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Raksha Bandhan

CBSE · Class 3 · Mathematics

NCERT Solutions for Raksha Bandhan — CBSE Class 3 Mathematics.

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57 Questions Solved · 16 Sections

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Let's Make Rakhis

1For each Rakhi, we need 1 flower, 2 threads and 4 beads. We need to make 5 Rakhis. Fill in: 5 × 1 = ?, 5 × 2 = ?, 5 × 4 = ?Show solution
Given: Each Rakhi needs 1 flower, 2 threads, and 4 beads. We need to make 5 Rakhis.

Flowers:
5×1=55 \times 1 = 5
For 5 Rakhis, we need 5 flowers.

Threads:
5×2=105 \times 2 = 10
For 5 Rakhis, we need 10 threads.

Beads:
4+4+4+4+4=204 + 4 + 4 + 4 + 4 = 20
5×4=205 \times 4 = 20
For 5 Rakhis, we need 20 beads.

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2Try it Out! For making 10 such Rakhis, we need ... flowers, ... threads and ... beads.Show solution
Given: Each Rakhi needs 1 flower, 2 threads, and 4 beads. We need to make 10 Rakhis.

Flowers:
10×1=1010 \times 1 = 10
We need 10 flowers.

Threads:
10×2=2010 \times 2 = 20
We need 20 threads.

Beads:
10×4=4010 \times 4 = 40
We need 40 beads.

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3There are 30 flowers, 30 threads and 30 beads. How many Rakhis can you make with this material?Show solution
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÷2=1530 \div 2 = 15 Rakhis.
- With 30 beads (4 per Rakhi): 30÷4=730 \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).

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Jagannath Sweet Shop

1How would we count the laddoos in the box? 3 + 3 + 3 = ? or 3 × 3 = ?Show solution
Given: Laddoos are arranged in 3 rows of 3 each.

3+3+3=93 + 3 + 3 = 9
3×3=93 \times 3 = 9

There are 9 laddoos in the box.

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2When 18 laddoos are shared equally among nine people, how many laddoos does each person get?Show solution
Given: 18 laddoos, 9 people.

We distribute equally:
18÷9=218 \div 9 = 2

Each person gets 2 laddoos.

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Try it Out! (Kaju Katlis)

1Look at the figure carefully. Count the number of kaju katlis. Distribute all kaju katlis equally among 5 people. How many will each get?Show solution
Note: Based on the context and the solution shown in the book (20÷5=420 \div 5 = 4), the tray contains 20 kaju katlis.

Total number of kaju katlis = 20

Distributing equally among 5 people:
20÷5=420 \div 5 = 4

Each person will get 4 kaju katlis.

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Let us Do (Page 90–91)

1Distribute all the kaju katlis equally among 4 people. How many kaju katlis will each get? Each will get ... kaju katlis. 16÷4=?16 \div 4 = ?Show solution
Given: 16 kaju katlis, 4 people.

Distributing equally:
16÷4=416 \div 4 = 4

Each person will get 4 kaju katlis.

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2Distribute 15 pedas in plates equally among 5 people. How many pedas will each get? 15÷5=?15 \div 5 = ?Show solution
Given: 15 pedas, 5 people.

15÷5=315 \div 5 = 3

15 equally shared by 5 is 3 each.

Each person will get 3 pedas.

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3Each cycle needs 2 wheels. How many cycles can be fitted with 12 wheels? 12÷2=?12 \div 2 = ?Show solution
Given: 12 wheels, each cycle needs 2 wheels.

12÷2=612 \div 2 = 6

6 cycles can be fitted with 12 wheels.

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4Look at the picture carefully. Count the number of jalebis. There are ... jalebis. Counting in groups: 4 + 4 + 4 + 4 + 4 + 4 = ? or ... × 4 = ? jalebis.Show solution
Given: Jalebis are arranged in 6 groups of 4 each.

4+4+4+4+4+4=244 + 4 + 4 + 4 + 4 + 4 = 24
6×4=246 \times 4 = 24

There are 24 jalebis.

