CG Board Class 9 Mathematics Chapter 2 Solutions: Real Numbers (वास्तविक संख्याएँ) – 2026-27

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CG SCERT (Chhattisgarh Board) Class 9 Mathematics Solutions | 2026-27 Official Edition

Complete 100% textbook-grounded CG Board Class 9 Mathematics Chapter 2 Solutions: Real Numbers (वास्तविक संख्याएँ) in Bilingual Format (English & Hindi Medium).

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Q1

Write those numbers which you have in Q…


📝 Question:

Write those numbers which you have in Q but not in N

💡 Solution:

Rational numbers in Q that are not Natural numbers N include 0, negative integers, proper fractions, and decimal numbers. Examples: 0, -3, 3/7, -0.37, 10/9, 22/7, -55, 3.2323, -7.5.

Question 2

Write those numbers which you have in Z…


📝 Question:

Write those numbers which you have in Z but not in N

💡 Solution:

Integers in Z not present in Natural numbers N are zero and negative integers. Examples: 0, -3, -55, -108.

Question 3

Write those numbers which you have in W…


📝 Question:

Write those numbers which you have in W but not in N

💡 Solution:

The only Whole number in W not present in Natural numbers N is 0.

Q4

Show on number line 2 3 4


📝 Question:

Show on number line: -2, +3, +4

💡 Solution:

  • Steps (English): Draw a horizontal line. Mark the origin O as 0. Points to the right are positive (+1, +2, +3, +4…) and to the left are negative (-1, -2, -3…). Mark point -2 two units to the left of 0, and points +3 and +4 at 3 and 4 units to the right of 0 respectively.
  • : O 0 0 (+1, +2, +3, +4…) (-1, -2, -3…) -2 0 2 +3 +4 3 4

Question 2

How many steps ahead 2 is from 3


📝 Question:

How many steps ahead +2 is from -3?

💡 Solution:

Distance = (+2) - (-3) = 2 + 3 = 5 steps.

+2 is 5 steps ahead of -3.

Q6

What will you have to do to show…


📝 Question:

/ :** What will you have to do to show -3/5, 2/5, 8/5, 12/5 on number line?

💡 Solution:

Divide each unit length on the number line into 5 equal parts. Then count sub-divisions from 0: 2 parts right for 2/5, 8 parts right for 8/5, 12 parts right for 12/5, and 3 parts left for -3/5.

Q7

How many integers lie between 5 15


📝 Question:

How many integers lie between 5 & 15?

💡 Solution:

  • List of integers / : 6, 7, 8, 9, 10, 11, 12, 13, 14.
  • Count / : 9 integers.

Question 2

How many integers lie between 3 8


📝 Question:

How many integers lie between -3 & 8?

💡 Solution:

  • List of integers / : -2, -1, 0, 1, 2, 3, 4, 5, 6, 7.
  • Count / : 10 integers.

Q9

Find any 5 rational numbers between 27 and…


📝 Question:

Find any 5 rational numbers between 2/7 and 4/7

💡 Solution:

  • Multiply numerator and denominator of both numbers by 6 (since we need 5 numbers, multiplier = 5 + 1 = 6)
  • 2/7 = (2 6) / (7 6) = 12/42
  • 4/7 = (4 6) / (7 6) = 24/42
  • 5 Rational Numbers / 5 : 13/42, 14/42, 15/42, 16/42, 17/42 (or 25/42, 26/42, 27/42, 28/42, 29/42 by using multiplier 6 with 4/7 and 5/7 equivalent forms).

Question 2

Find 3 rational numbers between 15 and 17


📝 Question:

Find 3 rational numbers between 1/5 and 1/7

💡 Solution:

  • Equating denominators (LCM of 5 and 7 = 35)
  • 1/5 = 7/35 and 1/7 = 5/35.
  • Multiply numerator and denominator by 4 (since multiplier = 3 + 1 = 4)
  • 5/35 = (5 4) / (35 4) = 20/140
  • 7/35 = (7 4) / (35 4) = 28/140
  • 3 Rational Numbers / 3 : 21/140, 22/140, 23/140.

Question 3

Find 11 rational numbers between 13 and 12


📝 Question:

Find 11 rational numbers between -1/3 and 1/2

💡 Solution:

  • Make denominators equal (LCM of 3 and 2 = 6, expand denominator to 24)
  • -1/3 = -8/24
  • 1/2 = 12/24
  • 11 Rational Numbers / 11 : -7/24, -6/24, -5/24, -4/24, -3/24, -2/24, -1/24, 0, 1/24, 2/24, 3/24.

Q12

Change the following decimals into pq form 1…


📝 Question:

/ : Change the following decimals into p/q form: 1. 1.2333..... (1.23 bar on 3) 2. 3.88..... (3.8 bar on 8) 3. 3.204343..... (3.2043 bar on 43) Solutions / : 1. For 1.2333…..: Let x = 1.2333..... (eq i) 10x = 12.333..... (eq ii) 100x = 123.333..... (eq iii) * Subtracting eq (ii) from eq (iii): 100x - 10x = 123.333... - 12.333... -> 90x = 111 -> x = 111/90 = 37/30

💡 Solution:

37/30

  1. For 3.88…..:
  • Let x = 3.888..... (eq i)
  • 10x = 38.888..... (eq ii)
  • Subtracting eq (i) from eq (ii): 9x = 35 -> x = 35/9.

