Raksha Bandhan

Given: Each Rakhi needs 1 flower, 2 threads, and 4 beads. We need to make 5 Rakhis.

Flowers:
$$5 \times 1 = 5$$
For 5 Rakhis, we need 5 flowers.

Threads:
$$5 \times 2 = 10$$
For 5 Rakhis, we need 10 threads.

Beads:
$$4 + 4 + 4 + 4 + 4 = 20$$
$$5 \times 4 = 20$$
For 5 Rakhis, we need 20 beads.

Given: Each Rakhi needs 1 flower, 2 threads, and 4 beads. We need to make 10 Rakhis.

Flowers:
$$10 \times 1 = 10$$
We need 10 flowers.

Threads:
$$10 \times 2 = 20$$
We need 20 threads.

Beads:
$$10 \times 4 = 40$$
We need 40 beads.

Given: 30 flowers, 30 threads, 30 beads. Each Rakhi needs 1 flower, 2 threads, 4 beads.

- With 30 flowers: can make 30 Rakhis.
- With 30 threads (2 per Rakhi): $30 \div 2 = 15$ Rakhis.
- With 30 beads (4 per Rakhi): $30 \div 4 = 7$ Rakhis (with 2 beads left over).

The limiting material is beads. We can make only 7 Rakhis (with 2 beads remaining, 2 threads remaining, and 23 flowers remaining).

Given: Laddoos are arranged in 3 rows of 3 each.

$$3 + 3 + 3 = 9$$
$$3 \times 3 = 9$$

There are 9 laddoos in the box.

Given: 18 laddoos, 9 people.

We distribute equally:
$$18 \div 9 = 2$$

Each person gets 2 laddoos.

Note: Based on the context and the solution shown in the book ($20 \div 5 = 4$), the tray contains 20 kaju katlis.

Total number of kaju katlis = 20

Distributing equally among 5 people:
$$20 \div 5 = 4$$

Each person will get 4 kaju katlis.

Given: 16 kaju katlis, 4 people.

Distributing equally:
$$16 \div 4 = 4$$

Each person will get 4 kaju katlis.

Given: 15 pedas, 5 people.

$$15 \div 5 = 3$$

15 equally shared by 5 is 3 each.

Each person will get 3 pedas.

Given: 12 wheels, each cycle needs 2 wheels.

$$12 \div 2 = 6$$

6 cycles can be fitted with 12 wheels.

Given: Jalebis are arranged in 6 groups of 4 each.

$$4 + 4 + 4 + 4 + 4 + 4 = 24$$
$$6 \times 4 = 24$$

There are 24 jalebis.

Dhara's family has 9 members. If each person gets 4 jalebis:
$$9 \times 4 = 36 \text{ jalebis needed}$$
But Dhara has only 24 jalebis, so there are NOT enough jalebis for everyone to have 4 each.

Dhara should buy $36 - 24 = 12$ more jalebis so that everyone can get 4 each.

Given: A flower bed with 8 rows of 6 plants (or 6 columns of 8 plants).

Dhara's way: 8 groups of 6
$$8 \times 6 = 48$$

Gopal's way: 6 groups of 8
$$6 \times 8 = 48$$

Both are correct! $8 \times 6 = 6 \times 8 = 48$. The total number of plants is 48. This shows that multiplication can be done in any order (commutative property).

Using repeated addition (counting chickens in pairs):

$$1 \times 2 = 2$$
$$2 \times 2 = 4$$
$$3 \times 2 = 6$$
$$4 \times 2 = 8$$
$$5 \times 2 = 10$$
$$6 \times 2 = 12$$
$$7 \times 2 = 14$$
$$8 \times 2 = 16$$
$$9 \times 2 = 18$$
$$10 \times 2 = 20$$

Dhara starts at 0 and jumps 3 each time:

| Number of jumps | Number reached |
|---|---|
| 1 jump | $3 = 1 \times 3$ |
| 2 jumps | $3+3=6 = 2 \times 3$ |
| 3 jumps | $3+3+3=9 = 3 \times 3$ |
| 4 jumps | $3+3+3+3=12 = 4 \times 3$ |
| 5 jumps | $3+3+3+3+3=15 = 5 \times 3$ |
| 6 jumps | $18 = 6 \times 3$ |
| 7 jumps | $21 = 7 \times 3$ |
| 8 jumps | $24 = 8 \times 3$ |
| 9 jumps | $27 = 9 \times 3$ |
| 10 jumps | $30 = 10 \times 3$ |}

Dhara is skip jumping by 3. After 9, the next numbers are:
$$9 + 3 = 12, \quad 12 + 3 = 15, \quad 15 + 3 = 18, \ldots$$
The next number after 9 is 12.

