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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 1 - Number Systems

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Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 1 - Number Systems.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 1 - Number Systems Page/Excercise MCQ

Solution 1

Correct option: (d)

0 can be written as   where p and q are integers and q ≠ 0.

Solution 2

Correct option: (a)

On a number line, 0 is a rational number that lies between -3 and 3. 

Solution 3

Correct option: (c)

Two rational numbers between    

Solution 4

Correct option: (d)

Every point on number line represents a unique number. 

Solution 5

Solution 6

Solution 7

Solution 8

  

Solution 9

Solution 10

Correct option: (d)

The decimal expansion of a rational number is either terminating or non-terminating recurring.

The decimal expansion of 0.5030030003…. is non-terminating non-recurring, which is not a property of a rational number. 

Solution 11

Correct option: (d)

The decimal expansion of an irrational number is non-terminating non-recurring.

Hence, 3.141141114….. is an irrational number. 

Solution 12

Correct option: (d)

  

Solution 13

Correct option: (b)

Given two rational numbers are negative and   is a positive rational number.

So, it does not lie between   

Solution 14

Correct option: (c)

Π = 3.14159265359…….., which is non-terminating non-recurring.

Hence, it is an irrational number.

Solution 15

Correct option: (d)

The decimal expansion of  , which is non-terminating, non-recurring. 

Solution 16

Correct option: (a)

The decimal expansion of  , which is non-terminating, non-recurring.

Hence, it is an irrational number. 

Solution 17

Correct option: (b)

  

Solution 18

Correct option: (c)

The decimal expansion of  , which is non-terminating, non-recurring.

Hence, it is an irrational number. 

Solution 19

Solution 20

Solution 21

Solution 22

Correct option: (c)

  

  

Solution 23

Correct option: (d)

The decimal expansion of a rational number is either terminating or non-terminating recurring.

Hence, 0.853853853... is a rational number. 

Solution 24

Correct option: (a)

The product of a non-zero rational number with an irrational number is always an irrational number. 

Solution 25

Correct option: (b)

  

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Correct option: (b)

Let x =

i.e. x = 0.3333…. ….(i)

10x = 3.3333…. ….(ii)

On subtracting (i) from (ii), we get

9x = 3

  

Let y =

i.e. y = 0.4444…. ….(i)

10y = 4.4444…. ….(ii)

On subtracting (i) from (ii), we get

9y = 4

  

Solution 34

Correct option: (c)

Let x =

i.e. x = 2.4545….  ….(i)

100x = 245.4545…….  ….(ii)

On subtracting (i) from (ii), we get

99x = 243

  

Let y =   

i.e. y = 0.3636….  ….(iii)

100y = 36.3636….  ….(iv)

On subtracting (iii) from (iv), we get

99y = 36

  

  

Solution 35

Correct option: (b)

  

Solution 36

Correct option: (b)

  

Which is positive and rational number. 

Solution 37

Correct option: (b)

  

Which is positive and rational number. 

Solution 38

Correct option: (c)

  

Solution 39

Correct option: (a)

  

Solution 40

Correct option: (b)

  

Solution 41

Correct option: (b)

  

Solution 42

Correct option: (b)

  

Solution 43

Correct option: (c)

Solution 44

Correct option: (b)

  

Solution 45

Correct option: (d)

  

Solution 46

Correct option: (a)

  

Solution 47

Correct option: (b)

  

Solution 48

Correct option: (a)

Solution 49

Correct option: (c)

93 + (-3)3 - 63 = 729 - 27 - 216 = 486 

Solution 50

Correct option: (b)

  

Solution 51

Correct option: (c)

  

Solution 52

Correct option: (d)

Solution 53

Correct option: (a)

  

Solution 54

Correct option: (d)

  

Solution 55

Correct option: (b)

  

Solution 56

Correct option: (b)

  

Solution 57

Correct option: (b)

xp-q xq - r xr - p

= xp - q + q - r + r - p

= x0

= 1 

Solution 58

Correct option: (c)

  

Solution 59

Correct option: (a)

