What is the HCF of 24 and 36?

12. 24 = 2^3 3 and 36 = 2^2 3^2. The common factors are 2^2 and 3, so the HCF is 4 3 = 12.

What is the LCM of 8, 12 and 15?

120. Take the highest power of each prime: 2^3 (from 8), 3 (from 12 or 15) and 5 (from 15). The LCM is 8 3 5 = 120.

The product of two numbers is 2160 and their HCF is 12. What is their LCM?

180. For any two numbers, HCF LCM = product of the numbers. So LCM = 2160 12 = 180.

Find the greatest number that divides 43, 91 and 183 leaving the same remainder in each case.

4. Such a number divides the differences 91 - 43 = 48, 183 - 91 = 92 and 183 - 43 = 140. The HCF of 48, 92 and 140 is 4.

Three bells ring at intervals of 18, 24 and 32 minutes. If they ring together at 8:00 a.m., when will they next ring together?

12:48 p.m.. They coincide every LCM(18, 24, 32) = 288 minutes = 4 hours 48 minutes. Adding this to 8:00 a.m. gives 12:48 p.m.

AI-guided practice

Quantitative Aptitude · Numbers & Simplification

HCF and LCM — Questions & Answers

HCF (greatest common divisor) and LCM (least common multiple) drive a large family of word problems: synchronising repeating events, cutting the largest equal pieces, and finding numbers that leave fixed remainders.

5 solved questions
·
Step-by-step AI walkthroughs
·
Key formulas included

Key formulas

For any two numbers: $\text{HCF} \times \text{LCM} = \text{product of the numbers}$
HCF of fractions $= \dfrac{\text{HCF of numerators}}{\text{LCM of denominators}}$
LCM of fractions $= \dfrac{\text{LCM of numerators}}{\text{HCF of denominators}}$
Events repeating every $a$ , $b$ , $c$ units coincide again after $\operatorname{LCM}(a,b,c)$ units

5 solved HCF and LCM questions

Try each question first — tap an option to check your answer, or open Learn step by step for a guided walkthrough.

01
What is the HCF of 24 and 36?
4 steps · reveal one at a time
1. Step 1: Factorize the first number
  To find the HCF, we first find the prime factorization of each number. For 24, we can write it as $24 = 2 \times 12 = 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3$ .
2. Step 2: Factorize the second number
  Next, we do the same for 36. The prime factorization of 36 is $36 = 2 \times 18 = 2 \times 2 \times 9 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$ . We now have both numbers in their prime factor forms.
3. Step 3: Calculate the HCF
  To find the HCF, we multiply the common prime factors raised to their lowest powers. The common factors are $2$ and $3$ . The lowest power of $2$ is $2^2$ and for $3$ is $3^1$ . So, HCF is $2^2 \times 3 = 4 \times 3 = 12$ .
4. Step 4: The answer
  So the answer is (C) 12. This is the largest number that divides both 24 and 36 without leaving a remainder.
02
What is the LCM of 8, 12 and 15?
5 steps · reveal one at a time
1. Step 1: Prime Factorize Numbers
  First, find the prime factorization for each number: $8$ , $12$ , and $15$ . This helps us see all the building blocks. We get: $8 = 2^3$ , $12 = 2^2 \times 3$ , and $15 = 3 \times 5$ .
2. Step 2: Find Highest Powers
  To find the LCM, we take the highest power of each unique prime factor that appears across all numbers. The unique prime factors are $2$ , $3$ , and $5$ .
3. Step 3: Select Highest Powers
  From $8$ , we get $2^3$ . From $12$ or $15$ , we get $3^1$ . From $15$ , we get $5^1$ . So, the highest powers are $2^3$ , $3$ , and $5$ .
4. Step 4: Calculate the LCM
  Now, multiply these highest powers together to find the LCM. This gives us $2^3 \times 3 \times 5 = 8 \times 3 \times 5$ . Calculating this product gives $24 \times 5 = 120$ .
5. Step 5: The answer
  So the answer is (C) 120. This is the smallest number that is a multiple of $8$ , $12$ , and $15$ .
03
The product of two numbers is 2160 and their HCF is 12. What is their LCM?
3 steps · reveal one at a time
1. Step 1: Read the facts
  We are given the product of two numbers, which is $2160$ . We also know their Highest Common Factor (HCF) is $12$ . Our goal is to find their Least Common Multiple (LCM).
2. Step 2: Recall the key property
  For any two positive integers, there's a fundamental relationship: the product of the numbers is always equal to the product of their HCF and LCM. This is written as $\text{HCF} \times \text{LCM} = \text{Product of numbers}$ .
3. Step 3: The answer
  So the answer is (B) 180. Using the property, we can find the LCM by dividing the product of the numbers by their HCF: $\text{LCM} = 2160 \div 12 = 180$ .
04
Find the greatest number that divides 43, 91 and 183 leaving the same remainder in each case.
5 steps · reveal one at a time
1. Step 1: Understand the problem
  We need to find the greatest number that divides 43, 91, and 183, leaving the same remainder in each division. This is a classic HCF problem variant.
2. Step 2: Find the key property
  If a number divides $a$ , $b$ , and $c$ leaving the same remainder, then it must also divide the differences between these numbers. Let's calculate these differences.
3. Step 3: Calculate differences
  The differences are: $91 - 43 = 48$ , $183 - 91 = 92$ , and $183 - 43 = 140$ . The greatest number we seek must divide all these differences.
4. Step 4: Find the HCF
  We need to find the Highest Common Factor (HCF) of 48, 92, and 140. Factoring them: $48 = 2^4 \times 3$ , $92 = 2^2 \times 23$ , and $140 = 2^2 \times 5 \times 7$ . The common factors are $2^2 = 4$ .
5. Step 5: The answer
  So the answer is (A) 4. The HCF of 48, 92, and 140 is 4, which is the greatest number that divides all three original numbers leaving the same remainder.
05
Three bells ring at intervals of 18, 24 and 32 minutes. If they ring together at 8:00 a.m., when will they next ring together?
5 steps · reveal one at a time
1. Step 1: Understand the Goal
  To find when the bells will ring together again, we need to determine the least common multiple (LCM) of their individual ringing intervals. This LCM will be the time in minutes until they all coincide.
2. Step 2: Calculate the LCM
  Let's find the LCM of $18$ , $24$ , and $32$ . The prime factorizations are $18 = 2 \times 3^2$ , $24 = 2^3 \times 3$ , and $32 = 2^5$ . The LCM is $2^5 \times 3^2 = 32 \times 9 = 288$ minutes.
3. Step 3: Convert Time
  The bells will next ring together after $288$ minutes. To convert this to hours and minutes, we divide by $60$ . $288 \div 60 = 4$ with a remainder of $48$ . So, this is $4$ hours and $48$ minutes.
4. Step 4: Find Next Ring Time
  The bells initially rang together at 8:00 a.m. Adding $4$ hours and $48$ minutes to this time: $8:00 \text{ a.m.} + 4 \text{ hours } 48 \text{ minutes} = 12:48 \text{ p.m.}$
5. Step 5: The answer
  So the answer is (B) 12:48 p.m. This is the time when the bells will next ring together after their initial synchronization at 8:00 a.m.

Free to start

Get exam-ready on HCF and LCM

Generate fresh hcf and lcm practice sets in Test Studio, get step-by-step AI explanations the moment you slip, and close the gaps in Socrative Mode.

Start practising free Browse all topics

Step-by-step AI solutions
Fresh practice sets on demand
Guided Socrative Mode drills

Key formulas

5 solved HCF and LCM questions

Get exam-ready on HCF and LCM

More in Numbers & Simplification