RS Aggarwal Class 9 Real Numbers

RS Aggarwal Class 9 Chapter 1 Solutions

Question 1:

What is the rational numbers? Give 10 examples of rational numbers?

Solution:

The number that can be written in the pq form, where p and q are integers and q 0 are known as rational numbers.

Examples of rational numbers:

(1) 12 (2) 15 (3) 54 (4) 41 = 1 (5) 52 (6) 17

(7) 01 = 0

 

Question 2: Represent Rational Numbers on Number Lines.

Solution 2:

Question 2:

Represent each of the following rational numbers on the number line:

I – 5

II — 3

III — 57

IV– 83

V – 1.3

VI — -24

VII — 236

SOLUTION :

(i) Let

X = 14 and y = 13

Rational number lying between x and y:

12 ( x + y) = 12(14+13)

= 724

(ii) Let:

X = 38 and y = 25

Rational number lying between x and y:

12 ( x + y) = 12(38+25)

= 12(15+1640) = 3180

(iii) Let:

X = 1.3 and y = 1.4

Rational number lying between x and y:

12 (x + y) = 12 (1.3 + 1.4)

= 12 (2.7) = 1.35

(iv) Let:

X = 0.75 and y = 1.2

Rational number lying between x and y;

12 ( x + y ) = 12 ( 0.75 + 1.2)

= 12 (1.95) = 0.975

(v) Let:

X= -1 and y = 12

Rational number lying between x and y:

12 ( x + y ) = 12 (1+12)

= 14

(vi) Let:

X =  34 and y =  25

Rational number lying between x and y:

12 ( x + y ) = 12 (3425)

= 12(15820)=2340

Question 3:

Find three rational numbers lying between 15  and 14

Solution:

Let:

X = 15 , Y = 14 and n = 3

We know:

d=yxn+1=14153+1=1204=180

So, three rational numbers lying between x and y are:

(x + d), (x + 2d) and ( x + 3d)

= (15+180) , (15+280) and (15+380)

= 1780 , 1880 and 1980

Question 4:

Find five rational numbers lying between 25 and  34

Solution:

Let:

X= 25, y = 34 and n= 5

We know:

d=yxn+1=34255+1=7206=7120

So, five rational numbers between x and y are:

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

= (25+7120)(25+14120)(25+21120)(25+28120)and(25+35120)

= 55120 , 62120 , 69120 , 76120 and 83120

Question 5:

Insert 6 rational numbers between 3 and 4.

Solution:

Let:

X= 3, y =4 and n=6

We know:

d = yxn+1 = 436+1 = 17

So, six rational numbers between x and y are:

(x + d) ,(x + 2d), (x + 3d), (x + 4d), (x + 5d) and (x + 6d)

= (3+17), (3+27), (3+37) , (3+47), (3+57), (3+67)

= 227, 237, 247, 257, 267 and 277

Question 6: Insert 16 rational numbers between 2.1 and 2.2 .

Solution:

Let x = 2.1, y = 2.2 and n= 16

We know,

d = yxn+1 =2.22.116+1= 0.117 = 1170 = 0.005 (approx)

So, 16 rational numbers between 2.1 and 2.2 are:

(x + d), (x + 2d), . . . . . (x + 16d)

= [ 2.1 + 0.005], [ 2.1 +2( 0.005)]. . . . . . . . . [2.1 + 16(0.005)]

= 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 and 2.18

Question 7: Without actual division find which of the following rational are terminating decimals

  1. 1380
  2. 724
  3. 512
  4. 835
  5. 16125

Solution:

(i) Denominator of 1380 is 80.

And,

80 = 24 x 5

Therefore, 80 has no other factors than 2 and 5.

Thus, 1380 is a terminating decimal.

(ii) Denominator of 724 is 24

And 24 = 23x 3

So, 24 has a prime factor 3, which is other than 2 and 5.

Thus, 724 is not a terminating decimal.

(iii) Denominator of 512 is 12.

And,

12 = 22x 3

So, 12 has a prime factor 3, which is other than 2 and 5.

Thus, 512 is not a terminating decimal.

(iv) Denominator of 835 is 35

And,

35 = 7 x 5

So, 35 has a prime factor 7, which is other than 2 and 5.

Thus, 835 is not a terminating decimal.

(v) Denominator of 16125 is 125.

And,

125 = 53

Therefore, 125 has no other factors than 2 and 5.

Thus, 16125 is a terminating decimal.

Question 8: Convert each of the following into a decimal.

  1. 58
  2. 916
  3. 725
  4. 1124
  5. 2512

Solution:

(i) 58 = 0.625

By actual division, we have:

(ii) 916 = 0.5625

By actual division, we have:

(iii) 725 = 0.28

By actual division, we have:

(iv) 1124 = 0.4583333

By actual division, we have:

(v) 2512 = 2912 = 2.416666

By actual division, we have:

Question 9: Express each of the following as a fraction in simplest form.

