HCF (Highest Common Factor) and LCM (Least Common Factor) are the basics of many mathematical operations. These help solve the various complex mathematical problems at wide variety of competitive exams like **RBI Assistant Exam**, **GATE**, **NEET**, and more. To ace the performance at the examination, it is important to develop a grip with these questions. Scroll below to find practice problems on H.C.F. and L.C.M.

## Formulas to Find HCF and LCM✏️

There are different formulas to find HCF and LCM individually. Let’s explore them one by one:

**Division Method**

You need to follow three steps to find the HCF by the division method. The steps are below.

- Step 1- Divide the larger numbers by smaller numbers.
- Step 2- You will get a remainder from Step 1. Assume this remainder as the divisor, and then the last divisor will be taken as the dividend.
- Step 3- You need to repeat the Step 2 until 0 is obtained in the remainder. The last divisor which will be obtained will be the HCF.

**Prime Factorisation Method**

IN this method, every number must be treated as the product of prime factors. The product of the least common prime factor will give the HCF of those numbers.

*HCF and LCM Formula = Product of two numbers = LCM of two numbers x HCF of two numbers.*

**Also Read:** **Average Value and Calculation**

## Questions and Answers on H.C.F and L.C.M⬇️

**Q1: Two numbers are in the ratio of 5:11. If their HCF is 7, find the numbers.**

**Solution: **Let the numbers be 5m and 11m. Since 5:11 is already the reduced ratio, ‘m’ has to be the HCF. So, the numbers are 5 x 7 = 35 and 11 x 7 = 77.

**Q2: Find the length of the plank which can be used to measure exactly the lengths 4 m 50 cm, 9 m 90 cm, and 16 m 20 cm in the least time.**

**Solution: **Let us first convert each length to cm. So, the lengths are 450 cm, 990 cm, and 1620 cm. Now, we need to find the length of the largest plank that can be used to measure these lengths as the largest plank will take the least time. For this, we need to take the HCF of 450, 990, and 1620. 450 = 2 x 3 x 3 x 5 x 5 = 2 x 3^{2} x 5^{2} 990 = 2 x 3 x 3 x 5 x 11 = 2 x 3^{2} x 5 x 11 1620 = 2 x 2 x 3 x 3 x 3 x 3 x 5 = 2^{2} x 3^{4} x 5 Therefore, HCF (450, 990, 1620) = 2 x 3 x 3 x 5 = 90 Thus, we need a plank of length 90 cm to measure the given lengths in the least time.

**Q3: Find the greatest number which on dividing 70 and 50 leaves remainders 1 and 4 respectively.**

**Solution: **The required number leaves remainders 1 and 4 on dividing 70 and 50 respectively. This means that the number exactly divides 69 and 46. So, we need to find the HCF of 69 (3 x 23) and 46 (2 x 23). HCF (69, 46) = 23 Thus, 23 is the required number.

**Q4: Find the largest number which divides 64, 136, and 238 to leave the same remainder in each case.**

**Solution: **To find the required number, we need to find the HCF of (136-64), (238-136), and (238-64), i.e., HCF (72, 102, 174). 72 = 2^{3} x 3^{2} 102 = 2 x 3 x 17 174 = 2 x 3 x 29 Therefore, HCF (72, 102, 174) = 2 x 3 = 6 hence, 6 is the required number.

**Q5: Find the least number which when divided by 5,7,9 and 12, leaves the same remainder 3 in each case.**

**Solution: **In these types of questions, we need to find the LCM of the divisors and add the common remainder (3) to it. So, LCM (5, 7, 9, 12) = 1260 Therefore, required number = 1260 + 3 = 1263

**6. What is the LCM of 6 and 8?****Solution: **To find: LCM (6, 8).

The multiples of 6 are 6, 12, 18, **24**, 30, 36, 42, 48, …

The multiples of 8 are 8, 16, **24**, 32, 40, 48, ….

Thus, the smallest common multiple of 6 and 8 is 24.

