Class 5
Mathematics
Federal Board of Intermediate and Secondary Education (FBISE) (FBISE)
📚 Model Textbook of Mathematics Grade 5 (FBISE / NBF)
Mastery Guide: Prime Factorization, HCF, LCM & Real-World Word Problems
📖 Chapter 2: HCF and LCM
📅 Updated: Sep 09, 2026
Teacher Pedagogical Roadmap
Grade 5 Mathematics • FBISE / SNC Latest Curriculum
Instructional Blueprint: Unit 2 — Highest Common Factor (HCF) & Least Common Multiple (LCM)
Target Learning Outcomes
- Master divisibility tests for $7$ and $11$ on numbers up to $5$ digits without manual division.
- Distinguish between prime, composite, twin primes, and prime triplets up to $100$ using the Sieve of Eratosthenes.
- Construct prime factorizations using Factor Trees and Continuous Division in index notation ($a^b$).
- Calculate HCF using Prime Factorization and Continued Long Division.
- Calculate LCM using Prime Factorization and Common Division algorithms.
- Solve complex real-life scheduling, grouping, and measurement word problems.
Pacing & Time Budget (60 Min)
- 00-15m: Divisibility Superheroes ($7$ & $11$) & Sieve of Eratosthenes.
- 15-30m: Prime Factorization & Dual HCF Algorithms (Trees vs Long Division).
- 30-45m: Dual LCM Algorithms & The Golden Product Formula.
- 45-60m: Word Problem Detective: "Greatest Shared Cut" vs "Next Meeting Time".
Differentiation Strategies
- Struggling: Use color-coded factor loops and Venn diagrams for common primes.
- Advanced: 3-number long division ladders and multi-step clock/bus synchronization problems.
1. The Divisibility Superheroes: Rules for 7 and 11
Imagine you are given a giant number like $756$ or $4\,653$. Do you need to do long division just to see if it divides evenly with no remainder? No! You can use mathematical secret codes called Divisibility Rules!
🦸 Super Rule for 7: The "Double & Subtract" Magic
Step 1: Take the very last digit (ones place) and double it ($\times 2$).
Step 2: Subtract that doubled number from the remaining truncated part.
Step 3: If the answer is $0$ or a multiple of $7$ ($7, 14, 21, 28, 35, \dots$), then the whole number is divisible by $7$!
Example 1: Check $756$
• Last digit is $6 \implies 6 \times 2 = 12$.
• Remaining part is $75 \implies 75 - 12 = 63$.
• Since $63 = 7 \times 9$ (multiple of $7$), $\mathbf{756}$ is divisible by 7! 🎉
Example 2: Check $2\,975$
• Last digit is $5 \implies 5 \times 2 = 10$.
• Remaining part is $297 \implies 297 - 10 = 287$.
• Repeat on $287$: Last digit $7 \times 2 = 14 \implies 28 - 14 = 14$ (multiple of $7$!). So $\mathbf{2\,975}$ is divisible by 7!
• Last digit is $6 \implies 6 \times 2 = 12$.
• Remaining part is $75 \implies 75 - 12 = 63$.
• Since $63 = 7 \times 9$ (multiple of $7$), $\mathbf{756}$ is divisible by 7! 🎉
Example 2: Check $2\,975$
• Last digit is $5 \implies 5 \times 2 = 10$.
• Remaining part is $297 \implies 297 - 10 = 287$.
• Repeat on $287$: Last digit $7 \times 2 = 14 \implies 28 - 14 = 14$ (multiple of $7$!). So $\mathbf{2\,975}$ is divisible by 7!
🦸 Super Rule for 11: The "Hopscotch Alternate Sum"
Step 1: Add up all digits in the odd positions (1st, 3rd, 5th, ...).
Step 2: Add up all digits in the even positions (2nd, 4th, 6th, ...).
Step 3: Find the difference between the two sums. If the difference is $\mathbf{0}$ or a multiple of $\mathbf{11}$ ($11, 22, 33, \dots$), the number is divisible by $11$!
Example 1: Check $4\,653$
• Sum of alternate digits: $(4 + 5) = 9$ and $(6 + 3) = 9$.
• Difference: $9 - 9 = \mathbf{0}$. Therefore, $\mathbf{4\,653}$ is divisible by 11!
Example 2: Check $918\,291$
• Odd places: $9 + 8 + 9 = 26$.
• Even places: $1 + 2 + 1 = 4$.
• Difference: $26 - 4 = \mathbf{22}$ (multiple of $11$, $11 \times 2 = 22$). So $\mathbf{918\,291}$ is divisible by 11!
• Sum of alternate digits: $(4 + 5) = 9$ and $(6 + 3) = 9$.
• Difference: $9 - 9 = \mathbf{0}$. Therefore, $\mathbf{4\,653}$ is divisible by 11!
Example 2: Check $918\,291$
• Odd places: $9 + 8 + 9 = 26$.
• Even places: $1 + 2 + 1 = 4$.
• Difference: $26 - 4 = \mathbf{22}$ (multiple of $11$, $11 \times 2 = 22$). So $\mathbf{918\,291}$ is divisible by 11!
⚡ Quick Reference Table: Divisibility Rules at a Glance
| Divisor | The Superhero Condition | Quick Example |
|---|---|---|
| 2 | Ends in $0, 2, 4, 6, 8$ (even number) | $1\,438 \implies \text{Ends in } 8$ ✓ |
| 3 | Sum of all digits is a multiple of $3$ | $231 \implies 2+3+1=6$ ✓ |
| 5 | Last digit is $0$ or $5$ | $785 \implies \text{Ends in } 5$ ✓ |
| 7 | $(\text{Truncated part}) - 2 \times (\text{Units digit}) = 0 \text{ or } 7k$ | $147 \implies 14 - (2 \times 7) = 0$ ✓ |
| 10 | Ends in $0$ | $990 \implies \text{Ends in } 0$ ✓ |
| 11 | $|\text{Sum of odd positions} - \text{Sum of even positions}| = 0 \text{ or } 11k$ | $2\,816 \implies (2+1)-(8+6) = -11$ ✓ |
2. Prime Numbers, Composite Numbers & The Sieve of Eratosthenes
Every whole number greater than $1$ belongs to one of two big kingdoms:
🌟 Prime Numbers (The Unbreakables)
A natural number that has exactly two distinct factors: $1$ and itself. They cannot be broken into smaller whole number rectangles!
Examples: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, \dots$
• $3 = 1 \times 3$ (only 2 factors)
• $17 = 1 \times 17$ (only 2 factors)
• $3 = 1 \times 3$ (only 2 factors)
• $17 = 1 \times 17$ (only 2 factors)
🧱 Composite Numbers (The Builders)
A natural number that has more than two factors. They can be broken down into products of smaller numbers.
Examples: $4, 6, 8, 9, 10, 12, 14, 15, 18, \dots$
• $4 = 1, 2, 4$ ($3$ factors → Smallest composite!)
• $18 = 1, 2, 3, 6, 9, 18$ ($6$ factors)
• $4 = 1, 2, 4$ ($3$ factors → Smallest composite!)
• $18 = 1, 2, 3, 6, 9, 18$ ($6$ factors)
💡 Critical Key Facts Every 5th Grader Must Know
- Why is $1$ neither Prime nor Composite? Because a prime number must have two distinct factors ($1$ and itself). For $1$, the only factor is $1$. So $1$ is a unique unit number!
- The King of Primes: $\mathbf{2}$ is the smallest prime number and the only even prime number in the universe! Every other even number ($4, 6, 8, 10, \dots$) is divisible by $2$ and thus composite.
- The Smallest Composite Number: $\mathbf{4}$ is the smallest composite number.
- Twin Primes: Pairs of prime numbers that have a difference of $2$ ($p_2 - p_1 = 2$). Examples: $(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)$.
- The Prime Triplet: Three consecutive prime numbers that differ by $2$: $\mathbf{(3, 5, 7)}$ is the only known prime triplet!
- Goldbach's Secret: Every even number greater than $2$ can be written as the sum of two prime numbers! For example: $4 = 2+2$, $12 = 5+7$, $36 = 17+19$.
📜 The Sieve of Eratosthenes (Primes Between 1 and 100)
Over $2\,200$ years ago, the Greek mathematician Eratosthenes invented this genius filtering method. There are exactly 25 prime numbers between $1$ and $100$ (exactly one-fourth of the first 100 natural numbers):
2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97
3. Prime Factorization: The DNA of Numbers
By the Fundamental Theorem of Arithmetic, every composite number can be uniquely factored into prime numbers. We have two great methods:
🌳 Method 1: The Factor Tree
Split the number into two branches. If a branch is composite, keep splitting until all branch tips (leaves) are prime numbers!
$$24 = 2 \times 12 = 2 \times (2 \times 6) = 2 \times 2 \times 2 \times 3$$
In index notation: $$24 = \mathbf{2^3 \times 3}$$
🪜 Method 2: The Continuous Division Ladder
Divide by the smallest prime number ($2, 3, 5, 7, \dots$) repeatedly until the quotient at the bottom becomes $1$.
$$72 \div 2 = 36 \to 36 \div 2 = 18 \to 18 \div 2 = 9 \to 9 \div 3 = 3 \to 3 \div 3 = 1$$
$$72 = 2 \times 2 \times 2 \times 3 \times 3 = \mathbf{2^3 \times 3^2}$$
4. Highest Common Factor (HCF) — The Great Equal Divider
The Highest Common Factor (HCF) of two or more numbers is the greatest natural number that divides each of the numbers exactly without leaving any remainder.