Yes, there is a pattern. Each number is 3 more than the previous number. The numbers are multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, …

Dhara is jumping 3 steps forward each time.

Starting from 0, skip jumping by 6:
$$0 \to 6 \to 12 \to 18 \to 24 \to 30 \to 36 \to 42 \to 48 \to 54 \to 60$$
The numbers reached are: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.

Yes! Each jump of 6 gives the next multiple of 6. The times-6 table is:
$$1 \times 6 = 6$$
$$2 \times 6 = 12$$
$$3 \times 6 = 18$$
$$4 \times 6 = 24$$
$$5 \times 6 = 30$$
$$6 \times 6 = 36$$
$$7 \times 6 = 42$$
$$8 \times 6 = 48$$
$$9 \times 6 = 54$$
$$10 \times 6 = 60$$

Using repeated addition (skip jumping by 4):
$$1 \times 4 = 4$$
$$2 \times 4 = 4+4 = 8$$
$$3 \times 4 = 4+4+4 = 12$$
$$4 \times 4 = 4+4+4+4 = 16$$
$$5 \times 4 = 4+4+4+4+4 = 20$$
$$6 \times 4 = 24$$
$$7 \times 4 = 28$$
$$8 \times 4 = 32$$
$$9 \times 4 = 36$$
$$10 \times 4 = 40$$

Note: Based on the figure (not visible), Gopal appears to be skip jumping by 9 (since 27 is a multiple of 9 and the pattern fits).

Gopal is doing skip jumping of 9 steps.

After 27:
$$27 + 9 = 36$$
$$36 + 9 = 45$$

After 27 he will jump on 36, 45.

Since Gopal is skip jumping by 9, we can construct the times-9 table:
$$1 \times 9 = 9$$
$$2 \times 9 = 18$$
$$3 \times 9 = 27$$
$$4 \times 9 = 36$$
$$5 \times 9 = 45$$
$$6 \times 9 = 54$$
$$7 \times 9 = 63$$
$$8 \times 9 = 72$$
$$9 \times 9 = 81$$
$$10 \times 9 = 90$$

Given numbers: 32, 40, 48, 56.

Finding the difference between consecutive numbers:
$$40 - 32 = 8, \quad 48 - 40 = 8, \quad 56 - 48 = 8$$

Dhara was skip jumping by 8.

Times-8 table:
$$1 \times 8 = 8$$
$$2 \times 8 = 16$$
$$3 \times 8 = 24$$
$$4 \times 8 = 32$$
$$5 \times 8 = 40$$
$$6 \times 8 = 48$$
$$7 \times 8 = 56$$
$$8 \times 8 = 64$$
$$9 \times 8 = 72$$
$$10 \times 8 = 80$$

To reach 12 in the fewest jumps, choose the largest factor of 12.

Factors of 12: 1, 2, 3, 4, 6, 12.

Choose skip jumping by 6 → reach 12 in just 2 jumps (0→6→12). This is the fewest jumps (other than jumping directly, which is not allowed).

Numbers reachable only by skip jumping by 1 are numbers that have no factors other than 1 and themselves — these are prime numbers.

Three such numbers: 7, 11, 13 (prime numbers — they cannot be reached by any skip jump other than 1).

Pattern in times-5 table: Every answer ends in either 0 or 5. When an even number is multiplied by 5, the answer ends in 0. When an odd number is multiplied by 5, the answer ends in 5.

$$11 \times 5: \text{ 11 is odd, so the answer ends in } \mathbf{5}. \quad (11 \times 5 = 55)$$
$$12 \times 5: \text{ 12 is even, so the answer ends in } \mathbf{0}. \quad (12 \times 5 = 60)$$

(a) Numbers that give an answer ending in 0 when multiplied by 5 (even numbers):
$$2 \times 5 = 10, \quad 4 \times 5 = 20, \quad 6 \times 5 = 30$$
Examples: 2, 4, 6

(b) Numbers that give an answer ending in 5 when multiplied by 5 (odd numbers):
$$1 \times 5 = 5, \quad 3 \times 5 = 15, \quad 7 \times 5 = 35$$
Examples: 1, 3, 7

Using the pattern (even × 5 ends in 0; odd × 5 ends in 5):

- $18 \times 5$: 18 is even → last digit is 0 (Answer: 90)
- $23 \times 5$: 23 is odd → last digit is 5 (Answer: 115)
- $32 \times 5$: 32 is even → last digit is 0 (Answer: 160)
- $50 \times 5$: 50 is even → last digit is 0 (Answer: 250)

Given: 5 jars, 4 cookies in each jar.

$$5 \times 4 = 20$$

There are 20 cookies in total.