  

Solution 60

Correct option: (d)

  

Solution 61

Correct option: (d)

(33)2 = 9x

(32)3 = (32)x

x = 3

Then 5x = 53 = 125 

Solution 62

Correct option: (b)

  

Solution 63

Correct option: (d)

  

Thus, the simplest rationalisation factr of    

Solution 64

Correct option: (b)

The simplest rationalisation factor of   is   

Solution 65

Correct option: (d)

  

Solution 66

Correct option: (d)

  

Solution 67

Correct option: (c)

  

Solution 68

Correct option: (c)

  

Solution 69

Correct option: (b)

  

Solution 70

Correct option: (c)

  

Solution 71

Correct option: (b)

  

Solution 72

Correct option: (d)

  

Solution 73

Correct option: (c)

  

Solution 74

Correct option: (c)

  

Solution 75

Correct option: (a)

  

Solution 76

Solution 77

Solution 78

Solution 79

Solution 80

Solution 81

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 1 - Number Systems Page/Excercise VSAQ

Solution 1

The sum of a rational number and an irrational number is irrational.

Example: 5 +   is irrational. 

Solution 2

  

Solution 3

Thus, the given number will terminate after 3 decimal places. 

Solution 4

(1296)0.17× (1296)0.08

Solution 5

  

Solution 6

An irrational number between 5 and 6 =   

Solution 7

  

Solution 8

  

Solution 9

  

Solution 10

Solution 11

  

Solution 12

  

Solution 13

Given, a = 1 and b = 2

  

Solution 14

  

Solution 15

Solution 16

Yes, the product of a rational and irrational numbers is always irrational.

For example,

  

Solution 17

Solution 18

The reciprocal of ( )

Solution 19

  

Solution 20

  

Solution 21

  

Solution 22

  

Solution 23

  

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 1 - Number Systems Page/Excercise 1A

Solution 1

A number which can be expressed as  , where 'a' and 'b' both are integers and b ≠ 0, is called a rational number.

Since, 0 can be expressed as  , it is a rational number.

Solution 2(iii)

  

Solution 2(i)

(i) 

 

Solution 2(ii)

(ii) 

Solution 2(iv)

(iv) 1.3

Solution 2(v)

(v) -2.4

Solution 3(v)

  

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 4

Infinite rational numbers can be determined between given two rational numbers.

Solution 5

We have

  

We know that 9 < 10 < 11 < 12 < 13 < 14 < 15

  

Solution 6

2 and 3 can be represented as respectively.

Now six rational numbers between 2 and 3 are 

 . 

Solution 7

  

Solution 8

Let x = 2.1 and y = 2.2

Then, x < y because 2.1 < 2.2

Or we can say that,

Or,

That is, we have,

We know that,

Therefore, we can have,

Therefore, 16 rational numbers between, 2.1 and 2.2 are:


So, 16 rational numbers between 2.1 and 2.2 are:

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, 2.18



Solution 9(iii)

False, integers include negative of natural numbers as well, which are clearly not whole numbers. For example -1 is an integer but not a whole number.

Solution 9(v)

False, integers are counting numbers on both sides of the number line i.e. they are both positive and negative while rational numbers are of the form  . Hence, Every rational number is not an integer but every integer is a rational number.

Solution 9(i)

True. Since the collection of natural number is a sub collection of whole numbers, and every element of natural numbers is an element of whole numbers

Solution 9(ii)

False. Since 0 is whole number but it is not a natural number.

Solution 9(iv)

True. Every integer can be represented in a fraction form with denominator 1.

Solution 9(vi)

False. Since division of whole numbers is not closed under division, the value of , may not be a whole number.

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 1 - Number Systems Page/Excercise 1B

Solution 1(iv)

  

Since the denominator of a given rational number is not of the form 2m × 2n, where m and n are whole numbers, it has non-terminating decimal. 

Solution 1(i)

 

If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.

Since, 80 has prime factors 2 and 5,  is a terminating decimal.