Dual Methods to Calculate HCF:
Method A: Prime Factorization
1. Find prime factors of each number.
2. Identify the shared (common) prime factors.
3. Multiply the common factors:
$$\mathbf{\text{HCF} = \text{Product of Common Prime Factors}}$$ Example: $18 = 2 \times 3 \times 3$ and $24 = 2 \times 2 \times 2 \times 3$.
Common primes: $2 \times 3 \implies \mathbf{\text{HCF} = 6}$.
1. Find prime factors of each number.
2. Identify the shared (common) prime factors.
3. Multiply the common factors:
$$\mathbf{\text{HCF} = \text{Product of Common Prime Factors}}$$ Example: $18 = 2 \times 3 \times 3$ and $24 = 2 \times 2 \times 2 \times 3$.
Common primes: $2 \times 3 \implies \mathbf{\text{HCF} = 6}$.
Method B: Continued Long Division
1. Divide the larger number by the smaller number.
2. If remainder $> 0$, the remainder becomes the new divisor, and the previous divisor becomes the new dividend.
3. Repeat until remainder is $0$. The last divisor is the HCF!
Example: HCF of $36$ and $54$:
$54 \div 36 = 1 \text{ rem } 18 \implies 36 \div 18 = 2 \text{ rem } 0$.
Last divisor is $\mathbf{18}$. $\text{HCF} = \mathbf{18}$.
1. Divide the larger number by the smaller number.
2. If remainder $> 0$, the remainder becomes the new divisor, and the previous divisor becomes the new dividend.
3. Repeat until remainder is $0$. The last divisor is the HCF!
Example: HCF of $36$ and $54$:
$54 \div 36 = 1 \text{ rem } 18 \implies 36 \div 18 = 2 \text{ rem } 0$.
Last divisor is $\mathbf{18}$. $\text{HCF} = \mathbf{18}$.
5. Least Common Multiple (LCM) — The Great Meeting Point
The Least Common Multiple (LCM) of two or more numbers is the smallest non-zero number that is a common multiple of all given numbers.
Dual Methods to Calculate LCM:
Method A: Prime Factorization Formula
1. Find prime factors of each number.
2. Separate into Common Prime Factors and Non-Common Prime Factors.
3. Use the fundamental LCM formula:
$$\mathbf{\text{LCM} = (\text{Product of Common Factors}) \times (\text{Product of Non-Common Factors})}$$ Example for $8$ and $12$:
$8 = 2 \times 2 \times 2$, $12 = 2 \times 2 \times 3$.
Common $= 2 \times 2 = 4$; Non-common $= 2 \times 3 = 6$.
$$\text{LCM} = 4 \times 6 = \mathbf{24}$$
1. Find prime factors of each number.
2. Separate into Common Prime Factors and Non-Common Prime Factors.
3. Use the fundamental LCM formula:
$$\mathbf{\text{LCM} = (\text{Product of Common Factors}) \times (\text{Product of Non-Common Factors})}$$ Example for $8$ and $12$:
$8 = 2 \times 2 \times 2$, $12 = 2 \times 2 \times 3$.
Common $= 2 \times 2 = 4$; Non-common $= 2 \times 3 = 6$.
$$\text{LCM} = 4 \times 6 = \mathbf{24}$$
Method B: Common Division Method
1. Write all numbers in a row separated by commas.
2. Divide by any prime number that divides at least one of the numbers.
3. Bring down unchanged any number that is not divisible.
4. Continue until all quotients are $1$.
5. Multiply all divisors on the left to get the LCM!
1. Write all numbers in a row separated by commas.
2. Divide by any prime number that divides at least one of the numbers.
3. Bring down unchanged any number that is not divisible.
4. Continue until all quotients are $1$.
5. Multiply all divisors on the left to get the LCM!
6. Word Problem Detective: How to Choose Between HCF and LCM
| 🔍 When to Use HCF (Dividing & Sharing) | ⏰ When to Use LCM (Repeating & Syncing) |
|---|---|
|
Look for clue words: • "Greatest possible length" • "Maximum capacity of container" • "Cut into equal pieces without leftover" • "Largest number of equal kits / rows / groups" Goal: Breaking down bigger quantities into the largest possible equal smaller portions. |
Look for clue words: • "Smallest / Minimum length / Least number" • "When will they meet / ring / toll together again?" • "Buses leaving station simultaneously next time" • "Shortest height where stacks match" Goal: Finding the earliest or smallest common cycle where different repeating events align. |
📝 Complete Solved Textbook Exercises (Unit 2: HCF & LCM)
Ex 1, 2, 3, 4 & Review Ex 2 Fully SolvedExercise 1: Divisibility Rules for 7 and 11
Q1. Which of the following numbers are divisible by 7?
(a) $280$: Truncated part $= 28$, units digit $= 0$. $28 - (2 \times 0) = 28$. Since $28 = 7 \times 4$, $\mathbf{280}$ is Divisible by 7.
(b) $3\,500$: Truncated part $= 350$, units digit $= 0$. $350 - 0 = 350 = 7 \times 50$. $\mathbf{3\,500}$ is Divisible by 7.
(c) $5\,601$: Units digit $= 1 \implies 560 - (2 \times 1) = 558$. For $558$: $55 - (2 \times 8) = 55 - 16 = 39$. Since $39$ is not a multiple of $7$, $\mathbf{5\,601}$ is Not Divisible by 7.
(d) $7\,777$: Units digit $= 7 \implies 777 - 14 = 763 \implies 76 - 6 = 70 = 7 \times 10$. $\mathbf{7\,777}$ is Divisible by 7.
(e) $1\,414$: Units digit $= 4 \implies 141 - 8 = 133 \implies 13 - 6 = 7$. $\mathbf{1\,414}$ is Divisible by 7.
(f) $11\,111$: $1\,111 - 2 = 1\,109 \implies 110 - 18 = 92 \implies 9 - 4 = 5 \ne 7k$. $\mathbf{11\,111}$ is Not Divisible by 7.
(b) $3\,500$: Truncated part $= 350$, units digit $= 0$. $350 - 0 = 350 = 7 \times 50$. $\mathbf{3\,500}$ is Divisible by 7.
(c) $5\,601$: Units digit $= 1 \implies 560 - (2 \times 1) = 558$. For $558$: $55 - (2 \times 8) = 55 - 16 = 39$. Since $39$ is not a multiple of $7$, $\mathbf{5\,601}$ is Not Divisible by 7.
(d) $7\,777$: Units digit $= 7 \implies 777 - 14 = 763 \implies 76 - 6 = 70 = 7 \times 10$. $\mathbf{7\,777}$ is Divisible by 7.
(e) $1\,414$: Units digit $= 4 \implies 141 - 8 = 133 \implies 13 - 6 = 7$. $\mathbf{1\,414}$ is Divisible by 7.
(f) $11\,111$: $1\,111 - 2 = 1\,109 \implies 110 - 18 = 92 \implies 9 - 4 = 5 \ne 7k$. $\mathbf{11\,111}$ is Not Divisible by 7.
Q2. Which of the following numbers are divisible by 11?
(a) $484$: Alternate sums: $(4 + 4) - (8) = 8 - 8 = \mathbf{0}$. ⇒ Divisible by 11.
(b) $6\,000$: $(6 + 0) - (0 + 0) = 6 - 0 = \mathbf{6} \ne 11k$. ⇒ Not Divisible by 11.
(c) $2\,816$: Odd positions $(2 + 1) = 3$, Even positions $(8 + 6) = 14$. Difference $= 14 - 3 = \mathbf{11}$. ⇒ Divisible by 11.
(d) $7\,777$: $(7 + 7) - (7 + 7) = 14 - 14 = \mathbf{0}$. ⇒ Divisible by 11.
(e) $50\,187$: $(5 + 1 + 7) - (0 + 8) = 13 - 8 = \mathbf{5} \ne 11k$. ⇒ Not Divisible by 11.
(f) $7\,172$: $(7 + 7) - (1 + 2) = 14 - 3 = \mathbf{11}$. ⇒ Divisible by 11.
(b) $6\,000$: $(6 + 0) - (0 + 0) = 6 - 0 = \mathbf{6} \ne 11k$. ⇒ Not Divisible by 11.
(c) $2\,816$: Odd positions $(2 + 1) = 3$, Even positions $(8 + 6) = 14$. Difference $= 14 - 3 = \mathbf{11}$. ⇒ Divisible by 11.
(d) $7\,777$: $(7 + 7) - (7 + 7) = 14 - 14 = \mathbf{0}$. ⇒ Divisible by 11.
(e) $50\,187$: $(5 + 1 + 7) - (0 + 8) = 13 - 8 = \mathbf{5} \ne 11k$. ⇒ Not Divisible by 11.
