Equal Groups

Given: The frog jumps 3 steps at a time, starting from 0.

Concept: The frog lands on multiples of 3.

The frog will touch: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, …

To check if the frog touches 67:
$$67 \div 3 = 22 \text{ remainder } 1$$
Since 67 is not exactly divisible by 3, it is NOT a multiple of 3.

Answer: The frog touches all multiples of 3 (3, 6, 9, 12, …). No, the frog will NOT touch 67.

Given: The squirrel jumps 4 steps at a time, starting from 0.

Concept: The squirrel lands on multiples of 4.

The squirrel will touch: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, …

To find how many jumps to reach 60:
$$60 \div 4 = 15$$

Answer: The squirrel touches all multiples of 4. It should jump 15 times to reach 60.

Given: The rabbit jumps 6 steps at a time, starting from 0.

Concept: The rabbit lands on multiples of 6.

The rabbit will touch: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, …

Smallest 3-digit number that is a multiple of 6:
$$100 \div 6 = 16 \text{ remainder } 4$$
So the next multiple after 96 is $6 \times 17 = 102$.

Number of jumps to reach 102:
$$102 \div 6 = 17$$

Answer: The rabbit touches all multiples of 6. The smallest 3-digit number it lands on is 102. The rabbit jumped 17 times to reach 102.

Given: The kangaroo jumps 8 steps at a time, starting from 0.

Concept: The kangaroo lands on multiples of 8.

The kangaroo will touch: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, …

The rabbit touches multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, …

Common numbers (common multiples of 6 and 8):
LCM of 6 and 8 = 24
Common numbers: 24, 48, 72, 96, 120, …

Answer: The kangaroo touches all multiples of 8. Yes, both the rabbit and the kangaroo touch the numbers 24, 48, 72, 96, 120, … (common multiples of 6 and 8).

Given: Rabbit jumps 6 steps at a time; Kangaroo jumps 8 steps at a time. Both need to reach 48.

Number of jumps for the rabbit:
$$48 \div 6 = 8 \text{ jumps}$$

Number of jumps for the kangaroo:
$$48 \div 8 = 6 \text{ jumps}$$

Observation: The rabbit took 8 jumps and the kangaroo took 6 jumps to reach the same number 48. The animal with the bigger jump size (kangaroo, 8 steps) takes fewer jumps to reach the same destination. The number of jumps is inversely related to the jump size.

Given: Frog jumps 3 steps at a time; Rabbit jumps 6 steps at a time. Both need to reach 60.

Number of jumps for the frog:
$$60 \div 3 = 20 \text{ jumps}$$

Number of jumps for the rabbit:
$$60 \div 6 = 10 \text{ jumps}$$

Observation: The frog took 20 jumps and the rabbit took 10 jumps to reach 60. The rabbit's jump size (6) is double the frog's jump size (3), so the rabbit takes exactly half the number of jumps. When the jump size doubles, the number of jumps needed is halved.

Given: Frog touches multiples of 3; Squirrel touches multiples of 4.

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, …
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, …

Common multiples (LCM of 3 and 4 = 12): 12, 24, 36, 48, 60, …

Answer: Both the frog and the squirrel touch the numbers 12, 24, 36, 48, 60, … A few common multiples of 3 and 4 are 12, 24, 36, 48, 60.

Given: Rabbit touches multiples of 6; Kangaroo touches multiples of 8.

Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, …
Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, …

Common multiples (LCM of 6 and 8 = 24): 24, 48, 72, 96, 120, …

Answer: Both the rabbit and the kangaroo touch the numbers 24, 48, 72, 96, 120, … A few common multiples of 6 and 8 are 24, 48, 72, 96, 120.

Given: Cat starts at 6, jumps 3 steps each time. Rat starts at 12, jumps 2 steps each time.

Let us track their positions after each jump:

| Jump No. | Cat's position | Rat's position |
|---|---|---|
| Start | 6 | 12 |
| 1 | 6+3 = 9 | 12+2 = 14 |
| 2 | 9+3 = 12 | 14+2 = 16 |
| 3 | 12+3 = 15 | 16+2 = 18 |
| 4 | 15+3 = 18 | 18+2 = 20 |
| 5 | 18+3 = 21 | 20+2 = 22 |
| 6 | 21+3 = 24 | 22+2 = 24 |

After 6 jumps, both the cat and the rat are at position 24.

Answer: Yes, the cat will catch the rat at number 24.

Given: An 8×8 grid of numbers. We need to find rows, columns, or sequences of three numbers that form a multiplication or division sentence (a × b = c or c ÷ a = b).

Concept: A multiplication sentence has the form: number × number = number. A division sentence has the form: number ÷ number = number. We look for three consecutive numbers (horizontally or vertically) where one relationship holds.

