Important Questions Class 9 Maths Chapter 1

Important Questions Class 9 Mathematics Chapter 1 – Number System.

Mathematics is utilised worldwide as a critical component in numerous fields. Mathematics has several benefits, and having a solid foundation in the subject will benefit your future career prospects. Since Mathematics is completely established on numbers and is not a language-based subject, you can either get the answer right or wrong, which enhances your chance of getting full marks in this subject.

Chapter 1 of Class 9 Mathematics is about ‘Number System’. Number systems are approaches in Mathematics that are used to represent numbers in assorted forms. A number is a mathematical value utilised for calculating and measuring objects and for executing arithmetic calculations. Numbers have various classifications, like natural numbers, whole numbers, rational and irrational numbers, and so on. Besides, there are different types of number systems that have other properties, like the binary number system, the octal number system, the decimal number system, and the hexadecimal number system.

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Important Questions Class 9 Mathematics Chapter 1 – With Solution

Our in-house Mathematics specialists have collated a complete list of Important Questions Class 9 Mathematics Chapter 1 by referring to a variety of reliable sources such as NCERT textbook, exemplars, other reference books, and past years exam papers. For individual questions, the mathematics faculty  has designed a step-by-step explanation that will enable students to comprehend the concepts used in each question. Furthermore, the questions are selected in a way that would cover entire chapter topics. So by practising from our question bank, students will be competent to revise the chapter and understand their strong and weak points. And enhance their practice by also focusing on weaker sections of the chapter and improve their academic scores.

Given  below are some  of the questions and answers from our question bank of Important Questions Class 9 Mathematics Chapter 1:

Question 1: Find out five rational numbers between 1 and 2.

Answer 1: We have to find five rational numbers between 1 and 2.

So, let us write the numbers with the denominator = 6 

Therefore, 6/6 = 1, 12/6 = 2

Now, we can note the five rational numbers between 6/6 and 12/6 as:

7/6, 8/6, 9/6, 10/6, 11/6

Question 2: Express whether the following statement is true or false and give reasons for your answer.

 Every natural number is a whole number.

Answer 2: This statement is true

Natural numbers- Numbers beginning from 1 to infinity (without fractions or decimals)

That is, The required natural numbers= 1,2,3,4…

Whole numbers- Numbers beginning from 0 to infinity (without fractions or decimals)

That is, the required whole numbers= 0,1,2,3…

Or, we can express that whole numbers include all the elements of natural numbers and zero.

Each natural number is a whole number; nevertheless, each whole number is not a natural number.

Question 3: Find five rational numbers between ?/? and ?/?.

Answer 3: Consider 3/5 < 2/3

Now x = 3/5, y = 2/3 and n = 5

d=?−?/?+1 = (2/3−3/5)/5+1 = (10−9/15)/6 = 1/90

Rational numbers in between x as well as y are

x + d, x + 2d, x + 3d, x + 4d and x + 5d

Then we obtain,

= 3/5 + 1/90 , 3/5 + 2(1/90) , 3/5 + 3(1/90) , 3/5 + 4(1/90) as well as 3/5+ 5(1/90)

= 54+1/90 , 3/5 + 1/45 , 3/5 + 1/30 , 3/5 + 2/45 and 3/5+1/18

= 55/90, 27+1/45, 18+1/30 , 27+2/45 and 54+5/90

= 11/18 , 28/45 , 19/30 , 29/45 and 59/90

Question 4:The value of 1.999… in the form p/q, where p and q are integers and q ≠ 0, is

(A) 19/10

(B) 1999/1000

(C) 2

(D) 1/9

Answer 4: (C) 2

Explanation:

(A) 19/10 = 1.9

(B) 1999/1000= 1.999

(C) 2

(D) 1/9= 0.111….

Let x = 1.9999….. — ( 1 )

Multiply equation ( 1 ) with 10

10x = 19.9999….. — ( 2 )

Subtract equation (1) from equation(2) ,

We get,

9x = 18

x = 18 / 9

x = 2

Therefore,

x = 1.9999… = 2

Hence, (C) is the correct option.

Question 5: Find out the required six rational numbers between 3 and 4.

Answer 5:There are an infinite number of rational numbers between 3 and 4.

As we have to see 6 rational numbers between 3 and 4, multiplying both the given numbers 3 and 4, with 6+1 = 7 (or any other number greater than 6)

That is, 3 × (7/7) = 21/7

and, 4 × (7/7) = 28/7. The numbers between 21/7 and 28/7 will be rational and will fall between 3 and 4.

Accordingly, 22/7, 23/7, 24/7, 25/7, 26/7, and 27/7 are the required 6 rational numbers between 3 and 4.

Question 6: Express whether the subsequent statements are true or false. Give explanations for your answers.

                  (i) Every required natural number is a whole number.

                  (ii) Every required integer is a whole number.

                  (iii) Each and every required rational number is a whole number.

Answer 6: (i) Every natural number is a whole number.

This statement is true.

Natural numbers- Numbers beginning from 1 to infinity (without fractions or decimals)

i.e., Natural numbers= 1,2,3,4…

Whole numbers- Numbers beginning from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3…

Or, we can express that whole numbers have all the elements of natural numbers and zero.

Each natural number is a whole number; nevertheless, each whole number is not a natural number.

(ii) Every integer is a whole number.

This statement is false.

Integers: Integers are numbers containing positive, negative and 0, excluding fractional and decimal numbers.

that is, integers= {…-4,-3,-2,-1,0,1,2,3,4…}

Whole numbers- The numbers starting from 0 to infinity (without fractions or decimals)

that is, Whole numbers= 0,1,2,3….

Therefore, we can say that integers include whole and negative numbers.

Each whole number is an integer; however, every integer is not a whole number.

(iii) Each and every rational number is a whole number.

This statement is false.