35/9

  1. For 3.204343…..:
  • Let x = 3.204343..... (eq i)
  • 100x = 320.4343..... (eq ii)
  • 10000x = 32043.4343..... (eq iii)
  • Subtracting eq (ii) from eq (iii): 9900x = 31723 -> x = 31723/9900.

31723/9900

Q13

Identify rational and irrational numbers i sqrt6 ii…


📝 Question:

/ : Identify rational and irrational numbers: (i) sqrt(6) (ii) sqrt(7) (iii) -sqrt(25) (iv) -sqrt(8) (v) sqrt(9) (vi) sqrt(9/16) Solutions / : (i) sqrt(6): Irrational / (non-perfect square root). (ii) sqrt(7): Irrational / . (iii) -sqrt(25) = -5: Rational / (integer form -5/1). (iv) -sqrt(8) = -2sqrt(2): Irrational / . (v) sqrt(9) = 3: Rational / . (vi) sqrt(9/16) = 3/4: Rational / . Solve: (i) 3sqrt(2) + 3sqrt(2) = (3 + 3)sqrt(2) = 6sqrt(2). (ii) 3sqrt(6) . 2sqrt(3) = (3 2)sqrt(6 3) = 6sqrt(18) = 6 3sqrt(2) = 18sqrt(2). Subtract 5sqrt(3) + 7sqrt(5) from 3sqrt(5) + 7sqrt(3). (3sqrt(5) + 7sqrt(3) 5sqrt(3) + 7*sqrt(5) )

💡 Solution:

[3sqrt(5) + 7sqrt(3)] - [5sqrt(3) + 7sqrt(5)]

= (7 - 5)sqrt(3) + (3 - 7)sqrt(5) = 2sqrt(3) - 4sqrt(5).

Question 3

Multiply 6sqrt3 by 13sqrt3 6sqrt3 13sqrt3


📝 Question:

Multiply 6sqrt(3) by 13sqrt(3). (6sqrt(3) 13sqrt(3) )

💡 Solution:

(6 13) (sqrt(3) sqrt(3)) = 78 3 = 234.

Question 4

Solve i 3 sqrt32 sqrt2 6 3sqrt2 2sqrt3…


📝 Question:

Solve: (i) (3 + sqrt(3))(2 + sqrt(2)) = 6 + 3sqrt(2) + 2sqrt(3) + sqrt(6). (ii) 3(2 + sqrt(3))(2 - sqrt(3)) = 3[(2)2 - (sqrt(3))2] = 3 = 3 1 = 3. (iii) (sqrt(5) + sqrt(2))2 = (sqrt(5))2 + (sqrt(2))2 + 2sqrt(5)sqrt(2) = 5 + 2 + 2sqrt(10) = 7 + 2sqrt(10). (iv) (sqrt(5) - sqrt(2))(sqrt(5) + sqrt(2)) = (sqrt(5))2 - (sqrt(2))2 = 5 - 2 = 3. Rationalise the denominator: (i) 1 / (4 + sqrt(5))

💡 Solution:

Multiply numerator & denominator by (4 - sqrt(5))

[1 * (4 - sqrt(5))] / [(4 + sqrt(5))(4 - sqrt(5))] = (4 - sqrt(5)) / (16 - 5) = (4 - sqrt(5)) / 11.

  • (ii) 1 / (7 + 4*sqrt(3))

Multiply numerator & denominator by (7 - 4*sqrt(3))

[1 (7 - 4sqrt(3))] / [(7)2 - (4sqrt(3))2] = (7 - 4sqrt(3)) / (49 - 48) = 7 - 4*sqrt(3).

  • (iii) 1 / (7 + 4*sqrt(3)) + 1 / (2 + sqrt(5))

Rationalising terms separately

First term = 7 - 4*sqrt(3)

Second term = [1 * (2 - sqrt(5))] / (4 - 5) = (2 - sqrt(5)) / (-1) = sqrt(5) - 2

Sum = (7 - 4sqrt(3)) + (sqrt(5) - 2) = 5 - 4sqrt(3) + sqrt(5).

### Section 2: Exercise 2.1 Complete Step-by-Step Solutions / 2.1

Question 1

1 Give examples for the following statements StepbyStep…


📝 Question:

/ 1 Give examples for the following statements: * Step-by-Step Solution /

💡 Solution:

(i)** Any positive integer like 1 belongs to N, W, and Z.

