Chapter 13 of 15
Chapter 13: Animal Jumps
जानवरों की छलाँग (गुणज, सार्व गुणज एवं लघुत्तम समापवर्त्य - LCM) | CBSE Session 2026–27
Core Pedagogical Competencies & Concepts
- Multiples of a number: The numbers obtained by multiplying it by natural numbers (1, 2, 3, …).
- Common Multiples: Numbers that appear in the multiple lists of two or more given numbers.
- Lowest Common Multiple (LCM): The smallest non-zero common multiple of given numbers.
- Step/jump word problems on a number line representing periodic cycles.
- Frog lands on multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48...
- Rabbit lands on multiples of 7: 7, 14, 21, 28, 35, 42, 49...
- (a) Common steps under 50: 21 and 42.
- (b) First common step: $\text{LCM}(3, 7) = \mathbf{21}$.
All Textbook Exercises & In-Text Tasks (24 Questions)
100% Complete Ingestion
Q1
Number Line Jumps • Multiples
A frog jumps 3 steps at a time, and a rabbit jumps 5 steps at a time. Both start jumping from step 0.
(a) List the first 6 steps where the frog lands.
(b) List the first 6 steps where the rabbit lands.
(c) At which step will they both land on the same number for the first time? What is this number called?
[3 Marks]
(a) List the first 6 steps where the frog lands.
(b) List the first 6 steps where the rabbit lands.
(c) At which step will they both land on the same number for the first time? What is this number called?
✅ Step-by-Step Solution & Working
Multiples on Number Line:
• (a) Frog's Landing Steps: Multiples of 3: 3, 6, 9, 12, 15, 18.
• (b) Rabbit's Landing Steps: Multiples of 5: 5, 10, 15, 20, 25, 30.
• (c) First Common Landing: Both land on step 15.
This number is called the Lowest Common Multiple (LCM) of 3 and 5.
• (a) Frog's Landing Steps: Multiples of 3: 3, 6, 9, 12, 15, 18.
• (b) Rabbit's Landing Steps: Multiples of 5: 5, 10, 15, 20, 25, 30.
• (c) First Common Landing: Both land on step 15.
This number is called the Lowest Common Multiple (LCM) of 3 and 5.
CBSE Step-by-Step Marking Scheme
Frog: 3, 6, 9, 12, 15, 18
1.0 M
Rabbit: 5, 10, 15, 20, 25, 30
1.0 M
First common = 15 (LCM)
1.0 M
Q2
Finding LCM • Common Multiples
Find the Lowest Common Multiple (LCM) of:
(a) 4 and 6
(b) 6 and 8
(c) 5 and 10
[3 Marks]
(a) 4 and 6
(b) 6 and 8
(c) 5 and 10
✅ Step-by-Step Solution & Working
LCM Computations:
• (a) LCM(4, 6):
- Multiples of 4: 4, 8, 12, 16...
- Multiples of 6: 6, 12, 18...
- LCM = 12.
• (b) LCM(6, 8):
- Multiples of 6: 6, 12, 18, 24, 30...
- Multiples of 8: 8, 16, 24, 32...
- LCM = 24.
• (c) LCM(5, 10): Since 10 is a multiple of 5, LCM = 10.
• (a) LCM(4, 6):
- Multiples of 4: 4, 8, 12, 16...
- Multiples of 6: 6, 12, 18...
- LCM = 12.
• (b) LCM(6, 8):
- Multiples of 6: 6, 12, 18, 24, 30...
- Multiples of 8: 8, 16, 24, 32...
- LCM = 24.
• (c) LCM(5, 10): Since 10 is a multiple of 5, LCM = 10.
CBSE Step-by-Step Marking Scheme
LCM(4, 6) = 12
1.0 M
LCM(6, 8) = 24
1.0 M
LCM(5, 10) = 10
1.0 M
Q3
Factors & Common Factors • HCF
(a) Write all factors of 18.
(b) Write all factors of 24.
(c) What are their common factors? What is their Highest Common Factor (HCF)?
[3 Marks]
(b) Write all factors of 24.
(c) What are their common factors? What is their Highest Common Factor (HCF)?