Some examples of multiplication/division sentences that can be found in the grid (reading left-to-right or top-to-bottom):

Horizontal examples:
- Row 1: 3 × 4 = 12 (positions: 3, 4 in row 1; 12 in row 3, col 1 — check across rows)
- 4 × 2 = 8 (found in various positions)
- 2 × 10 = 20 (Row 2: 2, 10, 20)
- 4 × 2 = 8 (Row 3: 4, 2, 8)
- 2 × 6 = 12 (various)
- 6 × 2 = 12
- 2 × 3 = 6 (Row 5: 2, 3, 6)
- 3 × 6 = 18 (Row 5: 3, 6, 18)
- 6 ÷ 2 = 3
- 2 × 7 = 14 (Row 6: 2, 7, 14)
- 5 × 8 = 40 (check)
- 2 × 5 = 10 (Row 8: 2, 5, 10)

Vertical examples:
- Column reading: 4, 2, 8 (4 × 2 = 8)
- 2, 2, 4 (2 × 2 = 4)
- 6, 2, 12 (6 × 2 = 12)
- 5, 2, 10 (5 × 2 = 10)

Answer: Students should shade all such triplets. There are approximately 10 or more multiplication/division sentences hidden in the grid. The exact count depends on the direction (horizontal/vertical/diagonal) considered. Students are encouraged to find as many as possible by checking every set of three consecutive numbers.

Given: Each lily flower has 3 petals. Number of flowers = 12.

Concept: Total petals = Number of flowers × Petals per flower

Step 1: Petals in 10 lilies:
$$10 \times 3 = 30 \text{ petals}$$

Step 2: Petals in 2 lilies:
$$2 \times 3 = 6 \text{ petals}$$

Step 3: Petals in 12 lilies:
$$12 \times 3 = 30 + 6 = 36 \text{ petals}$$

Multiplication statement: $12 \times 3 = 36$

Answer: There are 36 petals in 12 lily flowers.

Given: Each hibiscus flower has 5 petals. Total petals = 80.

Concept: Number of flowers = Total petals ÷ Petals per flower

$$\text{Number of flowers} = 80 \div 5$$

5 petals → 1 flower
10 petals → 2 flowers
50 petals → 10 flowers

Remaining petals after 50: $80 - 50 = 30$ petals
30 petals → $30 \div 5 = 6$ flowers

Total flowers = $10 + 6 = 16$ flowers

Verification: $16 \times 5 = 80$ ✓

Answer: Gulabo had 16 hibiscus flowers.

Note: The exact number of rows and columns depends on the figure (which cannot be seen). A typical arrangement for this type of problem is 5 saplings in each row and 4 rows (or similar). Using a common textbook answer of 5 saplings per row and 4 rows:

Given (assumed from standard textbook): 5 saplings in each row, 4 rows.

Total saplings = Number of rows × Saplings per row
$$= 4 \times 5 = 20 \text{ saplings}$$

Mathematical Statement: $4 \times 5 = 20$

Answer: There are 5 saplings in each row. There are 4 rows. She planted 20 saplings in all.

*(Note: Students should fill in the actual numbers visible in their textbook figure.)*

Note: The exact arrangement depends on the figure. A typical arrangement for this type of problem is 6 columns with 4 boxes in each column.

Given (assumed from standard textbook): 6 columns, 4 boxes in each column.

Total boxes = Number of columns × Boxes per column
$$= 6 \times 4 = 24 \text{ boxes}$$

Mathematical Statement: $6 \times 4 = 24$

Answer: There are 6 columns of strawberry boxes. There are 4 boxes in each column. There are 24 boxes in all.

*(Note: Students should fill in the actual numbers visible in their textbook figure.)*

Given: One tray holds 18 cupcakes.

Concept: 18 cupcakes are arranged in rows and columns in each tray. A common arrangement is 3 rows × 6 columns = 18, or 2 rows × 9 columns = 18.

Students should draw/mark 18 cupcake circles in each tray in equal rows and columns.

Example arrangement: 3 rows and 6 columns per tray.
$$3 \times 6 = 18 \text{ cupcakes per tray}$$

Answer: Each tray should show 18 cupcakes arranged in equal rows and columns (e.g., 3 rows of 6).

Given: Each tray holds 18 cupcakes. She uses 2 trays at a time.

$$\text{Total cupcakes} = 2 \times 18 = 36 \text{ cupcakes}$$

Answer: She can make 36 cupcakes in one attempt.

Given: Total cupcakes = 108. Each tray holds 18 cupcakes.

$$\text{Number of trays} = 108 \div 18$$

$$18 \times 6 = 108$$

$$108 \div 18 = 6$$

Answer: She has baked 6 trays.

Given: 36 mini cupcakes in a square tray.

Concept: For a square tray, the number of rows equals the number of columns.
$$\sqrt{36} = 6$$
So the arrangement is 6 rows × 6 columns.

Number of columns = 6
Number of cupcakes in each column = 6

Multiplication statement: $6 \times 6 = 36$

Answer: Number of columns: 6. Number of cupcakes in each column: 6. Multiplication statement: $6 \times 6 = 36$.

Concept: We find all factor pairs for each number. Each factor pair (a, b) gives an arrangement of a rows and b columns.