(f) $7\,172$: $(7 + 7) - (1 + 2) = 14 - 3 = \mathbf{11}$. ⇒ Divisible by 11.
Q3. Check whether $4\,312$ is divisible by both $7$ and $11$? Write three numbers which are divisible by both $7$ and $11$.
Testing for 7: $431 - (2 \times 2) = 431 - 4 = 427 \implies 42 - (2 \times 7) = 42 - 14 = 28$ ($7 \times 4 = 28$). Divisible by 7! ✓
Testing for 11: $(4 + 1) - (3 + 2) = 5 - 5 = 0$. Divisible by 11! ✓
Conclusion: Yes, $\mathbf{4\,312}$ is divisible by both 7 and 11 ($4\,312 \div 77 = 56$).
Three numbers divisible by both 7 and 11 (multiples of $7 \times 11 = 77$):
• $77 \times 1 = \mathbf{77}$
• $77 \times 2 = \mathbf{154}$
• $77 \times 10 = \mathbf{770}$ (or $7\,777$)
Testing for 11: $(4 + 1) - (3 + 2) = 5 - 5 = 0$. Divisible by 11! ✓
Conclusion: Yes, $\mathbf{4\,312}$ is divisible by both 7 and 11 ($4\,312 \div 77 = 56$).
Three numbers divisible by both 7 and 11 (multiples of $7 \times 11 = 77$):
• $77 \times 1 = \mathbf{77}$
• $77 \times 2 = \mathbf{154}$
• $77 \times 10 = \mathbf{770}$ (or $7\,777$)
Q4. Write a 4-digit number which is: (i) divisible by both 5 and 7, (ii) divisible by 5, 7 and 11.
(i) Divisible by both 5 and 7: Must be a multiple of $5 \times 7 = 35$ with 4 digits.
$$\text{Example: } 35 \times 30 = \mathbf{1\,050} \quad (\text{or } 1\,400, 3\,500)$$
(ii) Divisible by 5, 7, and 11: Must be a multiple of $5 \times 7 \times 11 = 385$ with 4 digits.
$$\text{Example: } 385 \times 3 = \mathbf{1\,155} \quad (\text{or } 385 \times 4 = 1\,540, 385 \times 10 = 3\,850)$$
$$\text{Example: } 35 \times 30 = \mathbf{1\,050} \quad (\text{or } 1\,400, 3\,500)$$
(ii) Divisible by 5, 7, and 11: Must be a multiple of $5 \times 7 \times 11 = 385$ with 4 digits.
$$\text{Example: } 385 \times 3 = \mathbf{1\,155} \quad (\text{or } 385 \times 4 = 1\,540, 385 \times 10 = 3\,850)$$
Q5. A number is divisible by 7. Can we say that: (i) the number can only be even, (ii) the number can only be odd, (iii) the number can be even as well as odd.
Correct Statement: (iii) The number can be even as well as odd.
Proof: $7 \times 1 = 7$ (odd), $7 \times 2 = 14$ (even), $7 \times 3 = 21$ (odd), $7 \times 4 = 28$ (even). Multiples of $7$ alternate between odd and even!
Proof: $7 \times 1 = 7$ (odd), $7 \times 2 = 14$ (even), $7 \times 3 = 21$ (odd), $7 \times 4 = 28$ (even). Multiples of $7$ alternate between odd and even!
Exercise 2: Prime, Composite, Twin Primes & Primes Breakdown
Q1. List all prime numbers between 20 and 60.
The prime numbers between $20$ and $60$ are:
$$\mathbf{23, 29, 31, 37, 41, 43, 47, 53, 59}$$
$$\mathbf{23, 29, 31, 37, 41, 43, 47, 53, 59}$$
Q2. A prime number when added to 101 gives an odd prime number. Find it.
$101$ is an odd number. Adding an odd number to an odd number always gives an even number (which is composite). To get an odd prime number, we must add an even prime number.
The only even prime number is $\mathbf{2}$.
$$101 + 2 = 103 \quad (\text{which is an odd prime number!})$$
Answer: The prime number is $\mathbf{2}$.
The only even prime number is $\mathbf{2}$.
$$101 + 2 = 103 \quad (\text{which is an odd prime number!})$$
Answer: The prime number is $\mathbf{2}$.
Q3. One of the twin primes is 7. Find the other prime number if the sum of both numbers is an even number.
Twin primes have a difference of $2$. Since $7$ is given, the twin pair is either $(5, 7)$ or $(7, 9)$. But $9$ is not prime. So the other prime is $\mathbf{5}$.
Sum check: $5 + 7 = 12$ (even number ✓).
Answer: The other prime number is $\mathbf{5}$.
Sum check: $5 + 7 = 12$ (even number ✓).
Answer: The other prime number is $\mathbf{5}$.
Q4. Find the greatest two digit prime number which when added with 4, gives the smallest three digit prime number.
The smallest 3-digit prime number is $\mathbf{101}$.
Let the 2-digit prime be $p$.
$$p + 4 = 101 \implies p = 101 - 4 = \mathbf{97}$$
$97$ is indeed the greatest 2-digit prime number!
Answer: $\mathbf{97}$.
Let the 2-digit prime be $p$.
$$p + 4 = 101 \implies p = 101 - 4 = \mathbf{97}$$
$97$ is indeed the greatest 2-digit prime number!
Answer: $\mathbf{97}$.
Q5. Express each of the following as the sum of two prime numbers:
(a) $34$: $$34 = 3 + 31 \quad (\text{or } 5 + 29 = 11 + 23 = 17 + 17)$$
(b) $33$: $$33 = 2 + 31 \quad (\text{since 33 is odd, one prime must be 2})$$
(c) $42$: $$42 = 5 + 37 \quad (\text{or } 11 + 31 = 13 + 29 = 19 + 23)$$
(d) $60$: $$60 = 7 + 53 \quad (\text{or } 17 + 43 = 19 + 41 = 23 + 37 = 29 + 31)$$
(e) $54$: $$54 = 7 + 47 \quad (\text{or } 11 + 43 = 13 + 41 = 17 + 37 = 23 + 31)$$
(b) $33$: $$33 = 2 + 31 \quad (\text{since 33 is odd, one prime must be 2})$$
(c) $42$: $$42 = 5 + 37 \quad (\text{or } 11 + 31 = 13 + 29 = 19 + 23)$$
(d) $60$: $$60 = 7 + 53 \quad (\text{or } 17 + 43 = 19 + 41 = 23 + 37 = 29 + 31)$$
(e) $54$: $$54 = 7 + 47 \quad (\text{or } 11 + 43 = 13 + 41 = 17 + 37 = 23 + 31)$$
Q6. Find a pair of twin primes between 40 and 70.
The pairs of twin primes between $40$ and $70$ are: $\mathbf{(41, 43)}$ and $\mathbf{(59, 61)}$.
Q7. Write the following composite numbers as sum of twin primes:
(a) $36$: $$36 = 17 + 19 \quad (\text{since } 17 \text{ and } 19 \text{ are twin primes})$$
(b) $84$: $$84 = 41 + 43 \quad (\text{since } 41 \text{ and } 43 \text{ are twin primes})$$
(c) $144$: $$144 = 71 + 73 \quad (\text{since } 71 \text{ and } 73 \text{ are twin primes})$$
(d) $60$: $$60 = 29 + 31 \quad (\text{since } 29 \text{ and } 31 \text{ are twin primes})$$
(b) $84$: $$84 = 41 + 43 \quad (\text{since } 41 \text{ and } 43 \text{ are twin primes})$$
(c) $144$: $$144 = 71 + 73 \quad (\text{since } 71 \text{ and } 73 \text{ are twin primes})$$
(d) $60$: $$60 = 29 + 31 \quad (\text{since } 29 \text{ and } 31 \text{ are twin primes})$$
Q8. List ten composite numbers between 41 and 55.
The composite numbers strictly between $41$ and $55$ (excluding primes $43, 47, 53$) are:
$$\mathbf{42, 44, 45, 46, 48, 49, 50, 51, 52, 54}$$
$$\mathbf{42, 44, 45, 46, 48, 49, 50, 51, 52, 54}$$
Q9. List five consecutive composite numbers between 60 and 70.