Rational numbers: All the numbers in the form p/q, where p and q are integers and q≠0.

that is ., Rational numbers = 0, 19/30 , 2, 9/-3, -12/7…

Whole numbers- The numbers starting from 0 to infinity (without fractions or decimals)

that is Whole numbers= 0,1,2,3….

Consequently, we can express that integers contain whole and negative numbers.

Every whole number is rational. Nevertheless, every rational number is not a whole number.

Question 5: Find out five rational numbers between 3/5 and 4/5.

Answer 5: We have to find five rational numbers between 3/5 and 4/5.

So, let us write the given numbers by multiplying with 6/6 (here 6 = 5 + 1)

Now,

3/5 = (3/5) × (6/6) = 18/30

4/5 = (4/5) × (6/6) = 24/30

Thus, the required five rational numbers will be: 19/30, 20/30, 21/30, 22/30, 23/30

Question 6:The decimal representation of a given rational number cannot be

(A)It is terminating

(B)It is non-terminating

(C)It is a non-terminating repeating

(D)It is non-terminating non-repeating

Answer 6: (D) It is non-terminating non-repeating

Explanation:

The required decimal representation of a given rational number cannot be non-terminating and non-repeating.

Hence, (D) is the correct option.

Question 7: Express whether the following statement is true or false and give reasons for your answer.

 Every rational number is a whole number.

Answer 7: This is a false statement.

Rational numbers: All numbers in the form p/q, where p and q are integers and q≠0.

that is, Rational numbers = 0, 19/30 , 2, 9/-3, -12/7…

Whole numbers- The numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3….

Thus, we can say that integers include whole and negative numbers.

All the given whole numbers are rational. Regardless, all rational numbers are not whole numbers.

Question 8: Between two rational numbers

(A) there are no rational numbers

(B) there is exactly one rational number

(C) there are infinitely numerous rational numbers

(D) there is only a rational number and no irrational number

Answer 8: (C) there are infinitely many rational numbers

Explanation:

Between the two given rational numbers, there are infinitely many rational numbers.

Hence, (C) is the correct option.

Question 9: Value of (256)⁰˙¹⁶ x (256)⁰˙⁰⁹ is ______.

Answer 9:    (256)⁰˙¹⁶ x (256)⁰˙⁰⁹ = (256)⁰˙¹⁶ ⁺ ⁰˙⁰⁹

                                                     =(256)⁰˙²⁵

                                                     =(256) ⁽¹៸⁴⁾

                                                     =(4⁴) ⁽¹៸⁴⁾

                                                    =4

Question 10: 2√3 + √3 is equal to

(A) 2√6

(B) 6

(C) 3√3

(D) 4√6

Answer 10: (C) 3√3

Explanation:

2√3 + √3

Taking √3 commons,

We get,

√3(2+1) = √3(3) = 3√3

Hence, (C) is the correct option.

Question 11: The required value of 1.999… in the form p/q, where p and q are integers and q ≠ 0, is

(A) 19/10

(B) 1999/1000

(C) 2

(D) 1/9

Answer 11: (C) 2

Explanation:

(A) 19/10 = 1.9

(B) 1999/1000= 1.999

(C) 2

(D) 1/9= 0.111….

Let x = 1.9999….. — ( 1 )

Multiplying equation ( 1 ) with 10

10x = 19.9999….. — ( 2 )

Subtracting equation (1) from equation(2),

We obtain,

9x = 18

x = 18 / 9

x = 2

Thus,

x = 1.9999… = 2

Therefore, (C) is the correct option.

Question 12: Are the square roots of all the positive integers irrational? If not, give an illustration of the square root of a number that is rational.

Answer 12: No, since the square root of a positive integer 16 is equal to 4. Here, 4 is a rational number.

Question 13: Show that 0.333…. can be expressed in the form of p/q where p and q are integers and q ≠ 0.

Answer 13: Let x = 0.3333…. 

Multiply with 10,

10x = 3.3333…

Now, 3.3333… = 3 + x (as we assumed x = 0.3333…)

Thus, 10x = 3 + x

10x – x = 3

9x = 3

x = 1/3

Therefore, 0.3333… = 1/3. Here, 1/3 is in the form of p/q and q ≠ 0

Question 14: How many irrational numbers lie between √? and√?? Find any three irrational numbers lying between √? and√?.

Answer 14: We know that

√2 = 1.414213562 ….

√3= 1.7320508075 ….

Thus, the three irrational numbers which lie between √2 and√3 are

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

Question 15: The product of any of the two irrational numbers is

(A) It is always an irrational number

(B) It is always a rational number

(C) It is always an integer

(D) It is sometimes rational and sometimes irrational

Answer 15: (D)It is sometimes rational and sometimes irrational

Explanation:

The product of any of two irrational numbers is sometimes rational and sometimes irrational.

Accordingly, (D) is the correct option.

Question 16: Let x and y be the given rational and irrational numbers, respectively. Is x + y necessarily an irrational number? Provide an illustration in aid of your answer.

Answer 16: Yes, if x and y are rational and irrational numbers, respectively, then x+ y is an irrational number.

For example,

Let x = 5 and y = √2.

Here, x+y = 5 + √2 = 5 + 1.414… = 6.414…

Now, 6.414 is the required non-terminating and non-recurring decimal and hence is a given irrational number.

Thus, x + y is an irrational number.

Question 17: Express whether the following statement is true or false and give reasons for your answer.

 Every irrational number is a real number.

Answer 17: This statement is true.

Irrational Numbers – A given number is irrational if it cannot always be written in the p/q, where p, as well as q, are integers and q ≠ 0.

That is, The required irrational numbers = ‎π, e, √3, 5+√2, 6.23146…. , 0.101001001000….

Real numbers – The collection of rational and irrational numbers is known as real numbers.

i.e., Real numbers = √2, √5, 0.102…

Every irrational number is a real number. However, every real number is not an irrational number.

Question 18: Which of the following is irrational?

