001001...$

$\Rightarrow 1000x=1+x\text{(}using(i))$

$\Rightarrow 999x=1$

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

Therefore, $\frac{p}{q}$ form of $0.\overline{001}$ is $\frac{1}{999}$

Q4 Express 0.99999 .... in the form $\frac{p}{q}$ . Are you surprised by your answer?

Answer:

Let $x=0.\bar{9}=0.9999....$ -(i)

Now, multiply by 10 on both sides

$10x=9.999....$

$\Rightarrow 10x=9+x\text{(}using(i))$

$\Rightarrow 9x=9$

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

Therefore, $\frac{p}{q}$ form of $0.999....$ is 1

The difference between 1 and 0.999999 is o.000001 which is almost negligible.

Therefore, 0.999 is too much closer to 1. Hence, we can write 0.999999.... as 1

Q5 What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$ ? Perform the division to check your answer.

Answer:

We can rewrite $\frac{1}{17}$ as

$\Rightarrow \frac{1}{17}=0.05882352941176470588235294117647=0.\overline{0588235294117647}$
Therefore, there are total 16 number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$

Q6 Look at several examples of rational numbers in the form $\frac{p}{q}$ (q ≠ 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?

Answer:

We can observe that when q is 2, 4, 5, 8, 10… then the decimal expansion is terminating. For example:

$\frac{3}{2}=1.5$ , denominator $q=2^1$

$\frac{8}{5}=1.6$ , denominator $q=5^1$

$\frac{15}{10}=1.5$ , denominator $q=10=2\times 5=2^1,5^1$

Therefore,

It can be observed that the terminating decimal can be obtained in a condition where prime factorization of the denominator of the given fractions has the power of 2 only or 5 only or both.

Q7 Write three numbers whose decimal expansions are non-terminating non-recurring.

Answer:

Three numbers whose decimal expansions are non-terminating non-recurring are
1) 0.02002000200002......
2) 0.15115111511115.......
3) 0.27227222722227.......

Q8 Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$ .

Answer:

We can write $\frac{5}{7}$ as

$\Rightarrow \frac{5}{7}=0.714285714285....=0.\overline{714285}$

And $\frac{9}{11}$ as

$\Rightarrow \frac{9}{11}=0.818181....=0.\overline{81}$
Therefore, three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$ are

1) 0.72737475....
2) 0.750760770780...
3) 0.790780770760....

Q9 (i) Classify the following numbers as rational or irrational : $\sqrt{23}$

Answer:

We can rewrite $\sqrt{23}$ in decimal form as

$\Rightarrow \sqrt{23}=4.7958152....$

Now, as the decimal expansion of this number is non-terminating non-recurring.

Therefore, it is an irrational number.

Q9 (ii) Classify the following numbers as rational or irrational : $\sqrt{225}$

Answer:

We can rewrite $\sqrt{225}$ as

$\Rightarrow \sqrt{225}=15$
We can clearly see that it is a rational number because we can represent it in $\frac{p}{q}$ form

Q9 (iii) Classify the following numbers as rational or irrational : 0.3796

Answer:

We can rewrite 0.3796 as

$\Rightarrow 0.3796=\frac{3796}{10000}$
Now, we can clearly see that it is a rational number as the decimal expansion of this number is terminating and we can also write it in $\frac{p}{q}$ form.

Q9 (iv) Classify the following numbers as rational or irrational : 7.478478....

Answer:

We can rewrite 7.478478.... as

$\Rightarrow 7.478478....=7.\overline{478}$
Now, as the decimal expansion of this number is non-terminating recurring. Therefore, it is a rational number.

Q9 (v) Classify the following numbers as rational or irrational : 1.101001000100001...

Answer:

In the case of number 1.101001000100001...
As the decimal expansion of this number is non-terminating non-repeating. Therefore, it is an irrational number.

Class 9 maths chapter 1 NCERT solutions - Exercise: 1.4

Q1 Visualise 3.765 on the number line, using successive magnification.

