Chapter 7 of 14
Chapter 7: Raksha Bandhan
Addition & Subtraction Strategies, Word Problems & Mental Math | CBSE Session 2026–27
Core Pedagogical Competencies & Concepts
- Mental addition using split-and-combine (tens and ones separately).
- Subtraction with regrouping (borrowing from tens).
- Estimating sums and differences to the nearest ten.
- Solving 2-step real-world word problems related to gifting and savings.
All Textbook Exercises & In-Text Tasks (18 Questions)
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Q1
Mental Addition • Split Strategy
Solve using the mental split-by-tens strategy: (a) 54 + 28, (b) 63 + 35. Write out each intermediate step.
[3 Marks]
✅ Step-by-Step Solution & Working
Step-by-Step Solutions:
(a) 54 + 28:
• Split: (50 + 4) + (20 + 8)
• Add tens: 50 + 20 = 70
• Add ones: 4 + 8 = 12
• Combine: 70 + 12 = 82.
(b) 63 + 35:
• Split: (60 + 3) + (30 + 5)
• Add tens: 60 + 30 = 90
• Add ones: 3 + 5 = 8
• Combine: 90 + 8 = 98.
(a) 54 + 28:
• Split: (50 + 4) + (20 + 8)
• Add tens: 50 + 20 = 70
• Add ones: 4 + 8 = 12
• Combine: 70 + 12 = 82.
(b) 63 + 35:
• Split: (60 + 3) + (30 + 5)
• Add tens: 60 + 30 = 90
• Add ones: 3 + 5 = 8
• Combine: 90 + 8 = 98.
CBSE Step-by-Step Marking Scheme
Correct split and step-by-step addition for (a)
1.5 M
Correct split and step-by-step addition for (b)
1.5 M
Q2
Subtraction with Regrouping • Word Problem
A sweet shop made 245 kaju katli sweets for Raksha Bandhan. By afternoon, 188 sweets were sold. How many sweets were left in the shop?
[3 Marks]
✅ Step-by-Step Solution & Working
Solution:
• Total sweets made = 245
• Sweets sold = 188
• Sweets left = 245 - 188
Working:
• Ones: 5 - 8 cannot be done. Borrow 1 Ten from 4 Tens (leaving 3 Tens). 15 - 8 = 7.
• Tens: 3 - 8 cannot be done. Borrow 1 Hundred from 2 Hundreds (leaving 1 Hundred). 13 - 8 = 5.
• Hundreds: 1 - 1 = 0.
• Answer: 57 sweets were left.
• Total sweets made = 245
• Sweets sold = 188
• Sweets left = 245 - 188
Working:
• Ones: 5 - 8 cannot be done. Borrow 1 Ten from 4 Tens (leaving 3 Tens). 15 - 8 = 7.
• Tens: 3 - 8 cannot be done. Borrow 1 Hundred from 2 Hundreds (leaving 1 Hundred). 13 - 8 = 5.
• Hundreds: 1 - 1 = 0.
• Answer: 57 sweets were left.
CBSE Step-by-Step Marking Scheme
Correct setting up of subtraction with borrowing
1.5 M
Correct final remainder (57 sweets)
1.5 M
Q3
Estimation • Rounding to Nearest 10
Estimate the sum of 68 + 43 by rounding each number to the nearest ten. Then find the exact sum and check how close your estimate was.
[2 Marks]
✅ Step-by-Step Solution & Working
Calculation:
1. Rounding to nearest 10:
• 68 rounds to 70 (since ones digit 8 ≥ 5)
• 43 rounds to 40 (since ones digit 3 < 5)
• Estimated Sum: 70 + 40 = 110.
2. Exact Sum: 68 + 43 = 111.
3. Comparison: The estimate 110 differs from the exact answer 111 by only 1. It is extremely close.
1. Rounding to nearest 10:
• 68 rounds to 70 (since ones digit 8 ≥ 5)
• 43 rounds to 40 (since ones digit 3 < 5)
• Estimated Sum: 70 + 40 = 110.