✅ Step-by-Step Solution & Working
Factors & HCF:
• (a) Factors of 18: 1, 2, 3, 6, 9, 18.
• (b) Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24.
• (c) Common Factors: 1, 2, 3, 6.
- Highest Common Factor (HCF): 6.
• (a) Factors of 18: 1, 2, 3, 6, 9, 18.
• (b) Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24.
• (c) Common Factors: 1, 2, 3, 6.
- Highest Common Factor (HCF): 6.
CBSE Step-by-Step Marking Scheme
Factors of 18 and 24
1.5 M
Common factors: 1, 2, 3, 6 & HCF = 6
1.5 M
Q4
Church Bells Ringing • LCM Word Problem
Three school bells ring at intervals of 15, 20, and 30 minutes respectively. If they all ring together at 8:00 AM:
(a) After how many minutes will they all ring together again?
(b) At what clock time will they next toll together?
[3 Marks]
(a) After how many minutes will they all ring together again?
(b) At what clock time will they next toll together?
✅ Step-by-Step Solution & Working
LCM of 15, 20, 30:
• Multiples of 15: 15, 30, 45, 60, 75...
• Multiples of 20: 20, 40, 60, 80...
• Multiples of 30: 30, 60, 90...
• (a) LCM: 60 minutes (1 hour).
• (b) Clock Time: 8:00 AM + 1 hour = 9:00 AM.
• Multiples of 15: 15, 30, 45, 60, 75...
• Multiples of 20: 20, 40, 60, 80...
• Multiples of 30: 30, 60, 90...
• (a) LCM: 60 minutes (1 hour).
• (b) Clock Time: 8:00 AM + 1 hour = 9:00 AM.
CBSE Step-by-Step Marking Scheme
LCM = 60 minutes
1.5 M
Time = 9:00 AM
1.5 M
Q5
Tamarind Seeds Arrangement • Factor Rectangles
Sunita has 12 tamarind seeds (imli seeds). She arranges them in rectangular grids.
(a) List all the different rectangular arrays she can form (e.g. 1 × 12).
(b) How do these arrays relate to the factors of 12?
[3 Marks]
(a) List all the different rectangular arrays she can form (e.g. 1 × 12).
(b) How do these arrays relate to the factors of 12?
✅ Step-by-Step Solution & Working
Factor Rectangles:
• (a) Possible Arrays:
- 1 row of 12 (1 × 12)
- 2 rows of 6 (2 × 6)
- 3 rows of 4 (3 × 4)
- 4 rows of 3 (4 × 3)
- 6 rows of 2 (6 × 2)
- 12 rows of 1 (12 × 1).
• (b) Relation to Factors: The dimensions of the rectangles represent the factor pairs of 12: (1, 12), (2, 6), (3, 4). The complete set of factors is {1, 2, 3, 4, 6, 12}.
• (a) Possible Arrays:
- 1 row of 12 (1 × 12)
- 2 rows of 6 (2 × 6)
- 3 rows of 4 (3 × 4)
- 4 rows of 3 (4 × 3)
- 6 rows of 2 (6 × 2)
- 12 rows of 1 (12 × 1).
• (b) Relation to Factors: The dimensions of the rectangles represent the factor pairs of 12: (1, 12), (2, 6), (3, 4). The complete set of factors is {1, 2, 3, 4, 6, 12}.
CBSE Step-by-Step Marking Scheme
All rectangular arrays listed
1.5 M
Clear connection to factor pairs of 12
1.5 M
Q6
Circular Track Racers • Meeting Time
Two runners, Aman and Baman, start together at the starting line of a circular track. Aman completes a lap in 60 seconds, and Baman in 90 seconds.
(a) After how many seconds will they next cross the starting line together?
(b) How many laps will each runner have completed at that moment?
[3 Marks]
(a) After how many seconds will they next cross the starting line together?
(b) How many laps will each runner have completed at that moment?
✅ Step-by-Step Solution & Working
LCM of 60 and 90:
• 60 = 30 × 2; 90 = 30 × 3.
• (a) Meeting Time: LCM(60, 90) = 30 × 2 × 3 = 180 seconds (3 minutes).
