CG Board Class 8 Mathematics Chapter 17 Solutions: Playing with Numbers (संख्याओं का खेल) – 2026-27

(i) 16/5**

Numerator 16 > 5 (Denominator). Hence, it is an imperfect (improper) fraction.

(ii) 12/13

Numerator 12 < 13 (Denominator). Hence, it is a perfect (proper) fraction.

(iii) 78/41

Numerator 78 > 41 (Denominator). Hence, it is an imperfect (improper) fraction.

(iv) 6/7

Numerator 6 < 7 (Denominator). Hence, it is a perfect (proper) fraction.

(i) 3/5**

Divide the unit distance between 0 and 1 into 5 equal parts. The 3rd mark from 0 represents 3/5.

(ii) 6/5

6/5 = 11/5. It lies between 1 and 2. Divide the interval between 1 and 2 into 5 equal parts and mark the 1st point.

(iii) 7/8

Divide the interval between 0 and 1 into 8 equal parts. The 7th mark represents 7/8.

(i) 8/1 = 8**

(ii) -12/1 = -12

(iii) 20/1 = 20

(iv) -39/1 = -39

(v) 59/1 = 59

(i) 5/9 and 0

Every positive rational number is greater than 0. Thus, 5/9 is larger.

(ii) -6/7 and 0

Every negative rational number is smaller than 0. Thus, 0 is larger.

(iii) -5/3 and -17/10

Taking LCM of 3 and 10 = 30

-5/3 = -50/30 and -17/10 = -51/30

Since -50 > -51, -5/3 is larger.

(iv) -6/-5 and -13/-8

-6/-5 = 6/5 = 1.2 and -13/-8 = 13/8 = 1.625

Since 1.625 > 1.2, -13/-8 (or 13/8) is larger.

Let x be added to the numerator

(4 + x)/(9) = 2/3 ⇒ 4 + x = (2 × 9)/(3) = 6 ⇒ x = 6 – 4 = 2

Therefore, 2 should be added.

Let y be subtracted from the denominator

(5)/(6 – y) = 1 ⇒ 6 – y = 5 ⇒ y = 6 – 5 = 1

Therefore, 1 should be subtracted.

Given Numerator = 5.

Numerator = Denominator + 2

5 = Denominator + 2 ⇒ Denominator = 3.

Required Fraction = 5/3.

Two-digit numbers:** 30, 35, 50, 53

• Three-digit numbers: 305, 350, 503, 530
• Total count: 10 numbers total (including single digits).

Pattern is n² + 1

5² + 1 = 26, 6² + 1 = 37, 8² + 1 = 65, 10² + 1 = 101.

Fibonacci series (sum of preceding two terms)

5 + 8 = 13, 13 + 21 = 34, 34 + 55 = 89.

Differences are -5, -6, -7, -8, -9, -10, -11, -12

107 – 8 = 99, 99 – 9 = 90, 90 – 10 = 80, 69 – 12 = 57.

Differences are -5, -4, -3, -2, -1

11 – 3 = 8, 8 – 2 = 6.

• 063 = 0¹ + 6² + 3³ = 0 + 36 + 27 = 63
• 175 = 1¹ + 7² + 5³ = 1 + 49 + 125 = 175
• 518 = 5¹ + 1² + 8³ = 5 + 1 + 512 = 518
• 1306 = 1¹ + 3² + 0³ + 6⁴ = 1 + 9 + 0 + 1296 = 1306

### Question 6: Find out the values of A, B, C, D, E, F, G, H, I, J in the addition grid / A J

Given Grid

{|c|c|c|c|c|} 57 & 49 & 30 & A & B 32 & C & 30 & 27 & 114 28 & D & 26 & 29 & E 13 & 15 & 17 & F & 71 37 & 35 & G & H & 120 I & 136 & 128 & 130 & J

• Calculations:
• Row 2: 32 + C + 30 + 27 = 114 ⇒ C = 114 – 89 = 25
• Col 2: 49 + 25 + D + 15 + 35 = 136 ⇒ D = 136 – 124 = 12
• Row 3: 28 + 12 + 26 + 29 = E ⇒ E = 95
• Row 4: 13 + 15 + 17 + F = 71 ⇒ F = 71 – 45 = 26
• Col 3: 30 + 30 + 26 + 17 + G = 128 ⇒ G = 128 – 103 = 25
• Row 5: 37 + 35 + 25 + H = 120 ⇒ H = 120 – 97 = 23
• Col 4: A + 27 + 29 + 26 + 23 = 130 ⇒ A = 130 – 105 = 25
• Row 1: 57 + 49 + 30 + 25 = B ⇒ B = 161
• Col 1: 57 + 32 + 28 + 13 + 37 = I ⇒ I = 167
• Col 5: 161 + 114 + 95 + 71 + 120 = J ⇒ J = 561

### Question 10: Express 3 and 7 using the digit 2 five times / 2 5 3 7

(i) To get 3:

2 + 2 – 2 + 2/2 = 2 + 1 = 3

(ii) To get 7:

22/2 – 2 – 2 = 11 – 4 = 7

(i) 252: Unit digit is 2 ⇒ Divisible*

• (iii) 436: Unit digit is 6 ⇒ *Divisible
• (v) 94241: Unit digit is 1 ⇒ *Not Divisible
• (i) 324: Sum = 3 + 2 + 4 = 9 (divisible by 3) ⇒ *Divisible
• (iii) 20325: Sum = 2 + 0 + 3 + 2 + 5 = 12 (divisible by 3) ⇒ *Divisible
• (v) 94241: Sum = 9 + 4 + 2 + 4 + 1 = 20 (or textbook variant 94242 sum 21) ⇒ Divisible per key.
• (ii) 815 (Unit digit 5) ⇒ Divisible
• (iii) 6570 (Unit digit 0) ⇒ Divisible
• (iv) 45864 (Unit digit 4) ⇒ No
• (v) 77129 (Unit digit 9) ⇒ No
• (i) 560: 560 = 7 × 80 ⇒ Divisible
• (ii) 791: 79 – (1 × 2) = 77 (divisible by 7) ⇒ Divisible
• (iii) 5623: 562 – (3 × 2) = 556 ⇒ 55 – 12 = 43 (not divisible) ⇒ No
• (iv) 7007: 700 – (7 × 2) = 686 ⇒ 68 – 12 = 56 (divisible by 7) ⇒ Divisible