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Number Play Class 8 Maths Worksheet with Answers Chapter 5

Updated:6 Mins Read

Ganita Prakash Part-I, Grade 8, Textbook of Mathematics – The Chapter 5 from class 8 NCERT book is on Number Play. In this chapter we are going to cover –

  • Consecutive number sums
  • Alternate plus-minus patterns
  • Even-odd reasoning rules
  • Multiples of 4
  • Remainder form 5k+3

The worksheet is broken in 4 parts – Basic, Standard, Advance, and HOTS. Answers are there at the bottom.

Class 8 Maths Worksheet – Chapter 5: Number Play

Basic

  1. Write 21 as a sum of consecutive natural numbers in two different ways.
  2. Place + / − signs between 3, 4, 5, 6 to form this expression and evaluate:
    3 + 4 − 5 + 6
  3. Without dividing, decide if the number is divisible by 9:

    1. 123
    2. 405
    3. 8888
  4. Numbers that leave remainder 3 when divided by 5 can be written as ________.

    (Write in terms of k.)
  5. Find the remainder when 7309 is divided by 9 using the “sum of digits” idea.
  6. Find the digital root of 358095.

Standard

  1. For four consecutive numbers 5, 6, 7, 8 evaluate:

    5 + 6 − 7 + 8
  2. Without computing completely, decide which of these are even:

    1. 43 + 37
    2. 672 − 348
    3. 4 × 347 × 3
    4. 708 − 477
  3. Which expressions are always even for all integers? (Write Yes/No)

    1. 2a + 2b
    2. 3g + 5h
    3. 4m + 2n
    4. b² + 1
  4. Two even numbers are added. When will their sum be a multiple of 4?

    Choose the correct statement:

    1. Multiple of 4 + multiple of 4 → always multiple of 4
    2. (4p+2) + (4q+2) → always multiple of 4
    3. Multiple of 4 + (4q+2) → always multiple of 4
  5. If 661 leaves remainder 3 when divided by 7, and 4779 leaves remainder 5 when divided by 7,
    find the remainders (without full division):

    1. 4779 + 661
    2. 4779 − 661
  6. Find the smallest multiple of 9 that has no odd digits.

Advance

  1. Find the multiple of 9 that is closest to 6000.
  2. How many multiples of 9 are there between 4300 and 4400 (both inclusive)?
  3. Use the 11-test (alternating sum of digits) to check if 462 is divisible by 11.
    Write “Divisible” or “Not divisible” and also write the remainder if not divisible.
  4. The sum of four consecutive numbers is 34. Find the four numbers.
  5. If 31z5 is a multiple of 9 (z is a digit), find all possible values of z.
  6. If 48a23b is a multiple of 18, list all possible pairs (a, b).

HOTS

  1. Pebble Riddle: A number is:
    • 1 more than a multiple of 3
    • odd
    • 1 more than a multiple of 5
    • exactly divisible by 7
    • less than 100
    Find the number.
  2. Decide whether the statement is Always true / Sometimes true / Never true:
    “If a number is divisible by 7, then it is divisible by any multiple of 7.”
  3. Solve the cryptarithm (each letter is a digit, no two letters share a digit):
    AB + 37 = 6A
  4. Tathagat writes numbers that leave remainder 2 when divided by 6.
    He claims: “If you add any three such numbers, the sum will always be a multiple of 6.”
    Is the claim true? (Yes/No) Give a one-line algebra reason.
  5. What is the digital root of the expression 9a + 36b + 13 for any integers a and b?
  6. Explain in 2–3 lines:
    Why does the “sum of digits” test correctly tell divisibility by 9?

Answer Key

Basic – Answers

  1. Ans: 10 + 11 = 21 and 6 + 7 + 8 = 21
    Hint: Find different consecutive blocks that sum to 21.
  2. Ans: 3 + 4 − 5 + 6 = 8
    1. Ans: 123 → 1+2+3=6, not divisible by 9
    2. Ans: 405 → 4+0+5=9, divisible by 9
    3. Ans: 8888 → 8+8+8+8=32, not divisible by 9
  4. Ans: 5k + 3
    Hint: “3 more than a multiple of 5”.
  5. Ans: 7+3+0+9 = 19 → 1+9=10 → 1+0=1
    So remainder = 1.
  6. Ans: 3+5+8+0+9+5 = 30 → 3+0 = 3

Standard – Answers

  1. Ans: 5 + 6 − 7 + 8 = 12
    1. Ans: 43 + 37 = even (odd + odd = even)
    2. Ans: 672 − 348 = even (even − even = even)
    3. Ans: 4 × 347 × 3 = even (has factor 4)
    4. Ans: 708 − 477 = odd (even − odd = odd)
    1. Ans: 2a + 2b → Yes (factor 2)
    2. Ans: 3g + 5h → No (can be odd/even depending on g,h)
    3. Ans: 4m + 2n → Yes (2(2m+n))
    4. Ans: b² + 1 → No (if b is even, b² even → +1 odd)
  4. Ans: (1) and (2) are true, (3) is false.
    Hint: 4p + (4q+2) = 4(p+q) + 2 (remainder 2).
    1. Ans: (5 + 3) mod 7 = 8 mod 7 = 1
    2. Ans: (5 − 3) mod 7 = 2
  6. Ans: 288
    Hint: Only even digits; digit-sum must be a multiple of 9.

Advance – Answers

  1. Ans: 6003
    Hint: 9×666=5994, next is 6003 (closer).
  2. Ans: First multiple ≥4300 is 4302 (=9×478).
    Last multiple ≤4400 is 4392 (=9×488).
    Count = 488 − 478 + 1 = 11.
  3. Ans: 462: alternating sum from units → 2 − 6 + 4 = 0
    So Divisible by 11 (remainder 0).
  4. Ans: Let numbers be n, n+1, n+2, n+3.
    Sum = 4n + 6 = 34 → 4n = 28 → n = 7
    Numbers: 7, 8, 9, 10
  5. Ans: 3+1+z+5 = 9+z must be divisible by 9 → z = 0 or 9
  6. Ans: Divisible by 18 ⇒ divisible by 2 and 9.
    b must be even and 4+8+a+2+3+b = 17+a+b divisible by 9.
    So a+b ≡ 1 (mod 9).
    Possible pairs: (1,0), (8,2), (6,4), (4,6), (2,8)

HOTS – Answers

  1. Ans: 91
    Hint: 1 mod 3 and 1 mod 5 ⇒ 1 mod 15; check which is divisible by 7 under 100.
  2. Ans: Sometimes true
    Reason: If N is divisible by 7, it is not necessarily divisible by 14 or 21, etc.
    Example: 42 divisible by 7 but not divisible by 28.
  3. Ans: A=2, B=5
    So 25 + 37 = 62
  4. Ans: Yes
    Reason: Numbers are of form 6k+2. Sum of three: (6k₁+2)+(6k₂+2)+(6k₃+2)=6(k₁+k₂+k₃+1), a multiple of 6.
  5. Ans: 4
    Hint: 9a and 36b are multiples of 9, so only 13 matters → digital root of 13 is 4.
  6. Ans: Because 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, etc.
    So each place value contributes the same remainder as its digit when dividing by 9.
    Hence the number and the sum of its digits have the same remainder mod 9.

Worksheet From Other Chapters

Ganita Prakash Part-I

Ganita Prakash Part-II

Amit, a BE in Mechanical Engineering, is a math enthusiast dedicated to making math fun and accessible for kids in classes 1 to 10. With a knack for simplifying complex concepts, Amit offers easy-to-understand solutions, fostering a love for math in young minds across India.

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