• (b) Laps Completed:
- Aman: 180 ÷ 60 = 3 laps.
- Baman: 180 ÷ 90 = 2 laps.
• 60 = 30 × 2; 90 = 30 × 3.
• (a) Meeting Time: LCM(60, 90) = 30 × 2 × 3 = 180 seconds (3 minutes).
• (b) Laps Completed:
- Aman: 180 ÷ 60 = 3 laps.
- Baman: 180 ÷ 90 = 2 laps.
CBSE Step-by-Step Marking Scheme
LCM = 180 seconds (3 min)
1.5 M
Aman: 3 laps; Baman: 2 laps
1.5 M
Q7
Prime Factorization using Factor Trees
Draw or describe factor trees to find the prime factors of:
(a) 24
(b) 36
(c) 72
[3 Marks]
(a) 24
(b) 36
(c) 72
✅ Step-by-Step Solution & Working
Prime Factorizations:
• (a) 24: 24 = 2 × 12 = 2 × 2 × 6 = 2 × 2 × 2 × 3 (or 2³ × 3).
• (b) 36: 36 = 6 × 6 = (2 × 3) × (2 × 3) = 2 × 2 × 3 × 3 (or 2² × 3²).
• (c) 72: 72 = 8 × 9 = (2 × 2 × 2) × (3 × 3) = 2 × 2 × 2 × 3 × 3 (or 2³ × 3²).
• (a) 24: 24 = 2 × 12 = 2 × 2 × 6 = 2 × 2 × 2 × 3 (or 2³ × 3).
• (b) 36: 36 = 6 × 6 = (2 × 3) × (2 × 3) = 2 × 2 × 3 × 3 (or 2² × 3²).
• (c) 72: 72 = 8 × 9 = (2 × 2 × 2) × (3 × 3) = 2 × 2 × 2 × 3 × 3 (or 2³ × 3²).
CBSE Step-by-Step Marking Scheme
24 = 2 × 2 × 2 × 3
1.0 M
36 = 2 × 2 × 3 × 3
1.0 M
72 = 2 × 2 × 2 × 3 × 3
1.0 M
Q8
Bangles Grouping Activity
Meena has 18 bangles. She tries to make groups of equal size:
(a) What different size groups can she make without any bangles left over?
(b) How many groups can she make if she has a prime number of bangles, say 13?
[3 Marks]
(a) What different size groups can she make without any bangles left over?
(b) How many groups can she make if she has a prime number of bangles, say 13?
✅ Step-by-Step Solution & Working
Grouping & Factorization:
• (a) Groups for 18 Bangles: Group sizes correspond to factors of 18:
- Groups of 1, 2, 3, 6, 9, or 18 bangles.
• (b) For 13 Bangles (Prime): 13 has only two factors (1 and 13).
- She can only make 13 groups of 1 bangle each, or 1 group of 13 bangles.
• (a) Groups for 18 Bangles: Group sizes correspond to factors of 18:
- Groups of 1, 2, 3, 6, 9, or 18 bangles.
• (b) For 13 Bangles (Prime): 13 has only two factors (1 and 13).
- She can only make 13 groups of 1 bangle each, or 1 group of 13 bangles.
CBSE Step-by-Step Marking Scheme
Groups of 1, 2, 3, 6, 9, 18
1.5 M
Prime 13: only groups of 1 or 13
1.5 M
Q9
Venn Diagram of Multiples
In a diagram with two overlapping circles labeled 'Multiples of 3' and 'Multiples of 5':
(a) List the numbers from 1 to 30 that go in the common middle section.
(b) What is the smallest number in the common section?
[3 Marks]
(a) List the numbers from 1 to 30 that go in the common middle section.
(b) What is the smallest number in the common section?
✅ Step-by-Step Solution & Working
Common Multiples (1 to 30):
• Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
• Multiples of 5: 5, 10, 15, 20, 25, 30.
• (a) Common Section: 15, 30.
• (b) Smallest Common: 15 (which is LCM of 3 and 5).
• Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
• Multiples of 5: 5, 10, 15, 20, 25, 30.
• (a) Common Section: 15, 30.