Therefore, the LCM of 6 and 8 is 24.

**7. Determine the LCM of 4 and 12 using the prime factorisation method.****Solution: **To find the LCM of 4 and 12 using the prime factorisation method, follow the below steps.**Step 1: Find the prime factorization of given numbers:**

The prime factorisation of 4 is 2 × 2

The prime factorisation of 12 is 2 × 2 × 3.**Step 2: The LCM of given numbers is found by multiplying the product of all factors. (Note: The common factor is included only once)**

Hence, the product of prime factors = 2 × 2 × 3 = 12.

Therefore, the LCM of 4 and 12 is 12.

**8. What is the LCM of 54 and 60?****Solution: **The prime factorisation of 54 is 2 × 3 × 3 × 3.

The prime factorisation of 60 is 2 × 2 × 3 × 5.

Thus, the product of prime factors = 2 × 2 × 3 × 3 × 3 × 5 = 540

Hence, the LCM of 54 and 60 is 540.

**9. Prove that LCM(12, 7) × HCF(12, 7) = Product(12, 7).****Solution: **Given: LCM(12, 7) × HCF(12, 7) = Product(12, 7) …(1)**Finding LCM (12, 7):**

The multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, **84**, 91, 98, ..

The multiples of 12 are 12, 24, 36, 48, 60, 72, **84**, 96, 120, …

Hence, the LCM of 12 and 7 is 84.**Finding HCF (12, 7):**

The factors of 7 are **1** and 7.

The factors of 12 are **1**, 2, 3, 4, 6 and 12.

Hence, HCF of 12 and 7 is 1.**Finding Product of 12 and 7:**

The product of 12 and 7 = 12 × 7 = 84.

Now, substitute the obtained values in (1), we get

84 × 1 = 84

84 = 84

Hence, LHS = RHS

Therefore, LCM(12, 7) × HCF(12, 7) = Product(12, 7) is proved.

**10. Find the LCM of the fractions 2/5, 4/7 and 6/11.****Solution: **Given fractions: 2/5, 4/7 and 6/11

As we know, the formula to find the LCM of fractions is:

LCM of fractions = LCM of Numerators/HCF of Denominators .. (1)

Thus, the LCM of Numerators = LCM (2, 4, 6) = 12.

HCF of denominators = HCF (5, 7, 11) = 1

Now, substitute the values in (1), we get

LCM of fractions = 12/1 = 12

Hence, the LCM of the fractions 2/5, 4/7 and 6/11 is 12.

**Q.11.**What is the L.C.M. of 25, 30, 35 and 40?1. 3800

2. 4200

3. 4400

4. 3200

5. None of these**Solution: Option** 2

25 = 5 × 5, 30 = 5 × 3 × 2, 35 = 5 × 7, 40 = 2 × 2 × 2 × 5

Required LCM = 2 × 2 × 2 × 5 × 5 × 3 × 7 = 4200

**Q.12.**What is the greatest number which divides 639, 1065 and 1491 exactly?1. 193

2. 183

3. 223

4. 213

5. 233**Solution: Option** 4

H.C.F. of 639 and 1065 is 213. H.C.F. of 213 and 1491 is 213.

**Q.13.**What is the H.C.F. of 4/9, 10/21 and 20/63?1. 4/189

2. 6/63

3. 2/63

4. 20/21

5. None of these**Solution: Option** 3

H.C.F of 4/9, 10/21 and 20/63 = H.C.F of 4,10 and 20 / L.C.M of 9,21 and 63

= H.C.F of 4, 10 and 20 = 2 & L.C.M. of 9, 21 and 63 = 63. Required H.C.F. = 2/63

**Q.14.**The H.C.F. of two numbers is 12 and their difference is 12. Which of the following can be the numbers?1. 66, 77

2. 70, 84

3. 94, 108

4. 84, 96

5. 66, 106**Solution: Option 4**

The difference of requisite numbers must be 12 and each should be divisible by 12. Checking the options given, only the fourth option satisfies.