Solution 1(ii)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since, 24 has prime factors 2 and 3 and 3 is different from 2 and 5,

 is not a terminating decimal.

Solution 1(iii)

 

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since 12 has prime factors 2 and 3  and 3 is different from 2 and 5,

 is not a terminating decimal.

Solution 2(i)

  

Hence, it has terminating decimal expansion. 

Solution 2(ii)

Hence, it has terminating decimal expansion. 

Solution 2(iii)

  

Hence, it has non-terminating recurring decimal expansion. 

Solution 2(iv)

  

Hence, it has non-terminating recurring decimal expansion. 

Solution 2(v)

   

Hence, it has non-terminating recurring decimal expansion. 

Solution 2(vi)

   

Hence, it has terminating decimal expansion. 

Solution 2(vii)

   

Hence, it has terminating decimal expansion. 

Solution 2(viii)

 

 

Hence, it has non-terminating recurring decimal expansion. 

Solution 3(i)

Let x =

i.e. x = 0.2222…. ….(i)

10x = 2.2222…. ….(ii)

On subtracting (i) from (ii), we get

9x = 2

  

Solution 3(ii)

Let x =

i.e. x = 0.5353….  ….(i)

100x = 53.535353…. ….(ii)

On subtracting (i) from (ii), we get

99x = 53

  

Solution 3(iii)

Let x =

i.e. x = 2.9393….  ….(i)

100x = 293.939……. ….(ii)

On subtracting (i) from (ii), we get

99x = 291

  

Solution 3(iv)

Let x =

i.e. x = 18.4848….  ….(i)

100x = 1848.4848……. ….(ii)

On subtracting (i) from (ii), we get

99x = 1830

  

Solution 3(v)

Let x =

i.e. x = 0.235235..…   ….(i)

1000x = 235.235235……. ….(ii)

On subtracting (i) from (ii), we get

999x = 235

  

Solution 3(vi)

Let x =

i.e. x = 0.003232..…   

100x = 0.323232……. ….(i)

10000x = 32.3232…. ….(ii)

On subtracting (i) from (ii), we get

9900x = 32

  

Solution 3(vii)

Let x =

i.e. x = 1.3232323..… ….(i)   

100x = 132.323232……. ….(ii)

On subtracting (i) from (ii), we get

99x = 131

  

Solution 3(viii)

Let x =

i.e. x = 0.3178178..… 

10x = 3.178178…… ….(i)   

10000x = 3178.178……. ….(ii)

On subtracting (i) from (ii), we get

9990x = 3175

  

Solution 3(ix)

Let x =

i.e. x = 32.123535..… 

100x = 3212.3535…… ….(i)   

10000x = 321235.3535……. ….(ii)

On subtracting (i) from (ii), we get

9900x = 318023

  

Solution 3(x)

Let x =

i.e. x = 0.40777..… 

100x = 40.777…… ….(i)   

1000x = 407.777……. ….(ii)

On subtracting (i) from (ii), we get

900x = 367

  

Solution 4

Let x =

i.e. x = 2.3636….  ….(i)

100x = 236.3636……. ….(ii)

On subtracting (i) from (ii), we get

99x = 234

  

Let y =   

i.e. y = 0.2323….  ….(iii)

100y = 23.2323…. ….(iv)

On subtracting (iii) from (iv), we get

99y = 23

  

  

Solution 5

Let x =   

i.e. x = 0.3838….  ….(i)

100x = 38.3838….  ….(ii)

On subtracting (i) from (ii), we get

99x = 38

  

Let y =

i.e. y = 1.2727….  ….(iii)

100y = 127.2727…….  ….(iv)

On subtracting (iii) from (iv), we get

99y = 126

  

  

Solution 9(v)

 

If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.

 

Since 125 has prime factor 5 only

  is a terminating decimal.