• (b) Smallest Common: 15 (which is LCM of 3 and 5).
CBSE Step-by-Step Marking Scheme
Common section: 15, 30
1.5 M
Smallest number = 15
1.5 M
Q10
Cat and Mouse Game • Escape Steps
In the NCERT Cat and Mouse game: The mouse is on step 14 and can jump 2 steps at a time. The cat is on step 3 and can jump 3 steps at a time. There is a mouse hole at step 28.
(a) Can the mouse reach the safety hole before the cat catches it?
(b) Show the steps for both to prove your answer.
[3 Marks]
(a) Can the mouse reach the safety hole before the cat catches it?
(b) Show the steps for both to prove your answer.
✅ Step-by-Step Solution & Working
Step Tracking:
• Mouse (Starts at 14, jumps 2):
- Jumps: 14 → 16 → 18 → 20 → 22 → 24 → 26 → 28 (Safety hole reached in 7 jumps).
• Cat (Starts at 3, jumps 3):
- Jumps: 3 → 6 → 9 → 12 → 15 → 18 → 21 → 24.
• (a) Conclusion: In jump 7, the mouse reaches 28 while the cat is only at 24. The mouse reaches the hole safely and escapes!
• Mouse (Starts at 14, jumps 2):
- Jumps: 14 → 16 → 18 → 20 → 22 → 24 → 26 → 28 (Safety hole reached in 7 jumps).
• Cat (Starts at 3, jumps 3):
- Jumps: 3 → 6 → 9 → 12 → 15 → 18 → 21 → 24.
• (a) Conclusion: In jump 7, the mouse reaches 28 while the cat is only at 24. The mouse reaches the hole safely and escapes!
CBSE Step-by-Step Marking Scheme
Mouse steps to 28 (7 jumps)
1.5 M
Cat steps to 24 & positive escape proof
1.5 M
Q11
Tile Paving • HCF Room Problem
A courtyard floor is 18 metres long and 12 metres wide. The owner wants to pave it with the largest possible identical square tiles without cutting any tile.
(a) What is the side length of the largest square tile (HCF)?
(b) How many such tiles will be needed to cover the entire floor?
[3 Marks]
(a) What is the side length of the largest square tile (HCF)?
(b) How many such tiles will be needed to cover the entire floor?
✅ Step-by-Step Solution & Working
HCF Application:
• (a) Largest Tile Side: HCF(18, 12):
- Factors of 18: 1, 2, 3, 6, 9, 18.
- Factors of 12: 1, 2, 3, 4, 6, 12.
- HCF = 6 metres.
• (b) Number of Tiles:
- Along length: 18 ÷ 6 = 3 tiles.
- Along width: 12 ÷ 6 = 2 tiles.
- Total tiles = 3 × 2 = 6 square tiles.
• (a) Largest Tile Side: HCF(18, 12):
- Factors of 18: 1, 2, 3, 6, 9, 18.
- Factors of 12: 1, 2, 3, 4, 6, 12.
- HCF = 6 metres.
• (b) Number of Tiles:
- Along length: 18 ÷ 6 = 3 tiles.
- Along width: 12 ÷ 6 = 2 tiles.
- Total tiles = 3 × 2 = 6 square tiles.
CBSE Step-by-Step Marking Scheme
Largest side = HCF = 6 m
1.5 M
Tile count = 6 tiles
1.5 M
Q12
Ribbon Cutting • Equal Pieces (HCF)
A craft teacher has two rolls of ribbon: red ribbon of 36 cm and blue ribbon of 48 cm. She wants to cut both ribbons into pieces of equal length with nothing left over.
(a) What is the greatest possible length of each piece?
(b) How many pieces of red ribbon and blue ribbon will she get?
[3 Marks]
(a) What is the greatest possible length of each piece?
(b) How many pieces of red ribbon and blue ribbon will she get?
✅ Step-by-Step Solution & Working
HCF of 36 and 48:
• (a) Greatest Length: HCF(36, 48) = 12 cm.
• (b) Number of Pieces:
- Red ribbon: 36 ÷ 12 = 3 pieces.
- Blue ribbon: 48 ÷ 12 = 4 pieces.