**Q.15.**The HCF of two numbers is 29 & their sum is 174. The possible numbers are1. 1,174

2. 74,100

3. 29, 154

4. 29, 145

5. None of these**Solution: Option 4**

Let the numbers be 29a and 29b. Then, 29a + 29b= 174 or 29(a + b) = 174 or, a + b = 174/29 = 6. Values of co-primes (with sum 6) is(1, 5).

So, the possible pairs of numbers is (29 x 1, 29 x 5) i.e. 29 and 145, which will become the answer.

**Q16: Find the largest four-digit number exactly divisible by 15,21 and 28.**

**Solution: **The largest four-digit number is 9999. Now, LCM (15, 21, 28) = 420 On dividing 9999 by 420, we get 339 as the remainder. Therefore, the required number is 9999-339 = 9660

**Q17: The policemen at three different places on the ground blow a whistle after every 42 sec, 60 sec, and 78 sec respectively. If they all blow the whistle simultaneously at 9:30:00 hours, then at what time do they whistle again together.**

**Solution: **They all will whistle again at the same time after an interval that is equal to the LCM of their individual whistle-blowing cycles. So, LCM (42, 60, 78) = 2 x 3 x 7 x 10 x 13 = 5460 Therefore, they will blow the whistle again simultaneously after 5460 sec, i.e., after 1 hour 31 minutes, i.e., at 11:01:00 hours.

**Q18: Find the least number which when divided by 6,7,8 leaves a remainder of 3, but when divided by 9 leaves no remainder.**

**Solution: **LCM (6, 7, 8) = 168 So, the number is of the form 168m + 3. Now, 168m + 3 should be divisible by 9. We know that a number is divisible by 9 if the sum of its digits is a multiple of 9. For m = 1, the number is 168 + 3 = 171, the sum of whose digits is 9. Therefore, the required number is 171.

**Q19: Two numbers are in the ratio 2:3. If the product of their LCM and HCF is 294, find the numbers.**

**Solution: **Let the common ratio be ‘m’. So, the numbers are 2m and 3m. Now, we know that the Product of numbers is = Product of LCM and HCF. => 2m x 3m = 294 => m^{2} = 49 => m = 7 Therefore, the numbers are 14 and 21.

**Q20: A rectangular field of dimension 180m x 105m is to be paved by identical square tiles. Find the size of each tile and the number of tiles required.**

**Solution: **We need to find the size of a square tile such that a number of tiles cover the field exactly, leaving no area unpaved. For this, we find the HCF of the length and breadth of the field. HCF (180, 105) = 15 Therefore, size of each tile = 15m x 15m Also, number of tiles = area of field / area of each tile => Number of tiles = (180 x 105) / (15 x 15) => Number of tiles = 84 Hence, we need 84 tiles, each of size 15m x 15m.