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 1 - Number Systems Page/Excercise 1F

Solution 1

The rationalising factor of the denominator in   is   

Solution 2(vi)

  

Solution 2(vii)

  

Solution 2(viii)

  

Solution 2(ix)

  

Solution 2(i)

On multiplying the numerator and denominator of the given number by , we get



Solution 2(ii)

On multiplying the numerator and denominator of the given number by , we get


Solution 2(iii)

Solution 2(iv)

Solution 2(v)

Solution 3(i)

  

Solution 3(ii)

  

Solution 3(iii)

  

Solution 4(i)

  

Solution 4(ii)

  

Solution 4(iii)

  

Solution 4(iv)

  

Solution 5(i)

  

Solution 5(ii)

  

Solution 5(iii)

  

Solution 5(iv)

  

Solution 5(v)

  

Solution 5(vi)

  

Solution 6(i)

  

Solution 6(ii)

  

Solution 7(ii)

  

Solution 7(iii)

  

Solution 7(iv)

  

Solution 7(i)

 

Solution 8(i)

  

Solution 8(ii)

  

Solution 9

 *Back answer incorrect 

Solution 10

  

Solution 11

  

Thus, the given number is rational. 

Solution 12

  

Solution 13

  

Solution 14

  

Solution 15

  

Solution 16

  

Solution 17

  

Solution 18

  

Solution 19

  

Solution 20

  

Solution 21

  

Solution 22(i)

  

Solution 22(ii)

  

Solution 22(iii)

  

Solution 23

  

Solution 24

  

Solution 25

  

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 1 - Number Systems Page/Excercise 1C

Solution 1

Irrational number: A number which cannot be expressed either as a terminating decimal or a repeating decimal is known as irrational number. Rather irrational numbers cannot be expressed in the fraction form,

For example, 0.101001000100001 is neither a terminating nor a repeating decimal and so is an irrational number.

Also, etc. are examples of irrational numbers.

Solution 2(iii)


We know that, if n is a not a perfect square, then is an irrational number.

Here,  is a not a perfect square number.

So, is irrational.

Solution 2(i)

  

Since quotient of a rational and an irrational is irrational, the given number is irrational.  

Solution 2(v)

 

is the product of a rational number and an irrational number square root of 6.

Theorem: The product of a non-zero rational number and an irrational number is an irrational number.

Thus, by the above theorem, is an irrational number.

So, is an irrational number.

Solution 2(ii)

  

Solution 2(iv)

  

Solution 2(vi)

The given number 4.1276 has terminating decimal expansion.

Hence, it is a rational number. 

Solution 2(vii)

  

Since the given number has non-terminating recurring decimal expansion, it is a rational number. 

Solution 2(viii)

The given number 1.232332333.... has non-terminating and non-recurring decimal expansion.

Hence, it is an irrational number. 

Solution 2(ix)

The given number 3.040040004..... has non-terminating and non-recurring decimal expansion.

Hence, it is an irrational number. 

Solution 2(x)

The given number 2.356565656..... has non-terminating recurring decimal expansion.

Hence, it is a rational number. 

Solution 2(xi)

The given number 6.834834.... has non-terminating recurring decimal expansion.

Hence, it is a rational number. 

Solution 3

We know that the sum of a rational and an irrational is irrational.

Hence, if x is rational and y is irrational, then x + y is necessarily an irrational number.

For example,

Solution 4

We know that the product of a rational and an irrational is irrational.

Hence, if a is rational and b is irrational, then ab is necessarily an irrational number.

For example,

  

Solution 5

No, the product of two irrationals need not be an irrational.

For example,

Solution 6

(i) Difference is an irrational number:

(ii) Difference is a rational number:

(iii) Sum is an irrational number:

(iv) Sum is an rational number:

(v) Product is an irrational number:

(vi) Product is a rational number:

(vii) Quotient is an irrational number:

(viii) Quotient is a rational number:

  

Solution 7

  

Solution 8

Rational number between 2 and 2.5 =

Irrational number between 2 and 2.5 =

Solution 9

There are infinite irrational numbers between .

We have

  

Hence, three irrational numbers lying between   are as follows:

1.5010010001……., 1.6010010001…… and 1.7010010001……. 

