Raksha Bandhan — NCERT Solutions

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:

| 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.