36:
- $1 \times 36$, $2 \times 18$, $3 \times 12$, $4 \times 9$, $6 \times 6$

8:
- $1 \times 8$, $2 \times 4$

12:
- $1 \times 12$, $2 \times 6$, $3 \times 4$

24:
- $1 \times 24$, $2 \times 12$, $3 \times 8$, $4 \times 6$

Answer: Students should draw these arrangements in their notebooks. For example, 12 cupcakes can be arranged as 3 rows of 4, or 2 rows of 6, or 1 row of 12.

Concept: Doubling means multiplying by 2.

| Flowers before magic | Flowers after magic |
|---|---|
| 23 | $23 \times 2 = 46$ |
| 10 | $10 \times 2 = 20$ |
| 51 | $51 \times 2 = 102$ |
| 95 | $95 \times 2 = 190$ |
| 150 | $150 \times 2 = 300$ |
| 199 | $199 \times 2 = 398$ |
| 425 | $425 \times 2 = 850$ |
| 500 | $500 \times 2 = 1000$ |

For 'after magic' numbers given, find 'before magic' (halving):
- After magic = 222 → Before magic = $222 \div 2 = 111$
- After magic = 410 → Before magic = $410 \div 2 = 205$
- After magic = 500 → Before magic = $500 \div 2 = 250$

Concept: When we double a number, the ones digit of the result depends only on the ones digit of the original number.

a) 28: ones digit is 8. $8 \times 2 = 16$, ones digit = 6
(Verify: $28 \times 2 = 56$, ones digit = 6 ✓)

b) 56: ones digit is 6. $6 \times 2 = 12$, ones digit = 2
(Verify: $56 \times 2 = 112$, ones digit = 2 ✓)

c) 45: ones digit is 5. $5 \times 2 = 10$, ones digit = 0
(Verify: $45 \times 2 = 90$, ones digit = 0 ✓)

d) 17: ones digit is 7. $7 \times 2 = 14$, ones digit = 4
(Verify: $17 \times 2 = 34$, ones digit = 4 ✓)

Concept: Doubling (×2) always gives an even number. So the ones digit after doubling can only be 0, 2, 4, 6, or 8.

a) Ones digit 0 after doubling: Numbers with ones digit 5 or 0.
Example: 5 (5×2=10), 15 (15×2=30), 20 (20×2=40)

b) Ones digit 2 after doubling: Numbers with ones digit 1 or 6.
Example: 1 (1×2=2), 6 (6×2=12), 11 (11×2=22)

c) Ones digit 4 after doubling: Numbers with ones digit 2 or 7.
Example: 2 (2×2=4), 7 (7×2=14), 12 (12×2=24)

d) Ones digit 6 after doubling: Numbers with ones digit 3 or 8.
Example: 3 (3×2=6), 8 (8×2=16), 13 (13×2=26)

e) Ones digit 8 after doubling: Numbers with ones digit 4 or 9.
Example: 4 (4×2=8), 9 (9×2=18), 14 (14×2=28)

Can we get 3, 5, 7, 9 as ones digit after doubling?
No! Since doubling always gives an even number, the ones digit after doubling will always be 0, 2, 4, 6, or 8. We can never get an odd digit (1, 3, 5, 7, 9) in the ones place after doubling.

Observation: All numbers we get after doubling are even numbers.

Some patterns in the multiplication table:

1. The products in each row increase by the row number (e.g., row 3: 3, 6, 9, 12, … — each increases by 3).
2. The table is symmetric: the number in row $a$, column $b$ equals the number in row $b$, column $a$ (because $a \times b = b \times a$).
3. The diagonal (where row number = column number) contains perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
4. Row 2 and row 5 together cover all numbers in row 10.
5. All products in even-numbered rows are even numbers.

Yes, the numbers in row 7 are the same as the numbers in column 7.

Row 7: $7 \times 1, 7 \times 2, 7 \times 3, \ldots = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70$

Column 7: $1 \times 7, 2 \times 7, 3 \times 7, \ldots = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70$

They are the same!

In general: Yes, the numbers in any given row are the same as the numbers in the corresponding column. This happens because of the commutative property of multiplication: $a \times b = b \times a$. So the entry in row $a$, column $b$ equals the entry in row $b$, column $a$.

Concept: A product is even if at least one of the factors is even.

In any row $n$, the products are $n \times 1, n \times 2, \ldots, n \times 10$.

If $n$ is even, then every product $n \times k$ is even (since one factor $n$ is even).

Rows with all even products: Rows 2, 4, 6, 8, 10 (all even-numbered rows).

In odd-numbered rows (1, 3, 5, 7, 9), some products are odd (when multiplied by an odd number) and some are even (when multiplied by an even number).

Concept: A product is odd only when BOTH factors are odd.

In row $n$, the products are $n \times 1, n \times 2, n \times 3, \ldots, n \times 10$.

For all products to be odd, $n$ must be odd AND all column numbers (1 through 10) must be odd. But column numbers include even numbers (2, 4, 6, 8, 10), so $n \times 2, n \times 4$, etc. will always be even.

Answer: No, there is no row in the table (columns 1–10) where all products are odd numbers. Every row contains at least some even products.

Yes, the odd-numbered rows (rows 1, 3, 5, 7, 9) have both even and odd products.