Answer:

3.765 can be visualised as in the following steps.
First, we draw a number line and mark points on it after that we will divide the number line between points 3 and 4. And then we will divide the points between 3.7 and 3.8 as the number is between both of them.

Q2 Visualise $4.\overline{26}$ on the number line, up to 4 decimal places.

Answer:

We can rewrite as $\Rightarrow 4.\overline{26}=4.262626...$
Now, 4.2626 can be visualised as in the following steps.

Class 9 maths chapter 1 NCERT solutions - Exercise: 1.5

Q1 (i) Classify the following numbers as rational or irrational: $2-\sqrt{5}$

Answer:

Value of $\sqrt{5}$ is 2.23606798....
Now,
$\Rightarrow 2-\sqrt{5}=2-2.23606798...=-0.23606798...$
Now,
Since the number is in non-terminating non-recurring. Therefore, it is an irrational number.

Q1 (ii) Classify the following numbers as rational or irrational: $(3+\sqrt{23})-\sqrt{23}$

Answer:

Given number is $(3+\sqrt{23})-\sqrt{23}$

$\Rightarrow (3+\sqrt{23})-\sqrt{23}=3+\sqrt{23}-\sqrt{23}=3$
Now, it is clearly a rational number because we can represent it in the form of $\frac{p}{q}$

Q1 (iii) Classify the following numbers as rational or irrational: $\frac{2\sqrt{7}}{7\sqrt{7}}$

Answer:

Given number is $\frac{2\sqrt{7}}{7\sqrt{7}}$

$\Rightarrow \frac{2\sqrt{7}}{7\sqrt{7}}=\frac{2}{7}$
As we can clearly see that it can be represented in $\frac{p}{q}$ form. Therefore, it is a rational number.

Q1 (iv) Classify the following numbers as rational or irrational: $\frac{1}{\sqrt{2}}$

Answer:

Given number is $\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}=\frac{1.41421356..}{2}=0.7071068...$
Now,
Clearly, as the decimal expansion of this expression is non-terminating and non-recurring. Therefore, it is an irrational number.

Q1 (v) Classify the following numbers as rational or irrational: $2\pi$

Answer:

Given number is $2\pi$
We know that the value of $\pi =3.14159265...$
Now,
$\Rightarrow 2\pi =2\times 3.14159265...=6.2831853...$
Now,
Clearly, as the decimal expansion of this expression is non-terminating and non-recurring. Therefore, it is an irrational number.

Q2 (i) Simplify each of the following expressions: $(3+\sqrt{3})(2+\sqrt{2})$

Answer:

Given number is $(3+\sqrt{3})(2+\sqrt{2})$
Now, we will reduce it into
$\Rightarrow (3+\sqrt{3})(2+\sqrt{2})=3.2+3.\sqrt{2}+\sqrt{3}.2+\sqrt{3}.\sqrt{2}$
$=6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}$

Therefore, answer is $6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}$

Q2 (ii) Simplify each of the following expressions: $(3+\sqrt{3})(3-\sqrt{3})$

Answer:

Given number is

$(3+\sqrt{3})(3-\sqrt{3})$
Now, we will reduce it into
$\Rightarrow (3+\sqrt{3})(3-\sqrt{3})=((3{)}^2-(\sqrt{3}{)}^2)$ $(using(a+b)(a-b)=a^2-b^2)$
$=9-3=6$

Therefore, answer is 6

Q2 (iii) Simplify each of the following expressions: ${(\sqrt{5}+\sqrt{2})}^2$

Answer:

Given number is ${(\sqrt{5}+\sqrt{2})}^2$

Now, we will reduce it into
$\Rightarrow {(\sqrt{5}+\sqrt{2})}^2=((\sqrt{5}{)}^2+(\sqrt{2}{)}^2+2.\sqrt{5}.\sqrt{2})$ $(using(a+b{)}^2=a^2+b^2+2ab)$
$=5+2+2\sqrt{10}$
$=7+2\sqrt{10}$

Therefore, the answer is $7+2\sqrt{10}$

Q2 (iv) Simplify each of the following expressions: $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$

Answer:

Given number is $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$

Now, we will reduce it into
$\Rightarrow (\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=((\sqrt{5}{)}^2-(\sqrt{2}{)}^2)$ $(using(a+b)(a-b)=a^2-b^2)$
$=5-2$
$=3$

Therefore, the answer is 3 .

