Extra Questions: Class 8 Maths Chapter 5 Number Play

Practice questions beyond the textbook exercises, testing deeper understanding of divisibility rules and number puzzles for Class 8 Maths Chapter 5: Number Play. These Class 8 Mathematics Chapter 5 important questions are handy for last-minute exam practice.

Extra Questions: Class 8 Maths Chapter 5 Number Play

1. 1 (Assertion-Reason). Assertion: 4536 is divisible by 9. Reason: The sum of its digits is divisible by 9.
   Solution: Both true, and the reason correctly explains the assertion — 4+5+3+6=18, a multiple of 9.
2. 2 (Numerical). Find the digit x if 92×5 is divisible by 11.
   Solution: Alternating digit-sum (from the right): 5−x+2−9 = −x−2. For divisibility by 11 this must be 0 or a multiple of 11, giving x=9 (since −9−2=−11). Check: 9295÷11=845 exactly. x=9.
3. 3 (Short Answer). Is the sum of any 4 consecutive integers always divisible by 4?
   Solution: No — e.g. 1+2+3+4=10, which is not divisible by 4. (The sum of n consecutive integers starting at a is always divisible by n only when n is odd.)
4. 4 (Applied). Find the digital root of 123456789.
   Solution: Digit sum = 45, digit sum of 45 = 9. Digital root = 9.
5. 5 (Assertion-Reason). Assertion: The product of any 3 consecutive integers is always divisible by 6. Reason: Among any 3 consecutive integers, at least one is divisible by 2 and one is divisible by 3.
   Solution: Both true, and the reason correctly explains the assertion.
6. 6 (Numerical). Find the smallest 4-digit number divisible by both 6 and 9.
   Solution: LCM(6,9)=18; smallest 4-digit multiple of 18 is 1008 (18×56).
7. 7 (Synthesis). If a number leaves remainder 4 when divided by 9, what is its digital root?
   Solution: The digital root equals the remainder when divided by 9 (for non-multiples of 9), so the digital root is 4.
8. 8 (Applied). Solve the cryptarithm: AB × 3 = CAB (each letter is a distinct digit).
   Solution: Writing AB as 10A+B: (10A+B)×3 = 100C+10A+B ⇒ 20A+2B=100C ⇒ 10A+B=50C. Since AB is a 2-digit number, C must be 1, giving AB=50. Check: 50×3=150=CAB with A=5, B=0, C=1.
9. 9 (Short Answer). Are all multiples of 4 also multiples of 8?
   Solution: No — e.g. 4 and 12 are multiples of 4 but not of 8; every multiple of 8 is a multiple of 4, but not vice versa.
10. 10 (Assertion-Reason). Assertion: Reversing the digits of a multiple of 3 always gives another multiple of 3. Reason: Digit sum is unchanged by reversing the order of digits.
    Solution: Both true, and the reason correctly explains the assertion, since divisibility by 3 depends only on the digit sum.

Written by Satish

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