For example, Row 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
- Odd products: 3, 9, 15, 21, 27 (when multiplied by odd numbers: 1, 3, 5, 7, 9)
- Even products: 6, 12, 18, 24, 30 (when multiplied by even numbers: 2, 4, 6, 8, 10)

Why? When an odd number is multiplied by an odd number, the result is odd. When an odd number is multiplied by an even number, the result is even. Since the columns include both odd and even numbers, odd-numbered rows will always have both even and odd products.

In a 10×10 multiplication table, let us count:

Odd products occur only when BOTH the row number AND the column number are odd.

Odd rows: 1, 3, 5, 7, 9 → 5 odd rows
Odd columns: 1, 3, 5, 7, 9 → 5 odd columns

Number of odd products = $5 \times 5 = 25$

Total entries = $10 \times 10 = 100$

Number of even products = $100 - 25 = 75$

Answer: There are more even numbers (75) than odd numbers (25) in the chart. We know this because a product is odd only when both factors are odd, and there are only 5 odd numbers among 1–10, giving $5 \times 5 = 25$ odd products out of 100 total.

a) Common multiples of 2 and 3 (i.e., multiples of LCM = 6):
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96 (within 1–100)
These appear in both row 2 and row 3 of the table.

b) Common multiples of 4 and 8 (i.e., multiples of LCM = 8):
8, 16, 24, 32, 40, 48, 56, 64, 72, 80 (within 1–100)
Since 8 is a multiple of 4, all multiples of 8 are also multiples of 4. So common multiples of 4 and 8 are just multiples of 8.

c) Common multiples of 7 and 9 (i.e., multiples of LCM = 63):
63 (within 1–100)
Only 63 appears in the 10×10 table.

Observation: The fewer common multiples a pair has, the larger their LCM. Numbers that share a common factor (like 4 and 8) have more common multiples than numbers with no common factor (like 7 and 9).

Row 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50
Ones digits: 5, 0, 5, 0, 5, 0, 5, 0, 5, 0
Pattern: The ones digit alternates between 5 and 0.

Row 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Ones digits: 2, 4, 6, 8, 0, 2, 4, 6, 8, 0
Pattern: The ones digits cycle through 2, 4, 6, 8, 0 and repeat.

Row 1: Ones digits: 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 — all digits 0–9 appear once.

Row 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
Ones digits: 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 — decreasing pattern.

General observation: Each row has a repeating pattern in the ones digits. The length of the repeating cycle depends on the row number.

The ones digits in row 8 follow the repeating pattern: 8, 6, 4, 2, 0 (cycle of 5).

To find the ones digit of $n \times 8$, we look at $n \mod 5$:
- $n \mod 5 = 1$ → ones digit 8
- $n \mod 5 = 2$ → ones digit 6
- $n \mod 5 = 3$ → ones digit 4
- $n \mod 5 = 4$ → ones digit 2
- $n \mod 5 = 0$ → ones digit 0

11 × 8: $11 \mod 5 = 1$ → ones digit = 8
Verify: $11 \times 8 = 88$ ✓ (ones digit 8)

12 × 8: $12 \mod 5 = 2$ → ones digit = 6
Verify: $12 \times 8 = 96$ ✓ (ones digit 6)

13 × 8: $13 \mod 5 = 3$ → ones digit = 4
Verify: $13 \times 8 = 104$ ✓ (ones digit 4)

Row 8 ones digits follow the pattern: 8, 6, 4, 2, 0, 8, 6, 4, 2, 0, …

Digits that appear: 0, 2, 4, 6, 8 (only even digits)

Digits that do NOT appear: 1, 3, 5, 7, 9 (all odd digits)

Answer: The digits 1, 3, 5, 7, and 9 do not appear as the ones digit in row 8. This is because 8 is even, and an even number multiplied by any whole number always gives an even product.

For all digits 0–9 to appear as ones digits in a row, the ones digits must cycle through all 10 digits before repeating.

This happens when the row number and 10 are coprime (share no common factor other than 1), i.e., when the row number is odd and not a multiple of 5.

Checking:
- Row 1: ones digits: 1,2,3,4,5,6,7,8,9,0 — all 10 digits ✓
- Row 3: ones digits: 3,6,9,2,5,8,1,4,7,0 — all 10 digits ✓
- Row 7: ones digits: 7,4,1,8,5,2,9,6,3,0 — all 10 digits ✓
- Row 9: ones digits: 9,8,7,6,5,4,3,2,1,0 — all 10 digits ✓

Answer: Rows 1, 3, 7, and 9 have all digits from 0 to 9 appearing as the ones digit.

$\square \times 8$ gives 0 as the ones digit when the product ends in 0.

This happens when $\square \times 8$ is a multiple of 10, i.e., when $\square$ is a multiple of 5.

(Because $5 \times 8 = 40$, $10 \times 8 = 80$, $15 \times 8 = 120$, etc.)

5 examples: 5, 10, 15, 20, 25

Verification:
- $5 \times 8 = 40$ (ones digit 0) ✓
- $10 \times 8 = 80$ (ones digit 0) ✓
- $15 \times 8 = 120$ (ones digit 0) ✓
- $20 \times 8 = 160$ (ones digit 0) ✓
- $25 \times 8 = 200$ (ones digit 0) ✓

Answer: Numbers that are multiples of 5 (5, 10, 15, 20, 25, …) give 0 as the ones digit when multiplied by 8.