- Total pieces = 3 + 4 = 7 pieces.
• (a) Greatest Length: HCF(36, 48) = 12 cm.
• (b) Number of Pieces:
- Red ribbon: 36 ÷ 12 = 3 pieces.
- Blue ribbon: 48 ÷ 12 = 4 pieces.
- Total pieces = 3 + 4 = 7 pieces.
CBSE Step-by-Step Marking Scheme
Greatest length = 12 cm
1.5 M
Red: 3 pieces, Blue: 4 pieces
1.5 M
Q13
Even and Odd Multiples
(a) What is the only even prime number?
(b) Is the product of two odd numbers always odd or even? Give an example.
(c) Is the sum of two odd numbers always odd or even? Give an example.
[3 Marks]
(b) Is the product of two odd numbers always odd or even? Give an example.
(c) Is the sum of two odd numbers always odd or even? Give an example.
✅ Step-by-Step Solution & Working
Properties:
• (a) Even Prime: 2 is the only even prime number (all other even numbers are divisible by 2).
• (b) Product of Two Odds: Always Odd. Example: 3 × 5 = 15.
• (c) Sum of Two Odds: Always Even. Example: 3 + 5 = 8.
• (a) Even Prime: 2 is the only even prime number (all other even numbers are divisible by 2).
• (b) Product of Two Odds: Always Odd. Example: 3 × 5 = 15.
• (c) Sum of Two Odds: Always Even. Example: 3 + 5 = 8.
CBSE Step-by-Step Marking Scheme
2 is only even prime
1.0 M
Product of odds = Odd
1.0 M
Sum of odds = Even
1.0 M
Q14
Divisibility Rules • 2, 3, 5, 10
Test the number 4,260 for divisibility by:
(a) 2
(b) 3
(c) 5
(d) 10
State the rule used for each.
[3 Marks]
(a) 2
(b) 3
(c) 5
(d) 10
State the rule used for each.
✅ Step-by-Step Solution & Working
Divisibility Rules for 4,260:
• (a) By 2: Ones digit is 0 (even) → Divisible by 2.
• (b) By 3: Sum of digits = 4 + 2 + 6 + 0 = 12. Since 12 is divisible by 3 → Divisible by 3.
• (c) By 5: Ones digit is 0 → Divisible by 5.
• (d) By 10: Ones digit is 0 → Divisible by 10.
• (a) By 2: Ones digit is 0 (even) → Divisible by 2.
• (b) By 3: Sum of digits = 4 + 2 + 6 + 0 = 12. Since 12 is divisible by 3 → Divisible by 3.
• (c) By 5: Ones digit is 0 → Divisible by 5.
• (d) By 10: Ones digit is 0 → Divisible by 10.
CBSE Step-by-Step Marking Scheme
Divisible by 2 and 3 with rules
1.5 M
Divisible by 5 and 10 with rules
1.5 M
Q15
Kangaroo, Frog, and Hare Multi-Jump
A frog jumps 2 steps, a hare jumps 3 steps, and a kangaroo jumps 7 steps. All start from 0.
(a) At which step will all three animals land together for the first time?
(b) How many jumps will the kangaroo have made to reach that step?
[3 Marks]
(a) At which step will all three animals land together for the first time?
(b) How many jumps will the kangaroo have made to reach that step?
✅ Step-by-Step Solution & Working
LCM of 2, 3, 7 (All Prime Numbers):
• (a) Meeting Step: Since 2, 3, 7 have no common factors, LCM = 2 × 3 × 7 = 42.
• (b) Kangaroo Jumps: 42 ÷ 7 = 6 jumps.
• (a) Meeting Step: Since 2, 3, 7 have no common factors, LCM = 2 × 3 × 7 = 42.
• (b) Kangaroo Jumps: 42 ÷ 7 = 6 jumps.
CBSE Step-by-Step Marking Scheme
LCM(2, 3, 7) = 42
1.5 M
Kangaroo jumps = 6
1.5 M
Q16
Common Multiple Word Problem • Egg Packaging
A poultry farm packages eggs. When packed into cartons of 6, none are left over. When packed into cartons of 10, none are left over. If there are between 50 and 80 eggs:
(a) What is the exact number of eggs?