Given: 6 plates, 4 idlis per plate.

$$6 \times 4 = 24$$

24 idlis can be cooked in one go.

Given: 30 cookies, 5 children.

$$30 \div 5 = 6$$

Each child will get 6 cookies.

Given: Starts at 0, reaches 18 in 6 equal jumps.

$$18 \div 6 = 3$$

The size of Roro's jump is 3.

Toto needs to reach 18 in 6 jumps of different (or unequal) sizes. One possible way:
$$1 + 2 + 3 + 4 + 5 + 3 = 18$$
Another way:
$$5 + 4 + 3 + 2 + 2 + 2 = 18$$
Many answers are possible. The key is that the 6 jumps add up to 18 but are not all the same size.

Given: Saves ₹8 per day, total = ₹56.

$$56 \div 8 = 7$$

Suma will have ₹56 after 7 days.

Given: Mary has 63 sea-shells. She gives 7 to each of 5 friends.

Shells given away:
$$5 \times 7 = 35$$

Shells left:
$$63 - 35 = 28$$

Mary has 28 sea-shells left.

$$4 \times 9 = 36$$

Answer: 36

(Word problem example: There are 4 boxes with 9 apples each. Total apples = 36.)

$$32 \div 8 = 4$$

Answer: 4

(Word problem example: 32 chocolates shared equally among 8 children — each gets 4.)

$$6 \times 7 = 42$$

Answer: 42

(Word problem example: 6 weeks have 7 days each. Total days = 42.)

$$45 \div 5 = 9$$

Answer: 9

(Word problem example: 45 flowers arranged equally in 5 vases — each vase gets 9 flowers.)

Given: 20 wheels, 5 spokes per wheel.

Step 1: 10 wheels will need:
$$10 \times 5 = 50 \text{ spokes}$$

Step 2: Another 10 wheels will need:
$$10 \times 5 = 50 \text{ spokes}$$

Total spokes:
$$50 + 50 = 100 \text{ spokes}$$

Bhim will need 100 spokes.

Using the pattern (each group of 10 wheels needs 50 spokes):

$$30 \times 5 = 150 \quad (50+50+50)$$
$$40 \times 5 = 200$$
$$50 \times 5 = 250$$
$$60 \times 5 = 300$$
$$70 \times 5 = 350$$
$$80 \times 5 = 400$$
$$90 \times 5 = 450$$
$$100 \times 5 = 500$$

Pattern: As the number increases by 10, the answer increases by 50. All answers end in 0.

Each wheel needs 5 spokes.

With 45 spokes:
$$45 \div 5 = 9$$
Gopal can make 9 wheels.

Does he have enough for 10 wheels?
10 wheels need $10 \times 5 = 50$ spokes. Gopal has only 45. No, he does not have enough.

With 60 spokes:
$$60 \div 5 = 12$$
You can make 12 wheels with 60 spokes.

Each spider has 8 legs.

$$5 \text{ spiders}: 5 \times 8 = 40 \text{ legs}$$
$$10 \text{ spiders}: 10 \times 8 = 80 \text{ legs}$$
$$15 \text{ spiders}: 15 \times 8 = 120 \text{ legs}$$
$$23 \text{ spiders}: 23 \times 8 = 184 \text{ legs}$$

Given: 32 legs, each spider has 8 legs.

$$32 \div 8 = 4$$

There are 4 spiders in the group.

Each auto rickshaw has 3 wheels.

(a) 18 auto rickshaws:
$$18 \times 3 = 54 \text{ wheels}$$

(b) 34 auto rickshaws:
$$34 \times 3 = 102 \text{ wheels}$$

Given: 36 wheels, each auto rickshaw has 3 wheels.

$$36 \div 3 = 12$$

There are 12 auto rickshaws in the garage.

Given: 55 ants, each ant has 6 legs.

$$55 \times 6 = 330$$

The total number of legs is 330.

