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

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.

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

Correct option: (d)

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

Correct option: (a)

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

Correct option: (c)

Two rational numbers between

Correct option: (d)

Every point on number line represents a unique number.

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.

Correct option: (d)

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

Hence, 3.141141114….. is an irrational number.

Correct option: (d)

Correct option: (b)

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

So, it does not lie between

Correct option: (c)

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

Hence, it is an irrational number.

Correct option: (d)

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

Correct option: (a)

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

Hence, it is an irrational number.

Correct option: (b)

Correct option: (c)

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

Hence, it is an irrational number.

Correct option: (c)

Correct option: (d)

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

Hence, 0.853853853... is a rational number.

Correct option: (a)

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

Correct option: (b)

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

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

Correct option: (b)

Correct option: (b)

Which is positive and rational number.

Correct option: (b)

Which is positive and rational number.

Correct option: (c)

Correct option: (a)

Correct option: (b)

Correct option: (b)

Correct option: (b)

Correct option: (c)

Correct option: (b)

Correct option: (d)

Correct option: (a)

Correct option: (b)

Correct option: (a)

Correct option: (c)

9^{3} + (-3)^{3} -
6^{3} = 729 - 27 - 216 = 486

Correct option: (b)

Correct option: (c)

Correct option: (d)

Correct option: (a)

Correct option: (d)

Correct option: (b)

Correct option: (b)

Correct option: (b)

x^{p-q}⋅
x^{q - r}⋅ x^{r - p}

= x^{p - q + q - r + r - p}

= x^{0}

= 1

Correct option: (c)

Correct option: (a)

Correct option: (d)

Correct option: (d)

(3^{3})^{2} = 9^{x}

⇒
(3^{2})^{3} = (3^{2})^{x}

⇒ x = 3

Then 5^{x} = 5^{3}
= 125

Correct option: (b)

Correct option: (d)

Thus, the simplest rationalisation factr of

Correct option: (b)

The simplest rationalisation factor of is

Correct option: (d)

Correct option: (d)

Correct option: (c)

Correct option: (c)

Correct option: (b)

Correct option: (c)

Correct option: (b)

Correct option: (d)

Correct option: (c)

Correct option: (c)

Correct option: (a)

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

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

Example: 5 + is irrational.

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

(1296)^{0.17}× (1296)^{0.08}

An irrational number between 5 and 6 =

Given, a = 1 and b = 2

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

For example,

The reciprocal of ()

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

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.

(i)

(ii)

(iv) 1.3

(v) -2.4

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

We have

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

2 and 3 can be represented asrespectively.

Now six rational numbers between 2 and 3 are

.

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

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.

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.

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

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

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

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

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

_{}

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.

_{}

_{}

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.

_{}

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.

Hence, it has terminating decimal expansion.

Hence, it has terminating decimal expansion.

Hence, it has non-terminating recurring decimal expansion.

Hence, it has non-terminating recurring decimal expansion.

Hence, it has non-terminating recurring decimal expansion.

Hence, it has terminating decimal expansion.

Hence, it has terminating decimal expansion.

Hence, it has non-terminating recurring decimal expansion.

Let x =

i.e. x = 0.2222…. ….(i)

⇒ 10x = 2.2222…. ….(ii)

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

9x = 2

Let x =

i.e. x = 0.5353…. ….(i)

⇒ 100x = 53.535353…. ….(ii)

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

99x = 53

Let x =

i.e. x = 2.9393…. ….(i)

⇒ 100x = 293.939……. ….(ii)

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

99x = 291

Let x =

i.e. x = 18.4848…. ….(i)

⇒ 100x = 1848.4848……. ….(ii)

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

99x = 1830

Let x =

i.e. x = 0.235235..… ….(i)

⇒ 1000x = 235.235235……. ….(ii)

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

999x = 235

Let x =

i.e. x = 0.003232..…

⇒ 100x = 0.323232……. ….(i)

⇒ 10000x = 32.3232…. ….(ii)

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

9900x = 32

Let x =

i.e. x = 1.3232323..… ….(i)