(A) √4/√9

(B) √12/√3

(C) √7

(D) √81

Answer 18:  (C) √7

Explanation:

(A) √4/√9 = 2/3

(B) √12/√3 = 2√3/√3 = 2

(C) √7 = 2.64575131106

(D) √81 = 9

Here, (C) √7 = 2.64575131106 is a non-terminating decimal expansion.

Hence, (C) is the correct option.

Question 19: Let x and y be the required rational and irrational numbers, respectively. x + y is a given irrational number.

Answer 19:  Yes, if x and y are the required rational and irrational numbers, respectively, then x + y is an irrational number.

For example,

Let x = 5 and y = √2.

Now, x+y = 5 + √2 = 5 + 1.414… = 6.414…

Then, 6.414… is a non-terminating and non-recurring decimal and thus is an irrational number.

Accordingly, x + y is a given irrational number.

Question 20: Let x be the given rational, and y be irrational. Is xy necessarily irrational? Give a reason for your answer with an example.

Answer 20: No, if x is a rational number and y is an irrational number, then xy is not necessarily a given irrational number. It can be the given rational if x = 0, which is a rational number.

For Example:

Let y = √2, which is irrational.

Considering x = 2, which is rational.

Now, x × y = 2 × √2 = 2√2, which is irrational.

Considering x = 0, which is the given rational.

Then xy = 0 × √2 = 0, which is the given rational.

∴ we can also conclude that the product of a given rational and the required irrational number is still irrational, only if the rational number is not zero.

Question 21: Write down the correct answer in each of the following:

1. Every rational number is

(A) a natural number

(B) an integer

(C) a real number

(D) a whole number

Answer 21: C) a real number

Explanation:

We observe that rational and irrational numbers taken together are known as real numbers. Hence, every real number is either a rational number or an irrational number. Therefore, every rational number is a real number.

Hence, (C) is the correct option.

Question 22: Examine whether the following numbers are rational or irrational:

(i) (5 – √?) (5 +√?)

(ii) (√? + ?) ?

(iii) ?√??/(?√??−?√???)

(iv) √? + ?√?? − ?√?

Answer 22: (i) (5 – √5) (5 +√5)

As per the given formula,

 ?² − ?² = (? + ?)(? − ?)

= (5)²– (√5)²

So we acquire

= 25 – 5

= 20

Therefore, (5 – √5) (5 +√5) is rational.

(ii) (√3 + 2)²

As per the given formula (? + ?)2 = ?2 + ?2 + 2??

= (√3)² + 2 × 2 × √3 + (2)²

On further calculation

= 3 + 4√3 + 4

So we obtain

= 7 + 4√3

Therefore, (√3 + 2) 2 is irrational.

(iii)2√13/(3√52−4√117)

We can also write it as

=2√13/(3√4×13−4√9×13)

On further calculation

=2√13/(3×2√13−4×3√13)

So we obtain

=2√13/(6√13−12√13)

By taking out all the common terms

=2√13/−6√13

By the given division

= -1/3

Thus

2√13/(3√52−4√117) is rational.

(iv) √8 + 4√32 − 6√2

It can also be written as

= √4 × 2 + 4√16 × 2 – 6√2

 we acquire

= 2√2 + 16√2 – 6√2

On further calculation

= 12√2

Therefore √8 + 4√32 − 6√2 is irrational.

Question 23: The number of rational numbers between 15 and 18 is finite.

Answer 23: The given statement is false. There lie infinitely many rational numbers between any two rational numbers. Hence, a number of rational numbers between 15 and 18 are infinite.

Question 24:Find out three different irrational numbers between the rational numbers 5/7 and 9/11.

Answer 24: The given two rational numbers are 5/7 and 9/11.

5/7 = 0.714285714…..

9/11 = 0.81818181……

Hence, the three irrational numbers between 5/7 and 9/11 can be:

0.720720072000…

0.730730073000…

0.808008000…

Question 25: A rational number between the following numbers √2 and √3 is

(A) (√2+√3)/2

(B) (√2. √3)/2

(C) 1.5

(D) 1.8

Answer 25: (C) 1.5

Explanation:

√2 =1.4142135…. and √3 =1.732050807….

(A) (√2+√3)/2 = 1.57313218497… is a non-terminating as well as a non-recurring decimal, and hence it is an irrational number.

(B) (√2. √3)/2 = 1.22474487139… is a non-terminating as well as a non-recurring decimal and hence is an irrational number.

(C) 1.5 is a terminating decimal and hence is a rational number.

(D) 1.8 is a terminating decimal and hence is a rational number.

Now both 1.5 and 1.8 are rational numbers. However, 1.8 does not lie in between √2 =1.4142135…. and √3 =1.732050807…. Although 1.5 lies in between √2 =1.4142135…. and √3 =1.732050807….

Therefore, (C) is the correct option.

Question 26: There are infinitely multiple integers between any two integers.

Answer 26: The provided statement is false. Assume two integers, 3 and 4. There are no integers between 3 and 4.

Question 27: Find out which of the variables x, y, z and u represent rational numbers and which irrational numbers:

(i) x² = 5

(ii) y² = 9

(iii) z² = .04

(iv) ?² = 17/4

Answer 27: (i) x² = 5

On solving, we get

⇒ x = ± √5

Hence, x is an irrational number.

(ii) y² = 9

On solving, we get

⇒ y = ± 3

Hence, y is a rational number.

(iii) z² = .04

On solving, we get

⇒ z = ± 0.2

Hence, z is a rational number.

(iv) u² = 17/4

On solving, we get

⇒ u = ± √17/2

√17 is irrational.

Hence, u is an irrational number

Question 28: Are the square roots of all positive integers irrational? If not, give an example of the given square root of a number that is a rational number.

Answer 28: No, the square roots of all  positive integers are not irrational.

For example,

√4 = 2 is rational.

√9 = 3 is rational.