(b) How many cartons of 6 and cartons of 10 would be filled?
[3 Marks]
(a) What is the exact number of eggs?
(b) How many cartons of 6 and cartons of 10 would be filled?
✅ Step-by-Step Solution & Working
Working:
• Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78.
• Multiples of 10: 10, 20, 30, 40, 50, 60, 70, 80.
• (a) Exact Count: The only common multiple between 50 and 80 is 60 eggs.
• (b) Cartons:
- Cartons of 6 = 60 ÷ 6 = 10 cartons.
- Cartons of 10 = 60 ÷ 10 = 6 cartons.
• Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78.
• Multiples of 10: 10, 20, 30, 40, 50, 60, 70, 80.
• (a) Exact Count: The only common multiple between 50 and 80 is 60 eggs.
• (b) Cartons:
- Cartons of 6 = 60 ÷ 6 = 10 cartons.
- Cartons of 10 = 60 ÷ 10 = 6 cartons.
CBSE Step-by-Step Marking Scheme
Exact count = 60 eggs
1.5 M
10 cartons of 6; 6 cartons of 10
1.5 M
Q17
Mental Arithmetic • LCM & HCF Shortcuts
State the LCM and HCF of the following pairs by inspection:
(a) 7 and 14
(b) 8 and 9
(c) 15 and 30
[3 Marks]
(a) 7 and 14
(b) 8 and 9
(c) 15 and 30
✅ Step-by-Step Solution & Working
Inspection Shortcuts:
• (a) 7 and 14: Since 14 is a multiple of 7:
- HCF = 7, LCM = 14.
• (b) 8 and 9 (Consecutive numbers / Co-prime):
- HCF = 1, LCM = 8 × 9 = 72.
• (c) 15 and 30: Since 30 is a multiple of 15:
- HCF = 15, LCM = 30.
• (a) 7 and 14: Since 14 is a multiple of 7:
- HCF = 7, LCM = 14.
• (b) 8 and 9 (Consecutive numbers / Co-prime):
- HCF = 1, LCM = 8 × 9 = 72.
• (c) 15 and 30: Since 30 is a multiple of 15:
- HCF = 15, LCM = 30.
CBSE Step-by-Step Marking Scheme
Part (a): HCF=7, LCM=14
1.0 M
Part (b): HCF=1, LCM=72
1.0 M
Part (c): HCF=15, LCM=30
1.0 M
Q18
Planting Trees along Boundary (HCF)
A rectangular park of 75 m by 45 m is to have ornamental trees planted at equal intervals all along its boundary and at the corners.
(a) What is the maximum distance between two consecutive trees?
(b) What is the total number of trees needed?
[3 Marks]
(a) What is the maximum distance between two consecutive trees?
(b) What is the total number of trees needed?
✅ Step-by-Step Solution & Working
HCF & Perimeter Grid:
• (a) Maximum Distance: HCF(75, 45) = 15 metres.
• (b) Number of Trees:
- Perimeter = 2 × (75 + 45) = 2 × 120 = 240 metres.
- Trees = 240 m ÷ 15 m = 16 trees.
• (a) Maximum Distance: HCF(75, 45) = 15 metres.
• (b) Number of Trees:
- Perimeter = 2 × (75 + 45) = 2 × 120 = 240 metres.
- Trees = 240 m ÷ 15 m = 16 trees.
CBSE Step-by-Step Marking Scheme
Max distance = HCF = 15 m
1.5 M
Trees = 16
1.5 M
Q19
HOTS • Remainder with Multiple Jumpers
A number leaves a remainder of 1 when divided by 2, a remainder of 1 when divided by 3, and a remainder of 1 when divided by 4. What is the smallest such number greater than 1?
[3 Marks]
✅ Step-by-Step Solution & Working
LCM and Remainder Deduction:
1. The number is of the form: $\text{LCM}(2, 3, 4) + 1$.
2. LCM of 2, 3, 4:
• Multiples of 4: 4, 8, 12... (12 is divisible by both 2 and 3).
