Unit Test (Solutions): Number Play | Mathematics Class 8- New NCERT (Ganita Prakash) PDF Download

Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers 9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each

Q1: Which of the following expressions is always even for any integer values?  (1 Mark)
a) 2a + 2b
b) 3g + 5h
c) b²
d) x + 1

Answer: a) 2a + 2b

Q2: If two numbers are both divisible by 8, then which of the following is true?  (1 Mark)
a) Their sum is always divisible by 8
b) Their difference is always divisible by 8
c) Both (a) and (b)
d) Neither (a) nor (b)

Answer: c) Both (a) and (b)

Q3:  If a number is divisible by both 9 and 4, then it must also be divisible by:  (1 Mark)
a) 12
b) 18
c) 36
d) 72

Answer: c) 36

Q4: Which of the following is the digital root of 489710?  (1 Mark)
a) 1
b) 2
c) 9
d) 7

Answer: b) 2

Q5: The divisibility rule for 11 states that:  (1 Mark)
a) The sum of digits should be divisible by 11
b) The difference between the sum of digits at odd and even places should be divisible by 11
c) The unit digit should be 1 or 11
d) The number should end with digit 0 or 1

Answer: b) The difference between the sum of digits at odd and even places should be divisible by 11

Q6: Farmer Karan keeps eggs in crates of 4. If he has 236, 412, 768, and 995 eggs, which totals can be exactly divided into crates?  (2 Marks)

Solution: Divisible by 4 → check last two digits.

• 236: last two digits 36 → not divisible by 4

• 412: last two digits 12 → divisible by 4

• 768: last two digits 68 → divisible by 4

• 995: last two digits 95 → not divisible by 4

Divisible by 4: 412, 768

Q7: Aarav has one basket with an even number of bananas and another with an odd number. When he mixes them, will the total be even, odd, or both?  (2 Marks)

Solution:
Even + Odd = Odd.
Example: 8 (even) + 7 (odd) = 15 (odd).
So the total is always odd.

Q8: In a board game, each forward move = 4 steps, each backward move = 5 steps. Starting at 0, can you land on an odd square?  (2 Marks)

Solution:

• Forward move = +4 (even) → parity remains even.

• Backward move = –5 (odd) → parity flips.
  Start = 0 (even).

• First backward move → –5 (odd).

• After that, combinations of +4 and –5 can land on both odd and even numbers.

Yes, you can land on odd squares.

Q9: A fruit seller has 21 oranges. He wants to arrange them in baskets such that each basket has consecutive natural numbers of oranges.  Find two different ways.  (3 Marks)

Solution:
We need sums of consecutive numbers = 21.

• 1 + 2 + 3 + 4 + 5 + 6 = 21

• 6 + 7 + 8 = 21

Two possible ways found.

Q10: Aarav and Meera are solving a number puzzle. They discover the number 864 and want to check if it’s divisible by 2, 3, 4, 6, 8, and 9 without doing actual division. (3 Marks)

Solution:

• Divisibility by 2: Last digit = 4 (even). So, divisible by 2.

• Divisibility by 3: Sum of digits = 8 + 6 + 4 = 18. Since 18 is divisible by 3, 864 is divisible by 3.

• Divisibility by 4: Last two digits = 64. Since 64 is divisible by 4, the number is divisible by 4.

• Divisibility by 6: Number must be divisible by both 2 and 3. It is, so divisible by 6.

• Divisibility by 8: Last three digits = 864. Since 864 ÷ 8 = 108, divisible by 8.

• Divisibility by 9: Sum of digits = 18. Since 18 is divisible by 9, divisible by 9.

Final Answer: 864 is divisible by 2, 3, 4, 6, 8, and 9.

Q11: Four friends — Neha, Dev, Ritu, and Sam — are arranging tiles numbered consecutively starting from 11. They can add (+) or subtract (–) the numbers. No matter how they do it, the total is always even.
Tiles: 11, 12, 13, 14.
List all possible sums and check the pattern.  (3 Marks)

Solution:
Numbers: 11, 12, 13, 14

Possible sums:

• 11 + 12 + 13 + 14 = 50 (even)

• 11 + 12 + 13 – 14 = 22 (even)

• 11 + 12 – 13 + 14 = 24 (even)

• 11 – 12 + 13 + 14 = 26 (even)

• 11 + 12 – 13 – 14 = –4 (even)

• 11 – 12 + 13 – 14 = –2 (even)

• 11 – 12 – 13 + 14 = 0 (even)

• 11 – 12 – 13 – 14 = –28 (even)

All results are even → pattern holds true.

Q12: The school has 5 buses numbered: 132, 225, 450, 720, and 835.
The headmaster says:

• Buses divisible by 3 → red sticker

• Buses divisible by 5 → green sticker

Which buses get which stickers? Can any bus get both?  (5 Marks)

Solution:

• 132: Sum of digits = 6 → divisible by 3, not by 5 → Red

• 225: Sum of digits = 9 → divisible by 3, last digit 5 → divisible by 5 → Red + Green

• 450: Sum of digits = 9 → divisible by 3, last digit 0 → divisible by 5 → Red + Green

• 720: Sum of digits = 9 → divisible by 3, last digit 0 → divisible by 5 → Red + Green

• 835: Sum of digits = 16 (not divisible by 3), last digit 5 → divisible by 5 → Green

Red stickers: 132, 225, 450, 720
Green stickers: 225, 450, 720, 835
Both: 225, 450, 720