Therefore, the square roots of positive integers 4 and 9 are not irrational. ( 2 and 3, respectively).

Question 29: You comprehend that 17 = 0.142857¯. Can you expect the decimal expansion of 27, 137, 47, 57, and 67 without doing the long division? If so, how?

Answer 29: We are provided with

 17 = 0.142857¯.

 27 = 2 x 17 = 2 x (0.142857¯) = 0.285714¯

37 = 3 x 17 = 3 x (0.142857¯) = 0.428571¯

47 = 4 x 17 = 4 x (0.142857¯) = 0.571428¯

57 = 5 x 17 = 5 x (0.142857¯) = 0.714285¯

67 = 6 x 17 = 6 x (0.142857¯) = 0.857142¯

Accordingly, we can predict the decimal expansions of the given rational numbers without actually doing the long division.

Question 30: √12/√3 is not the required rational number, as √12 and √3 are not integers.

Answer 30: √12/√3 = √4​=2

2 is a rational number.

The given statement is false.

Question 31: The decimal expansion of the given number √2 is

(A)It is a finite decimal

(B) 1.41421

(C) It is a non-terminating recurring

(D) It is a non-terminating non-recurring

 

Answer 31: (D) It is a non-terminating non-recurring

Explanation:

The required decimal expansion of the number √2 = 1.41421356237…

Thus, (D) is the correct option.

Question 32: Find three rational numbers between

(i) –1 and –2

(ii) 0.1 and 0.11

(iii) 5/7 and 6/7

(iv) 1/4 and 1/5

Answer 32: (i) –1 and –2

The three rational numbers between –1 and –2 are –1.1, –1.2 and –1.3.

(ii) 0.1 and 0.11

The three given rational numbers between 0.1 and 0.11 are 0.101, 0.102 and 0.103.

(iii)5/7 and 6/7

5/7 can also be written as (5 × 10)/(7 × 10) = 50/70

Similarly,

6/7 can also be written as (6 × 10)/(7 × 10) = 60/70

The three given rational numbers between 5/7 and 6/7 = three rational numbers between 50/70 and 60/70.

The three given rational numbers between 5/7 and 6/7 are 51/70, 52/70, and 53/70.

(iv)1/4 and 1/5

Here, according to the question,

LCM of 4 and 5 is 20.

Let us make the denominators common, 80.

(4 × 20) = 80 and (5 × 16) = 80

Thus,

1/4 can also be written as (1 × 20)/(4 × 20) = 20/80

Likewise,

1/5 can be written as (1 × 16)/(5 × 16) = 16/80

The three given rational numbers between 1/4 and 1/5 = three rational numbers between 16/80 and 20/80.

Thus, the three given rational numbers are 17/80, 18/80 and 19/80.

Question 33: Add 2√2+ 5√3 and √2 – 3√3.

Answer 33: (2√2 + 5√3) + (√2 – 3√3)

= 2√2 + 5√3 + √2 – 3√3

= (2 + 1)√2 + (5 – 3)√3

= 3√2 + 2√3

Question 34: Insert a given rational number and an irrational number between the following:

(i) 2 and 3

(ii) 0 and 0.1

(iii) 1/3 and 1/2

(iv) – 2/5 and 1/2

(v) 0.15 and 0.16

(vi) √2 and √3

(vii) 2.357 and 3.121

(viii) .0001 and .001

(ix) 3.623623 and 0.484848

(x) 6.375289 and 6.375738.

Answer 34: (i) 2 and 3

So, a rational number between 2 and 3 = 2.5

And, the required irrational number between 2 and 3 = 2.040040004…

(ii) 0 and 0.1

Then, the given rational number between 0 and 0.1 = 0.05

And, the required irrational number between 0 and 0.1 = 0.007000700007…

(iii) 1/3 and 1/2

LCM of 3 and 2 is 6.

1/3 = 0.33

1/3 can also be written as (1 × 20)/(3 × 20) = 20/60

½ = 0.5

1/2 can also be written as (1 × 30)/(2 × 30) = 30/60

So, a given rational number between 1/3 and 1/2 = 25/60

And, the required irrational number between 1/3 and 1/2 = irrational number between 0.33 and 0.5 = 0.414114111…

(iv) – 2/5 and 1/2

LCM of 5 and 2 is 10.

-2/5 = -0.4

-2/5 can also be written as (-2 × 2)/(5 × 2) = -4/10

1/2 = 0.5

1/2 can also be written as (1 × 5)/(2 × 5) = 5/10

So, The given rational number between -2/5 and 1/2 = rational number between -4/10 and 5/10 = 1/10

And, The required irrational number between -2/5 and 1/2 = irrational number between -0.4 and 0.5 = 0.414114111…

(v) 0.15 and 0.16

The required rational number between 0.15 and 0.16 = 0.151

The required irrational number between 0.15 and 0.16 = 0.151551555…

(vi) √2 = 1.41 and √3 = 1.732

The required rational number between √2 and √3 = rational number between 1.41 and 1.732 = 1.5

The required irrational number between √2 and √3 = irrational number between 1.41 and 1.732 = 1.585585558…

(vii) 2.357 and 3.121

The required rational number between 2.357 and 3.121 = 3

The required irrational number between 2.357 and 3.121 = 3.101101110…

(viii) .0001 and .001

The required rational number between .0001 and .001 = 0.00011

The required irrational number between .0001 and .001 = 0.0001131331333…

(ix) 3.623623 and 0.484848

The rational number between 3.623623 and 0.484848 = 1

The irrational number between 3.623623 and 0.484848 = 1.909009000…

(x) 6.375289 and 6.375738.

The required rational number between 6.375289 and 6.375738 = 6.3753

The required irrational number between 6.375289 and 6.375738 = 6.375414114111…

Question 35: Express the following in the form p/q, where p and q are integers and q ≠ 0 :

 0.2

Answer 35:  0.2

We know that,

0/2 can be written as,

0.2 = 2/10 = 1/5

Question 36: Show 0.99999…. in the form p/q. 