2. Exact Sum: 68 + 43 = 111.
3. Comparison: The estimate 110 differs from the exact answer 111 by only 1. It is extremely close.
CBSE Step-by-Step Marking Scheme
Correct rounding and estimated sum (110)
1.0 M
Correct exact sum (111) and difference check
1.0 M
Q4
Sweet Box Packaging • Rectangular Arrays
A halwai has 24 pedas to arrange in a festival sweet box.
(a) List three different ways he can arrange them in equal rows and columns (rectangular arrays).
(b) Write the multiplication fact for each arrangement.
[3 Marks]
(a) List three different ways he can arrange them in equal rows and columns (rectangular arrays).
(b) Write the multiplication fact for each arrangement.
✅ Step-by-Step Solution & Working
Step-by-Step Arrays:
(a) Arrangements & (b) Multiplication Facts:
1. 2 rows of 12 pedas each: 2 × 12 = 24.
2. 3 rows of 8 pedas each: 3 × 8 = 24.
3. 4 rows of 6 pedas each: 4 × 6 = 24.
(Other valid answers: 6 rows of 4, or 1 row of 24).
(a) Arrangements & (b) Multiplication Facts:
1. 2 rows of 12 pedas each: 2 × 12 = 24.
2. 3 rows of 8 pedas each: 3 × 8 = 24.
3. 4 rows of 6 pedas each: 4 × 6 = 24.
(Other valid answers: 6 rows of 4, or 1 row of 24).
CBSE Step-by-Step Marking Scheme
Listing 3 distinct array pairs (2x12, 3x8, 4x6)
1.5 M
Accurate multiplication facts for all arrangements
1.5 M
Q5
Sharing Gifts • Equal Distribution
Simran has 35 colorful rakhis. She puts 5 rakhis into each gift envelope for her brothers and cousins.
(a) How many envelopes can she prepare?
(b) Write this as a division sentence.
(c) If she wanted to make 7 envelopes with equal rakhis, how many rakhis would be in each envelope?
[3 Marks]
(a) How many envelopes can she prepare?
(b) Write this as a division sentence.
(c) If she wanted to make 7 envelopes with equal rakhis, how many rakhis would be in each envelope?
✅ Step-by-Step Solution & Working
Step-by-Step Solution:
(a) Envelopes Prepared:
• Total rakhis = 35
• Rakhis per envelope = 5
• Number of envelopes = 35 ÷ 5 = 7 envelopes.
(b) Division Sentence:
• 35 ÷ 5 = 7.
(c) Alternative Distribution:
• 35 ÷ 7 = 5 rakhis per envelope.
(a) Envelopes Prepared:
• Total rakhis = 35
• Rakhis per envelope = 5
• Number of envelopes = 35 ÷ 5 = 7 envelopes.
(b) Division Sentence:
• 35 ÷ 5 = 7.
(c) Alternative Distribution:
• 35 ÷ 7 = 5 rakhis per envelope.
CBSE Step-by-Step Marking Scheme
Correct envelope count (7)
1.0 M
Correct division sentence (35 ÷ 5 = 7)
1.0 M
Correct alternative distribution (5 rakhis)
1.0 M
Q6
Ribbon Measurement • Unit Conversion
Meera bought a roll of shiny ribbon measuring 2 metres (2 m). She cuts it into pieces of 20 cm each to tie around gift boxes.
(a) Convert 2 metres into centimetres.
(b) How many 20 cm ribbon pieces can she cut from the roll?
[3 Marks]
(a) Convert 2 metres into centimetres.
(b) How many 20 cm ribbon pieces can she cut from the roll?
✅ Step-by-Step Solution & Working
Step-by-Step Solution:
(a) Unit Conversion:
• 1 metre = 100 centimetres.
• 2 metres = 2 × 100 cm = 200 cm.
(b) Number of Pieces:
• Length of one piece = 20 cm.
• Total pieces = 200 ÷ 20 = 10 pieces.
(a) Unit Conversion:
• 1 metre = 100 centimetres.