Dhara's family has 9 members. If each person gets 4 jalebis:
9×4=36 jalebis needed9 \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=1236 - 24 = 12 more jalebis so that everyone can get 4 each.

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Plants in the Garden

1Dhara says: 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 = ? (8 times 6 = 8 × 6 = ?). Gopal says: 8 + 8 + 8 + 8 + 8 + 8 = 6 × 8. Who is correct?Show solution
Given: A flower bed with 8 rows of 6 plants (or 6 columns of 8 plants).

Dhara's way: 8 groups of 6
8×6=488 \times 6 = 48

Gopal's way: 6 groups of 8
6×8=486 \times 8 = 48

Both are correct! 8×6=6×8=488 \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).

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Visit to a Farm — Times-2 Table

1Complete the times-2 table: 1×2, 2×2, 3×2, 4×2, 5×2, 6×2, 7×2, 8×2, 9×2, 10×2.Show solution
Using repeated addition (counting chickens in pairs):

1×2=21 \times 2 = 2
2×2=42 \times 2 = 4
3×2=63 \times 2 = 6
4×2=84 \times 2 = 8
5×2=105 \times 2 = 10
6×2=126 \times 2 = 12
7×2=147 \times 2 = 14
8×2=168 \times 2 = 16
9×2=189 \times 2 = 18
10×2=2010 \times 2 = 20

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Skip Jumping Game — Times-3 Table

1Dhara is skip jumping by 3. Complete the table of number of jumps and number reached.Show solution
Dhara starts at 0 and jumps 3 each time:

| Number of jumps | Number reached |
|---|---|
| 1 jump | 3=1×33 = 1 \times 3 |
| 2 jumps | 3+3=6=2×33+3=6 = 2 \times 3 |
| 3 jumps | 3+3+3=9=3×33+3+3=9 = 3 \times 3 |
| 4 jumps | 3+3+3+3=12=4×33+3+3+3=12 = 4 \times 3 |
| 5 jumps | 3+3+3+3+3=15=5×33+3+3+3+3=15 = 5 \times 3 |
| 6 jumps | 18=6×318 = 6 \times 3 |
| 7 jumps | 21=7×321 = 7 \times 3 |
| 8 jumps | 24=8×324 = 8 \times 3 |
| 9 jumps | 27=9×327 = 9 \times 3 |
| 10 jumps | 30=10×330 = 10 \times 3 |}

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Let us Do (Skip Jumping — Page 96–97)

1Guess and write the next number she will jump onto (after 9, skip jumping by 3).Show solution
Dhara is skip jumping by 3. After 9, the next numbers are:
9+3=12,12+3=15,15+3=18,…9 + 3 = 12, \quad 12 + 3 = 15, \quad 15 + 3 = 18, \ldots
The next number after 9 is 12.

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2Is there a pattern in these numbers: 3, 6, 9, ...?Show solution
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, …

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3How many steps forward is Dhara jumping each time?Show solution
Dhara is jumping 3 steps forward each time.

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4Continue skip jumping by 6, by drawing the jumps on the number track.Show solution
Starting from 0, skip jumping by 6:
0→6→12→18→24→30→36→42→48→54→600 \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.

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5Can this skip jumping be used to form times-6 table? Write times-6 table in your notebook.Show solution
Yes! Each jump of 6 gives the next multiple of 6. The times-6 table is:
1×6=61 \times 6 = 6
2×6=122 \times 6 = 12
3×6=183 \times 6 = 18
4×6=244 \times 6 = 24
5×6=305 \times 6 = 30
6×6=366 \times 6 = 36
7×6=427 \times 6 = 42
8×6=488 \times 6 = 48
9×6=549 \times 6 = 54
10×6=6010 \times 6 = 60

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6Make times-4 table using repeated addition.Show solution
Using repeated addition (skip jumping by 4):
1×4=41 \times 4 = 4
2×4=4+4=82 \times 4 = 4+4 = 8
3×4=4+4+4=123 \times 4 = 4+4+4 = 12
4×4=4+4+4+4=164 \times 4 = 4+4+4+4 = 16
5×4=4+4+4+4+4=205 \times 4 = 4+4+4+4+4 = 20
6×4=246 \times 4 = 24
7×4=287 \times 4 = 28
8×4=328 \times 4 = 32
9×4=369 \times 4 = 36
10×4=4010 \times 4 = 40

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7Gopal is doing skip jumping of ... steps. After 27 he will jump on ..., ...Show solution
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=3627 + 9 = 36
36+9=4536 + 9 = 45

After 27 he will jump on 36, 45.

