# Samacheer Kalvi Class 9 Maths Chapter 2 Real Numbers Solutions

## Samacheer Kalvi Class 9 Maths Chapter 2 Real Numbers Solutions

Welcome to NCTB Solutions. Here with this post we are going to help 9th class students by providing Solutions for Samacheer Kalvi Class 9 Maths chapter 2 Real Numbers. Here students can easily find all the solutions for Real Numbers Exercise 2.2, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7 and 2.8. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 2 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus. #### Real Numbers Exercise 2.1 Solutions :

(2) Find any three rational numbers between -7/11 and 2/11

Solution :

Rational numbers between -7/11 and 2/11 are -6/1, -5/11, -4/11

(3) Find any five rational numbers between

(i) 1/4 and 1/5

Solution :

Given, 1/4 and 1/5

1st case,

1/2 (1/4 + 1/5)

= 1/2 (5+4/20)

= 1/2 × 9/20

= 9/40

2nd case,

1/2 (1/4 + 9/40)

= 1/2 (10 + 9/40)

= 1/2 × 19/40

= 19/80

3rd case,

1/2 (1/5 + 9/40)

= 1/2 (8+9/40

= 1/2 × 17/40

= 17/80

4th case,

1/2 (1/5 + 17/80)

= 1/2 (16+17/80)

= 1/2 × 33/80

= 33/160

5th case,

1/2 (17/80 + 33/160 )

= 1/2 (34+33/160)

= 1/2 × 67/160

= 67/320

Thus, 5 rational numbers are –

9/40, 19,80, 17/80, 33/160, 67/320

(ii) 0.1 and 0.11

Solution :

0.1 = 1/10

0.11 = 11/100

1/10 = 1×10/10×10 = 10/100 = 10×10/100×101 = 100/1000

11/100 = 11×1/100×1 = 11/100 = 11×10/100×10 = 110/1000

Thus, Five rational number are,

101/1000, 102/1000, 103/1000, 104/1000, 105/1000

(iii) -1 and -2

Solution :

-1 = -1/1 = -1×10/1×10 = -10/10

-2 = -2/1 = -2×10/1×10 = -20/10

Thus, Five rational numbers are

-11/10 , -12/10, -13/10, -14/10, -15/10

#### Real Numbers Exercise 2.2 Solutions :

(1) Express the following rational numbers into decimal and state the kind of decimal expansion

(i) 2/7

(ii) -5 3/11

(iii) 22/3

(iv) 327/200

Solution :

(i) 2/7

= 0.28571

(ii) -5 3/11

= 5.72

(iii) 22/3

= 7.333

(iv) 327/200

= 1.635

(2) Express 1/13 in decimal form. find the length of the period of decimals

Solution :

Given, 1/13

= 0.07692

Length of the period of decimal is 5

(3) Express the rational number 1/33 in recurring decimal form by using the recurring decimal expansion of 1/11. hence write 71/33 in recurring decimal form.

Solution :

1/11 = 0.090909

Similarly,

1/33 = 0.030303

71/33 = 2.151515

(4) Express the following decimal expression into rational numbers.

(i) 0.24

(ii) 2.327

(iii) -5.132

(iv) 3.17

(v) 17.215

(vi) -21.2137

Solution :

(i) Let, x = 0.24

So, x = 0.212424 ….. (i)

Multiply both side by 100

100 x = 24.2424 …. (ii)

From (ii) and (iii) we get

100x – x = 24.2424 – 0.24

= 99 x = 24

= x = 24/99

(ii) Let, x = 2.327

x = 2.327327 …. (i)

Multiply 1000 both side

1000x = 2327.327327 …. (i)

1000x – x = 2327.3273

27 – 2.327327

= 999x = 2325

= x = 2325/999

(iii) -5.132

= -5132/1000

For the point 1 is come at the denominator and after point here are 3 digits so three zero.

(iv) Let, x = 3.17

x = 3.17777 ….. (i)

10x = 31.777….. (ii)

Again 100x = 317.777…. (iii)

From (ii) and (iii) we get

100x – 10x = 317.777 – 31.777

= 90x = 286

∴ x = 286/90

(v) Let, x = 17.215

x = 17.215215 …. (i)

1000x = 17215.215215 ….. (ii)

1000x – x = 17215.215215 – 17.215215

= 999x = 17198

= x = 17198/999

Next Chapter Solution :

Updated: August 1, 2023 — 4:05 pm