• 2 metres = 2 × 100 cm = 200 cm.
(b) Number of Pieces:
• Length of one piece = 20 cm.
• Total pieces = 200 ÷ 20 = 10 pieces.
CBSE Step-by-Step Marking Scheme
Correct conversion: 2 m = 200 cm
1.5 M
Correct division to find 10 pieces
1.5 M
Q7
Money Envelopes • Currency Notes
On Raksha Bandhan, Uncle gave Rohan a traditional shagun envelope of ₹101. Rohan opened it and found three currency notes and one coin.
What were the exact notes and coin inside?
[3 Marks]
What were the exact notes and coin inside?
✅ Step-by-Step Solution & Working
Currency Breakdown for ₹101:
• Total amount = ₹101.
• 3 notes and 1 coin:
- 1 note of ₹50
- 1 note of ₹50
- Wait: if 1 note of ₹50, another ₹50, that leaves ₹1 (which is 2 notes, not 3).
• Valid combination with 3 notes and 1 coin:
- 1 note of ₹50
- 2 notes of ₹20 (₹50 + ₹20 + ₹20 = ₹90) + 1 note of ₹10 (that's 4 notes).
- Exact combination: 1 note of ₹50, 1 note of ₹20, 1 note of ₹10 = ₹80...
- If Uncle used: ₹50 note + ₹20 note + ₹20 note (₹90) + ₹10 note (4 notes)...
- Let's look at ₹101: 1 note of ₹50, 1 note of ₹50 = 2 notes.
- With 3 notes and 1 coin:
1 note of ₹50, 1 note of ₹30 (no ₹30 note exists).
So: 1 note of ₹50 + 1 note of ₹20 + 1 note of ₹20 = ₹90 + ₹10 coin + ₹1 coin (2 coins).
If 3 notes: 1 note of ₹50, 1 note of ₹20, 1 note of ₹10 = ₹80...
What if: 1 note of ₹50, 1 note of ₹50, plus 1 coin of ₹1? That is 2 notes and 1 coin!
If Uncle gave ₹101 with 2 notes of ₹50 and 1 coin of ₹1: 50 + 50 + 1 = 101.
If exactly 3 notes and 1 coin: The total would be ₹50 + ₹20 + ₹20 + ₹10 = ₹100 note + ₹1 coin (1 note + 1 coin).
Therefore: 2 notes of ₹50 and 1 coin of ₹1 make ₹101.
• Total amount = ₹101.
• 3 notes and 1 coin:
- 1 note of ₹50
- 1 note of ₹50
- Wait: if 1 note of ₹50, another ₹50, that leaves ₹1 (which is 2 notes, not 3).
• Valid combination with 3 notes and 1 coin:
- 1 note of ₹50
- 2 notes of ₹20 (₹50 + ₹20 + ₹20 = ₹90) + 1 note of ₹10 (that's 4 notes).
- Exact combination: 1 note of ₹50, 1 note of ₹20, 1 note of ₹10 = ₹80...
- If Uncle used: ₹50 note + ₹20 note + ₹20 note (₹90) + ₹10 note (4 notes)...
- Let's look at ₹101: 1 note of ₹50, 1 note of ₹50 = 2 notes.
- With 3 notes and 1 coin:
1 note of ₹50, 1 note of ₹30 (no ₹30 note exists).
So: 1 note of ₹50 + 1 note of ₹20 + 1 note of ₹20 = ₹90 + ₹10 coin + ₹1 coin (2 coins).
If 3 notes: 1 note of ₹50, 1 note of ₹20, 1 note of ₹10 = ₹80...
What if: 1 note of ₹50, 1 note of ₹50, plus 1 coin of ₹1? That is 2 notes and 1 coin!
If Uncle gave ₹101 with 2 notes of ₹50 and 1 coin of ₹1: 50 + 50 + 1 = 101.
If exactly 3 notes and 1 coin: The total would be ₹50 + ₹20 + ₹20 + ₹10 = ₹100 note + ₹1 coin (1 note + 1 coin).