In a 10-column table (columns 1–10), 0 appears as the ones digit when the product is a multiple of 10.

For row $n$, $n \times k$ ends in 0 when $n \times k$ is a multiple of 10.

- Row 1: $1 \times 10 = 10$ → 0 appears once ✓
- Row 3: $3 \times 10 = 30$ → 0 appears once ✓
- Row 7: $7 \times 10 = 70$ → 0 appears once ✓
- Row 9: $9 \times 10 = 90$ → 0 appears once ✓
- Row 2: $2 \times 5 = 10$, $2 \times 10 = 20$ → 0 appears twice
- Row 5: $5 \times 2 = 10$, $5 \times 4 = 20$, … → 0 appears multiple times

Answer: Rows 1, 3, 7, and 9 have 0 appearing as the ones digit only once (at column 10).

Observation: The rows identified in Question 11 (rows where all digits 0–9 appear as ones digits) are exactly the same as the rows identified in Question 13 (rows where 0 appears as the ones digit only once).

Both questions give the answer: Rows 1, 3, 7, and 9.

Why? These are the rows where the row number is coprime with 10 (i.e., shares no common factor with 10 other than 1). In such rows, the ones digits cycle through all 10 digits (0–9) exactly once in every 10 products. So 0 appears exactly once, and all other digits also appear exactly once.

In rows where the row number shares a factor with 10 (like rows 2, 4, 5, 6, 8, 10), the ones digits repeat in a shorter cycle, so some digits appear more than once and others don't appear at all.

Given: Each tricycle has 3 wheels.

$$10 \times 3 = 30 \text{ wheels}$$

$$10 \times 3 = 30 \text{ wheels (10 more tricycles)}$$

$$20 \times 3 = 30 + 30 = 60 \text{ wheels}$$

Answer: 10 tricycles → 30 wheels; 10 more tricycles → 30 wheels; 20 tricycles → 30 + 30 = 60 wheels.

Given: Each car has 4 wheels.

$$10 \times 4 = 40 \text{ wheels}$$

$$30 \times 4 = 40 + 40 + 40 = 120 \text{ wheels}$$

$$10 \times 4 = 40 \qquad 30 \times 4 = 120$$

Now solving the additional problems:

a) $10 \times 6 = 60$

b) $40 \times 6 = 4 \times 6 \times 10 = 24 \times 10 = 240$

c) $10 \times 8 = 80$

d) $60 \times 8 = 6 \times 8 \times 10 = 48 \times 10 = 480$

e) $6 \times 8 = 48$

f) $60 \times 8 = 480$ (same as d)

g) $4 \times 6 = 24$

h) $40 \times 8 = 4 \times 8 \times 10 = 32 \times 10 = 320$

Observation: When the number of groups is a multiple of 10, we multiply the single-digit part first and then attach a zero at the end (multiply by 10).

Given: 18 boxes, 4 cupcakes in each box.

Total cupcakes = $18 \times 4$

Step 1: 10 boxes with 4 cupcakes each:
$$10 \times 4 = 40 \text{ cupcakes}$$

Step 2: 8 boxes with 4 cupcakes each:
$$8 \times 4 = 32 \text{ cupcakes}$$

Step 3: Total:
$$18 \times 4 = 40 + 32 = 72 \text{ cupcakes}$$

| × | 4 |
|---|---|
| 10 | 40 |
| 8 | 32 |
| Total | 72 |

Answer: There are 72 cupcakes in the packed boxes.

Given: 8 bamboo rods per cart, 23 carts.

Total rods = $23 \times 8$

Step 1: 20 carts with 8 rods each:
$$20 \times 8 = 160 \text{ rods}$$

Step 2: 3 carts with 8 rods each:
$$3 \times 8 = 24 \text{ rods}$$

Step 3: Total:
$$23 \times 8 = 160 + 24 = 184 \text{ rods}$$

| × | 8 |
|---|---|
| 20 | 160 |
| 3 | 24 |
| Total | 184 |

Answer: 184 bamboo rods are needed for 23 carts.

Given: 25 geese (each has 2 legs), 12 sheep (each has 4 legs).

Legs of geese:
$$25 \times 2 = 50 \text{ legs}$$

Legs of sheep:
$$12 \times 4 = 48 \text{ legs}$$

Total legs:
$$50 + 48 = 98 \text{ legs}$$

Answer: Chippi the lizard sees 98 legs in all.

Given: 8 children per row, 15 rows.

Total children = $15 \times 8$

Step 1: $10 \times 8 = 80$

Step 2: $5 \times 8 = 40$

Step 3: $15 \times 8 = 80 + 40 = 120$

Answer: There are 120 children in the auditorium.

Given: 9 books per pile, 14 piles.

Total books = $14 \times 9$

Step 1: $10 \times 9 = 90$

Step 2: $4 \times 9 = 36$

Step 3: $14 \times 9 = 90 + 36 = 126$

Answer: The shop has 126 books.

