Word Problems: Number Play | Mathematics Class 8- New NCERT (Ganita Prakash) PDF Download

Q1: Anshu and Priya are detectives solving a number mystery. They find a number, 672, and need to determine if it’s divisible by 2, 3, 4, 6, 8, and 9 without performing division.

Q5: The school has 5 buses numbered: 246, 375, 540, 816, and 999.
The headmaster says:

“Buses divisible by 3 will get a red sticker.”

“Buses divisible by 5 will get a green sticker.”

Without performing actual division state which buses get which stickers? Can a bus get both?

Solution: Divisibility Rules:

• Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.

• Divisibility by 5: A number is divisible by 5 if its last digit is either 0 or 5.

Step-by-Step Analysis:

Bus 246:

• Sum of digits: 2 + 4 + 6 = 12. Since 12 is divisible by 3, Bus 246 is divisible by 3.

• Last digit is 6, which is not 0 or 5, so Bus 246 is not divisible by 5.

• Sticker: Red (because it is divisible by 3).

Bus 375:

• Sum of digits: 3 + 7 + 5 = 15. Since 15 is divisible by 3, Bus 375 is divisible by 3.

• Last digit is 5, which means Bus 375 is divisible by 5.

• Sticker: Red and Green (because it is divisible by both 3 and 5).

Bus 540:

• Sum of digits: 5 + 4 + 0 = 9. Since 9 is divisible by 3, Bus 540 is divisible by 3.

• Last digit is 0, which means Bus 540 is divisible by 5.

• Sticker: Red and Green (because it is divisible by both 3 and 5).

Bus 816:

• Sum of digits: 8 + 1 + 6 = 15. Since 15 is divisible by 3, Bus 816 is divisible by 3.

• Last digit is 6, which is not 0 or 5, so Bus 816 is not divisible by 5.

• Sticker: Red (because it is divisible by 3).

Bus 999:

• Sum of digits: 9 + 9 + 9 = 27. Since 27 is divisible by 3, Bus 999 is divisible by 3.

• Last digit is 9, which is not 0 or 5, so Bus 999 is not divisible by 5.

• Sticker: Red (because it is divisible by 3).

Conclusion:

• Buses with red stickers (divisible by 3): 246, 375, 540, 816, 999.

• Buses with green stickers (divisible by 5): 375, 540.

• Buses that get both red and green stickers (divisible by both 3 and 5): 375, 540.

Q6: Farmer Leela puts eggs in crates of 4. If she has 328, 444, 812, and 973 eggs, which totals can be exactly divided into crates without leftovers?

Solution:
Divisible by 4 → last two digits divisible by 4

• 328 → last two digits 28 

• 444 → last two digits 44 

• 812 → last two digits 12 

• 973 → last two digits 73 

Q7: Ananya has one basket with an even number of apples and another with an odd number. She combines them into a big basket. Will the total always be odd, even, or can it be both? Explain.

Solution:  When you add an even number (which has no remainder) to an odd number (which leaves a remainder of 1), the total will always leave a remainder of 1, making the sum odd. According to parity rule:

Even + Odd = Always odd.

Example:

• 4 (even) + 5 (odd) = 9 (odd)

• 6 (even) + 3 (odd) = 9 (odd)

Q8: In a video game, each jump forward moves you 2 squares and each jump backward moves you 3 squares. If you start at square 0, can you ever land on an odd-numbered square? Why or why not?

Solution: Let's analyze the problem step-by-step:

• Forward move: Every jump forward moves you 2 squares. Since 2 is an even number, every forward move keeps your position's parity (odd or even) the same. If you start on an even square, you'll land on another even square after moving forward.

• Backward move: Every jump backward moves you 3 squares. Since 3 is an odd number, every backward move changes the parity of your position. If you start on an even square, you will land on an odd square after a backward move, and vice versa.

Starting at square 0 (even):

• First backward move: You move 3 squares back, so from square 0 (even), you will land on square -3 (odd).

• Subsequent moves:

  1. A forward move (+2) will take you from an odd square to an odd square (e.g., -3 + 2 = -1, which is odd).

  2. Another backward move (-3) will take you from an odd square to an even square (e.g., -1 - 3 = -4, which is even).