Answer 36: Assuming that x = 0.9999…..(1)

Multiplying both sides by 10,

10x = 9.9999…. (2)

Eq.(2) – Eq.(1), we obtain

10x = 9.9999…

-x = -0.9999…

___________

9x = 9

x = 1

The given difference between 1 and 0.999999 is 0.000001, which is also negligible.

Therefore, we can also conclude that 0.999 is too much near 1. 

Thus, 1 as the answer can be justified.

Question 37: Show the following in the form p/q, where p and q are integers and q ≠ 0:

 0.888…

Answer 37: 0.888…

Assume that ? = 0.888 …

⇒ ? = 0.8 ……………. Eq.(1)

Multiply L.H.S and R.H.S by 10,

We get

10 ? = 8.8 ……………. Eq.(2)

Subtracting equation (1) from (2),

We get

10 ? − ? = 8.8 − 0.8

⇒ 9? = 8

⇒ ? = 8/9

Question 38:Simplify each of the following expressions:

(i) (3+√3)(2+√2)

(ii) (3+√3)(3-√3 )

(iii) (√5+√2)2

(iv) (√5-√2)(√5+√2)

Answer 38: (i) (3+√3)(2+√2)

(3+√3)(2+√2 )

Open the required brackets. We obtain

 (3×2)+(3×√2)+(√3×2)+(√3×√2)

= 6+3√2+2√3+√6

(ii) (3+√3)(3-√3 )

(3+√3)(3-√3 ) = 32-(√3)2 = 9-3

= 6

(iii) (√5+√2)2

(√5+√2)2 = √52+(2×√5×√2)+ √22

= 5+2×√10+2 = 7+2√10

(iv) (√5-√2)(√5+√2)

(√5-√2)(√5+√2) = (√52-√22) = 5-2 = 3

Question 39: Simplify (√3+√7) (√3-√7). 

Answer 39: (√3 + √7)(√3 – √7)

Using the following identity (a + b)(a – b) = a2 – b2,

(√3 + √7)(√3 – √7) = (√3)2 – (√7)2

= 3 – 7

= -4

Question 40: Observe the example of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no common factors other than 1 and have terminating decimal representation (expansions). What property q must satisfy?

Answer 40: We see that when q is 2, 4, 5, 8, 10… Then the decimal expansion terminates. For example:

1/2 = 0. 5, denominator q = 21

7/8 = 0. 875, denominator q =23

4/5 = 0. 8, denominator q = 51

We can notice that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.

Question 41: Show the following in the form p/q, where p and q are integers and q ≠ 0:

 0.2555…

Answer 41:  0.2555…

Assume that ? = 0.2555 …

⇒ x = 0.25 ……………. Eq. (1)

Multiply L.H.S and R.H.S by 10,

We get

10 x = 2.5 ……………. Eq. (2)

Multiply L.H.S and R.H.S by 100,

We get

100 x = 25.5 …………. Eq. (3)

Subtracting equation (2) from (3),

We get

100 x-10x = 25.5 – 2.5

⇒ 90? = 23

⇒ ? = 23/90

Question 42: Remember that π is the ratio of a circle’s circumference (say c) to its diameter (say d). That is, π =c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

Answer 42: There is no contradiction. When we calculate a value with a scale, we only acquire an approximate value. We never get an exact value. Thus, we may not realise whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…

Question 43: Show the following in the form p/q, where p and q are integers and q ≠ 0:

.00323232…

Answer 43: .00323232…

Let ? = 0.00323232 …

 x = 0.0032 ………….…. (1)

Multiplying the required L.H.S and R.H.S by 100,

We acquire,

100x = 0.32 ……………. (2)

Multiplying the required L.H.S and R.H.S by 10000,

We acquired,

10000 x = 32.32 …………. Eq. (3)

Subtract the given equation (2) from (3),

We obtain

10000 x-100x = 32.32 – 0.32

9900? = 32

 ? = 32/9900

     = 8/2475

Question 44: Rationalise the denominator of 1/[7+3√3].

Answer 44: 1/(7 + 3√3)

By rationalising the denominator,

= [1/(7 + 3√3)] [(7 – 3√3)/(7 – 3√3)]

= (7 – 3√3)/[(7)2 – (3√3)2]

= (7 – 3√3)/(49 – 27)

= (7 – 3√3)/22

Question 45: Rationalise the denominators of the following:

(i) 1/√7

(ii) 1/(√7-√6)

(iii) 1/(√5+√2)

(iv) 1/(√7-2)

Answer 45: (i) 1/√7

Multiply and divide 1/√7 by √7

(1×√7)/(√7×√7) = √7/7

1. ii) 1/(√7-√6)

Multiplying and dividing 1/(√7-√6) by (√7+√6)

[1/(√7-√6)]×(√7+√6)/(√7+√6) = (√7+√6)/(√7-√6)(√7+√6)

= (√7+√6)/√72-√62 [The given denominator is acquired by the property, (a+b)(a-b) = a2-b2]

= (√7+√6)/(7-6)

= (√7+√6)/1

= √7+√6

(iii) 1/(√5+√2)

Multiplying and dividing 1/(√5+√2) by (√5-√2)

[1/(√5+√2)]×(√5-√2)/(√5-√2) = (√5-√2)/(√5+√2)(√5-√2)

= (√5-√2)/(√52-√22) [The given denominator is acquired by the property, (a+b)(a-b) = a2-b2]

= (√5-√2)/(5-2)

= (√5-√2)/3

(iv) 1/(√7-2)

Multiplying and dividing 1/(√7-2) by (√7+2)

1/(√7-2)×(√7+2)/(√7+2) = (√7+2)/(√7-2)(√7+2)

= (√7+2)/(√72-22) [The given denominator is acquired by the property, (a+b)(a-b) = a²-b²]

= (√7+2)/(7-4)

= (√7+2)/3

Question 46: Write the three numbers whose decimal expansions are non-terminating non-recurring.