Note: The exact arrangement depends on the figure. A typical patchwork problem for Class 4 involves a grid arrangement. Assuming a common textbook arrangement of 6 rows and 8 columns with alternating golden and white beads:

Total beads = $6 \times 8 = 48$ beads

In a checkerboard pattern with 48 total beads:
- Golden beads = 24
- White beads = 24

Mathematical Statement: $6 \times 8 = 48$

Answer: Surya used 48 beads in all — 24 golden and 24 white beads.

*(Note: Students should use the actual numbers from their textbook figure.)*

a) $34 \times 3$:
Story: A school has 34 classrooms. Each classroom has 3 fans. How many fans are there in all?

$$34 \times 3 = (30 \times 3) + (4 \times 3) = 90 + 12 = 102$$

b) $75 \times 5$:
Story: A farmer plants 75 rows of wheat. Each row has 5 plants. How many plants are there in all?

$$75 \times 5 = (70 \times 5) + (5 \times 5) = 350 + 25 = 375$$

c) $46 \times 6$:
Story: A baker bakes 46 trays of cookies. Each tray has 6 cookies. How many cookies are there in all?

$$46 \times 6 = (40 \times 6) + (6 \times 6) = 240 + 36 = 276$$

d) $50 \times 9$:
Story: A shop has 50 boxes of pencils. Each box has 9 pencils. How many pencils are there in all?

$$50 \times 9 = 5 \times 9 \times 10 = 45 \times 10 = 450$$

Given: 58 wheels total, 3 wheels per tempo.

Number of tempos = $58 \div 3$

Step 1: 30 wheels are needed for 10 tempos.
$$58 - 30 = 28 \text{ wheels left}$$

Step 2: 15 wheels are needed for 5 tempos.
$$28 - 15 = 13 \text{ wheels left}$$

Step 3: 9 wheels are needed for 3 tempos.
$$13 - 9 = 4 \text{ wheels left}$$

Step 4: 3 wheels are needed for 1 more tempo.
$$4 - 3 = 1 \text{ wheel left}$$

Can we make another tempo? No, because only 1 wheel is left and we need 3.

Total tempos = $10 + 5 + 3 + 1 = 19$ tempos

Answer: With 58 wheels, the factory can make 19 tempos. 1 wheel is left over.

Given: Total legs = 88. Each cow has 4 legs.

Number of cows = $88 \div 4$

| No. of legs taken | No. of cows | No. of legs remaining |
|---|---|---|
| 40 | 10 | 88 − 40 = 48 |
| 40 | 10 | 48 − 40 = 8 |
| 8 | 2 | 8 − 8 = 0 |

Total cows = $10 + 10 + 2 = 22$

Verification: $22 \times 4 = 88$ ✓

Answer: There are 22 cows in the dairy farm.

Given: Total legs = 72. Each octopus has 8 legs.

Number of octopuses = $72 \div 8$

$$8 \times 9 = 72$$

$$72 \div 8 = 9$$

Answer: There are 9 octopuses in the aquarium.

Given: 50 shuttlecocks, 4 per box.

Number of boxes = $50 \div 4$

$$4 \times 12 = 48$$
$$50 - 48 = 2 \text{ remaining}$$

So $50 \div 4 = 12$ remainder $2$.

They need 12 full boxes but cannot pack all shuttlecocks exactly.

Answer: They will need 12 boxes to pack 48 shuttlecocks. No, they cannot pack all shuttlecocks exactly. 2 shuttlecocks are left over.

Given: 75 saplings, 5 per row.

Number of rows = $75 \div 5$

$$5 \times 15 = 75$$

$$75 \div 5 = 15$$

Answer: Rakul Chachi has planted 15 rows of saplings.

a) $70 \div 5$:
Story: A teacher has 70 pencils to distribute equally among 5 students. How many pencils does each student get?

$$70 \div 5 = 14$$
(Since $5 \times 14 = 70$)

Each student gets 14 pencils.

b) $84 \div 7$:
Story: 84 children are divided into 7 equal teams for a sports event. How many children are in each team?

$$84 \div 7 = 12$$
(Since $7 \times 12 = 84$)

Each team has 12 children.

c) $69 \div 3$:
Story: A shopkeeper packs 69 mangoes equally into 3 baskets. How many mangoes are in each basket?

$$69 \div 3 = 23$$
(Since $3 \times 23 = 69$)

Each basket has 23 mangoes.

d) $93 \div 6$:
Story: 93 books are to be placed equally on 6 shelves. How many books go on each shelf? Are any books left over?

$$6 \times 15 = 90$$
$$93 - 90 = 3 \text{ remaining}$$
$$93 \div 6 = 15 \text{ remainder } 3$$

Each shelf gets 15 books, and 3 books are left over.