• LCM(2, 3, 4) = 12.
3. The number = 12 + 1 = 13.
4. Verification:
- 13 ÷ 2 = 6 with remainder 1.
- 13 ÷ 3 = 4 with remainder 1.
- 13 ÷ 4 = 3 with remainder 1. Solution verified!
1. The number is of the form: $\text{LCM}(2, 3, 4) + 1$.
2. LCM of 2, 3, 4:
• Multiples of 4: 4, 8, 12... (12 is divisible by both 2 and 3).
• LCM(2, 3, 4) = 12.
3. The number = 12 + 1 = 13.
4. Verification:
- 13 ÷ 2 = 6 with remainder 1.
- 13 ÷ 3 = 4 with remainder 1.
- 13 ÷ 4 = 3 with remainder 1. Solution verified!
CBSE Step-by-Step Marking Scheme
LCM(2, 3, 4) = 12
1.5 M
Number = 13 with verification
1.5 M
Q20
Traffic Light Synchronization
Three traffic signals turn green every 40 seconds, 60 seconds, and 80 seconds.
(a) After how many seconds will all three turn green simultaneously?
(b) Express this duration in minutes.
[3 Marks]
(a) After how many seconds will all three turn green simultaneously?
(b) Express this duration in minutes.
✅ Step-by-Step Solution & Working
LCM of 40, 60, 80:
• Prime factorizations:
- 40 = 2³ × 5
- 60 = 2² × 3 × 5
- 80 = 2⁴ × 5
• LCM = 2⁴ × 3 × 5 = 16 × 15 = 240 seconds.
• (b) In Minutes: 240 ÷ 60 = 4 minutes.
• Prime factorizations:
- 40 = 2³ × 5
- 60 = 2² × 3 × 5
- 80 = 2⁴ × 5
• LCM = 2⁴ × 3 × 5 = 16 × 15 = 240 seconds.
• (b) In Minutes: 240 ÷ 60 = 4 minutes.
CBSE Step-by-Step Marking Scheme
LCM = 240 seconds
1.5 M
Duration = 4 minutes
1.5 M
Q21
Factor Pairs • Table Formulation
Formulate all factor pairs for 40 and 50. Use them to prove whether 40 and 50 have more than two common factors.
[3 Marks]
✅ Step-by-Step Solution & Working
Factor Pairs:
• For 40: 1 × 40, 2 × 20, 4 × 10, 5 × 8. Factors = {1, 2, 4, 5, 8, 10, 20, 40}.
• For 50: 1 × 50, 2 × 25, 5 × 10. Factors = {1, 2, 5, 10, 25, 50}.
• Common Factors: 1, 2, 5, 10 (4 common factors).
• Conclusion: Yes, they have 4 common factors, which is more than two.
• For 40: 1 × 40, 2 × 20, 4 × 10, 5 × 8. Factors = {1, 2, 4, 5, 8, 10, 20, 40}.
• For 50: 1 × 50, 2 × 25, 5 × 10. Factors = {1, 2, 5, 10, 25, 50}.
• Common Factors: 1, 2, 5, 10 (4 common factors).
• Conclusion: Yes, they have 4 common factors, which is more than two.
CBSE Step-by-Step Marking Scheme
Factor pairs of 40 and 50
1.5 M
Common factors: 1, 2, 5, 10 (4 factors)
1.5 M
Q22
Multiples of 100 • Zero Tracking
(a) What is the LCM of 25 and 100?
(b) What is the LCM of 20 and 50?
(c) What is the HCF of 100 and 150?
[3 Marks]
(b) What is the LCM of 20 and 50?
(c) What is the HCF of 100 and 150?
✅ Step-by-Step Solution & Working
Calculations:
• (a) LCM(25, 100): 100 is a multiple of 25 → 100.
• (b) LCM(20, 50): Multiples of 50: 50, 100... (100 ÷ 20 = 5) → 100.
• (c) HCF(100, 150): 150 - 100 = 50. Both 100 and 150 are divisible by 50 → 50.
• (a) LCM(25, 100): 100 is a multiple of 25 → 100.
• (b) LCM(20, 50): Multiples of 50: 50, 100... (100 ÷ 20 = 5) → 100.
