Equal Groups — NCERT Solutions

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:

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.

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 = $500 \div 2 = 250$

$$32 \times 2 = 64$$
Answer: 64

$$14 \times 2 = 28$$
Answer: 28

$$26 \times 2 = 52$$
Answer: 52

$$17 \times 2 = 34$$
Answer: 34

$$39 \times 2 = 78$$
Answer: 78

$$45 \times 2 = 90$$
Answer: 90

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).