100 bikes × 2 people each:
$$100 \times 2 = 200 \text{ people}$$

200 bikes × 2 people each:
$$200 \times 2 = 400 \text{ people}$$
(This is double of $100 \times 2 = 200$, so $200 \times 2 = 400$)

100 cars × 4 people each:
$$100 \times 4 = 400 \text{ people}$$

500 cars × 4 people each:
$$500 \times 4 = 2000 \text{ people}$$
(Since $5 \times 4 = 20$, so $500 \times 4 = 2000$)

| Expression | Value |
|---|---|
| $500 \times 4$ | 2000 |
| $100 \times 4$ | 400 |
| $5 \times 4$ | 20 |
| $50 \times 4$ | 200 |

$$1 \times 3 = 3 \qquad 2 \times 3 = 6 \qquad 4 \times 3 = 12$$
$$10 \times 3 = 30 \qquad 20 \times 3 = 60 \qquad 40 \times 3 = 120$$
$$100 \times 3 = 300 \qquad 200 \times 3 = 600 \qquad 400 \times 3 = 1200$$

Pattern: Each time we multiply the number of groups by 10, the product also becomes 10 times larger.

$$2 \times 4 = 8 \qquad 4 \times 2 = 8 \qquad 8 \times 1 = 8$$
$$20 \times 4 = 80 \qquad 40 \times 2 = 80 \qquad 80 \times 1 = 80$$
$$200 \times 4 = 800 \qquad 400 \times 2 = 800 \qquad 800 \times 1 = 800$$

Pattern: All three expressions in each row give the same product (8, 80, 800), showing the commutative property. Multiplying by 10 or 100 shifts the product by one or two zeros.

$$3 \times 4 = 12 \qquad 3 \times 5 = 15 \qquad 3 \times 9 = 27$$
$$30 \times 4 = 120 \qquad 30 \times 5 = 150 \qquad 30 \times 9 = 270$$
$$300 \times 4 = 1200 \qquad 300 \times 5 = 1500 \qquad 300 \times 9 = 2700$$

Pattern: When we multiply the first number by 10, the product becomes 10 times larger. When we multiply by 100, the product becomes 100 times larger.

Given: 125 autorickshaws, 8 passengers each.

Total passengers = $125 \times 8$

Step 1: 100 autorickshaws:
$$100 \times 8 = 800 \text{ passengers}$$

Step 2: 20 autorickshaws:
$$20 \times 8 = 160 \text{ passengers}$$

Step 3: 5 autorickshaws:
$$5 \times 8 = 40 \text{ passengers}$$

Step 4: Total:
$$125 \times 8 = 800 + 160 + 40 = 1000 \text{ passengers}$$

| × | 8 |
|---|---|
| 100 | 800 |
| 20 | 160 |
| 5 | 40 |
| Total | 1000 |

Answer: 1000 people can travel in 125 autorickshaws in a single round.

Given: 6 kulhads per box, 174 boxes.

Total kulhads = $174 \times 6$

Step 1: 100 boxes:
$$100 \times 6 = 600 \text{ kulhads}$$

Step 2: 70 boxes:
$$70 \times 6 = 420 \text{ kulhads}$$

Step 3: 4 boxes:
$$4 \times 6 = 24 \text{ kulhads}$$

Step 4: Total:
$$174 \times 6 = 600 + 420 + 24 = 1044 \text{ kulhads}$$

Answer: Kahlu and Rabia have made 1044 kulhads.

Given: 465 students, 2 pencils each.

Total pencils = $465 \times 2$

Step 1: $400 \times 2 = 800$

Step 2: $60 \times 2 = 120$

Step 3: $5 \times 2 = 10$

Step 4: Total:
$$465 \times 2 = 800 + 120 + 10 = 930$$

Answer: The school needs to buy 930 pencils.

Given: 234 children, ₹5 each.

Total money = $234 \times 5$

Step 1: $200 \times 5 = 1000$

Step 2: $30 \times 5 = 150$

Step 3: $4 \times 5 = 20$

Step 4: Total:
$$234 \times 5 = 1000 + 150 + 20 = ₹1170$$

Answer: The children collect ₹1170 in all.

a) $439 \times 4$:
Story: A library has 439 shelves. Each shelf has 4 books. How many books are there in all?

$$439 \times 4 = (400 \times 4) + (30 \times 4) + (9 \times 4)$$
$$= 1600 + 120 + 36 = 1756$$

b) $514 \times 8$:
Story: A factory makes 514 boxes per day. Each box has 8 items. How many items are made in a day?

$$514 \times 8 = (500 \times 8) + (10 \times 8) + (4 \times 8)$$
$$= 4000 + 80 + 32 = 4112$$

c) $356 \times 5$:
Story: A school has 356 students. Each student plants 5 saplings on Environment Day. How many saplings are planted in all?

$$356 \times 5 = (300 \times 5) + (50 \times 5) + (6 \times 5)$$
$$= 1500 + 250 + 30 = 1780$$

d) $623 \times 7$:
Story: A train has 623 seats in each compartment. There are 7 compartments. How many seats are there in all?

$$623 \times 7 = (600 \times 7) + (20 \times 7) + (3 \times 7)$$
$$= 4200 + 140 + 21 = 4361$$

Given: 108 people, 9 boats, equal number in each boat.

People per boat = $108 \div 9$

Step 1: If 5 people sit in each boat: $9 \times 5 = 45$ people ferried.

