HCF (Highest Common Factor) Explained with Examples | 20 MCQs for Competitive Exams
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HCF (Highest Common Factor) Explained with Examples | 20 MCQs for Competitive Exams

⭐ Highest Common Factor (HCF) — Detailed Explanation

HCF (Highest Common Factor) is the greatest number that divides two or more numbers without leaving any remainder.

It is also called:

  • GCD (Greatest Common Divisor)
  • GCM (Greatest Common Measure)

HCF is widely used in:

  • Number system
  • Simplifying ratios
  • LCM–HCF word problems
  • Fractions
  • Aptitude and competitive exams

🧩 How to Find HCF

Method 1: Prime Factorization

  1. Write prime factors of each number.
  2. Take only common prime factors.
  3. Multiply them.

Example:
HCF(36, 60)
36 = 2² × 3²
60 = 2² × 3 × 5
Common factors = 2² × 3 = 12

Method 2: Division Method (Euclid’s Method)

Use repeated division:

HCF(a, b) = HCF(b, remainder)

Continue until remainder becomes 0.
The last divisor is the HCF.

Method 3: Listing Common Factors

  • List all factors → choose the highest.
  • Useful for small numbers only.

⭐ 5 Solved Examples of HCF (With Explanation)

Example 1: Find HCF of 24 and 36

24 = 2³ × 3
36 = 2² × 3²
Common = 2² × 3 = 12

HCF = 12

Example 2: Find HCF of 18, 27 and 45

18 = 2 × 3²
27 = 3³
45 = 3² × 5
Common = 3² = 9

HCF = 9

Example 3: Find HCF of 72 and 120 (Euclid Method)

120 ÷ 72 = 48
72 ÷ 48 = 24
48 ÷ 24 = 0

HCF = 24

Example 4: Find HCF of 14 and 35

14 = 2 × 7
35 = 5 × 7
Common = 7

HCF = 7

Example 5: Find HCF of 48, 64 and 80

48 = 2⁴ × 3
64 = 2⁶
80 = 2⁴ × 5
Common = 2⁴ = 16

HCF = 16

MCQ's For Exam

Q.1. What is the HCF of 8 and 12?

  1. 2
  2. 4
  3. 6
  4. 12

2) 4

📝Explanation: Prime factors:
8 = 2³
12 = 2² × 3
Common factor is 2² = 4.
Thus HCF = 4 because this is the largest number that divides both 8 and 12 without leaving any remainder.

Q.2. HCF of 9 and 27 is:

  1. 3
  2. 9
  3. 27
  4. 1

2) 9

📝 Description:

9 = 3²
27 = 3³
Common = 3² = 9.
9 divides both numbers exactly, while no larger number does. Hence the correct HCF is 9.

Q.3. The HCF of 16 and 20 is:

  1. 2
  2. 4
  3. 8
  4. 16

2) 4

📝 Description: Prime factors:
16 = 2⁴
20 = 2² × 5
Common = 2² = 4.
Thus the largest common divisor of both numbers is 4.

Q.4. What is the HCF of 15 and 25?

  1. 5
  2. 10
  3. 15
  4. 25

1) 5

📝 Description:

15 = 3 × 5
25 = 5 × 5
Common = 5.
5 is the greatest number dividing both without remainder. Hence HCF = 5.

Q.5. Find the HCF of 12, 18 and 30.

  1. 3
  2. 6
  3. 9
  4. 12

2) 6

📝 Description:

512 = 2² × 3
18 = 2 × 3²
30 = 2 × 3 × 5
Common = 2 × 3 = 6
Thus HCF = 6, the largest number dividing all three.

Q.6. HCF of 11 and 121 is:

  1. 1
  2. 11
  3. 121
  4. 33

2) 11

📝 Description:

11 = 11
121 = 11²
Common = 11
So the highest factor common to both is 11.

Q.7. HCF of 40 and 56 is:

  1. 4
  2. 8
  3. 16
  4. 24

2) 8

📝 Description:

40 = 2³ × 5
56 = 2³ × 7
Common = 2³ = 8
Thus HCF = 8.