Solution 10

Since 0.5 < 0.55

Let x = 0.5, y = 0.55 and y = 2

Two irrational numbers between 0.5 and 0.55 are 0.5151151115……. and 0.5353553555…. 

Solution 11

  

Thus, three different irrational numbers between the rational numbers   are as follows:

0.727227222….., 0.757557555….. and 0.808008000….. 

Solution 12

Let a and b be two rational numbers between the numbers 0.2121121112... and 0.2020020002......

Now, 0.2020020002......

Solution 13

Two irrational numbers between 0.16 and 0.17 are as follows:

0.1611161111611111611111…… and 0.169669666……. 

Solution 14(i)

True

Solution 14(ii)

False

Solution 14(iii)

True

Solution 14(iv)

False

Solution 14(v)

True

Solution 14(vi)

False

Solution 14(vii)

False

Solution 14(viii)

True

Solution 14(ix)

True

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 1 - Number Systems Page/Excercise 1D

Solution 1(i)

We have:

Solution 1(ii)

We have:

Solution 1(iii)

Solution 2(i)

 

 

 

Solution 2(ii)

Solution 2(iii)

Solution 2(iv)

 

 

 

Solution 2(v)

 

 

 

Solution 2(vi)

 

 

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 4(i)

  

= 9 - 11

= -2  

Solution 4(ii)

  

= 9 - 5

= 4 

Solution 4(v)

  

Solution 4(iii)

Solution 4(iv)

Solution 4(vi)

Solution 5

  

Solution 6(i)

  

Thus, the given number is rational. 

Solution 6(ii)

  

Clearly, the given number is irrational. 

Solution 6(iii)

  

Thus, the given number is rational. 

Solution 6(iv)

  

Thus, the given number is irrational. 

Solution 7

(i) Number of chocolates distributed by Reema

  

 

(ii) Loving, helping and caring attitude towards poor and needy children.

Solution 8(i)

  

Solution 8(ii)

  

Solution 8(iii)

  

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 1 - Number Systems Page/Excercise 1E

Solution 1

Draw a number line as shown.

On the number line, take point O corresponding to zero.

Now take point A on number line such that OA = 2 units.

Draw perpendicular AZ at A on the number line and cut-off arc AB = 1 unit.

By Pythagoras Theorem,

OB2 = OA2 + AB2 = 22 + 12 = 4 + 1 = 5

OB =

Taking O as centre and OB =   as radius draw an arc cutting real line at C.

Clearly, OC = OB =

  

Hence, C represents   on the number line.

Solution 2

Draw a number line as shown.

On the number line, take point O corresponding to zero.

Now take point A on number line such that OA = 1 unit.

Draw perpendicular AZ at A on the number line and cut-off arc AB = 1 unit.

By Pythagoras Theorem,

OB2 = OA2 + AB2 = 12 + 12 = 1 + 1 = 2

OB =

Taking O as centre and OB =   as radius draw an arc cutting real line at C.

Clearly, OC = OB =

Thus, C represents   on the number line.

Now, draw perpendicular CY at C on the number line and cut-off arc CE = 1 unit.

By Pythagoras Theorem,

OE2 = OC2 + CE2 =  2 + 12 = 2 + 1 = 3

OE =

Taking O as centre and OE =   as radius draw an arc cutting real line at D.

Clearly, OD = OE =

  

Hence, D represents   on the number line. 

Solution 3

Draw a number line as shown.

On the number line, take point O corresponding to zero.

Now take point A on number line such that OA = 3 units.

Draw perpendicular AZ at A on the number line and cut-off arc AB = 1 unit.

By Pythagoras Theorem,

OB2 = OA2 + AB2 = 32 + 12 = 9 + 1 = 10

OB =

Taking O as centre and OB =   as radius draw an arc cutting real line at C.

Clearly, OC = OB =

  

Hence, C represents   on the number line. 

Solution 4

Draw a number line as shown.

On the number line, take point O corresponding to zero.

Now take point A on number line such that OA = 2 units.

Draw perpendicular AZ at A on the number line and cut-off arc AB = 2 units.