Between $61$ (prime) and $67$ (prime), we have five consecutive composite numbers:
$$\mathbf{62, 63, 64, 65, 66}$$
$$\mathbf{62, 63, 64, 65, 66}$$
Exercise 3: Highest Common Factor (HCF) Calculation & Word Problems
Q1. Find HCF of the following numbers using prime factorization method:
(a) $58, 72$: $58 = 2 \times 29$, $72 = 2^3 \times 3^2$. Common factor: $2$. ⇒ $\mathbf{\text{HCF} = 2}$
(b) $21, 48$: $21 = 3 \times 7$, $48 = 2^4 \times 3$. Common factor: $3$. ⇒ $\mathbf{\text{HCF} = 3}$
(c) $56, 70$: $56 = 2^3 \times 7$, $70 = 2 \times 5 \times 7$. Common factors: $2 \times 7$. ⇒ $\mathbf{\text{HCF} = 14}$
(d) $45, 90$: $45 = 3^2 \times 5$, $90 = 2 \times 3^2 \times 5$. Common factors: $3^2 \times 5$. ⇒ $\mathbf{\text{HCF} = 45}$
(e) $42, 49$: $42 = 2 \times 3 \times 7$, $49 = 7^2$. Common factor: $7$. ⇒ $\mathbf{\text{HCF} = 7}$
(f) $15, 18, 56$: $15 = 3 \times 5$, $18 = 2 \times 3^2$, $56 = 2^3 \times 7$. No common prime factor across all three. ⇒ $\mathbf{\text{HCF} = 1}$
(g) $42, 54, 64$: $42 = 2 \times 3 \times 7$, $54 = 2 \times 3^3$, $64 = 2^6$. Common factor: $2$. ⇒ $\mathbf{\text{HCF} = 2}$
(h) $18, 30, 90$: $18 = 2 \times 3^2$, $30 = 2 \times 3 \times 5$, $90 = 2 \times 3^2 \times 5$. Common factors: $2 \times 3$. ⇒ $\mathbf{\text{HCF} = 6}$
(i) $12, 24, 36$: $12 = 2^2 \times 3$, $24 = 2^3 \times 3$, $36 = 2^2 \times 3^2$. Common factors: $2^2 \times 3$. ⇒ $\mathbf{\text{HCF} = 12}$
(j) $18, 36, 76$: $18 = 2 \times 3^2$, $36 = 2^2 \times 3^2$, $76 = 2^2 \times 19$. Common factor: $2$. ⇒ $\mathbf{\text{HCF} = 2}$
(k) $5, 35, 40$: $5 = 5$, $35 = 5 \times 7$, $40 = 2^3 \times 5$. Common factor: $5$. ⇒ $\mathbf{\text{HCF} = 5}$
(l) $13, 52, 78$: $13 = 13$, $52 = 2^2 \times 13$, $78 = 2 \times 3 \times 13$. Common factor: $13$. ⇒ $\mathbf{\text{HCF} = 13}$
(b) $21, 48$: $21 = 3 \times 7$, $48 = 2^4 \times 3$. Common factor: $3$. ⇒ $\mathbf{\text{HCF} = 3}$
(c) $56, 70$: $56 = 2^3 \times 7$, $70 = 2 \times 5 \times 7$. Common factors: $2 \times 7$. ⇒ $\mathbf{\text{HCF} = 14}$
(d) $45, 90$: $45 = 3^2 \times 5$, $90 = 2 \times 3^2 \times 5$. Common factors: $3^2 \times 5$. ⇒ $\mathbf{\text{HCF} = 45}$
(e) $42, 49$: $42 = 2 \times 3 \times 7$, $49 = 7^2$. Common factor: $7$. ⇒ $\mathbf{\text{HCF} = 7}$
(f) $15, 18, 56$: $15 = 3 \times 5$, $18 = 2 \times 3^2$, $56 = 2^3 \times 7$. No common prime factor across all three. ⇒ $\mathbf{\text{HCF} = 1}$
(g) $42, 54, 64$: $42 = 2 \times 3 \times 7$, $54 = 2 \times 3^3$, $64 = 2^6$. Common factor: $2$. ⇒ $\mathbf{\text{HCF} = 2}$
(h) $18, 30, 90$: $18 = 2 \times 3^2$, $30 = 2 \times 3 \times 5$, $90 = 2 \times 3^2 \times 5$. Common factors: $2 \times 3$. ⇒ $\mathbf{\text{HCF} = 6}$
(i) $12, 24, 36$: $12 = 2^2 \times 3$, $24 = 2^3 \times 3$, $36 = 2^2 \times 3^2$. Common factors: $2^2 \times 3$. ⇒ $\mathbf{\text{HCF} = 12}$
(j) $18, 36, 76$: $18 = 2 \times 3^2$, $36 = 2^2 \times 3^2$, $76 = 2^2 \times 19$. Common factor: $2$. ⇒ $\mathbf{\text{HCF} = 2}$
(k) $5, 35, 40$: $5 = 5$, $35 = 5 \times 7$, $40 = 2^3 \times 5$. Common factor: $5$. ⇒ $\mathbf{\text{HCF} = 5}$
(l) $13, 52, 78$: $13 = 13$, $52 = 2^2 \times 13$, $78 = 2 \times 3 \times 13$. Common factor: $13$. ⇒ $\mathbf{\text{HCF} = 13}$
Q2. Find HCF of the following numbers using division method:
(a) $13, 65$: $65 \div 13 = 5 \text{ rem } 0$. Last divisor: $\mathbf{13}$. ⇒ $\mathbf{\text{HCF} = 13}$
(b) $25, 75$: $75 \div 25 = 3 \text{ rem } 0$. Last divisor: $\mathbf{25}$. ⇒ $\mathbf{\text{HCF} = 25}$
(c) $42, 98$: $98 \div 42 = 2 \text{ rem } 14 \implies 42 \div 14 = 3 \text{ rem } 0$. Last divisor: $\mathbf{14}$. ⇒ $\mathbf{\text{HCF} = 14}$
(d) $16, 20, 70$: $20 \div 16 = 1 \text{ rem } 4 \implies 16 \div 4 = 4 \text{ rem } 0$ (HCF of 16, 20 is 4). Now $70 \div 4 = 17 \text{ rem } 2 \implies 4 \div 2 = 2 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 2}$
(e) $56, 84, 88$: $84 \div 56 = 1 \text{ rem } 28 \implies 56 \div 28 = 2 \text{ rem } 0$ (HCF of 56, 84 is 28). Now $88 \div 28 = 3 \text{ rem } 4 \implies 28 \div 4 = 7 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 4}$
(f) $57, 76, 95$: $76 \div 57 = 1 \text{ rem } 19 \implies 57 \div 19 = 3 \text{ rem } 0$. Now $95 \div 19 = 5 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 19}$
(g) $16, 32, 96$: $32 \div 16 = 2 \text{ rem } 0$. Now $96 \div 16 = 6 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 16}$
(h) $20, 40, 80$: $40 \div 20 = 2 \text{ rem } 0$. Now $80 \div 20 = 4 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 20}$
(i) $48, 76, 96$: $76 \div 48 = 1 \text{ rem } 28 \implies 48 \div 28 = 1 \text{ rem } 20 \implies 28 \div 20 = 1 \text{ rem } 8 \implies 20 \div 8 = 2 \text{ rem } 4 \implies 8 \div 4 = 2 \text{ rem } 0$ (HCF of 48, 76 is 4). Now $96 \div 4 = 24 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 4}$
(j) $24, 48, 72$: $48 \div 24 = 2 \text{ rem } 0$. Now $72 \div 24 = 3 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 24}$
(k) $51, 65, 75$: $65 \div 51 = 1 \text{ rem } 14 \implies 51 \div 14 = 3 \text{ rem } 9 \implies 14 \div 9 = 1 \text{ rem } 5 \implies 9 \div 5 = 1 \text{ rem } 4 \implies 5 \div 4 = 1 \text{ rem } 1 \implies 4 \div 1 = 4 \text{ rem } 0$ (HCF of 51, 65 is 1). Since $\text{HCF}(1, 75) = 1$. ⇒ $\mathbf{\text{HCF} = 1}$
(l) $13, 39, 78$: $39 \div 13 = 3 \text{ rem } 0$. Now $78 \div 13 = 6 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 13}$
(b) $25, 75$: $75 \div 25 = 3 \text{ rem } 0$. Last divisor: $\mathbf{25}$. ⇒ $\mathbf{\text{HCF} = 25}$
(c) $42, 98$: $98 \div 42 = 2 \text{ rem } 14 \implies 42 \div 14 = 3 \text{ rem } 0$. Last divisor: $\mathbf{14}$. ⇒ $\mathbf{\text{HCF} = 14}$
(d) $16, 20, 70$: $20 \div 16 = 1 \text{ rem } 4 \implies 16 \div 4 = 4 \text{ rem } 0$ (HCF of 16, 20 is 4). Now $70 \div 4 = 17 \text{ rem } 2 \implies 4 \div 2 = 2 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 2}$
(e) $56, 84, 88$: $84 \div 56 = 1 \text{ rem } 28 \implies 56 \div 28 = 2 \text{ rem } 0$ (HCF of 56, 84 is 28). Now $88 \div 28 = 3 \text{ rem } 4 \implies 28 \div 4 = 7 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 4}$
(f) $57, 76, 95$: $76 \div 57 = 1 \text{ rem } 19 \implies 57 \div 19 = 3 \text{ rem } 0$. Now $95 \div 19 = 5 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 19}$
(g) $16, 32, 96$: $32 \div 16 = 2 \text{ rem } 0$. Now $96 \div 16 = 6 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 16}$
(h) $20, 40, 80$: $40 \div 20 = 2 \text{ rem } 0$. Now $80 \div 20 = 4 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 20}$
(i) $48, 76, 96$: $76 \div 48 = 1 \text{ rem } 28 \implies 48 \div 28 = 1 \text{ rem } 20 \implies 28 \div 20 = 1 \text{ rem } 8 \implies 20 \div 8 = 2 \text{ rem } 4 \implies 8 \div 4 = 2 \text{ rem } 0$ (HCF of 48, 76 is 4). Now $96 \div 4 = 24 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 4}$
(j) $24, 48, 72$: $48 \div 24 = 2 \text{ rem } 0$. Now $72 \div 24 = 3 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 24}$
(k) $51, 65, 75$: $65 \div 51 = 1 \text{ rem } 14 \implies 51 \div 14 = 3 \text{ rem } 9 \implies 14 \div 9 = 1 \text{ rem } 5 \implies 9 \div 5 = 1 \text{ rem } 4 \implies 5 \div 4 = 1 \text{ rem } 1 \implies 4 \div 1 = 4 \text{ rem } 0$ (HCF of 51, 65 is 1). Since $\text{HCF}(1, 75) = 1$. ⇒ $\mathbf{\text{HCF} = 1}$
(l) $13, 39, 78$: $39 \div 13 = 3 \text{ rem } 0$. Now $78 \div 13 = 6 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 13}$
Q3. The lengths of two ropes is 24 metres and 14 metres. Ali wants to cut the ropes into pieces of equal lengths completely. What will be the maximum length of each piece?