Q.8. HCF of 48 and 180 is:

  1. 6
  2. 12
  3. 24
  4. 36

2) 12

📝 Description:

48 = 2⁴ × 3
180 = 2² × 3² × 5
Common = 2² × 3 = 12
So HCF = 12.

Q.9. HCF of 32, 96 and 160 is:

  1. 16
  2. 32
  3. 64
  4. 8

1) 16

📝 Description: 32 = 2⁵
96 = 2⁵ × 3
160 = 2⁵ × 5
Common factor = 2⁵ = 32
But all three share only 2⁴ = 16.
Thus HCF = 16.

Q.10. HCF of two prime numbers is always:

  1. 1
  2. The smaller prime
  3. The larger prime
  4. Their product

1) 1

📝 Description: Prime numbers have no common factors except 1, unless they are the same prime. Therefore, the HCF of any two different primes is always 1.

Q.11. HCF of 81 and 108 is:

  1. 9
  2. 18
  3. 27
  4. 36

3) 27

📝 Description:

81 = 3⁴
108 = 2² × 3³
Common = 3³ = 27
Thus HCF = 27.

Q.12. Which pair has HCF = 1?

  1. 8 & 12
  2. 10 & 25
  3. 14 & 21
  4. 16 & 27

4) 16 & 27

📝 Description:

16 = 2⁴
27 = 3³
No common prime factors → co-primes.
Hence HCF = 1.

Q.13. HCF of 50 and 75 is:

  1. 5
  2. 10
  3. 15
  4. 25

4) 25

📝 Description: 

50 = 2 × 5²
75 = 3 × 5²
Common = 5² = 25
Thus HCF = 25.

Q.14. Using Euclid’s method, HCF of 96 and 144 is:

  1. 24
  2. 32
  3. 48
  4. 96

3) 48

📝 Description:

144 ÷ 96 = 48
96 ÷ 48 = 0
Last non-zero divisor = 48
Thus HCF = 48.

Q.15. If HCF of two numbers is 7, which pair is possible?

  1. 7 & 49
  2. 14 & 21
  3. 35 & 70
  4. 42 & 63

1) 7 & 49

📝 Description:

HCF(7, 49) = 7
Other pairs have HCF > 7.
Thus the only correct pair is 7 and 49.

Q.16. HCF of 72, 108 and 180 is:

  1. 18
  2. 36
  3. 12
  4. 24

2) 36

📝 Description: 

72 = 2³ × 3²
108 = 2² × 3³
180 = 2² × 3² × 5
Common = 2² × 3² = 4 × 9 = 36?
But 36 doesn’t divide 72 evenly? Yes it does.
Actually dividing:
72 ÷ 36 = 2
108 ÷ 36 = 3
180 ÷ 36 = 5
So HCF = 36

Q.17. HCF of 101 and 103 is:

  1. 1
  2. 101
  3. 103
  4. 2

1) 1

📝 Description: Both are prime numbers. Different primes have no common factors except 1. Hence HCF = 1.

Q.18. HCF of 120 and 150 is:

  1. 10
  2. 20
  3. 30
  4. 60

3) 30

📝 Description:

120 = 2³ × 3 × 5
150 = 2 × 3 × 5²
Common = 2 × 3 × 5 = 30
But 30 must divide 120 and 150 → Yes.
Thus HCF = 30.

Q.19. HCF of 3 consecutive integers is always:

  1. 1
  2. 2
  3. 3
  4. Cannot be determined

1) 1

📝 Description: Any 3 consecutive integers always have different prime factor structures. They never share any common factor except 1. Thus their HCF is always 1.

Q.20. HCF of 84, 140 and 196 is:

  1. 14
  2. 28
  3. 42
  4. 7

2) 28

📝 Description:

84 = 2² × 3 × 7
140 = 2² × 5 × 7
196 = 2² × 7²
Common = 2² × 7 = 28
Thus HCF = 28.

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