By Pythagoras Theorem,

OB2 = OA2 + AB2 = 22 + 22 = 4 + 4 = 8

OB =

Taking O as centre and OB =   as radius draw an arc cutting real line at C.

Clearly, OC = OB =

  

Hence, C represents   on the number line. 

Solution 5

Draw a line segment AB = 4.7 units and extend it to C such that BC = 1 unit.

Find the midpoint O of AC.

With O as centre and OA as radius, draw a semicircle.

Now, draw BD AC, intersecting the semicircle at D.

Then, BD =   units.

With B as centre and BD as radius, draw an arc, meeting AC produced at E.

 Then, BE = BD =   units. 

Solution 6

Draw a line segment OB = 10.5 units and extend it to C such that BC = 1 unit.

Find the midpoint D of OC.

With D as centre and DO as radius, draw a semicircle.

Now, draw BE AC, intersecting the semicircle at E.

Then, BE =   units.

With B as centre and BE as radius, draw an arc, meeting AC produced at F.

 

  

Then, BF = BE =   units.

Solution 7

Draw a line segment AB = 7.28 units and extend it to C such that BC = 1 unit.

Find the midpoint O of AC.

With O as centre and OA as radius, draw a semicircle.

Now, draw BD AC, intersecting the semicircle at D.

Then, BD = units.

With D as centre and BD as radius, draw an arc, meeting AC produced at E.

Then, BE = BD = units.

Solution 8

Draw a line segment OB = 9.5 units and extend it to C such that BC = 1 unit.

Find the midpoint D of OC.

With D as centre and DO as radius, draw a semicircle.

Now, draw BE AC, intersecting the semicircle at E.

Then, BE =   units.

With B as centre and BE as radius, draw an arc, meeting AC produced at F.

Then, BF = BE =   units.

Extend BF to G such that FG = 1 unit.

Then, BG =   

  

 

Solution 9

 

  

Solution 10

  

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 1 - Number Systems Page/Excercise 1G

Solution 1(i)

  

Solution 1(ii)

  

Solution 1(iv)

  

Solution 1(iii)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 4(i)

Solution 4(ii)

Solution 4(iii)

Solution 5(i)

Solution 5(ii)

 

Solution 5(iii)

Solution 5(iv)

Solution 5(v)

Solution 5(vi)

Solution 6(i)

Given, a = 2 and b = 3

Solution 6(ii)

Given, a = 2 and b = 3

  

Solution 7(i)

  

Solution 7(ii)

(14641)0.25

Solution 7(iii)

  

Solution 7(iv)

  

Solution 8(i)

  

Solution 8(ii)

  

Solution 8(iii)

  

Solution 8(iv)

  

Solution 9(i)

  

Solution 9(ii)

  

Solution 9(iii)

  

Solution 9(iv)

  

Solution 10(i)

  

Solution 10(ii)

  

Solution 10(iii)

  

Solution 11

  

Solution 12

  

Solution 13(i)

  

Solution 13(ii)

  

Solution 13(iii)

  

Solution 14(i)

  

Solution 14(ii)

  

Solution 14(iii)

  

Solution 14(iv)

5x - 3 × 32x - 8 = 225

5x - 3× 32x - 8 = 52 × 32

x - 3 = 2 and 2x - 8 = 2

x = 5 and 2x = 10

x = 5 

Solution 14(v)

  

Solution 15(i)

  

Solution 15(ii)

  

Solution 15(iii)

  

Solution 15(iv)

  

Solution 16

  

Solution 17

  

Solution 18

  

TopperLearning provides step-by-step solutions for each question in each chapter in the RS Aggarwal & V Aggarwal Textbook. Access the CBSE Class 9 Mathematics Chapter 1 - Number Systems for free. The questions have been solved by our subject matter experts to help you understand how to answer them. Our RS Aggarwal Solutions for class 9 will help you to study and revise the whole chapter, and you can easily clear your fundamentals in Chapter 1 - Number Systems now.

Text Book Solutions

CBSE IX - Mathematics

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