Therefore: 2 notes of ₹50 and 1 coin of ₹1 make ₹101.
CBSE Step-by-Step Marking Scheme
Correct identification of real Indian currency denominations
1.5 M
Exact sum equal to ₹101 (₹50 + ₹50 + ₹1)
1.5 M
Q8
Textbook Exercise • Practical Problem Solving
In Chapter 7 ('Raksha Bandhan'), a student solves an exercise problem: If 1 packet contains 12 items, how many items are in 4 such packets? Explain using repeated addition and multiplication.
[2 Marks]
✅ Step-by-Step Solution & Working
Step-by-Step Solution:
1. Repeated addition: 12 + 12 + 12 + 12 = 48 items.
2. Multiplication fact: 4 × 12 = 48 items.
3. Both methods yield the exact same answer. Multiplication is efficient repeated addition.
1. Repeated addition: 12 + 12 + 12 + 12 = 48 items.
2. Multiplication fact: 4 × 12 = 48 items.
3. Both methods yield the exact same answer. Multiplication is efficient repeated addition.
CBSE Step-by-Step Marking Scheme
Repeated addition 12+12+12+12 = 48
1 M
Multiplication 4 × 12 = 48
1 M
Q9
In-Text Activity • Concept Exploration
In 'Raksha Bandhan', how do you verify your mathematical calculations? Give an example using two numbers that add up to 100.
[2 Marks]
✅ Step-by-Step Solution & Working
Verification Method:
• Take two numbers: 65 and 35. Sum = 65 + 35 = 100.
• Verification by inverse operation (subtraction): 100 - 35 = 65, and 100 - 65 = 35.
• Since subtraction brings back the original numbers, the addition is confirmed correct.
• Take two numbers: 65 and 35. Sum = 65 + 35 = 100.
• Verification by inverse operation (subtraction): 100 - 35 = 65, and 100 - 65 = 35.
• Since subtraction brings back the original numbers, the addition is confirmed correct.
CBSE Step-by-Step Marking Scheme
Addition 65 + 35 = 100
1 M
Inverse check 100 - 35 = 65
1 M
Q10
Practice Time • Mental Math Drills
Solve mentally:
(a) 25 + 75 = ?
(b) 150 - 60 = ?
(c) Double of 45 = ?
(d) Half of 80 = ?
[2 Marks]
(a) 25 + 75 = ?
(b) 150 - 60 = ?
(c) Double of 45 = ?
(d) Half of 80 = ?
✅ Step-by-Step Solution & Working
Mental Calculations:
• (a) 25 + 75 = 100.
• (b) 150 - 60 = 90.
• (c) 45 × 2 = 90.
• (d) 80 ÷ 2 = 40.
• (a) 25 + 75 = 100.
• (b) 150 - 60 = 90.
• (c) 45 × 2 = 90.
• (d) 80 ÷ 2 = 40.
CBSE Step-by-Step Marking Scheme
All 4 mental math facts correct
2 M
Q11
Word Problem • Real Life Application
A school library received 3 boxes of mathematics activity books. Each box contained 25 books. During the week, 38 books were distributed to Grade 3 children. How many books remain in the library?
[3 Marks]
✅ Step-by-Step Solution & Working
Step-by-Step Solution:
1. Total books received = 3 × 25 = 75 books.
2. Books distributed = 38 books.
3. Books remaining = 75 - 38 = 37 books.
1. Total books received = 3 × 25 = 75 books.
2. Books distributed = 38 books.
3. Books remaining = 75 - 38 = 37 books.
CBSE Step-by-Step Marking Scheme
Total 3 × 25 = 75
1.5 M
Remainder 75 - 38 = 37
1.5 M
Q12
Critical Thinking • Finding the Pattern
Observe the sequence related to 'Raksha Bandhan': 12, 24, 36, 48, ___, ___. Find the next two terms and state the mathematical rule.