Answer 46: We know that all irrational numbers are non-terminating and non-recurring. The three given numbers with decimal expansions that are non-terminating and non-recurring are:

1. √3 = 1.732050807568
2. √26 =5.099019513592
3. √101 = 10.04987562112

Question 48: Find (i) 64

                            (ii) 32 

                            (iii) 125

Answer 48: (i) 64 = 8 x 8 = 8²

∴ (64)¹៸² = (82)¹៸² = 8² x ¹៸² = 8 [(a^m)ⁿ = a^m x ⁿ]

(ii) 32 = 2 x 2x 2 x 2 x 2 = 2⁵

∴ (32)¹៸⁵= (2⁵)¹៸⁵= 2⁵ x ¹៸⁵ = 2 [(a^m)ⁿ= a^m x ⁿ]

(iii) 125 = 5 x 5 x 5 = 5³

∴ (125)¹៸³ = (5³)¹៸³ = 5³ x ¹៸³ = 5 [(a^m)ⁿ = a^m x ⁿ]

Question 49: Classify the following numbers as rational or irrational according to their type:

                        (i)√23

                         (ii)√225

                        (iii) 0.3796

                        (iv) 7.478478

                        (v) 1.101001000100001…

Answer 49:

 (i)√23 = 4.79583152331…

Since the number is non-terminating and non-recurring, thus, it is an irrational number.

(ii)√225

√225 = 15 = 15/1

The number can also be represented in p/q form. It is a rational number.

(iii) 0.3796

The number 0.3796 is terminating. It is the required rational number.

(iv) 7.478478

The number,7.478478, is non-terminating and, at the same time, recurring; it is the required rational number.

(v) 1.101001000100001…

The number 1.101001000100001… is non-terminating non-repeating (non-recurring). It is an irrational number.

Question 50: What is the actual product of a rational and an irrational number?

1. a) It is always an integer
2. b) It is always a rational number
3. c) It is always an irrational number
4. d) It is sometimes rational and sometimes irrational

Answer 50:  Option (c)

Explanation:

The product of any rational and irrational number is always an irrational number.

For example, if 2 is a rational number and √3 is irrational. Thus, 2√3 is always an irrational number.

Question 51: Show the following in the form p/q, where p and q are integers and q ≠ 0:

 .404040…

Answer 51:  .404040…

Let ? = 0.404040 …

⇒ ? = 0. 40 ………..….…. (1)

Multiplying the given L.H.S and R.H.S by 100,

We obtain

100 ? = 40.40 ……….…. (2)

Subtract the given equation (1) from (2),

We acquire

100 ? − ? = 40.40 − 0.40

 99? = 40

 ? = 40/99

Question 52: Express whether the following statements are true or false. Explain your answers

                   (i) Each and every irrational number is a real number.

                   (ii) Each and every point on the number line is of the form √m, where m is a natural number.

                   (iii) Each real number is an irrational number.

Answer 52: (i) Each and every irrational number is a real number.

This statement is true.

Irrational Numbers – A given number is irrational if it cannot be written in the p/q, where p, as well as q, are integers and q ≠ 0.

That is, Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….

Real numbers – The collection of rational and irrational numbers is known as real numbers.

that is, Real numbers = √2, √5, , 0.102…

Each and every irrational number is a real number. However, every real number is not an irrational number.

(ii) Each and every point on the number line is of the form √m, where m is a natural number.

The statement is false.

Since, as per the law, a negative number cannot be represented as square roots.

For example, √9 =3 is a natural number.

While √2 = 1.414 is not a natural number.

Besides, we learn that there are negative numbers on the number line, but when we carry the root of a negative number, it evolves into a complex number and not a natural number.

Example., √-7 = 7i, where i = √-1

The view that every point on the number line is of the form √m, where m is a natural number, is false.

(iii) Each real number is an irrational number.

The statement is false.

The real numbers contain both irrational and rational numbers. Accordingly, every real number cannot be an irrational number.

Real numbers – The collection of rational and irrational numbers is known as real numbers.

that is, Real numbers = √2, √5, , 0.102…

Irrational Numbers – A given number is irrational if it cannot be written in the p/q, where p, as well as q, are integers and q ≠ 0.

That is, Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….

Each and every irrational number is a real number. Nonetheless, every real number is not irrational.

Question 53:Classify the number as rational or irrational:

 2 –√5

Answer 53: We know that √5 = 2.2360679…

Now, 2.2360679…is non-terminating and non-recurring.

Then, substitute the value of √5 in 2 –√5, and we get,

2-√5 = 2-2.2360679… = -0.2360679

The required number – 0.2360679…, is non-terminating non-recurring, and 2 –√5 is an irrational number.

Question 55: What is the value of (256)⁰˙¹⁶ X (256)⁰˙⁰⁹?

1. a) 4
2. b) 16
3. c) 64
4. d) 256.25

Answer 55:  Option (a)

(256)⁰˙¹⁶ x (256)⁰˙⁰⁹ = (256)(⁰˙¹⁶ ⁺⁰˙⁰⁹)

= (256)⁰˙²⁵

= (256)(25/100)

= (256)(1/4)

= (44)(1/4)

= 44(1/4)

= 4

Question 56: Classify the number as rational or irrational:

 (3 +√23)- √23

Answer 56: (3 +√23) –√23 = 3+√23–√23

= 3

= 3/1

The number 3/1 is in the given p/q form, (3 +√23)- √23 is rational.

Question 57: Represent √ 9.3 on the number line.

Answer 57: Mark a line segment AB = 9.3 units and expand it to C such that BC = 1 unit.

Locate the midpoint of AC and label it as O.

Drawing a semicircle taking O as the centre and AO as the radius. Draw BD ⊥ AC.

