HCF and LCM Word Problems
HCF and LCM word problems help students apply the concepts of Highest Common Factor and Lowest Common Multiple to real-life situations. In Class 5, students move beyond just computing HCF and LCM — they learn to identify which concept to use based on the wording of a problem.
Understanding when to use HCF and when to use LCM is the key skill tested in NCERT/CBSE exams. This page covers both types with step-by-step solutions and plenty of practice.
Many students can calculate HCF and LCM correctly but struggle to decide which one to use in a word problem. The trick is to read the problem carefully and look for key words:
- If the problem asks for the largest group, maximum pieces, or equal distribution with nothing left over → use HCF
- If the problem asks about events happening together again, the smallest common quantity, or when things coincide → use LCM
This page teaches you to recognise these patterns through 10 solved examples covering all common problem types — from distributing items equally to finding when events repeat together.
What is HCF and LCM Word Problems - Class 5 Maths (Factors and Multiples)?
The Highest Common Factor (HCF) of two or more numbers is the largest number that divides all of them exactly. The Lowest Common Multiple (LCM) of two or more numbers is the smallest number that is a multiple of all of them.
When to use HCF:
- Dividing things into equal groups with nothing left over
- Finding the largest piece that fits evenly
- Key words: "largest", "maximum", "greatest", "divide equally", "cut into equal pieces"
When to use LCM:
- Finding when events that repeat will happen together again
- Finding the smallest quantity that can be divided by all given numbers
- Key words: "together again", "at the same time", "smallest number divisible by", "minimum"
HCF and LCM Word Problems Formula
HCF × LCM = Product of the two numbers
HCF(a, b) × LCM(a, b) = a × b
Quick Decision Rule:
Equal groups / largest piece → HCF
Repeat together / smallest common → LCM
Types and Properties
Type 1: Equal Distribution Problems (Use HCF)
These problems ask you to divide items into equal groups with no remainder. You find the HCF of the given quantities.
Type 2: Repeating Events Problems (Use LCM)
These problems involve events that happen at different intervals. You find the LCM to know when they coincide.
Type 3: Minimum Quantity Problems (Use LCM)
These ask for the smallest number divisible by all given numbers. You find the LCM of those numbers.
Type 4: Combined HCF-LCM Problems
Some problems require using the relationship: HCF × LCM = Product of the two numbers.
Solved Examples
Example 1: Example 1: Equal Groups (HCF)
Problem: Ria has 24 mangoes and 36 guavas. She wants to pack them into bags so that each bag has the same number of fruits and no fruits are left over. What is the maximum number of fruits she can put in each bag?
Solution:
Step 1: We need the largest number that divides both 24 and 36 → find HCF.
Step 2: Prime factorisation of 24 = 2 × 2 × 2 × 3
Step 3: Prime factorisation of 36 = 2 × 2 × 3 × 3
Step 4: Common factors = 2 × 2 × 3 = 12
Answer: Ria can put a maximum of 12 fruits in each bag.
Example 2: Example 2: Repeating Events (LCM)
Problem: Aman goes to a library every 6 days. Dev goes to the same library every 8 days. If they both visited today, after how many days will they visit together again?
Solution:
Step 1: We need the smallest number divisible by both 6 and 8 → find LCM.
Step 2: Prime factorisation of 6 = 2 × 3
Step 3: Prime factorisation of 8 = 2 × 2 × 2
Step 4: LCM = 2 × 2 × 2 × 3 = 24
Answer: They will visit together again after 24 days.
Example 3: Example 3: Cutting into Equal Pieces (HCF)
Problem: A cloth merchant has two rolls of fabric — one 48 m long and the other 60 m long. He wants to cut them into pieces of equal length with no fabric wasted. What is the greatest possible length of each piece?
Solution:
Step 1: We need the greatest length that divides both 48 and 60 → find HCF.
Step 2: Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3
Step 3: Prime factorisation of 60 = 2 × 2 × 3 × 5
Step 4: Common factors = 2 × 2 × 3 = 12
Answer: The greatest possible length of each piece is 12 m.
Example 4: Example 4: Bells Ringing Together (LCM)
Problem: In a school, one bell rings every 15 minutes and another bell rings every 20 minutes. If both bells ring together at 8:00 AM, when will they ring together again?