To find the maximum equal length, we find $\text{HCF}(24, 14)$:
$$24 = 2 \times 2 \times 2 \times 3$$ $$14 = 2 \times 7$$ $$\text{Common Factor} = 2 \implies \text{HCF} = \mathbf{2\text{ meters}}$$ Answer: The maximum length of each piece will be $\mathbf{2\text{ meters}}$.
$$24 = 2 \times 2 \times 2 \times 3$$ $$14 = 2 \times 7$$ $$\text{Common Factor} = 2 \implies \text{HCF} = \mathbf{2\text{ meters}}$$ Answer: The maximum length of each piece will be $\mathbf{2\text{ meters}}$.
Q4. For the Independence day celebrations, 52 students in white, 65 students in green and 39 students in golden dress are to be arranged in equal rows such that students of the same colour dress are in each row. What is the greatest number of students that could be in each row?
To find the greatest number of students in each row, we find $\text{HCF}(52, 65, 39)$:
$$52 = 2 \times 2 \times 13$$ $$65 = 5 \times 13$$ $$39 = 3 \times 13$$ $$\text{Common Prime Factor} = 13 \implies \text{HCF} = \mathbf{13}$$ Answer: The greatest number of students in each row is $\mathbf{13\text{ students}}$.
$$52 = 2 \times 2 \times 13$$ $$65 = 5 \times 13$$ $$39 = 3 \times 13$$ $$\text{Common Prime Factor} = 13 \implies \text{HCF} = \mathbf{13}$$ Answer: The greatest number of students in each row is $\mathbf{13\text{ students}}$.
Q5. Find the greatest number that divides 16, 24 and 48 completely.
We calculate $\text{HCF}(16, 24, 48)$:
$$16 = 2^4 = 2 \times 2 \times 2 \times 2$$ $$24 = 2^3 \times 3 = 2 \times 2 \times 2 \times 3$$ $$48 = 2^4 \times 3 = 2 \times 2 \times 2 \times 2 \times 3$$ $$\text{Common Factors} = 2 \times 2 \times 2 = 2^3 = \mathbf{8}$$ Answer: The greatest number is $\mathbf{8}$.
$$16 = 2^4 = 2 \times 2 \times 2 \times 2$$ $$24 = 2^3 \times 3 = 2 \times 2 \times 2 \times 3$$ $$48 = 2^4 \times 3 = 2 \times 2 \times 2 \times 2 \times 3$$ $$\text{Common Factors} = 2 \times 2 \times 2 = 2^3 = \mathbf{8}$$ Answer: The greatest number is $\mathbf{8}$.
Q6. Ibrahim and Marwa are preparing first aid kits for the students. They have 30 perforated adhesive bandages, 60 triangular bandages and 75 rectangular bandages. They must distribute these equally in the kits, with nothing left over. What is the greatest number of kits they can be made with this quantity of bandages?
We find the $\text{HCF}(30, 60, 75)$:
$$30 = 2 \times 3 \times 5$$ $$60 = 2 \times 2 \times 3 \times 5$$ $$75 = 3 \times 5 \times 5$$ $$\text{Common Prime Factors} = 3 \times 5 = \mathbf{15}$$ Answer: The greatest number of first aid kits that can be made is $\mathbf{15\text{ kits}}$ (Each kit will contain $2$ adhesive, $4$ triangular, and $5$ rectangular bandages).
$$30 = 2 \times 3 \times 5$$ $$60 = 2 \times 2 \times 3 \times 5$$ $$75 = 3 \times 5 \times 5$$ $$\text{Common Prime Factors} = 3 \times 5 = \mathbf{15}$$ Answer: The greatest number of first aid kits that can be made is $\mathbf{15\text{ kits}}$ (Each kit will contain $2$ adhesive, $4$ triangular, and $5$ rectangular bandages).
Exercise 4: Least Common Multiple (LCM) Calculation & Real-Life Applications
Q1. Find LCM of the following numbers using prime factorization:
(a) $3, 21$: $3 = 3$, $21 = 3 \times 7$. $\text{LCM} = 3 \times 7 = \mathbf{21}$
(b) $12, 80$: $12 = 2^2 \times 3$, $80 = 2^4 \times 5$. $\text{LCM} = 2^4 \times 3 \times 5 = 16 \times 15 = \mathbf{240}$
(c) $20, 15$: $20 = 2^2 \times 5$, $15 = 3 \times 5$. $\text{LCM} = 2^2 \times 3 \times 5 = 4 \times 15 = \mathbf{60}$
(d) $4, 10, 16$: $4 = 2^2$, $10 = 2 \times 5$, $16 = 2^4$. $\text{LCM} = 2^4 \times 5 = 16 \times 5 = \mathbf{80}$
(e) $9, 18, 27$: $9 = 3^2$, $18 = 2 \times 3^2$, $27 = 3^3$. $\text{LCM} = 2 \times 3^3 = 2 \times 27 = \mathbf{54}$
(f) $10, 20, 35$: $10 = 2 \times 5$, $20 = 2^2 \times 5$, $35 = 5 \times 7$. $\text{LCM} = 2^2 \times 5 \times 7 = 4 \times 35 = \mathbf{140}$
(g) $20, 60, 75$: $20 = 2^2 \times 5$, $60 = 2^2 \times 3 \times 5$, $75 = 3 \times 5^2$. $\text{LCM} = 2^2 \times 3 \times 5^2 = 4 \times 3 \times 25 = \mathbf{300}$
(h) $30, 45, 90$: $30 = 2 \times 3 \times 5$, $45 = 3^2 \times 5$, $90 = 2 \times 3^2 \times 5$. $\text{LCM} = 2 \times 3^2 \times 5 = \mathbf{90}$
(i) $16, 24, 36$: $16 = 2^4$, $24 = 2^3 \times 3$, $36 = 2^2 \times 3^2$. $\text{LCM} = 2^4 \times 3^2 = 16 \times 9 = \mathbf{144}$
(j) $18, 60, 75$: $18 = 2 \times 3^2$, $60 = 2^2 \times 3 \times 5$, $75 = 3 \times 5^2$. $\text{LCM} = 2^2 \times 3^2 \times 5^2 = 4 \times 9 \times 25 = \mathbf{900}$
(k) $49, 51, 56$: $49 = 7^2$, $51 = 3 \times 17$, $56 = 2^3 \times 7$. $\text{LCM} = 2^3 \times 3 \times 7^2 \times 17 = 8 \times 3 \times 49 \times 17 = \mathbf{19\,992}$
(l) $13, 65, 71$: $13 = 13$, $65 = 5 \times 13$, $71 = 71$ (prime). $\text{LCM} = 5 \times 13 \times 71 = 65 \times 71 = \mathbf{4\,615}$
(b) $12, 80$: $12 = 2^2 \times 3$, $80 = 2^4 \times 5$. $\text{LCM} = 2^4 \times 3 \times 5 = 16 \times 15 = \mathbf{240}$
(c) $20, 15$: $20 = 2^2 \times 5$, $15 = 3 \times 5$. $\text{LCM} = 2^2 \times 3 \times 5 = 4 \times 15 = \mathbf{60}$
(d) $4, 10, 16$: $4 = 2^2$, $10 = 2 \times 5$, $16 = 2^4$. $\text{LCM} = 2^4 \times 5 = 16 \times 5 = \mathbf{80}$
(e) $9, 18, 27$: $9 = 3^2$, $18 = 2 \times 3^2$, $27 = 3^3$. $\text{LCM} = 2 \times 3^3 = 2 \times 27 = \mathbf{54}$
(f) $10, 20, 35$: $10 = 2 \times 5$, $20 = 2^2 \times 5$, $35 = 5 \times 7$. $\text{LCM} = 2^2 \times 5 \times 7 = 4 \times 35 = \mathbf{140}$
(g) $20, 60, 75$: $20 = 2^2 \times 5$, $60 = 2^2 \times 3 \times 5$, $75 = 3 \times 5^2$. $\text{LCM} = 2^2 \times 3 \times 5^2 = 4 \times 3 \times 25 = \mathbf{300}$
(h) $30, 45, 90$: $30 = 2 \times 3 \times 5$, $45 = 3^2 \times 5$, $90 = 2 \times 3^2 \times 5$. $\text{LCM} = 2 \times 3^2 \times 5 = \mathbf{90}$
(i) $16, 24, 36$: $16 = 2^4$, $24 = 2^3 \times 3$, $36 = 2^2 \times 3^2$. $\text{LCM} = 2^4 \times 3^2 = 16 \times 9 = \mathbf{144}$
(j) $18, 60, 75$: $18 = 2 \times 3^2$, $60 = 2^2 \times 3 \times 5$, $75 = 3 \times 5^2$. $\text{LCM} = 2^2 \times 3^2 \times 5^2 = 4 \times 9 \times 25 = \mathbf{900}$