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8What times table can you construct from Gopal's jumps? Make it in your notebook.Show solution
Since Gopal is skip jumping by 9, we can construct the times-9 table:
1×9=91 \times 9 = 9
2×9=182 \times 9 = 18
3×9=273 \times 9 = 27
4×9=364 \times 9 = 36
5×9=455 \times 9 = 45
6×9=546 \times 9 = 54
7×9=637 \times 9 = 63
8×9=728 \times 9 = 72
9×9=819 \times 9 = 81
10×9=9010 \times 9 = 90

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9Dhara also skip jumps. Gopal notes down the jumps but misses the first few numbers: 32, 40, 48, 56. By what numbers was Dhara skip jumping? Construct the times table of this number in your notebook.Show solution
Given numbers: 32, 40, 48, 56.

Finding the difference between consecutive numbers:
40−32=8,48−40=8,56−48=840 - 32 = 8, \quad 48 - 40 = 8, \quad 56 - 48 = 8

Dhara was skip jumping by 8.

Times-8 table:
1×8=81 \times 8 = 8
2×8=162 \times 8 = 16
3×8=243 \times 8 = 24
4×8=324 \times 8 = 32
5×8=405 \times 8 = 40
6×8=486 \times 8 = 48
7×8=567 \times 8 = 56
8×8=648 \times 8 = 64
9×8=729 \times 8 = 72
10×8=8010 \times 8 = 80

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10Let us Play: Atya places a flower on 12. Skip jump with equal steps to reach the flower. What skip jumping number will you choose? Are there numbers that can be reached only through skip jumping by 1? Find 3 such numbers.Show solution
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).

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Seeing Patterns in Multiplication Tables

1Look at the times-5 table. What patterns do you see? Guess what will be the last digits of 11 × 5 and 12 × 5.Show solution
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.

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

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2Give 3 examples of numbers that when taken 5 times gives an answer ending with (a) 0 (b) 5.Show solution
(a) Numbers that give an answer ending in 0 when multiplied by 5 (even numbers):
2×5=10,4×5=20,6×5=302 \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):
1×5=5,3×5=15,7×5=351 \times 5 = 5, \quad 3 \times 5 = 15, \quad 7 \times 5 = 35
Examples: 1, 3, 7

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3Without finding the answer, can you tell the last digits of 18 × 5, 23 × 5, 32 × 5, 50 × 5?Show solution
Using the pattern (even × 5 ends in 0; odd × 5 ends in 5):

- 18×518 \times 5: 18 is even → last digit is 0 (Answer: 90)
- 23×523 \times 5: 23 is odd → last digit is 5 (Answer: 115)
- 32×532 \times 5: 32 is even → last digit is 0 (Answer: 160)
- 50×550 \times 5: 50 is even → last digit is 0 (Answer: 250)

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Let us Do (Word Problems — Page 100)

1aThere are 5 jars with 4 cookies in each jar. How many cookies are there?Show solution
Given: 5 jars, 4 cookies in each jar.

5×4=205 \times 4 = 20

There are 20 cookies in total.

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1bAn idli vessel contains 6 idli plates. In each plate we can make 4 idlis. How many idlis can be cooked in one go?Show solution
Given: 6 plates, 4 idlis per plate.

6×4=246 \times 4 = 24

24 idlis can be cooked in one go.

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1c30 cookies are to be distributed among 5 children equally. How many cookies will each child get?Show solution
Given: 30 cookies, 5 children.

30÷5=630 \div 5 = 6

Each child will get 6 cookies.

