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.

Last Updated: September 23, 2026

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.

Frequently Asked Questions

How can the divisibility rule for 9 help you quickly check whether a large number is divisible by 9?
A number is divisible by 9 if the sum of all its digits is also divisible by 9, so instead of long division, you can simply add up its digits and check whether that smaller sum is a multiple of 9.

What is the significance of finding common factors when simplifying a fraction to its lowest terms?
Finding the highest common factor of the numerator and denominator allows both to be divided by that same number, reducing the fraction to its simplest equivalent form without changing its actual value.

Written by Satish

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