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).
Write those numbers which you have in Q…
हिंदी परिमेय संख्या Q की ऐसी संख्याएँ जो…
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.
Write those numbers which you have in Z…
हिंदी पूर्णांक Z की ऐसी संख्याएँ जो N…
Integers in Z not present in Natural numbers N are zero and negative integers. Examples: 0, -3, -55, -108.
Write those numbers which you have in W…
हिंदी पूर्ण संख्या W की एकमात्र संख्या जो…
The only Whole number in W not present in Natural numbers N is 0.
Show on number line 2 3 4
Show on number line 2 3 4
- 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
-2two units to the left of 0, and points+3and+4at 3 and 4 units to the right of 0 respectively. - : O 0 0 (+1, +2, +3, +4…) (-1, -2, -3…)
-20 2+3+43 4
How many steps ahead 2 is from 3
How many steps ahead 2 is from 3
Distance = (+2) - (-3) = 2 + 3 = 5 steps.
+2 is 5 steps ahead of -3.
What will you have to do to show…
हिंदी संख्या रेखा पर प्रत्येक 1 इकाई दूरी…
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.
How many integers lie between 5 15
How many integers lie between 5 15
- List of integers / :
6, 7, 8, 9, 10, 11, 12, 13, 14. - Count / :
9 integers.
How many integers lie between 3 8
How many integers lie between 3 8
- List of integers / :
-2, -1, 0, 1, 2, 3, 4, 5, 6, 7. - Count / :
10 integers.
Find any 5 rational numbers between 27 and…
Find any 5 rational numbers between 27 and…
- Multiply numerator and denominator of both numbers by
6(since we need 5 numbers, multiplier = 5 + 1 = 6) 2/7 = (2 6) / (7 6) = 12/424/7 = (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).
Find 3 rational numbers between 15 and 17
Find 3 rational numbers between 15 and 17
- Equating denominators (LCM of 5 and 7 = 35)
1/5 = 7/35and1/7 = 5/35.- Multiply numerator and denominator by
4(since multiplier = 3 + 1 = 4) 5/35 = (5 4) / (35 4) = 20/1407/35 = (7 4) / (35 4) = 28/140- 3 Rational Numbers / 3 :
21/140, 22/140, 23/140.
Find 11 rational numbers between 13 and 12
Find 11 rational numbers between 13 and 12
- Make denominators equal (LCM of 3 and 2 = 6, expand denominator to 24)
-1/3 = -8/241/2 = 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.
Change the following decimals into pq form 1…
Change the following decimals into pq form 1…
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 = 37/3037/30
- 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
- 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
Identify rational and irrational numbers i sqrt6 ii…
Identify rational and irrational numbers i sqrt6 ii…
(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) )[3sqrt(5) + 7sqrt(3)] - [5sqrt(3) + 7sqrt(5)]
= (7 - 5)sqrt(3) + (3 - 7)sqrt(5) = 2sqrt(3) - 4sqrt(5).
Multiply 6sqrt3 by 13sqrt3 6sqrt3 13sqrt3
Multiply 6sqrt3 by 13sqrt3 6sqrt3 13sqrt3
6sqrt(3) by 13sqrt(3). (6sqrt(3) 13sqrt(3) )(6 13) (sqrt(3) sqrt(3)) = 78 3 = 234.
Solve i 3 sqrt32 sqrt2 6 3sqrt2 2sqrt3…
Solve i 3 sqrt32 sqrt2 6 3sqrt2 2sqrt3…
(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))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
1 Give examples for the following statements StepbyStep…
i A number which is natural number whole…
(i)** Any positive integer like 1 belongs to N, W, and Z.
- (ii)
0is 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)(or16 - 3*sqrt(8)for print representation variant).