• (c) HCF(100, 150): 150 - 100 = 50. Both 100 and 150 are divisible by 50 → 50.
CBSE Step-by-Step Marking Scheme
Part (a): 100
1.0 M
Part (b): 100
1.0 M
Part (c): 50
1.0 M
Q23
True or False with Factor/Multiple Reason
State True or False with reasoning:
(a) Every number is both a factor and a multiple of itself.
(b) The HCF of two numbers is always greater than their LCM.
(c) A number that has more than two factors is called a composite number.
[3 Marks]
(a) Every number is both a factor and a multiple of itself.
(b) The HCF of two numbers is always greater than their LCM.
(c) A number that has more than two factors is called a composite number.
✅ Step-by-Step Solution & Working
Justifications:
• (a) True: For any number $n$, $n \times 1 = n$ (so $n$ divides itself, making it a factor), and $n = 1 \times n$ (so it is in its own table of multiples).
• (b) False: HCF is a divisor (less than or equal to the numbers), while LCM is a multiple (greater than or equal to the numbers). HCF is always less than or equal to LCM.
• (c) True: By definition, numbers with 3 or more factors are composite.
• (a) True: For any number $n$, $n \times 1 = n$ (so $n$ divides itself, making it a factor), and $n = 1 \times n$ (so it is in its own table of multiples).
• (b) False: HCF is a divisor (less than or equal to the numbers), while LCM is a multiple (greater than or equal to the numbers). HCF is always less than or equal to LCM.
• (c) True: By definition, numbers with 3 or more factors are composite.
CBSE Step-by-Step Marking Scheme
Part (a): True with self-factor reasoning
1.0 M
Part (b): False (HCF ≤ LCM)
1.0 M
Part (c): True with definition
1.0 M
Q24
Comprehensive Chapter Review • Step Challenge
Three children, Anita (step length 40 cm), Bablu (step length 50 cm), and Chintu (step length 60 cm), take part in a walkathon from the same starting line.
(a) At what minimum distance from the start will their steps align exactly again?
(b) How many steps will each child have taken at that distance?
[3 Marks]
(a) At what minimum distance from the start will their steps align exactly again?
(b) How many steps will each child have taken at that distance?
✅ Step-by-Step Solution & Working
LCM of 40, 50, 60:
• Prime factorizations:
- 40 = 2³ × 5
- 50 = 2 × 5²
- 60 = 2² × 3 × 5
• (a) Minimum Distance: LCM = 2³ × 3 × 5² = 8 × 3 × 25 = 600 cm (6 metres).
• (b) Steps Taken:
- Anita: 600 ÷ 40 = 15 steps.
- Bablu: 600 ÷ 50 = 12 steps.
- Chintu: 600 ÷ 60 = 10 steps.
• Prime factorizations:
- 40 = 2³ × 5
- 50 = 2 × 5²
- 60 = 2² × 3 × 5
• (a) Minimum Distance: LCM = 2³ × 3 × 5² = 8 × 3 × 25 = 600 cm (6 metres).
• (b) Steps Taken:
- Anita: 600 ÷ 40 = 15 steps.
- Bablu: 600 ÷ 50 = 12 steps.
- Chintu: 600 ÷ 60 = 10 steps.
CBSE Step-by-Step Marking Scheme
LCM = 600 cm = 6 m
1.5 M
Anita: 15, Bablu: 12, Chintu: 10 steps
1.5 M
🌟 NEP 2020 Competency-Based Case Study
Competency Question: Traffic Light Synchronization
Three traffic signals at a busy junction change their lights after every 40 seconds, 60 seconds, and 75 seconds respectively. If they all change simultaneously at 8:00 AM, when will they change together next?
Analytical Task: Find the time at which all three traffic lights will change together again.
Evaluation & Model Answer:
Suggested Model Solution:
LCM(40, 60, 75) = 600 seconds = 10 minutes.
Next change time = 8:00 AM + 10 min = 8:10 AM.
Next change time = 8:00 AM + 10 min = 8:10 AM.
Rubric: 1M concept identification + 1M step execution + 1M conclusion