Drawing an arc taking B as centre and BD as radius met AC produced at E, so BE = BD = √ 9.3 units.

image source: https://www.learncbse.in/wp-content/uploads/2020/10/NCERT-Solutions-for-Class-9-Maths-Chapter-1-Number-Systems-Ex-1.5-Q4.png

Question 58: Show 0.00323232… in the form p/q, where p and q are integers and q ≠ 0 and select the correct option.

Answer 58: Let ? = 0.00323232…

⇒ x = 0.003232… Eq. (1)

Multiplying L.H.S and R.H.S by 100,

We get,

100x = 0.32…… Eq. (2)

Multiplying L.H.S and R.H.S in Eq 1 by 10000,

We get

10000 x = 32.3232……. Eq. (3)

Subtract equation (2) from (3),

We obtain

10000 x-100x = 32.32.. – 0.32..

⇒ 9900? = 32

⇒ ? = 32/9900 

        = 8/2478

Question 59: Which is the rational number between 0.15 and 0.16?

Answer 59: 0.15 can be written as 0.150, and 0.16 can be written as 0.160

The rational number between 0.15 and 0.16 is 0.151.

Question 60: Classify the number as rational or irrational:

 (i)2√7/7√7

 (ii)1/√2

Answer 60:

 (i)2√7/7√7 = ( 2/7)× (√7/√7)

We understand that (√7/√7) = 1

Therefore, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7

Considering the number 2/7 is in p/q form, 2√7/7√7 is rational.

(ii) 1/√2

Multiplying and dividing the given numerator and denominator by √2, we acquire,

(1/√2) × (√2/√2) = √2/2 ( after √2×√2 = 2)

We know that √2 = 1.4142…

Then, √2/2 = 1.4142/2 = 0.7071..

Since the number 0.7071..is no.

Question 61: Find a rational number between

(i)?/? and ?/?

(ii) ?. ? ??? ?. ?

(iii) -1 and ?/?

(iv) -3/? and – ?/?

(v) ?/? and ?/?

Answer 61: (i) Consider the following x = 3/8 and y = 2/5

so

3/8 < 2/5

The rational number that lies in-between x and y = 1/2 (x+y)

Substitute the values of x and y =

1/2(3/8+2/5)

On further calculating = 1/2[(15+16)/40]

So we obtain =

1/2 x 31/40 = 31/80

Thus the rational number which lies between 3/8 and 2/5 is 31/80

Consider the following x = 1.3 and y = 1.4

So 1.3 < 1.4

The rational number in-between x and y = 1/2 (1.3+1.4)

Thus we obtain

= 1/2 x 2.7= 1.35

Hence the rational number which lies in-between 1.3 and 1.4 is 1.35.

(ii) Consider the following x = -1 and y = ?/?

Then -1 < 1/2

The rational number in-between x and y = 1/2(-1+1/2)

Thus on further calculating, we obtain = 1/2(−2+1/2)

Hence we obtain, = 1/2 x -1/2 = -1/4

Thus the rational number which lies in between -1 and 1/2 is – ¼

(iii) Consider x = −3/4 and y = −2/5

Then − 3/4 < −2/5

The rational number in-between x and y = 1/2((−3/4) +(−2/5))

On further calculation=1/2(−15−18/20)

So we get

=1/2x−23/20= −23/40

Thus the rational number which lies in between – 3/4 and -2/5 is −23/40

(iv) Consider the following x = 1/9 and y =2/9 

So,

1/9 < /2/9

Rational number in-between x and y =

1/2(1/9+2/9)

So we get, =

1/2 x 3/9 = 1/6

Thus the rational number which lies in between 1/9 and 2/9 is 1/6.

Question 62: Imagine 3.765 on the number line, using subsequent magnification.

Answer 62: 3.765 lies between 3 and 4.

Question 63: A glance at some examples of rational numbers in the required form p/q (q ≠ 0), where p as well as q are integers with no given common factors other than 1 and have terminating decimal representations (expansions). Can you assume what property q must satisfy?

Answer 63: When q is 2, 4, 5, 8, 10… Then the decimal expansion terminates. For example:

1/2 = 0. 5, denominator q = 21

7/8 = 0. 875, denominator q =23

4/5 = 0. 8, denominator q = 51

We can see that the terminating decimal may be acquired in the situation where prime factorization of the given denominator of the provided fractions has the power of only 2 or only 5 or both.

Question 64: Are the square roots of all the positive integers irrational? If not, provide an example of the square root of a number that is a rational number.

Answer 64: No, the square roots of all the positive integers are not irrational.

For example,

√4 = 2 is rational.

√9 = 3 is rational.

Therefore, the square roots of the given positive integers 4 and 9 are not irrational. ( 2 and 3, respectively).

Question 65: Represent the following in the form p/q, where p and q are integers and q is not equal to 0.

        (i)   

        (ii)  

Answer 65:

(i) Assume that  x = 0.666…

Then,10x = 6.666…

10x = 6 + x

9x = 6

x = 2/3

(ii) Assume that  x = 0.001001…

Then, 1000x = 1.001001…

1000x = 1 + x

999x = 1

x = 1/999

Question 66:Classify the following number as rational or irrational.

(i) 2 – √5 

(ii)    (3 + √23) – √23 

(iii)  2√7/7√7 

 (iv)      √12

 (v)2π = 2 x π 

 Answer 66:

(i)  There  is a difference between a rational and an irrational number.

∴ 2 – √5 is an irrational number.

(ii) 3 + √23 –√ 23= 3 + √23– √23= 3

which is a rational number.

(iii) Since, 2√7/7√7 = 2×√7/7×7√ = 2/7 , which is a rational number.

(iv) ∵ The quotient of a rational and irrational number is an irrational number.

∴ √12 is an irrational number.

(v) ∵ 2π = 2 x π = The product of a given rational and a given irrational number is an irrational number.

∴ 2π is the required irrational number.

