CG SCERT (Chhattisgarh Board) Class 9 Mathematics Solutions | 2026-27 Official Edition
Complete 100% textbook-grounded CG Board Class 9 Mathematics Chapter 6 Solutions: Playing with Numbers (संख्याओं में भी है खेल). These solutions cover all official exercises (प्रश्नावली 6.1 एवं 6.2) directly from the official CG SCERT textbook in full unparaphrased bilingual format.
End-of-Chapter Exercises (Solutions)
अभ्यास के सभी प्रश्नों के हल
Exercise 6.1 – Cryptarithmetic Puzzles
प्रश्नावली 6.1 – अक्षरों के मान ज्ञात करना
1. Find out value of letters A, B, X, Y, Z, L, M, N as used in the following puzzles:
(i)
B A
+ 1 2 B
——-
2 2 5
(ii)
3 X Y
+ Y Y 2
——-
1 0 1 8
(iii)
M N
× 6
—–
M L N
(iv)
1 Z
× Z
—–
7 Z
(v)
X X
+ Y Y Y
+ 6
——-
4 6 1
(vi)
2 P Q
+ P Q 1
——-
Q 1 8
(vii)
M L
× 6
—–
L L L
(i) A = 6, B = 9
Step: In the units column, A + B ends in 5. Looking at the hundreds place, B + 1 + carry = 2 (with carry 1 from tens), which gives B = 9. Then 9 + A = 15, so A = 6. Check: 96 + 129 = 225.
(ii) X = 5, Y = 6
Step: Units column: Y + 2 = 8 ⇒ Y = 6. Tens column: X + Y = X + 6 = 11 (carry 1 to hundreds) ⇒ X = 5. Check: 356 + 662 = 1018.
(iii) L = 0, M = 1, N = 8
Step: In units place, 6 × N ends in N. Testing digits: 6 × 8 = 48 (ends in 8, carry 4). Then 6 × M + 4 = 10 × M + L. For M = 1: 6(1) + 4 = 10 ⇒ M = 1, L = 0. Check: 18 × 6 = 108.
(iv) Z = 5
Step: (10 + Z) × Z = 70 + Z ⇒ 10Z + Z² = 70 + Z ⇒ Z² + 9Z – 70 = 0 ⇒ (Z + 14)(Z – 5) = 0 ⇒ Z = 5. Check: 15 × 5 = 75.
(v) X = 1, Y = 4
Step: 11X + 111Y + 6 = 461 ⇒ 11X + 111Y = 455. If Y = 4: 111(4) = 444, leaving 11X = 455 – 444 = 11 ⇒ X = 1. Check: 11 + 444 + 6 = 461.
(vi) P = 4, Q = 7
Step: Units column: Q + 1 = 8 ⇒ Q = 7. Tens column: P + Q = P + 7 = 11 (carry 1) ⇒ P = 4. Hundreds column: 2 + P + 1 = 2 + 4 + 1 = 7 = Q. Check: 247 + 471 = 718.
(vii) M = 7, L = 4
Step: (10M + L) × 6 = 111L ⇒ 60M + 6L = 111L ⇒ 60M = 105L ⇒ 4M = 7L. Since M and L are single non-zero digits, M = 7 and L = 4. Check: 74 × 6 = 444.
Exercise 6.2 – Multiples of 5 and 10
प्रश्नावली 6.2 – 5 और 10 के गुणज
2. Which of the following numbers are multiples of 5 and 10:
316, 9560, 205, 311, 800, 7936
Multiples of 5: 9560, 205, 800 (Numbers whose unit digit is either 0 or 5).
Multiples of 10: 9560, 800 (Numbers whose unit digit is 0).
Exercise 6.2 – Divisibility by 3
प्रश्नावली 6.2 – 3 से विभाज्यता
3. If a number divisible by 3 has 8 in its units place, what can be the possible digits in its tens place?
Possible tens digits: 1, 4, 7
Reason: Let the tens digit be x. The sum of digits is x + 8, which must be divisible by 3.
– If x = 1: 1 + 8 = 9 (divisible by 3)
– If x = 4: 4 + 8 = 12 (divisible by 3)
– If x = 7: 7 + 8 = 15 (divisible by 3)
Thus, the possible digits are 1, 4, or 7.
Exercise 6.2 – Divisibility by 5 (Finding P)
प्रश्नावली 6.2 – 5 से विभाज्यता (P का मान)
4. If the number 35P is a multiple of 5, find the value of P.
Value of P: 0 or 5
Reason: Any integer is divisible by 5 if and only if its units digit is either 0 or 5. Since P is the units digit of 35P, P must be 0 or 5.
Exercise 6.2 – Divisibility by 9 (Finding A and B)
प्रश्नावली 6.2 – 9 से विभाज्यता (A तथा B के मान)
5. If 6A3B is a number divisible by 9, find the possible values of A and B.
Possible pairs of (A, B): (0, 0), (0, 9), (9, 0), (1, 8), (8, 1), (2, 7), (7, 2), (3, 6), (6, 3), (4, 5), (5, 4)
Reason: A number is divisible by 9 if the sum of its digits is a multiple of 9. Here, sum = 6 + A + 3 + B = 9 + (A + B). Therefore, (A + B) must be either 0, 9, or 18 (if A=9, B=9).
Exercise 6.2 – Divisibility by 7
प्रश्नावली 6.2 – 7 से विभाज्यता की जाँच
6. Check the divisibility by 7 for the following numbers:
(i) 672 (ii) 905 (iii) 2205 (iv) 9751
(i) 672: Divisible by 7 (Yes)
Rule: Double the unit digit and subtract from remaining number: 67 – 2(2) = 67 – 4 = 63. Since 63 is divisible by 7 (63 ÷ 7 = 9), 672 is divisible by 7.
(ii) 905: Not divisible by 7 (No)
Rule: 90 – 2(5) = 90 – 10 = 80. Since 80 is not divisible by 7, 905 is not divisible by 7.
(iii) 2205: Divisible by 7 (Yes)
Rule: 220 – 2(5) = 220 – 10 = 210. 210 ÷ 7 = 30. Therefore, 2205 is divisible by 7.
(iv) 9751: Not divisible by 7 (No)
Rule: 975 – 2(1) = 973. Repeating: 97 – 2(3) = 97 – 6 = 91. 91 ÷ 7 = 13. Since 91 is divisible by 7, 9751 is divisible by 7 (Yes).