⇒ 100x = 132.323232……. ….(ii)

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

99x = 131

Let x =

i.e. x = 0.3178178..…

⇒ 10x = 3.178178…… ….(i)

⇒ 10000x = 3178.178……. ….(ii)

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

9990x = 3175

Let x =

i.e. x = 32.123535..…

⇒ 100x = 3212.3535…… ….(i)

⇒ 10000x = 321235.3535……. ….(ii)

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

9900x = 318023

Let x =

i.e. x = 0.40777..…

⇒ 100x = 40.777…… ….(i)

⇒ 1000x = 407.777……. ….(ii)

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

900x = 367

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

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

_{}

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

The rationalising factor of the denominator in is

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

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

_{}

*Back answer incorrect

Thus, the given number is rational.

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

**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.

_{}

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.

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

is the product of a rational number and an irrational number .

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.

The given number 4.1276 has terminating decimal expansion.

Hence, it is a rational number.

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

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

Hence, it is an irrational number.

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

Hence, it is an irrational number.

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

Hence, it is a rational number.

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

Hence, it is a rational number.

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,

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,

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

For example,

(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:

Rational number between 2 and 2.5 =

Irrational number between 2 and 2.5 =

There are infinite irrational numbers between.

We have

Hence, three irrational numbers lying between are as follows:

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

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….

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

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

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

Now, 0.2020020002......

Two irrational numbers between 0.16 and 0.17 are as follows:

0.1611161111611111611111…… and 0.169669666…….

True

False

True

False

True

False

False

True

True

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

_{}

We have:

_{}

_{}

We have:

_{}

_{}

_{}

_{}

_{}

_{}

= 9 - 11

= -2

= 9 - 5

= 4

_{}

_{}

_{}

_{}

_{}

Thus, the given number is rational.

Clearly, the given number is irrational.

Thus, the given number is rational.

Thus, the given number is irrational.

(i) Number of chocolates distributed by Reema

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

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

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,

OB^{2} = OA^{2}
+ AB^{2} = 2^{2} + 1^{2 }= 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.

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,

OB^{2} = OA^{2}
+ AB^{2} = 1^{2} + 1^{2 }= 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,

OE^{2} = OC^{2}
+ CE^{2} = ^{2} + 1^{2 }= 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.

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,

OB^{2} = OA^{2}
+ AB^{2} = 3^{2} + 1^{2 }= 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.

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,

OB^{2} = OA^{2}
+ AB^{2} = 2^{2} + 2^{2 }= 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.

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.

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.

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.

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 =

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

_{}

_{}

_{}

_{}

_{}

Given, a = 2 and b = 3

Given, a = 2 and b = 3

(14641)^{0.25}

5^{x - 3} × 3^{2x - 8}
= 225

⇒ 5^{x - 3}× 3^{2x - 8} = 5^{2} × 3^{2}

⇒ x - 3 = 2 and 2x - 8 = 2

⇒ x = 5 and 2x = 10

⇒ x = 5

## R. S. Aggarwal and V. Aggarwal - Mathematics - IX Class 9 Chapter Solutions

- Chapter 1 - Number Systems
- Chapter 2 - Polynomials
- Chapter 3 - Factorisation of Polynomials
- Chapter 4 - Linear Equations in Two Variables
- Chapter 5 - Coordinate Geometry
- Chapter 6 - Introduction to Euclid's Geometry
- Chapter 7 - Lines And Angles
- Chapter 8 - Triangles
- Chapter 9 - Congruence of Triangles and Inequalities in a Triangle
- Chapter 10 - Quadrilaterals
- Chapter 11 - Areas of Parallelograms and Triangles
- Chapter 12 - Circles
- Chapter 13 - Geometrical Constructions
- Chapter 14 - Areas of Triangles and Quadrilaterals
- Chapter 15 - Volume and Surface Area of Solids
- Chapter 16 - Presentation of Data in Tabular Form
- Chapter 17 - Bar Graph, Histogram and Frequency Polygon
- Chapter 18 - Mean, Median and Mode of Ungrouped Data
- Chapter 19 - Probability

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