Question 67: 10.124124… is an irrational number.

Answer 67:

10.124124……

10.124124… is a decimal expansion which is non-terminating but recurring. Hence, it is a rational number.

Question 68: Classify the following numbers as rational or irrational according to their type:

                (i)√23

                 (ii)√225

                 (iii) 0.3796

                (iv) 7.478478

                (v) 1.101001000100001…

Answer 68: (i)√23

√23 = 4.79583152331…

The number is non-terminating and non-recurring. Therefore, it is an irrational number.

(ii)√225

√225 = 15 = 15/1

The number can also be represented in p/q form. It is a rational number.

(iii) 0.3796

The number 0.3796 is terminating. It is a rational number.

(iv) 7.478478

The number 7.478478 is non-terminating but recurring. It is a rational number.

(v) 1.101001000100001…

The number 1.101001000100001… is non-terminating non-repeating (non-recurring). It is an irrational number.

Question 69: Let x be the required rational number, and y be the required irrational number. Is xy necessarily irrational? Justify your answer with an example.

Answer 69: A number is rational if it can be written as a/b, where both the numerator and the denominator are integers.

It is said to be irrational if a number cannot be readily furthered to any fraction of a natural number and an integer. Irrational numbers’ decimal expansion is neither finite nor recurrent.

If x is a rational number and y is an irrational number, then xy is not necessarily a given irrational number. It can also be rational if x = 0, which is also a rational number.

For Example:

Let _y_ = √2, which is also irrational.

Taking into consideration x = 2, which is also rational.

Now, x × y = 2 × √2 = 2√2, which is also irrational.

Taking into consideration x = 0, which is also rational.

Now, _xy_ = 0 × √2 = 0, which is rational.

∴, we can suppose that the given product of a rational and an irrational number is always irrational,

just if the rational number is not zero.

Question 70: Simplify y² = 9

 The value of y is an r

Answer 70:

 y² = 9

On solving, we get

⇒ y = ± 3

Hence, y is a rational number.

Question 71: Show 0.888… in the form p/q, where p and q are integers and q ≠ 0 and select the correct option.

Answer 71: 0.888…

Suppose that ? = 0.888…

⇒ ? = 0.888 ……………. Eq I

Multiplying L.H.S and R.H.S by 10,

We get

10 ? = 8.888……………. Eq II

Subtract equation I from II,

We obtain

10 ? − ? = 8.888.. − 0.888..

⇒ 9? = 8

⇒ ? = 8/9

Question 72: See which of the variables x, y, z and u represent rational numbers and which irrational

numbers:

(i) x² = 5

(ii) y² = 9

(iii) z² = .04

(iv) ?² = 17/4

Answer 72: (i) x² = 5

On solving, we get

⇒ x = ± √5

Hence, x is an irrational number.

(ii) y² = 9

On solving, we get

⇒ y = ± 3

Hence, y is a rational number.

(iii) z² = .04

On solving, we get

⇒ z = ± 0.2

Hence, z is a rational number.

(iv) u² = 17/4

On solving, we get

⇒ u = ± √17/2

√17 is irrational.

Hence, u is an irrational number

Question 73: Put a rational number and an irrational number between the following:

(i) 2 and 3

(ii) 0 and 0.1

(iii) 1/3 and 1/2

(iv) – 2/5 and 1/2

(v) 0.15 and 0.16

(vi) √2 and √3

(vii) 2.357 and 3.121

(viii) .0001 and .001

(ix) 3.623623 and 0.484848

(x) 6.375289 and 6.375738.

Answer 73: (i) 2 and 3

So, the required rational number between 2 and 3 = 2.5

And, the required irrational number between 2 and 3 = 2.040040004…

(ii) 0 and 0.1

So,The required rational number between 0 and 0.1 = 0.05

And,The required irrational number between 0 and 0.1 = 0.007000700007…

(iii) 1/3 and 1/2

LCM of 3 and 2 is 6.

1/3 = 0.33

1/3 can also be written as (1 × 20)/(3 × 20) = 20/60

½ = 0.5

1/2 can also be written as (1 × 30)/(2 × 30) = 30/60

So, the required rational number between 1/3 and 1/2 = 25/60

And,The required irrational number between 1/3 and 1/2 = irrational number between 0.33 and 0.5 = 0.414114111…

(iv) – 2/5 and 1/2

The Least Common Multiple of the numbers 5 and 2 is 10.

-2/5 = -0.4

-2/5 can also be written as (-2 × 2)/(5 × 2) = -4/10

1/2 = 0.5

1/2 can also be written as (1 × 5)/(2 × 5) = 5/10

now,The required rational number between -2/5 and 1/2 = rational number between -4/10 and 5/10 = 1/10

And,The required irrational number between -2/5 and 1/2 = irrational number between -0.4 and 0.5 = 0.414114111…

(v) 0.15 and 0.16

The required rational number between 0.15 and 0.16 = 0.151

The required irrational number between 0.15 and 0.16 = 0.151551555

(vi) √2 = 1.41 and √3 = 1.732

The required rational number between √2 and √3 = rational number between 1.41 and 1.732 = 1.5

The required irrational number between √2 and √3 = irrational number between 1.41 and 1.732 = 1.585585558…

(vii) 2.357 and 3.121

The required rational number between 2.357 and 3.121 = 3

The required irrational number between 2.357 and 3.121 = 3.101101110…

(viii) .0001 and .001

The required rational number between .0001 and .001 = 0.00011

The required irrational number between .0001 and .001 = 0.0001131331333…

(ix) 3.623623 and 0.484848

The required rational number between 3.623623 and 0.484848 = 1

The required irrational number between 3.623623 and 0.484848 = 1.909009000…

(x) 6.375289 and 6.375738.

The required rational number between 6.375289 and 6.375738 = 6.3753

The required irrational number between 6.375289 and 6.375738 = 6.375414114111…

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