(k) $49, 51, 56$: $49 = 7^2$, $51 = 3 \times 17$, $56 = 2^3 \times 7$. $\text{LCM} = 2^3 \times 3 \times 7^2 \times 17 = 8 \times 3 \times 49 \times 17 = \mathbf{19\,992}$
(l) $13, 65, 71$: $13 = 13$, $65 = 5 \times 13$, $71 = 71$ (prime). $\text{LCM} = 5 \times 13 \times 71 = 65 \times 71 = \mathbf{4\,615}$
Q2. Find LCM of the following numbers using division method:
(a) $14, 70$: Common prime division: $2 \times 5 \times 7 = \mathbf{70}$
(b) $15, 30$: Common prime division: $2 \times 3 \times 5 = \mathbf{30}$
(c) $45, 90$: Common prime division: $2 \times 3 \times 3 \times 5 = \mathbf{90}$
(d) $35, 60, 75$: Common division: $2^2 \times 3 \times 5^2 \times 7 = 4 \times 3 \times 25 \times 7 = \mathbf{2\,100}$
(e) $7, 21, 49$: Common division: $3 \times 7 \times 7 = \mathbf{147}$
(f) $25, 45, 95$: Common division: $3^2 \times 5^2 \times 19 = 9 \times 25 \times 19 = \mathbf{4\,275}$
(g) $16, 32, 48$: Common division: $2^5 \times 3 = 32 \times 3 = \mathbf{96}$
(h) $28, 32, 40$: Common division: $2^5 \times 5 \times 7 = 32 \times 35 = \mathbf{1\,120}$
(i) $12, 14, 26$: Common division: $2^2 \times 3 \times 7 \times 13 = 12 \times 91 = \mathbf{1\,092}$
(j) $10, 20, 25$: Common division: $2^2 \times 5^2 = 4 \times 25 = \mathbf{100}$
(k) $7, 14, 21$: Common division: $2 \times 3 \times 7 = \mathbf{42}$
(l) $8, 32, 42$: Common division: $2^5 \times 3 \times 7 = 32 \times 21 = \mathbf{672}$
(b) $15, 30$: Common prime division: $2 \times 3 \times 5 = \mathbf{30}$
(c) $45, 90$: Common prime division: $2 \times 3 \times 3 \times 5 = \mathbf{90}$
(d) $35, 60, 75$: Common division: $2^2 \times 3 \times 5^2 \times 7 = 4 \times 3 \times 25 \times 7 = \mathbf{2\,100}$
(e) $7, 21, 49$: Common division: $3 \times 7 \times 7 = \mathbf{147}$
(f) $25, 45, 95$: Common division: $3^2 \times 5^2 \times 19 = 9 \times 25 \times 19 = \mathbf{4\,275}$
(g) $16, 32, 48$: Common division: $2^5 \times 3 = 32 \times 3 = \mathbf{96}$
(h) $28, 32, 40$: Common division: $2^5 \times 5 \times 7 = 32 \times 35 = \mathbf{1\,120}$
(i) $12, 14, 26$: Common division: $2^2 \times 3 \times 7 \times 13 = 12 \times 91 = \mathbf{1\,092}$
(j) $10, 20, 25$: Common division: $2^2 \times 5^2 = 4 \times 25 = \mathbf{100}$
(k) $7, 14, 21$: Common division: $2 \times 3 \times 7 = \mathbf{42}$
(l) $8, 32, 42$: Common division: $2^5 \times 3 \times 7 = 32 \times 21 = \mathbf{672}$
Q3. Find minimum length of the ribbon which can be completely cut into pieces of lengths 45 cm, 75 cm and 85 cm without any leftover.
We find the $\text{LCM}(45, 75, 85)$:
$$45 = 3^2 \times 5$$ $$75 = 3 \times 5^2$$ $$85 = 5 \times 17$$ $$\text{LCM} = 3^2 \times 5^2 \times 17 = 9 \times 25 \times 17 = 225 \times 17 = \mathbf{3\,825\text{ cm}}$$ Answer: The minimum length of the ribbon is $\mathbf{3\,825\text{ cm}}$ (or $38.25\text{ meters}$).
$$45 = 3^2 \times 5$$ $$75 = 3 \times 5^2$$ $$85 = 5 \times 17$$ $$\text{LCM} = 3^2 \times 5^2 \times 17 = 9 \times 25 \times 17 = 225 \times 17 = \mathbf{3\,825\text{ cm}}$$ Answer: The minimum length of the ribbon is $\mathbf{3\,825\text{ cm}}$ (or $38.25\text{ meters}$).
Q4. Find the smallest number that is completely divisible by 42, 38 and 16.
We calculate $\text{LCM}(42, 38, 16)$:
$$42 = 2 \times 3 \times 7$$ $$38 = 2 \times 19$$ $$16 = 2^4$$ $$\text{LCM} = 2^4 \times 3 \times 7 \times 19 = 16 \times 21 \times 19 = 336 \times 19 = \mathbf{6\,384}$$ Answer: The smallest number is $\mathbf{6\,384}$.
$$42 = 2 \times 3 \times 7$$ $$38 = 2 \times 19$$ $$16 = 2^4$$ $$\text{LCM} = 2^4 \times 3 \times 7 \times 19 = 16 \times 21 \times 19 = 336 \times 19 = \mathbf{6\,384}$$ Answer: The smallest number is $\mathbf{6\,384}$.
Q5. The tour buses for Badshahi Masjid leave the station after every 25 minutes, for the interior city after every 15 minutes, and for the zoo after every 30 minutes. If the three buses leave the station simultaneously at 11:05 am, find the time when the three buses will leave the next station simultaneously.
We find the $\text{LCM}(25, 15, 30)$ in minutes:
$$25 = 5^2$$ $$15 = 3 \times 5$$ $$30 = 2 \times 3 \times 5$$ $$\text{LCM} = 2 \times 3 \times 5^2 = 6 \times 25 = \mathbf{150\text{ minutes}}$$ Converting minutes to hours: $150\text{ minutes} = 2\text{ hours and } 30\text{ minutes}$.
Adding to initial departure time ($11:05\text{ am}$):
$$11:05\text{ am} + 2\text{ hr } 30\text{ min} = \mathbf{1:35\text{ pm}}$$ Answer: The buses will leave simultaneously again at $\mathbf{1:35\text{ pm}}$.
$$25 = 5^2$$ $$15 = 3 \times 5$$ $$30 = 2 \times 3 \times 5$$ $$\text{LCM} = 2 \times 3 \times 5^2 = 6 \times 25 = \mathbf{150\text{ minutes}}$$ Converting minutes to hours: $150\text{ minutes} = 2\text{ hours and } 30\text{ minutes}$.
Adding to initial departure time ($11:05\text{ am}$):
$$11:05\text{ am} + 2\text{ hr } 30\text{ min} = \mathbf{1:35\text{ pm}}$$ Answer: The buses will leave simultaneously again at $\mathbf{1:35\text{ pm}}$.
Q6. Boxes having heights of 22 cm, 35 cm and 50 cm respectively are to be stacked next to each other. What is the shortest possible height at which the three types of boxes will be at the same height?
We find $\text{LCM}(22, 35, 50)$:
$$22 = 2 \times 11$$ $$35 = 5 \times 7$$ $$50 = 2 \times 5^2$$ $$\text{LCM} = 2 \times 5^2 \times 7 \times 11 = 50 \times 77 = \mathbf{3\,850\text{ cm}}$$ Answer: The shortest possible matching height is $\mathbf{3\,850\text{ cm}}$ (or $38.5\text{ meters}$).
$$22 = 2 \times 11$$ $$35 = 5 \times 7$$ $$50 = 2 \times 5^2$$ $$\text{LCM} = 2 \times 5^2 \times 7 \times 11 = 50 \times 77 = \mathbf{3\,850\text{ cm}}$$ Answer: The shortest possible matching height is $\mathbf{3\,850\text{ cm}}$ (or $38.5\text{ meters}$).
Q7. Students of grade 5 have a Mathematics test after every 3 days, English test after every 6 days and Science test after every 9 days. If all the three tests were conducted today, find when will the three tests be conducted together again?
