NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

 

NCERT solutions for class 9 maths chapter 1 Number Systems- Are you looking for in-depth knowledge of numbers and its applications? Then you must read this chapter. You will have a great conceptual clarity related to number system after going through it. Solutions of NCERT class 9 maths chapter 1 Number Systems are also there to provide assistance whenever you feel trouble while facing any problem related to this particular chapter. In this particular chapter, there is a total of 6 exercises which consist of 27 questions. It has a weightage of 8 marks in the CBSE class 9 final examination. CBSE NCERT solutions for class 9 maths chapter 1 Number Systems are covering the solutions to each and every question present in the practice exercises. Apart from school exams, the solutions are also beneficial if you are aiming for exams like- National Talent Search Examination (NTSE), Indian National Olympiad (INO), SSC, CAT, etc. In NCERT class 9 maths chapter 1 Number Systems, you will learn the extended form of the number line and learn how to represent various types of the number on it. The Chapter starts with an introduction which includes rational numbers, whole numbers, and integers followed by important topics like irrational numbers, real numbers, how to represent real numbers on the number line, operations on real numbers and many more. With the understanding of rational numbers, we will also study how to represent square roots of 2, 3, 5 and other non-rational numbers. NCERT solutions for class 9 maths chapter 1 Number Systems are written keeping important basics in mind so that a student can get 100% out of it. NCERT solutions are also available for different subjects and classes which you can get by clicking on the link.

NCERT solutions for class 9 maths chapter 1 Number Systems Excercise: 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}  (where \ q \neq 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} \ and \ 4 = \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} \ ( where \ q \neq 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 

NCERT solutions for class 9 maths chapter 1 Number Systems Excercise: 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.

NCERT solutions for class 9 maths chapter 1 Number Systems Excercise: 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.

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

Answer:

Let x = 0.\overline6= 0.6666....             -(i)

Now, multiply by 10 on both sides

10x= 6.6666...

\Rightarrow 10x = 6 + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(using \ (i))

\Rightarrow 9x = 6

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

Therefore,  \frac{p}{q}  form of  0.\bar{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\bar{7}

Answer:

We can write 0.4\overline7 as

\Rightarrow 0.4\overline7 = \frac{4}{10}+ \frac{0.777..}{10}             -(i)

Now,

Let x = 0.\overline7= 0.7777....             -(ii)

Now, multiply by 10 on both sides

10x= 7.7777...

\Rightarrow 10x = 7 + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(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\overline7 = \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\overline7  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 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(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.\overline{9}= 0.9999....             -(i)

Now, multiply by 10 on both sides

10x= 9.999....

\Rightarrow 10x = 9 + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(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.

NCERT solutions for class 10 maths chapter 1 Number Systems Excercise: 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 4.\overline{26}  as
\Rightarrow 4.\overline{26}= 4.262626...
Now,  4.2626 can be visualised as in the following steps.

NCERT solutions for class 9 maths chapter 1 Number Systems Excercise: 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: \left ( 3+\sqrt{23} \right )-\sqrt{23}

Answer:

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

\Rightarrow \left ( 3+\sqrt{23} \right )-\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 reduces 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: \left (3+\sqrt{3} \right )\left ( 3-\sqrt{3} \right )

Answer:

Given number is 

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

Therefore, answer is  6

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

Answer:

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

Now, we will reduce it into
\Rightarrow \left (\sqrt{5}+\sqrt{2} \right )^{2} = \left ( (\sqrt{5})^2+(\sqrt{2})^2+2.\sqrt{5}.\sqrt{2} \right )            \left ( using \ (a+b)^2=a^2+b^2 +2ab\right )
                                   =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: \left (\sqrt{5}-\sqrt{2} \right )\left ( \sqrt{5}+\sqrt{2} \right )

Answer:

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

Now, we will reduce it into
\Rightarrow \left (\sqrt{5}-\sqrt{2} \right )\left ( \sqrt{5}+\sqrt{2} \right )= \left ( (\sqrt{5})^2-(\sqrt{2})^2 \right )           \left ( using \ (a+b)(a-b)=a^2-b^2\right )
                                                           =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{\sqrt7}{7}

Therefore, the answer is  \frac{\sqrt7}{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-\sqrt6} = \frac{1}{\sqrt 7-\sqrt 6}\times \frac{\sqrt 7+\sqrt 6 }{\sqrt 7+\sqrt 6 } = \frac{\sqrt7+\sqrt 6}{(\sqrt7)^2-(\sqrt6)^2} = \frac{\sqrt7+\sqrt6}{7-6} = \sqrt7+\sqrt6
                                                                                                                          

Therefore, the answer is  \sqrt7+\sqrt6

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 

\\= \frac{1}{\sqrt 5+\sqrt2} \\\\= \frac{1}{\sqrt 5+\sqrt 2}\times \frac{\sqrt 5-\sqrt 2 }{\sqrt 5-\sqrt 2 }\\\\ = \frac{\sqrt5-\sqrt 2}{(\sqrt5)^2-(\sqrt2)^2} \\\\= \frac{\sqrt5-\sqrt2}{5-2}
 = \frac{\sqrt5-\sqrt2}{3}

Therefore, the answer is  \frac{\sqrt5-\sqrt2}{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{\sqrt7+ 2}{(\sqrt7)^2-(2)^2} = \frac{\sqrt7+2}{7-4}
                                                                                                        = \frac{\sqrt7+2}{3}

Therefore, answer is  \frac{\sqrt7+2}{3}

NCERT solutions for class 9 maths chapter 1 Number Systems Excercise: 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}            \left ( \because a^n.a^m = a^{n+m} \right )

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\times7}} = \frac{1}{3^{21}} = 3^{-21}                   \left ( \because (a^n)^m = a^{n.m} \ and \ \frac{1}{a^m}= a^{-m}\right )

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}                \left ( \because \frac{a^n}{a^m}= a^{n-m}\right )
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\times8)^{\frac{1}{2}} = 56^{\frac{1}{2}}                                                          \left ( \because a^n.b^n=(a.b)^n\right )
Therefore, the answer is  56^{\frac{1}{2}} 

NCERT solutions for class 9 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 9 maths chapter 1 Number Systems

Chapter 2

CBSE NCERT solutions for class 9 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry

Chapter 4

NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables

Chapter 5

CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry

Chapter 6

Solutions of NCERT class 9 maths chapter 6 Lines And Angles

Chapter 7

NCERT solutions for class 9 maths chapter 7 Triangles

Chapter 8

CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals

Chapter 9

Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles

Chapter 10

NCERT solutions for class 9 maths chapter 10 Circles

Chapter 11

CBSE NCERT solutions for class 9 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 9 maths chapter 12 Heron’s Formula

Chapter 13

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes

Chapter 14

CBSE NCERT solutions for class 9 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 9 maths chapter 15 Probability

NCERT solutions for class 9 subject wise

CBSE NCERT Solutions for Class 9 Maths

Solutions of NCERT Solutions for Class 9 Science

How to use NCERT Solutions for class 9 maths chapter 1 Number Systems?

  • First of all, go through the conceptual text given in the book before the exercises.

  • After covering the conceptual theory, you must go through some examples to understand the application of that particular concept. 

  • After observing the application, come to the practice exercises available in the textbook

  • While solving the practice exercises, you can take the help of NCERT solutions for class 9 maths chapter 1 Number Systems to boost your preparation.

Keep working hard & happy learning!

 

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