Given: 48 legs, each cow has 4 legs.

$$48 \div 4 = 12$$

There are 12 cows in the shed.

Given: 24 horns, each cow has 2 horns.

Number of cows:
$$24 \div 2 = 12 \text{ cows}$$

Total legs (each cow has 4 legs):
$$12 \times 4 = 48 \text{ legs}$$

The total number of legs in the shed is 48.

The frog jumps: 0, 7, 14, 21, 28, 35, 42, 49, 56, …

7 \times 7 = 49 < 50
8 \times 7 = 56 > 50

The largest number the frog reaches before crossing 50 is 49.

The frog jumps backwards from 50: 50, 43, 36, 29, 22, 15, 8, 1.

From 1, the frog cannot jump back 7 (would reach $1 - 7 = -6$, which is below 0).

The number after which it is not possible to make a jump of 7 is 1 (since 1 - 7 < 0).

The frog reaches 0 by jumping backwards in steps of 7. Starting numbers that eventually reach 0:
$$7 \to 0 \quad (1 \text{ jump})$$
$$14 \to 7 \to 0 \quad (2 \text{ jumps})$$
$$21 \to 14 \to 7 \to 0 \quad (3 \text{ jumps})$$
$$28, 35, 42, 49, \ldots$$

These are all multiples of 7: 7, 14, 21, 28, 35, 42, 49, …

Observation: Only multiples of 7 can reach exactly 0 by taking jumps of 7.

Given: One wall-hanging = ₹42.

$$2 \times ₹42 = ₹42 + ₹42 = ₹84$$

The cost of two wall hangings is ₹84.

Given: One cup = ₹75, Preeti buys 5 cups.

Total cost:
$$5 \times ₹75 = ₹375$$

Number of ₹100 notes needed:
She needs at least ₹375. 3 \times 100 = 300 < 375 and 4 \times 100 = 400 > 375.
She should give 4 notes of ₹100 (= ₹400).

Change returned:
$$₹400 - ₹375 = ₹25$$

The shopkeeper will return ₹25 to Preeti.

Given: 112 shells, each necklace needs 28 shells.

After 1st necklace:
$$112 - 28 = 84 \text{ shells left}$$

After 2nd necklace:
$$84 - 28 = 56 \text{ shells left}$$

After 3rd necklace:
$$56 - 28 = 28 \text{ shells left}$$

Yes, the shells are enough for 3 necklaces (one for each friend).

After 4th necklace:
$$28 - 28 = 0 \text{ shells left}$$

Dhruv can make 4 necklaces from 112 shells.

Given: 100 shells, each necklace needs 17 shells.

Using repeated subtraction:
$$100 - 17 = 83$$
$$83 - 17 = 66$$
$$66 - 17 = 49$$
$$49 - 17 = 32$$
$$32 - 17 = 15$$
15 is less than 17, so we stop.

5 necklaces can be made (with 15 shells remaining).

Given: 127 pebbles, 3 friends.

Using repeated subtraction or division:
$$127 \div 3 = 42 \text{ remainder } 1$$

Each friend will get 42 pebbles (and 1 pebble will be left over).

Given: ₹500 note.

(a) All ₹50 notes:
$$500 \div 50 = 10$$
She will get 10 notes of ₹50.

(b) All ₹20 notes:
$$500 \div 20 = 25$$
She will get 25 notes of ₹20.

(c) All ₹10 notes:
$$500 \div 10 = 50$$
She will get 50 notes of ₹10.

Given: Cards 1–10, each card used exactly once, two cards per envelope. The fifth envelope has 5 and 9 (product = 45). The remaining cards are 1, 2, 3, 4, 6, 7, 8, 10.

Note: The products on the other four envelopes are not visible in the OCR. However, using the remaining cards {1, 2, 3, 4, 6, 7, 8, 10}, we need to pair them into 4 pairs. One logical set of pairings (based on typical puzzle design):

- Envelope 1: Cards 1 and 6 → $1 \times 6 = 6$
- Envelope 2: Cards 2 and 10 → $2 \times 10 = 20$
- Envelope 3: Cards 3 and 8 → $3 \times 8 = 24$
- Envelope 4: Cards 4 and 7 → $4 \times 7 = 28$
- Envelope 5: Cards 5 and 9 → $5 \times 9 = 45$ (given)

Note: The exact pairing depends on the products written on the envelopes in the figure, which is not visible in the OCR. Students should match the product on each envelope to the correct pair of cards from 1–10.