[2 Marks]
✅ Step-by-Step Solution & Working
Pattern Analysis:
• 12 + 12 = 24
• 24 + 12 = 36
• 36 + 12 = 48
• Rule: Successive addition of 12 (Multiples of 12).
• Next terms: 48 + 12 = 60, 60 + 12 = 72.
• 12 + 12 = 24
• 24 + 12 = 36
• 36 + 12 = 48
• Rule: Successive addition of 12 (Multiples of 12).
• Next terms: 48 + 12 = 60, 60 + 12 = 72.
CBSE Step-by-Step Marking Scheme
Next terms 60, 72
1 M
Rule: multiples of 12 / +12 step
1 M
Q13
Estimation & Measurement • Practical Judgment
Estimate whether each measurement is greater than or less than 1 metre:
(a) Length of your classroom pencil.
(b) Height of your classroom teacher.
(c) Width of your mathematics textbook.
(d) Length of the school blackboard.
[2 Marks]
(a) Length of your classroom pencil.
(b) Height of your classroom teacher.
(c) Width of your mathematics textbook.
(d) Length of the school blackboard.
✅ Step-by-Step Solution & Working
Estimation:
• (a) Pencil: Less than 1 metre (approx. 15 cm).
• (b) Teacher height: Greater than 1 metre (approx. 1.6 to 1.7 m).
• (c) Textbook width: Less than 1 metre (approx. 20 cm).
• (d) Blackboard length: Greater than 1 metre (approx. 2 to 3 m).
• (a) Pencil: Less than 1 metre (approx. 15 cm).
• (b) Teacher height: Greater than 1 metre (approx. 1.6 to 1.7 m).
• (c) Textbook width: Less than 1 metre (approx. 20 cm).
• (d) Blackboard length: Greater than 1 metre (approx. 2 to 3 m).
CBSE Step-by-Step Marking Scheme
Correct estimation for all 4 items
2 M
Q14
Data Interpretation • Reading Simple Tables
A table shows student preferences for fruit: Apple (15), Banana (22), Mango (35), Orange (18).
(a) Which fruit is preferred by the maximum number of students?
(b) How many more students prefer Mango over Banana?
(c) What is the total number of students surveyed?
[3 Marks]
(a) Which fruit is preferred by the maximum number of students?
(b) How many more students prefer Mango over Banana?
(c) What is the total number of students surveyed?
✅ Step-by-Step Solution & Working
Table Interpretation:
• (a) Maximum preference: Mango with 35 students.
• (b) Difference: 35 - 22 = 13 more students prefer Mango.
• (c) Total students: 15 + 22 + 35 + 18 = 90 students.
• (a) Maximum preference: Mango with 35 students.
• (b) Difference: 35 - 22 = 13 more students prefer Mango.
• (c) Total students: 15 + 22 + 35 + 18 = 90 students.
CBSE Step-by-Step Marking Scheme
Mango identified
1 M
Difference = 13
1 M
Total = 90
1 M
Q15
Let us Do • Shapes and Symmetry
Draw a line of symmetry through each shape:
(a) A rectangle
(b) An equilateral triangle
(c) A circle.
How many lines of symmetry does a circle have?
[3 Marks]
(a) A rectangle
(b) An equilateral triangle
(c) A circle.
How many lines of symmetry does a circle have?
✅ Step-by-Step Solution & Working
Symmetry Properties:
• (a) A rectangle has 2 lines of symmetry (horizontal and vertical through centers of opposite sides).
• (b) An equilateral triangle has 3 lines of symmetry (from each vertex to opposite midpoint).
• (c) A circle has infinite lines of symmetry (any diameter passing through the center).
• (a) A rectangle has 2 lines of symmetry (horizontal and vertical through centers of opposite sides).
• (b) An equilateral triangle has 3 lines of symmetry (from each vertex to opposite midpoint).
• (c) A circle has infinite lines of symmetry (any diameter passing through the center).