Q3 Recall, $\pi$ is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, $\pi =\frac{c}{d}$ ⋅ This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?

Answer:

There is no contradiction.
When we measure a length with scale or any other instrument, we only obtain an approximate rational value. We never obtain an exact value.
For this reason, we cannot say that either c or d is irrational.
Therefore, the fraction $\frac{c}{d}$ is irrational. Hence, the value of $\pi$ is approximately equal to $\frac{22}{7}=3.142857....$

Therefore, $\pi$ is irrational.

Q4 Represent $\sqrt{9.3}$ on the number line.

Answer:

Draw a line segment OB of 9.3 unit. Then, extend it to C so that BC = 1 unit. Find the mid-point D of OC and draw a semi-circle on OC while taking D as its centre and OD as the radius. Now, Draw a perpendicular to line OC passing through point B and intersecting the semi-circle at E. Now, Take B as the centre and BE as radius, draw an arc intersecting the number line at F. the length BF is $9\sqrt{3}$ units.

Q5 (i) Rationalise the denominators of the following: $\frac{1}{\sqrt{7}}$

Answer:

Given number is $\frac{1}{\sqrt{7}}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$

Therefore, the answer is $\frac{\sqrt{7}}{7}$

Q5 (ii) Rationalise the denominators of the following: $\frac{1}{\sqrt{7}-\sqrt{6}}$

Answer:

Given number is $\frac{1}{\sqrt{7}-\sqrt{6}}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7}{)}^2-(\sqrt{6}{)}^2}=\frac{\sqrt{7}+\sqrt{6}}{7-6}$ $=\sqrt{7}+\sqrt{6}$


Therefore, the answer is $\sqrt{7}+\sqrt{6}$

Q5 (iii) Rationalise the denominators of the following: $\frac{1}{\sqrt{5}+\sqrt{2}}$

Answer:

Given number is $\frac{1}{\sqrt{5}+\sqrt{2}}$

Now, on rationalisation, we will get

$$\begin{gathered} =\frac{1}{\sqrt{5}+\sqrt{2}} \\ =\frac{1}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ =\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}{)}^2-(\sqrt{2}{)}^2} \\ =\frac{\sqrt{5}-\sqrt{2}}{5-2} \end{gathered}$$
$=\frac{\sqrt{5}-\sqrt{2}}{3}$

Therefore, the answer is $\frac{\sqrt{5}-\sqrt{2}}{3}$

Q5 (iv) Rationalise the denominators of the following: $\frac{1}{\sqrt{7}-2}$

Answer:

Given number is $\frac{1}{\sqrt{7}-2}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2}=\frac{\sqrt{7}+2}{(\sqrt{7}{)}^2-(2{)}^2}=\frac{\sqrt{7}+2}{7-4}$
$=\frac{\sqrt{7}+2}{3}$

Therefore, answer is $\frac{\sqrt{7}+2}{3}$

Class 9 number systems ncert solutions - Exercise: 1.6

Q1 (i) Find : ${64}^{\frac{1}{2}}$

Answer:

Given number is ${64}^{\frac{1}{2}}$

Now, on simplifying it we will get

$\Rightarrow {64}^{\frac{1}{2}}=(8^2{)}^{\frac{1}{2}}=8$

Therefore, answer is 8

Q1 (ii) Find : ${32}^{\frac{1}{5}}$

Answer:

Given number is ${32}^{\frac{1}{5}}$

Now, on simplifying it we will get

$\Rightarrow {32}^{\frac{1}{5}}=(2^5{)}^{\frac{1}{5}}=2$

Therefore, the answer is 2

Q1 (iii) Find : ${125}^{\frac{1}{3}}$

Answer:

Given number is ${125}^{\frac{1}{3}}$

Now, on simplifying it we will get

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

Therefore, the answer is 5

Q2 (i) Find : $9^{\frac{3}{2}}$

Answer:

Given number is $9^{\frac{3}{2}}$

Now, on simplifying it we will get

$\Rightarrow 9^{\frac{3}{2}}=(3^2{)}^{\frac{3}{2}}=3^3=27$

Therefore, the answer is 27

Q2 (ii) Find : ${32}^{\frac{2}{5}}$

Answer:

Given number is ${32}^{\frac{2}{5}}$

Now, on simplifying it we will get

$\Rightarrow {32}^{\frac{2}{5}}=(2^5{)}^{\frac{2}{5}}=2^2=4$

Therefore, the answer is 4

Q2 (iii) Find : ${16}^{\frac{3}{4}}$

Answer:

Given number is ${16}^{\frac{3}{4}}$

Now, on simplifying it we will get

$\Rightarrow {16}^{\frac{3}{4}}=(2^4{)}^{\frac{3}{4}}=2^3=8$

Therefore, the answer is 8

Q2 (iv) Find : ${125}^{\frac{-1}{3}}$

Answer:

Given number is ${125}^{\frac{-1}{3}}$

Now, on simplifying it we will get

$\Rightarrow {125}^{\frac{-1}{3}}=(5^3{)}^{\frac{-1}{3}}=5^{-1}=\frac{1}{5}$
Therefore, the answer is $\frac{1}{5}$

Q3 (i) Simplify : $2^{\frac{2}{3}}{.2}^{\frac{1}{5}}$

Answer:

Given number is $2^{\frac{2}{3}}{.2}^{\frac{1}{5}}$

Now, on simplifying it we will get

$\Rightarrow 2^{\frac{2}{3}}{.2}^{\frac{1}{5}}=2^{\frac{2}{3}+\frac{1}{5}}=2^{\frac{10+3}{15}}=2^{\frac{13}{15}}$ $(\because a^n.a^m=a^{n+m})$

Therefore, the answer is $2^{\frac{13}{15}}$

Q3 (ii) Simplify : ${\left(\frac{1}{3^3}\right)}^7$

Answer:

Given number is ${\left(\frac{1}{3^3}\right)}^7$

Now, on simplifying it we will get

$\Rightarrow {\left(\frac{1}{3^3}\right)}^7=\frac{1^7}{3^{3\times 7}}=\frac{1}{3^{21}}=3^{-21}$ $(\because (a^n{)}^m=a^{n.m}and\frac{1}{a^m}=a^{-m})$

Therefore, the answer is $3^{-21}$

Q3 (iii) Simplify : $\frac{{11}^{\frac{1}{2}}}{{11}^{\frac{1}{4}}}$

Answer:

Given number is $\frac{{11}^{\frac{1}{2}}}{{11}^{\frac{1}{4}}}$

Now, on simplifying it we will get

$\Rightarrow \frac{{11}^{\frac{1}{2}}}{{11}^{\frac{1}{4}}}={11}^{\frac{1}{2}-\frac{1}{4}}={11}^{\frac{2-1}{4}}={11}^{\frac{1}{4}}$ $(\because \frac{a^n}{a^m}=a^{n-m})$
Therefore, the answer is ${11}^{\frac{1}{4}}$

Q3 (iv) Simplify : $7^{\frac{1}{2}}{.8}^{\frac{1}{2}}$

Answer:

Given number is $7^{\frac{1}{2}}{.8}^{\frac{1}{2}}$

Now, on simplifying it we will get

$\Rightarrow 7^{\frac{1}{2}}{.8}^{\frac{1}{2}}=(7\times 8{)}^{\frac{1}{2}}={56}^{\frac{1}{2}}$ $(\because a^n.b^n=(a.b{)}^n)$
Therefore, the answer is ${56}^{\frac{1}{2}}$