Solution:
Step 1: We need the LCM of 15 and 20.
Step 2: Prime factorisation of 15 = 3 × 5
Step 3: Prime factorisation of 20 = 2 × 2 × 5
Step 4: LCM = 2 × 2 × 3 × 5 = 60 minutes = 1 hour
Answer: Both bells will ring together again at 9:00 AM.
Example 5: Example 5: Arranging in Rows (HCF)
Problem: Priya has 45 red roses and 75 white roses. She wants to make bouquets with the same number of red and the same number of white roses in each bouquet, using all flowers. What is the maximum number of bouquets she can make?
Solution:
Step 1: The maximum number of bouquets = HCF of 45 and 75.
Step 2: Prime factorisation of 45 = 3 × 3 × 5
Step 3: Prime factorisation of 75 = 3 × 5 × 5
Step 4: HCF = 3 × 5 = 15
Step 5: Each bouquet gets 45 ÷ 15 = 3 red roses and 75 ÷ 15 = 5 white roses.
Answer: Priya can make 15 bouquets (3 red + 5 white roses each).
Example 6: Example 6: Minimum Sweets (LCM)
Problem: Meera wants to buy the minimum number of sweets so she can distribute them equally among groups of 6, 8, or 12 children with none left over. How many sweets should she buy?
Solution:
Step 1: We need the smallest number divisible by 6, 8, and 12 → find LCM.
Step 2: Prime factorisation: 6 = 2 × 3, 8 = 2 × 2 × 2, 12 = 2 × 2 × 3
Step 3: LCM = 2 × 2 × 2 × 3 = 24
Answer: Meera should buy at least 24 sweets.
Example 7: Example 7: Tiling a Floor (HCF)
Problem: A rectangular room is 16 m long and 12 m wide. Square tiles of the largest possible size are used to tile the floor without cutting any tile. What is the side length of each tile?
Solution:
Step 1: The largest square tile side = HCF of 16 and 12.
Step 2: Prime factorisation of 16 = 2 × 2 × 2 × 2
Step 3: Prime factorisation of 12 = 2 × 2 × 3
Step 4: HCF = 2 × 2 = 4
Answer: Each tile has a side length of 4 m.
Example 8: Example 8: Using the HCF-LCM Relationship
Problem: The HCF of two numbers is 6 and their LCM is 60. If one number is 12, find the other number.
Solution:
Step 1: Use the formula: HCF × LCM = Product of the two numbers
Step 2: 6 × 60 = 12 × other number
Step 3: 360 = 12 × other number
Step 4: Other number = 360 ÷ 12 = 30
Answer: The other number is 30.
Example 9: Example 9: Traffic Lights (LCM)
Problem: At a crossing, one traffic light turns red every 40 seconds and another turns red every 50 seconds. If both turned red at the same moment, after how many seconds will they turn red together again?
Solution:
Step 1: Find the LCM of 40 and 50.
Step 2: Prime factorisation of 40 = 2 × 2 × 2 × 5
Step 3: Prime factorisation of 50 = 2 × 5 × 5
Step 4: LCM = 2 × 2 × 2 × 5 × 5 = 200
Answer: Both lights will turn red together after 200 seconds.
Example 10: Example 10: Distributing Equally (HCF)
Problem: Arjun has 56 pencils and 98 erasers to distribute equally among students. Each student must get the same number of pencils and the same number of erasers. What is the maximum number of students?
Solution:
Step 1: Maximum students = HCF of 56 and 98.
Step 2: Prime factorisation of 56 = 2 × 2 × 2 × 7
Step 3: Prime factorisation of 98 = 2 × 7 × 7
Step 4: HCF = 2 × 7 = 14
Step 5: Each student gets 56 ÷ 14 = 4 pencils and 98 ÷ 14 = 7 erasers.
Answer: Maximum 14 students (4 pencils + 7 erasers each).