CBSE Step-by-Step Marking Scheme
Rectangle = 2 lines
1 M
Triangle = 3 lines
1 M
Circle = infinite lines
1 M
Q16
Puzzle Time • Missing Number Grid
In a 3x3 magic square, the sum of each row, column, and diagonal is 15. The center number is 5. What numbers go in the opposite corners if one corner is 2?
[2 Marks]
✅ Step-by-Step Solution & Working
Magic Square Logic:
• In a 3x3 magic square with sum 15 and center 5, opposite corners must sum to 10 (since 2 + 5 + x = 15 → x = 8).
• If one corner is 2, the diagonally opposite corner must be 8.
• Other corner pairs: 4 and 6.
• In a 3x3 magic square with sum 15 and center 5, opposite corners must sum to 10 (since 2 + 5 + x = 15 → x = 8).
• If one corner is 2, the diagonally opposite corner must be 8.
• Other corner pairs: 4 and 6.
CBSE Step-by-Step Marking Scheme
Opposite corner = 8 with reasoning
2 M
Q17
Money & Shopping • Cash Memo Bill
Rani visits a stationery shop and buys: 2 Pencils at ₹5 each, 1 Eraser for ₹3, and 1 Notebook for ₹25. Prepare a simple cash memo showing total bill and change returned from a ₹50 note.
[3 Marks]
✅ Step-by-Step Solution & Working
Cash Memo Preparation:
• 2 Pencils @ ₹5 = ₹10
• 1 Eraser @ ₹3 = ₹3
• 1 Notebook @ ₹25 = ₹25
• Total Bill Amount = ₹10 + ₹3 + ₹25 = ₹38.
• Amount paid = ₹50.
• Change returned = ₹50 - ₹38 = ₹12.
• 2 Pencils @ ₹5 = ₹10
• 1 Eraser @ ₹3 = ₹3
• 1 Notebook @ ₹25 = ₹25
• Total Bill Amount = ₹10 + ₹3 + ₹25 = ₹38.
• Amount paid = ₹50.
• Change returned = ₹50 - ₹38 = ₹12.
CBSE Step-by-Step Marking Scheme
Correct item breakdown and total ₹38
2 M
Change = ₹12
1 M
Q18
Chapter Review • Comprehensive Concept Check
True or False:
(a) 1 kilometre is equal to 1000 metres.
(b) 500 millilitres is equal to half a litre.
(c) A square has 4 equal straight edges and 4 right-angled corners.
[3 Marks]
(a) 1 kilometre is equal to 1000 metres.
(b) 500 millilitres is equal to half a litre.
(c) A square has 4 equal straight edges and 4 right-angled corners.
✅ Step-by-Step Solution & Working
Review Answers:
• (a) True: 1 km = 1000 m.
• (b) True: 1 Litre = 1000 mL, so half a litre = 500 mL.
• (c) True: A square has 4 equal sides and 4 right angles.
• (a) True: 1 km = 1000 m.
• (b) True: 1 Litre = 1000 mL, so half a litre = 500 mL.
• (c) True: A square has 4 equal sides and 4 right angles.
CBSE Step-by-Step Marking Scheme
All 3 statements correctly evaluated as True
3 M
🌟 NEP 2020 Competency-Based Case Study
Rakhi Gift Hamper Budgeting
Simran has ₹250 savings. She wants to buy a rakhi for ₹45, a box of chocolates for ₹120, and a greeting card for ₹35 for her brother.
Analytical Task: Can Simran afford all three items with her savings? How much money will be left with her or how much more will she need?
Evaluation & Model Answer:
1. Total cost of gifts = ₹45 + ₹120 + ₹35 = ₹(45 + 35 + 120) = ₹80 + ₹120 = ₹200.
2. Simran's savings = ₹250.
3. Since ₹250 > ₹200, she can easily afford all three items.
4. Remaining money = ₹250 - ₹200 = ₹50 left.
2. Simran's savings = ₹250.
3. Since ₹250 > ₹200, she can easily afford all three items.
4. Remaining money = ₹250 - ₹200 = ₹50 left.
Rubric: 1.5M addition of 3 gift items + 1.5M comparison with savings and balance.