  • (ii) 0 is a whole number (W) but not a natural number (N).
  • (iii) 0 (or -2, 1/2) is a rational number (Q) but not a natural number (N).
  • (i) 126/5:
  • (ii) 335/16:
  • (iii) 22/7:
  • (iv) -118/3:
  • (i) 0.53: 53 / 100.
  • (ii) 16.8: 168 / 10 = 84 / 5.
  • (iii) 105.25: 10525 / 100 = 421 / 4.
  • (iv) 7.36: 736 / 100 = 184 / 25.
  • (i) 0.70 bar (0.707070…..):

Let x = 0.7070..... -> 100x = 70.7070.....

99x = 70 -> x = 70/99.

  • (ii) 8.39 bar (8.393939…..):

Let x = 8.3939..... -> 100x = 839.3939.....

99x = 831 -> x = 831/99 = 277/33.

  • (iii) 5.63180 bar:

Expressing repeating block yields fraction = 563180 / 99900 = 56318 / 9990.

  • (iv) 5.125 bar (5.125125125…..):

Let x = 5.125125..... -> 1000x = 5125.125125.....

999x = 5120 -> x = 5120/999.

  • (i) 3sqrt(3) + 15sqrt(3) = (3 + 15)sqrt(3) = 18sqrt(3).
  • (ii) 5sqrt(7) - 2sqrt(7) = (5 - 2)sqrt(7) = 3sqrt(7).
  • (i) Using (a + b)(a - b) = a2 - b2

(sqrt(3))2 - (sqrt(2))2 = 3 - 2 = 1.

  • (ii) Using (a + b)(a - b) = a2 - b2

(5)2 - (sqrt(5))2 = 25 - 5 = 20.

  • (iii) Expanding terms

sqrt(2)sqrt(3) + sqrt(2)sqrt(8) + sqrt(3)sqrt(3) + sqrt(3)sqrt(8)

= sqrt(6) + sqrt(16) + 3 + sqrt(24)

= sqrt(6) + 4 + 3 + 2sqrt(6) = 7 + 3sqrt(6).

  • (iv) Using (a - b)2 = a2 + b2 - 2ab

(sqrt(7))2 + (sqrt(3))2 - 2sqrt(7)sqrt(3) = 7 + 3 - 2sqrt(21) = 10 - 2sqrt(21).

  • (i) Multiply numerator and denominator by sqrt(5)

[1 sqrt(5)] / [sqrt(5) sqrt(5)] = sqrt(5) / 5.

  • (ii) Multiply numerator and denominator by sqrt(6)

[2 sqrt(6)] / [sqrt(6) sqrt(6)] = 2*sqrt(6) / 6 = sqrt(6) / 3.

  • (iii) Simplify radical fraction

sqrt(21 / 3) = sqrt(7) = sqrt(7) / 1.

  • (i) Rationalising LHS

[(6 + sqrt(3))(6 + sqrt(3))] / [(6 - sqrt(3))(6 + sqrt(3))]

= [(6)2 + (sqrt(3))2 + 2(6)(sqrt(3))] / [(6)2 - (sqrt(3))2]

= (36 + 3 + 12*sqrt(3)) / (36 - 3)

= (39 + 12sqrt(3)) / 33 = 39/33 + (12/33)sqrt(3) = 13/11 + (4/11)*sqrt(3).

Comparing with a + b*sqrt(3) gives

a = 13/11, b = 4/11.

  • (ii) Rationalising LHS

[(sqrt(5) + sqrt(3))(sqrt(5) + sqrt(3))] / [(sqrt(5) - sqrt(3))(sqrt(5) + sqrt(3))]

= [(sqrt(5))2 + (sqrt(3))2 + 2sqrt(5)sqrt(3)] / [(sqrt(5))2 - (sqrt(3))2]

= (5 + 3 + 2sqrt(15)) / (5 - 3) = (8 + 2sqrt(15)) / 2 = 4 + 1*sqrt(15).

Comparing with a + b*sqrt(15) gives

a = 4, b = 1.

  • (i) Rationalise each term
  • First term: [(5 + sqrt(3))(2 + sqrt(3))] / (4 - 3) = 10 + 5sqrt(3) + 2sqrt(3) + 3 = 13 + 7*sqrt(3)
  • Second term: [(2 - sqrt(3))(2 - sqrt(3))] / (4 - 3) = 4 + 3 - 4sqrt(3) = 7 - 4sqrt(3)
  • Sum = (13 + 7sqrt(3)) + (7 - 4sqrt(3)) = 20 + 3*sqrt(3).
  • (ii) Rationalise each term
  • First term: [(2 + sqrt(3))2] / (4 - 3) = 4 + 3 + 4sqrt(3) = 7 + 4sqrt(3)
  • Second term: [(2 - sqrt(3))2] / (4 - 3) = 4 + 3 - 4sqrt(3) = 7 - 4sqrt(3)
  • Third term: [(sqrt(3) - 1)2] / (3 - 1) = (3 + 1 - 2sqrt(3)) / 2 = (4 - 2sqrt(3)) / 2 = 2 - sqrt(3)
  • Total Sum = (7 + 4sqrt(3)) + (7 - 4sqrt(3)) + (2 - sqrt(3)) = 16 - sqrt(3) (or 16 - 3*sqrt(8) for print representation variant).

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