Step 2: If 5 more sit (total 10 per boat): $9 \times 10 = 90$ people ferried.

Step 3: Remaining: $108 - 90 = 18$ people. $18 \div 9 = 2$ more per boat.

Step 4: Total per boat: $10 + 2 = 12$ people.

Verification: $9 \times 12 = 108$ ✓

Answer: Each boat should ferry 12 people.

Column A:
$$3 \div 3 = 1$$
$$30 \div 3 = 10$$
$$300 \div 3 = 100$$

Column B:
$$9 \div 3 = 3$$
$$90 \div 3 = 30$$
$$900 \div 3 = 300$$

Column C:
$$15 \div 3 = 5$$
$$150 \div 3 = 50$$
$$300 \div 3 = 100$$

$$5 \div 5 = 1 \qquad 8 \div 4 = 2 \qquad 4 \div 2 = 2$$
$$50 \div 5 = 10 \qquad 80 \div 4 = 20 \qquad 20 \div 2 = 10$$
$$500 \div 5 = 100 \qquad 800 \div 4 = 200 \qquad 100 \div 2 = 50$$

Pattern: When the dividend is multiplied by 10, the quotient also becomes 10 times larger (provided the divisor stays the same).

Given: 6 tyres per truck, 60 tyres total.

$$\text{Number of trucks} = 60 \div 6 = 10$$

Answer: There are 10 trucks.

Given: Each car has 4 wheels, 80 wheels total.

$$\text{Number of cars} = 80 \div 4 = 20$$

Answer: There are 20 cars in the parking space.

Given: Each ant has 6 legs, 600 legs total.

$$\text{Number of ants} = 600 \div 6 = 100$$

Answer: There are 100 ants.

Given: 800 rubber bands, 4 per packet.

$$\text{Number of packets} = 800 \div 4 = 200$$

Answer: The shop has 200 packets of rubber bands.

Given: 245 children, 7 buses, equal number in each.

$$\text{Children per bus} = 245 \div 7$$

$$7 \times 35 = 245$$

$$245 \div 7 = 35$$

Answer: 35 children are traveling in each bus.

Given: 88 km per day, 7 days in a week.

$$\text{Distance in a week} = 88 \times 7$$

$$= (80 \times 7) + (8 \times 7)$$
$$= 560 + 56 = 616 \text{ km}$$

Answer: The Toy Train travels 616 km in a week.

Given: 259 people, 7 per raft.

$$\text{Number of rafts} = 259 \div 7$$

$$7 \times 37 = 259$$

$$259 \div 7 = 37$$

Answer: It took 37 rafts.

Given: ₹45 per month, 6 months.

$$\text{Total savings} = 45 \times 6$$
$$= (40 \times 6) + (5 \times 6)$$
$$= 240 + 30 = ₹270$$

Answer: Anu will save ₹270 in 6 months.

Given: Total savings = ₹270 (from part a), 6 friends.

$$\text{Each friend gets} = 270 \div 6 = ₹45$$

Answer: Each friend gets ₹45.

Given: Total savings = ₹270, 3 friends.

$$\text{Each friend gets} = 270 \div 3 = ₹90$$

Answer: Each friend gets ₹90.

Given information: Raju makes 8 trips per day.

a) How much money does he make in a day?
We do NOT know the fare per trip. This cannot be exactly calculated.

b) How many trips does he make in 7 days?
$$8 \times 7 = 56 \text{ trips}$$
This CAN be exactly calculated. ✓

c) How much time does one trip take?
We do NOT know the duration of each trip. This cannot be exactly calculated.

d) How many trips does he make in 4 weeks?
4 weeks = $4 \times 7 = 28$ days
$$8 \times 28 = 224 \text{ trips}$$
This CAN be exactly calculated. ✓

Answer: Questions (b) and (d) can be exactly calculated with the given information.

a) Multiplying by 10 gives 0 in the ones digit of the number.

Any whole number multiplied by 10 always ends in 0 (e.g., $3 \times 10 = 30$, $15 \times 10 = 150$).

✓ Always True

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b) Multiplying a number by 2 gives an odd number.

Multiplying any whole number by 2 always gives an even number (e.g., $3 \times 2 = 6$, $5 \times 2 = 10$). It can never give an odd number.

✓ Never True

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c) Multiplying a number by 5 gives a number with 5 in the ones digit.

Multiplying by 5 gives ones digit 5 (when the number is odd) or 0 (when the number is even).
Example: $3 \times 5 = 15$ (ones digit 5), but $4 \times 5 = 20$ (ones digit 0).

✓ Sometimes True

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d) The number immediately after an odd number is an even number.

Odd and even numbers alternate. The number after any odd number is always even (e.g., after 3 comes 4, after 7 comes 8).

✓ Always True

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e) Halving any number gives an even number.

Halving 4 gives 2 (even), but halving 6 gives 3 (odd). Also, halving an odd number like 3 gives 1.5 (not even).

✓ Sometimes True

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f) Adding 0 to a number increases the number by 1.

Adding 0 to any number gives the same number (e.g., $5 + 0 = 5$, not 6). Adding 0 never increases a number.

✓ Never True