## Practice Set on on H.C.F and L.C.M⬇️

**1. What is the H.C.F of 18 and 24?**

- A) 4
- B) 6
- C) 8
- D) 12

**2. The L.C.M of 15 and 20 is:**

- A) 40
- B) 50
- C) 60
- D) 75

**3. Find the H.C.F of 36 and 48:**

- A) 6
- B) 8
- C) 12
- D) 18

**4. If the H.C.F of two numbers is 5 and their L.C.M is 60, what are the numbers?**

- A) 5, 60
- B) 10, 30
- C) 15, 20
- D) 20, 30

**5. The L.C.M of two co-prime numbers is always:**

- A) A prime number
- B) A composite number
- C) An even number
- D) Odd number

**6. Find the H.C.F of 56 and 72:**

- A) 8
- B) 12
- C) 16
- D) 24

**7. If the L.C.M of two numbers is 120 and their H.C.F is 5, find the product of the numbers:**

- A) 100
- B) 120
- C) 150
- D) 200

**8. What is the smallest number which leaves a remainder of 4 when divided by 6, 9, and 10?**

- A) 64
- B) 94
- C) 124
- D) 154

**9. The H.C.F of 72, 120, and 144 is:**

- A) 12
- B) 18
- C) 24
- D) 36

**10. If the L.C.M of two numbers is 45 and their H.C.F is 3, find the numbers:**

- A) 15, 30
- B) 12, 36
- C) 9, 45
- D) 6, 54

**11. The L.C.M of two numbers is 36, and one of the numbers is 12. Find the other number:**

- A) 2
- B) 3
- C) 4
- D) 6

**12. Find the H.C.F of 36, 54, and 72:**

- A) 12
- B) 18
- C) 24
- D) 36

**13. If the H.C.F of two numbers is 7 and their L.C.M is 105, what are the numbers?**

- A) 7, 105
- B) 14, 15
- C) 21, 35
- D) 28, 30

**14. The product of two numbers is 1200, and their H.C.F is 5. What is their L.C.M?**

- A) 200
- B) 240
- C) 300
- D) 360

**15. The L.C.M of two numbers is 80, and their H.C.F is 8. If one number is 16, find the other:**

- A) 24
- B) 32
- C) 40
- D) 48

**16. What is the H.C.F of 42 and 63?**

- A) 7
- B) 14
- C) 21
- D) 28

**17. The sum of two numbers is 80, and their H.C.F is 10. What is their L.C.M?**

- A) 40
- B) 80
- C) 120
- D) 160

**18. If the L.C.M of two numbers is 72 and their H.C.F is 9, find the numbers:**

- A) 18, 36
- B) 24, 48
- C) 27, 45
- D) 36, 72

**19. The H.C.F of three numbers is 3, and their L.C.M is 72. If two of the numbers are 9 and 12, find the third number:**

- A) 3
- B) 4
- C) 6
- D) 8

**20. If the H.C.F of two numbers is 9, and one of the numbers is 45, find the other number:**

- A) 5
- B) 9
- C) 15
- D) 25

**21. The L.C.M of three numbers is 120. If two of the numbers are 15 and 20, find the third number:**

- A) 4
- B) 6
- C) 8
- D) 10

**22. Find the H.C.F of 84, 126, and 168:**

- A) 12
- B) 14
- C) 18
- D) 21

**23. If the L.C.M of two numbers is 56 and their H.C.F is 8, find the numbers:**

- A) 16, 24
- B) 20, 28
- C) 24, 32
- D) 28, 36

**24. What is the H.C.F of 64 and 96?**

- A) 16
- B) 24
- C) 32
- D) 48

**25. The L.C.M of two numbers is 120, and their H.C.F is 6. If one number is 24, find the other:**

- A) 36
- B) 48
- C) 60
- D) 72

**26. Find the H.C.F of 56, 70, and 84:**

- A) 14
- B) 18
- C) 21
- D) 28

**27. If the H.C.F of two numbers is 4, and their L.C.M is 120, what are the numbers?**

- A) 8, 30
- B) 12, 40
- C) 16, 60
- D) 24, 80

**28. The sum of two numbers is 45, and their H.C.F is 9. What is their L.C.M?**

- A) 45
- B) 81
- C) 90
- D) 135

**29. If the L.C.M of two numbers is 72, and one of the numbers is 9, find the other number:**

- A) 8
- B) 12
- C) 18
- D) 24

**30. The H.C.F of 56, 98, and 126 is:**

- A) 14
- B) 18
- C) 21
- D) 28

**Answer Key for Practice Set**

1.A) | 2.B) | 3.C) | 4.A) | 5.B) |

6.B) | 7.A) | 8.A) | 9.C) | 10.A) |

11.B) | 12.D) | 13.A) | 14.D) | 15.C) |

16.C) | 17.B) | 18.A) | 19.A) | 20.C) |

21.D) | 22.A) | 23.B) | 24.D) | 25.A) |

26.D) | 27.C) | 28.A) | 29.D) | 30.B) |

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