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1dRoro starts from 0 and takes 6 jumps to reach 18. All his jumps are of the same size. What is the size of Roro's jump?
1eToto does not take jumps of the same size and still reaches 18 in 6 jumps. How did Toto jump?
1fSuma saves ₹8 every day. After how many days will she have ₹56?
1gMary has 63 sea-shells. She gives 7 sea-shells to each of her 5 friends. How many does she have left?
2aSolve: 4×94 \times 9
2bSolve: 32÷832 \div 8
2cSolve: 6×76 \times 7
2dSolve: 45÷545 \div 5

Bhim's Rath — Spokes Problem

1Bhim needs to make spokes for 20 wheels of the rath. Each wheel needs 5 spokes. How many spokes will Bhim need?
2Try these: 30×5=?30 \times 5 = ?, 40×5=?40 \times 5 = ?, 50×5=?50 \times 5 = ?, 60×5=?60 \times 5 = ?, 70×5=?70 \times 5 = ?, 80×5=?80 \times 5 = ?, 90×5=?90 \times 5 = ?, 100×5=?100 \times 5 = ?
3Gopal collected 45 spokes. How many wheels can he make? Does Gopal have enough spokes to make 10 wheels? How many wheels can you make with 60 spokes?

Let us Do (Legs and Wheels — Page 102–103)

1A spider has 8 legs. 5 spiders will have ... legs. 10 spiders will have ... legs. 15 spiders will have ... legs. 23 spiders will have ... legs.
2A group of spiders have 32 legs. How many spiders are there in the group?
3Here is a 3-wheeled auto rickshaw. How many wheels are there in: (a) 18 auto rickshaws? (b) 34 auto rickshaws?
4Auto rickshaws in a garage have a total of 36 wheels. How many auto rickshaws are there in the garage?
5There is a line of 55 ants (one ant has 6 legs). What is the total number of legs in the line?
6Micky, the mouse, can see 48 legs of cows in the shed. How many cows are there in the shed?
7Karry, the crow, can see 24 horns of cows in the shed. What is the total number of legs in the shed?

Frog Jump Problems

1A frog is at 0. It takes jumps of only 7. What would be the largest number that the frog will reach before crossing 50?
2A frog wants to jump backwards from 50. It continues to take jumps of 7. What is the number after which it is not possible for the frog to make a jump of 7?
3What numbers should the frog start from to reach 0, taking jumps of 7 each time? What do you observe?

Puri Beach

1One wall-hanging costs ₹42. How much do two wall hangings cost?
2One rabdi cup costs ₹75. Preeti buys 5 cups of rabdi. She has only ₹100 notes. How many ₹100 notes should she give? How much will the shopkeeper return? What is the total cost of 5 cups?

Sea-Shells

1Dhruv has 112 sea-shells. He takes 28 for one necklace. 112 − 28 = 84. Then takes 28 more. How many shells are left? Then takes 28 for the third necklace. How many are left? Are the shells enough? How many necklaces can Dhruv make from 112 shells?
2Try these: 1. Kannu makes a necklace of 17 sea-shells. How many such necklaces can be made using 100 sea-shells?
3Dhruv finds 127 shiny pebbles. He distributes them equally to his 3 friends. How many will each get?
4Preeti has a ₹500 note and wants to exchange it for lower denomination notes. How many notes will she get if she wants: (a) All ₹50 notes? (b) All ₹20 notes? (c) All ₹10 notes?

LaLue Explorer — Number Cards in Envelopes

1There are ten number cards from 1–10 in five sealed envelopes (two cards each). The fifth envelope contains cards 5 and 9, with 5 × 9 = 45 written on it. Identify the number cards inside each of the other envelopes (products given on envelopes).

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Frequently Asked Questions

What are the important topics in Raksha Bandhan for CBSE Class 3 Mathematics?
Raksha Bandhan covers several key topics that are frequently asked in CBSE Class 3 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Raksha Bandhan — CBSE Class 3 Mathematics?
Understand the core concepts first, then work through the 43 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
Where can I get free NCERT Solutions for Raksha Bandhan Class 3 Mathematics?
This page has free step-by-step NCERT Solutions for every exercise question in Raksha Bandhan (CBSE Class 3 Mathematics) — written the way examiners award marks: given, formula, working, answer.

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