Real-World Applications
Real-life uses of HCF and LCM:
- Packing and distribution: Dividing items equally into groups with no leftovers (HCF)
- Event scheduling: Finding when two repeating events happen at the same time (LCM)
- Tiling and carpeting: Finding the largest tile that fits perfectly (HCF)
- Shopping: Finding the minimum quantity to buy for equal sharing (LCM)
- Bus and train schedules: Finding when two buses depart together from the same stop (LCM)
Key Points to Remember
- HCF = largest number that divides all given numbers exactly. Use when dividing into equal groups.
- LCM = smallest number that is a multiple of all given numbers. Use when finding when events repeat together.
- Key words for HCF: "largest", "maximum", "greatest", "divide equally", "no remainder".
- Key words for LCM: "together again", "at the same time", "smallest number divisible by", "minimum".
- HCF × LCM = Product of the two numbers (works for any pair of numbers).
- HCF of two numbers is always a factor of their LCM.
- Always use prime factorisation to find HCF and LCM accurately.
- Read word problems carefully — underline key words before deciding HCF or LCM.
Practice Problems
- Kavi has 32 apples and 48 oranges. He wants to pack them into baskets with the same number of apples and the same number of oranges in each basket, using all fruits. What is the maximum number of baskets?
- Two buses leave a bus stand at the same time. One comes back every 12 minutes and the other every 18 minutes. After how many minutes will both buses be at the bus stand together again?
- A rope 72 m long and another rope 90 m long are to be cut into pieces of equal length. What is the greatest possible length of each piece?
- Three children start cycling around a track. They take 10, 12, and 15 minutes to complete one round. If they start together, after how many minutes will they meet at the starting point again?
- Neha has 84 red beads and 60 blue beads. She wants to make identical necklaces using all beads. What is the maximum number of necklaces she can make?
- The HCF of two numbers is 8 and their LCM is 96. If one number is 32, find the other number.
- A room is 18 m long and 24 m wide. What is the side of the largest square tile that can tile the floor completely?
- Aditi visits a park every 4 days and Rahul visits the same park every 6 days. They met at the park today. After how many days will they meet again at the park?
Frequently Asked Questions
Q1. How do I know whether to use HCF or LCM in a word problem?
Look for key words. If the problem says "largest", "maximum", "divide equally", or "no remainder", use HCF. If it says "together again", "at the same time", "minimum number divisible by", or "repeat", use LCM.
Q2. What is the relationship between HCF and LCM of two numbers?
For any two numbers a and b: HCF(a, b) × LCM(a, b) = a × b. This means if you know the HCF and one number, you can find the LCM, and vice versa.
Q3. Can the HCF of two numbers be equal to one of the numbers?
Yes. If one number is a factor of the other, the HCF equals the smaller number. For example, HCF of 6 and 18 is 6.
Q4. Can the LCM of two numbers be equal to one of the numbers?
Yes. If one number is a multiple of the other, the LCM equals the larger number. For example, LCM of 5 and 15 is 15.
Q5. What is the HCF and LCM of two co-prime numbers?
For co-prime numbers (like 8 and 15), the HCF is always 1, and the LCM is their product. So HCF(8, 15) = 1 and LCM(8, 15) = 120.
Q6. How do I find the LCM of three numbers?
Write the prime factorisation of all three numbers. For each prime factor, take the highest power that appears. Multiply them together. For example, LCM of 4, 6, 10: 4 = 2², 6 = 2 × 3, 10 = 2 × 5. LCM = 2² × 3 × 5 = 60.
Q7. Is HCF always smaller than LCM?
Yes, for two different numbers, HCF is always less than or equal to LCM. They are equal only when both numbers are the same (e.g., HCF(5, 5) = LCM(5, 5) = 5).
Q8. Why is prime factorisation useful for HCF and LCM?
Prime factorisation breaks numbers into their building blocks. For HCF, you pick common factors. For LCM, you pick the highest power of each factor. This method works reliably for all numbers, including large ones.
Q9. What are some common mistakes in HCF and LCM word problems?
The most common mistake is confusing HCF and LCM — using LCM when HCF is needed, or vice versa. Another error is incorrect prime factorisation. Always double-check by multiplying the factors back together.
Q10. Is this topic in the NCERT Class 5 syllabus?
Yes. HCF and LCM word problems are part of the Factors and Multiples chapter in Class 5 NCERT Mathematics. Students are expected to solve application-based problems using both concepts.