We find $\text{LCM}(3, 6, 9)$:
$$3 = 3$$ $$6 = 2 \times 3$$ $$9 = 3^2$$ $$\text{LCM} = 2 \times 3^2 = 2 \times 9 = \mathbf{18\text{ days}}$$ Answer: All three tests will be conducted together again after $\mathbf{18\text{ days}}$.
$$3 = 3$$ $$6 = 2 \times 3$$ $$9 = 3^2$$ $$\text{LCM} = 2 \times 3^2 = 2 \times 9 = \mathbf{18\text{ days}}$$ Answer: All three tests will be conducted together again after $\mathbf{18\text{ days}}$.
Review Exercise 2: Comprehensive Review & Mastery Assessment
Q1. Choose the correct option:
(a) The HCF of 20, 48 and 56 is _____:
• $20 = 2^2 \times 5$, $48 = 2^4 \times 3$, $56 = 2^3 \times 7 \implies \text{HCF} = 2^2 = \mathbf{4}$. ⇒ Correct Option: (i) 4
(b) Which of the following is a prime number?
• $33 = 3 \times 11$, $35 = 5 \times 7$, $37 = \text{Prime}$, $39 = 3 \times 13$. ⇒ Correct Option: (iii) 37
(c) The prime factorization of 16 is _____:
• $16 = 2 \times 2 \times 2 \times 2$. ⇒ Correct Option: (iii) $2 \times 2 \times 2 \times 2$
(d) The LCM of 33, 66 and 81 is _____:
• $33 = 3 \times 11$, $66 = 2 \times 3 \times 11$, $81 = 3^4 \implies \text{LCM} = 2 \times 3^4 \times 11 = 2 \times 81 \times 11 = \mathbf{1\,782}$. ⇒ Correct Option: (iii) 1 782
(e) The LCM of two or more prime numbers is always equal to their _____:
• Since prime numbers have no common factors other than $1$, their LCM is their product. ⇒ Correct Option: (iv) product
(f) Which of the following is a composite number?
• $1$ (neither), $2$ (prime), $3$ (prime), $4$ (composite, factors $1, 2, 4$). ⇒ Correct Option: (iv) 4
• $20 = 2^2 \times 5$, $48 = 2^4 \times 3$, $56 = 2^3 \times 7 \implies \text{HCF} = 2^2 = \mathbf{4}$. ⇒ Correct Option: (i) 4
(b) Which of the following is a prime number?
• $33 = 3 \times 11$, $35 = 5 \times 7$, $37 = \text{Prime}$, $39 = 3 \times 13$. ⇒ Correct Option: (iii) 37
(c) The prime factorization of 16 is _____:
• $16 = 2 \times 2 \times 2 \times 2$. ⇒ Correct Option: (iii) $2 \times 2 \times 2 \times 2$
(d) The LCM of 33, 66 and 81 is _____:
• $33 = 3 \times 11$, $66 = 2 \times 3 \times 11$, $81 = 3^4 \implies \text{LCM} = 2 \times 3^4 \times 11 = 2 \times 81 \times 11 = \mathbf{1\,782}$. ⇒ Correct Option: (iii) 1 782
(e) The LCM of two or more prime numbers is always equal to their _____:
• Since prime numbers have no common factors other than $1$, their LCM is their product. ⇒ Correct Option: (iv) product
(f) Which of the following is a composite number?
• $1$ (neither), $2$ (prime), $3$ (prime), $4$ (composite, factors $1, 2, 4$). ⇒ Correct Option: (iv) 4
Q2. Find HCF of the following numbers using prime factorization:
(a) $15, 18$: $15 = 3 \times 5$, $18 = 2 \times 3^2 \implies \mathbf{\text{HCF} = 3}$
(b) $10, 20$: $10 = 2 \times 5$, $20 = 2^2 \times 5 \implies \mathbf{\text{HCF} = 10}$
(c) $25, 40$: $25 = 5^2$, $40 = 2^3 \times 5 \implies \mathbf{\text{HCF} = 5}$
(d) $56, 88$: $56 = 2^3 \times 7$, $88 = 2^3 \times 11 \implies \mathbf{\text{HCF} = 8}$
(e) $10, 18, 22$: $10 = 2 \times 5$, $18 = 2 \times 3^2$, $22 = 2 \times 11 \implies \mathbf{\text{HCF} = 2}$
(f) $20, 40, 82$: $20 = 2^2 \times 5$, $40 = 2^3 \times 5$, $82 = 2 \times 41 \implies \mathbf{\text{HCF} = 2}$
(g) $16, 38, 98$: $16 = 2^4$, $38 = 2 \times 19$, $98 = 2 \times 7^2 \implies \mathbf{\text{HCF} = 2}$
(h) $39, 51, 75$: $39 = 3 \times 13$, $51 = 3 \times 17$, $75 = 3 \times 5^2 \implies \mathbf{\text{HCF} = 3}$
(b) $10, 20$: $10 = 2 \times 5$, $20 = 2^2 \times 5 \implies \mathbf{\text{HCF} = 10}$
(c) $25, 40$: $25 = 5^2$, $40 = 2^3 \times 5 \implies \mathbf{\text{HCF} = 5}$
(d) $56, 88$: $56 = 2^3 \times 7$, $88 = 2^3 \times 11 \implies \mathbf{\text{HCF} = 8}$
(e) $10, 18, 22$: $10 = 2 \times 5$, $18 = 2 \times 3^2$, $22 = 2 \times 11 \implies \mathbf{\text{HCF} = 2}$
(f) $20, 40, 82$: $20 = 2^2 \times 5$, $40 = 2^3 \times 5$, $82 = 2 \times 41 \implies \mathbf{\text{HCF} = 2}$
(g) $16, 38, 98$: $16 = 2^4$, $38 = 2 \times 19$, $98 = 2 \times 7^2 \implies \mathbf{\text{HCF} = 2}$
(h) $39, 51, 75$: $39 = 3 \times 13$, $51 = 3 \times 17$, $75 = 3 \times 5^2 \implies \mathbf{\text{HCF} = 3}$
Q3. Find HCF of the following numbers using the division method:
(a) $20, 50$: $50 \div 20 = 2 \text{ rem } 10 \implies 20 \div 10 = 2 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 10}$
(b) $15, 45$: $45 \div 15 = 3 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 15}$
(c) $60, 70, 80$: $70 \div 60 = 1 \text{ rem } 10 \implies 60 \div 10 = 6 \text{ rem } 0$. $80 \div 10 = 8 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 10}$
(d) $22, 28, 32$: $28 \div 22 = 1 \text{ rem } 6 \implies 22 \div 6 = 3 \text{ rem } 4 \implies 6 \div 4 = 1 \text{ rem } 2 \implies 4 \div 2 = 2 \text{ rem } 0$. $32 \div 2 = 16 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 2}$
(e) $44, 55, 99$: $55 \div 44 = 1 \text{ rem } 11 \implies 44 \div 11 = 4 \text{ rem } 0$. $99 \div 11 = 9 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 11}$
(f) $34, 48, 62$: $48 \div 34 = 1 \text{ rem } 14 \implies 34 \div 14 = 2 \text{ rem } 6 \implies 14 \div 6 = 2 \text{ rem } 2 \implies 6 \div 2 = 3 \text{ rem } 0$. $62 \div 2 = 31 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 2}$
(g) $30, 45, 70$: $45 \div 30 = 1 \text{ rem } 15 \implies 30 \div 15 = 2 \text{ rem } 0$. $70 \div 15 = 4 \text{ rem } 10 \implies 15 \div 10 = 1 \text{ rem } 5 \implies 10 \div 5 = 2 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 5}$
(h) $26, 52, 65$: $52 \div 26 = 2 \text{ rem } 0$. $65 \div 26 = 2 \text{ rem } 13 \implies 26 \div 13 = 2 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 13}$
(b) $15, 45$: $45 \div 15 = 3 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 15}$
(c) $60, 70, 80$: $70 \div 60 = 1 \text{ rem } 10 \implies 60 \div 10 = 6 \text{ rem } 0$. $80 \div 10 = 8 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 10}$
(d) $22, 28, 32$: $28 \div 22 = 1 \text{ rem } 6 \implies 22 \div 6 = 3 \text{ rem } 4 \implies 6 \div 4 = 1 \text{ rem } 2 \implies 4 \div 2 = 2 \text{ rem } 0$. $32 \div 2 = 16 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 2}$
(e) $44, 55, 99$: $55 \div 44 = 1 \text{ rem } 11 \implies 44 \div 11 = 4 \text{ rem } 0$. $99 \div 11 = 9 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 11}$
(f) $34, 48, 62$: $48 \div 34 = 1 \text{ rem } 14 \implies 34 \div 14 = 2 \text{ rem } 6 \implies 14 \div 6 = 2 \text{ rem } 2 \implies 6 \div 2 = 3 \text{ rem } 0$. $62 \div 2 = 31 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 2}$
(g) $30, 45, 70$: $45 \div 30 = 1 \text{ rem } 15 \implies 30 \div 15 = 2 \text{ rem } 0$. $70 \div 15 = 4 \text{ rem } 10 \implies 15 \div 10 = 1 \text{ rem } 5 \implies 10 \div 5 = 2 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 5}$