  1. 0. 3¯¯¯
  2. 1. 3¯¯¯
  3. 0. 34¯¯¯¯¯
  4. 3. 14¯¯¯¯¯
  5. 0. 324¯¯¯¯¯¯¯¯
  6. 0. 17¯¯¯
  7. 0.54¯¯¯
  8. 0.163¯¯¯¯¯

Solution:  

(i) 0. 3¯¯¯

Let x = 0. 3¯¯¯

Therefore, X = 0.3333 – – – – – – – – – – – – – – – (1)

10x = 3.3333 – – – – – – – – – – – – – – (2)

On subtracting (1) from (2) we get:

9x = 3

= x = 13

Therefore, 0. 3¯¯¯ = 13

(ii) 1. 3¯¯¯

Let x = 1. 3¯¯¯

Therefore, X = 1.33333 – – – – – – – – – – – – – (i)

10x = 13.33333 – – – – – – – – – – – (ii)

On subtracting (i) and (ii) we get:

9x = 12

= x = 43

Therefore, 1. 3¯¯¯ = 43

(iii) 0. 34¯¯¯¯¯

Let x = 0. 34¯¯¯¯¯

Therefore, X = 0.3434… – – – – – – – – – – – – (1)

100x = 34.343434… – – – – – – – – – – (2)

On Subtracting (1) and (2) we get:

99x = 34

= x = 3499

Therefore, 0. 34¯¯¯¯¯ = 3499

(iv) 3. 14¯¯¯¯¯

Let x = 3. 34¯¯¯¯¯

Therefore, X = 3.1414.. – – – – – – – – – – – (1)

100x = 314.1414… – – – – – – – – – – (2)

On subtracting (1) from (2), we get:

99x = 311

= x = 31199

Therefore, 3. 34¯¯¯¯¯ = 31199

(v) 0. 324¯¯¯¯¯¯¯¯

Let x = 0. 324¯¯¯¯¯¯¯¯

Therefore, X = 0.324324… – – – – – – – – – – – – – – – (1)

1000x = 324.324324.. – – – – – – – – – – – – (2)

On subtracting (1) from (2), we get:

999x = 324

= x = 324999 = 1237

Therefore, 0. 324¯¯¯¯¯¯¯¯ = 1237

(vi) 0. 17¯¯¯

Let x = 0.1 7¯¯¯

Therefore, X = 0.1777.. – – – – – – – – – – – – – (1)

10x = 1.7777.. – – – – – – – – – – – – – (2)

On subtracting (1) from (2) we get:

90x = 16

= x = 845

Therefore, 0.17¯¯¯ = 845

(vii) 0.54¯¯¯

Let x = 0.54¯¯¯

Therefore, X = 0.54444…

10x = 5.4444.. – – – – – – – – – – – – – – – (1)

100x = 54.4444… – – – – – – – – – – – – – – (2)

On subtracting (1) from (2) we get:

90x = 49

= x = 4990

Therefore, 0.54¯¯¯ = 4990

(viii) 0.163¯¯¯¯¯

Let x = 0.163¯¯¯¯¯

Therefore, X = 0.1636363…

10x = 1.6363.. – – – – – – – – – – – – – – – (1)

1000x = 163.6363… – – – – – – – – – – – – – – (2)

On subtracting (1) from (2) we get:

990x = 162

= x = 162990= 955

Therefore, 0.163¯¯¯¯¯ = 955

Question 10: Write, whether the given statement is true or false. Give reason.

  1. Every natural number is a whole number.
  2. Every whole number is a natural number.
  3. Every integer is a rational numbers.
  4. Every rational number is a whole number.
  5. Every terminating decimal is a rational number.
  6. Every repeating decimal is a rational number.
  7. 0 is a rational number

Solution:

(i) True

Natural numbers start from 1 to infinity and whole numbers start from 0 to infinity; hence, every natural number is a whole number.

(ii) False

0 is a whole number not a natural number, so every whole number is not a natural number.

(iii) True

Every integer can be expressed in the pq form.

(iv) False

Because whole numbers consist only of number of the form p1 , where p is a positive number. On the other hand, rational numbers are the numbers whose denominator can be anything except 0.

(v) True

Every terminating decimal can be easily expressed in the pq form.

(vi) True

Every terminating decimal can be easily expressed in the pq form.

(vii) True

0 can be expressed in the form pq, so it is a rational number.


Practise This Question

For the graph given below choose the correct option :

Table ATable Ba) Line Ai) Accelerated Motionb) Line Bii) Constant velocity (high magnitude)c) Curve Ciii) At restd) Line Eiv) Constant Velocity (low magnitude)