(h) $26, 52, 65$: $52 \div 26 = 2 \text{ rem } 0$. $65 \div 26 = 2 \text{ rem } 13 \implies 26 \div 13 = 2 \text{ rem } 0$. ⇒ $\mathbf{\text{HCF} = 13}$
Q4. Find LCM of the following numbers using prime factorization method:
(a) $2, 5$: $\text{LCM} = 2 \times 5 = \mathbf{10}$
(b) $3, 7$: $\text{LCM} = 3 \times 7 = \mathbf{21}$
(c) $5, 8$: $\text{LCM} = 5 \times 2^3 = \mathbf{40}$
(d) $4, 10, 16$: $4 = 2^2$, $10 = 2 \times 5$, $16 = 2^4 \implies \text{LCM} = 2^4 \times 5 = \mathbf{80}$
(e) $20, 25, 50$: $20 = 2^2 \times 5$, $25 = 5^2$, $50 = 2 \times 5^2 \implies \text{LCM} = 2^2 \times 5^2 = \mathbf{100}$
(f) $45, 90, 95$: $45 = 3^2 \times 5$, $90 = 2 \times 3^2 \times 5$, $95 = 5 \times 19 \implies \text{LCM} = 2 \times 3^2 \times 5 \times 19 = 90 \times 19 = \mathbf{1\,710}$
(g) $32, 70, 80$: $32 = 2^5$, $70 = 2 \times 5 \times 7$, $80 = 2^4 \times 5 \implies \text{LCM} = 2^5 \times 5 \times 7 = 32 \times 35 = \mathbf{1\,120}$
(h) $33, 66$: $33 = 3 \times 11$, $66 = 2 \times 3 \times 11 \implies \text{LCM} = 2 \times 3 \times 11 = \mathbf{66}$
(b) $3, 7$: $\text{LCM} = 3 \times 7 = \mathbf{21}$
(c) $5, 8$: $\text{LCM} = 5 \times 2^3 = \mathbf{40}$
(d) $4, 10, 16$: $4 = 2^2$, $10 = 2 \times 5$, $16 = 2^4 \implies \text{LCM} = 2^4 \times 5 = \mathbf{80}$
(e) $20, 25, 50$: $20 = 2^2 \times 5$, $25 = 5^2$, $50 = 2 \times 5^2 \implies \text{LCM} = 2^2 \times 5^2 = \mathbf{100}$
(f) $45, 90, 95$: $45 = 3^2 \times 5$, $90 = 2 \times 3^2 \times 5$, $95 = 5 \times 19 \implies \text{LCM} = 2 \times 3^2 \times 5 \times 19 = 90 \times 19 = \mathbf{1\,710}$
(g) $32, 70, 80$: $32 = 2^5$, $70 = 2 \times 5 \times 7$, $80 = 2^4 \times 5 \implies \text{LCM} = 2^5 \times 5 \times 7 = 32 \times 35 = \mathbf{1\,120}$
(h) $33, 66$: $33 = 3 \times 11$, $66 = 2 \times 3 \times 11 \implies \text{LCM} = 2 \times 3 \times 11 = \mathbf{66}$
Q5. Find LCM of the following numbers using division method:
(a) $4, 9$: $2^2 \times 3^2 = \mathbf{36}$
(b) $7, 11$: $7 \times 11 = \mathbf{77}$
(c) $14, 26$: $2 \times 7 \times 13 = \mathbf{182}$
(d) $20, 40$: $2^3 \times 5 = \mathbf{40}$
(e) $6, 24, 42$: $2^3 \times 3 \times 7 = 24 \times 7 = \mathbf{168}$
(f) $10, 20, 30$: $2^2 \times 3 \times 5 = \mathbf{60}$
(g) $12, 18, 38$: $2^2 \times 3^2 \times 19 = 36 \times 19 = \mathbf{684}$
(h) $6, 15, 21$: $2 \times 3 \times 5 \times 7 = \mathbf{210}$
(b) $7, 11$: $7 \times 11 = \mathbf{77}$
(c) $14, 26$: $2 \times 7 \times 13 = \mathbf{182}$
(d) $20, 40$: $2^3 \times 5 = \mathbf{40}$
(e) $6, 24, 42$: $2^3 \times 3 \times 7 = 24 \times 7 = \mathbf{168}$
(f) $10, 20, 30$: $2^2 \times 3 \times 5 = \mathbf{60}$
(g) $12, 18, 38$: $2^2 \times 3^2 \times 19 = 36 \times 19 = \mathbf{684}$
(h) $6, 15, 21$: $2 \times 3 \times 5 \times 7 = \mathbf{210}$
Q6. 84 apples, 56 bananas and 21 oranges were distributed equally among some children. If the same combination of all kinds of fruits is distributed among all the children, find the maximum possible number of children who can receive the fruits?
We find the $\text{HCF}(84, 56, 21)$:
$$84 = 2^2 \times 3 \times 7$$ $$56 = 2^3 \times 7$$ $$21 = 3 \times 7$$ $$\text{Common Prime Factor} = \mathbf{7}$$ Answer: The maximum possible number of children is $\mathbf{7\text{ children}}$ (Each child receives $12$ apples, $8$ bananas, and $3$ oranges).
$$84 = 2^2 \times 3 \times 7$$ $$56 = 2^3 \times 7$$ $$21 = 3 \times 7$$ $$\text{Common Prime Factor} = \mathbf{7}$$ Answer: The maximum possible number of children is $\mathbf{7\text{ children}}$ (Each child receives $12$ apples, $8$ bananas, and $3$ oranges).
Q7. Three water containers contain 12 litres, 24 litres and 42 litres of water. (a) Find the maximum capacity of a measuring container that can fully measure the amount of water in all three containers. (b) Find how many times the container needs to be filled to empty each container.
(a) Maximum capacity of measuring container:
$$\text{HCF}(12, 24, 42) \implies 12 = 2^2 \times 3, \quad 24 = 2^3 \times 3, \quad 42 = 2 \times 3 \times 7$$ $$\text{Common Factors} = 2 \times 3 = \mathbf{6\text{ litres}}$$ (b) Number of fillings to empty each container:
• 1st container ($12\text{ L}$): $12 \div 6 = \mathbf{2\text{ times}}$
• 2nd container ($24\text{ L}$): $24 \div 6 = \mathbf{4\text{ times}}$
• 3rd container ($42\text{ L}$): $42 \div 6 = \mathbf{7\text{ times}}$
• Total fillings across all containers: $2 + 4 + 7 = \mathbf{13\text{ times}}$.
$$\text{HCF}(12, 24, 42) \implies 12 = 2^2 \times 3, \quad 24 = 2^3 \times 3, \quad 42 = 2 \times 3 \times 7$$ $$\text{Common Factors} = 2 \times 3 = \mathbf{6\text{ litres}}$$ (b) Number of fillings to empty each container:
• 1st container ($12\text{ L}$): $12 \div 6 = \mathbf{2\text{ times}}$
• 2nd container ($24\text{ L}$): $24 \div 6 = \mathbf{4\text{ times}}$
• 3rd container ($42\text{ L}$): $42 \div 6 = \mathbf{7\text{ times}}$
• Total fillings across all containers: $2 + 4 + 7 = \mathbf{13\text{ times}}$.
Q8. Find the smallest number that is completely divisible by 32 and 55.
We calculate $\text{LCM}(32, 55)$:
$$32 = 2^5$$ $$55 = 5 \times 11$$ Since they are co-prime (no shared prime factors):
$$\text{LCM} = 32 \times 55 = \mathbf{1\,760}$$ Answer: The smallest number is $\mathbf{1\,760}$.
$$32 = 2^5$$ $$55 = 5 \times 11$$ Since they are co-prime (no shared prime factors):
$$\text{LCM} = 32 \times 55 = \mathbf{1\,760}$$ Answer: The smallest number is $\mathbf{1\,760}$.
Q9. Find the least number of stickers which can be equally distributed among 15, 12 and 10 children.
We calculate $\text{LCM}(15, 12, 10)$:
$$15 = 3 \times 5$$ $$12 = 2^2 \times 3$$ $$10 = 2 \times 5$$ $$\text{LCM} = 2^2 \times 3 \times 5 = 4 \times 15 = \mathbf{60}$$ Answer: The least number of stickers is $\mathbf{60\text{ stickers}}$.
$$15 = 3 \times 5$$ $$12 = 2^2 \times 3$$ $$10 = 2 \times 5$$ $$\text{LCM} = 2^2 \times 3 \times 5 = 4 \times 15 = \mathbf{60}$$ Answer: The least number of stickers is $\mathbf{60\text{ stickers}}$.
🎯 Unit 2 Synthesis Summary
HCF is the largest shared factor that evenly divides numbers (used for cutting, sharing, and finding maximum portions). LCM is the smallest common multiple divisible by numbers (used for matching cycles, schedules, and minimum matching quantities). Prime numbers are numbers with exactly 2 factors, $2$ is the only even prime, and $1$ is unique (neither prime nor composite). For any two numbers: $\mathbf{\text{Number}_1 \times \text{Number}_2 = \text{HCF} \times \text{LCM}}$.
More Chapter Notes for Class 5 (FBISE)
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Mastery Guide: Large Numbers, Place Value Systems, Arithmetic Operations & BODMAS Rule
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Class 5 Mathematics - Ch 2: HCF and LCM Chapter Mock Test
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