Question 1

Find the LCM of 24, 36, and 48.

Accepted Answer

Step 1: Find prime factorization of each number. 24 = 2³ × 3, 36 = 2² × 3², 48 = 2⁴ × 3. Step 2: LCM = highest power of each prime factor = 2⁴ × 3² = 16 × 9 = 144. Therefore, LCM is 144.

Question 2

Three ropes of lengths 48 m, 60 m, and 72 m are to be cut into equal pieces without any waste. What is the maximum length of each piece?

Accepted Answer

Step 1: The maximum length of each piece is the HCF of 48, 60, and 72. Step 2: Prime factorization: 48 = 2⁴ × 3, 60 = 2² × 3 × 5, 72 = 2³ × 3². Step 3: HCF = 2² × 3 = 4 × 3 = 12. Therefore, the maximum length of each piece is 12 m.

Question 3

Find the HCF of 144 and 180.

Accepted Answer

Step 1: Prime factorization of 144 = 2⁴ × 3². Step 2: Prime factorization of 180 = 2² × 3² × 5. Step 3: HCF = product of lowest powers of common prime factors = 2² × 3² = 4 × 9 = 36. Therefore, HCF is 36.

Question 4

The HCF of two numbers is 18. If the numbers are in the ratio 2:3, find the numbers.

Accepted Answer

Step 1: Let the two numbers be 2x and 3x, where x is a positive integer. Step 2: HCF of 2x and 3x is x (since HCF(2,3) = 1). Step 3: Given HCF = 18, so x = 18. Step 4: The numbers are 2 × 18 = 36 and 3 × 18 = 54. Therefore, the numbers are 36 and 54.

Question 5

The HCF of two numbers is 15. If the numbers are in the ratio 3:5, what are the two numbers?

Accepted Answer

Step 1: If HCF is 15 and numbers are in ratio 3:5, express numbers as 3k and 5k where k is a constant. Step 2: HCF(3k, 5k) = k × HCF(3, 5) = k × 1 = k. Step 3: Since HCF = 15, we have k = 15. Step 4: First number = 3 × 15 = 45, Second number = 5 × 15 = 75. Therefore, the answer is 45 and 75.

Question 6

Find the HCF of 144 and 180 using the Euclidean algorithm.

Accepted Answer

Step 1: 180 = 144 × 1 + 36. Step 2: 144 = 36 × 4 + 0. Step 3: When remainder is 0, the HCF is the last non-zero remainder = 36.

Question 7

The product of two numbers is 2880 and their HCF is 12. Find their LCM.

Accepted Answer

Step 1: Use the formula HCF × LCM = Product of two numbers. Step 2: 12 × LCM = 2880. Step 3: LCM = 2880 ÷ 12 = 240.

Question 8

The HCF of two numbers is 15. If the numbers are 60 and 75, verify the HCF and find their LCM.

Accepted Answer

Step 1: Verify HCF of 60 and 75 using prime factorization. 60 = 2² × 3 × 5, 75 = 3 × 5². Step 2: HCF = 3 × 5 = 15 ✓ (verified). Step 3: LCM = 2² × 3 × 5² = 4 × 3 × 25 = 300.

Question 9

Find the LCM of 24, 36, and 48.

Accepted Answer

Step 1: Prime factorize each number. 24 = 2³ × 3, 36 = 2² × 3², 48 = 2⁴ × 3. Step 2: LCM = highest power of each prime = 2⁴ × 3² = 16 × 9 = 144.

Question 10

Three bells ring at intervals of 8, 12, and 16 minutes respectively. If they all ring together at 9:00 AM, at what time will they ring together again?

Accepted Answer

Step 1: Find LCM of 8, 12, and 16. Step 2: Prime factorisation: 8 = 2³, 12 = 2² × 3, 16 = 2⁴. Step 3: LCM = 2⁴ × 3 = 48 minutes. Step 4: They will ring together after 48 minutes. Step 5: 9:00 AM + 48 minutes = 9:48 AM. Therefore, the answer is 9:48 AM.

Question 11

Three bells ring at intervals of 12, 18, and 24 minutes respectively. If they all ring together at 9:00 AM, at what time will they ring together again?

Accepted Answer

Step 1: Find LCM of 12, 18, and 24. Prime factorizations: 12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3. Step 2: LCM = 2³ × 3² = 8 × 9 = 72 minutes. Step 3: 72 minutes = 1 hour 12 minutes. Step 4: 9:00 AM + 1 hour 12 minutes = 10:12 AM.

Question 12

Two numbers have an HCF of 18. Their LCM is 1080. How many such pairs of numbers are possible?

Accepted Answer

Step 1: If HCF = 18, then both numbers can be written as 18a and 18b where HCF(a,b) = 1. Step 2: LCM(18a, 18b) = 18 × LCM(a,b) = 18 × a × b (since HCF(a,b) = 1). Step 3: 18 × a × b = 1080 → a × b = 60. Step 4: Find pairs (a,b) where HCF(a,b) = 1 and a × b = 60. Step 5: 60 = 2² × 3 × 5. Step 6: Coprime factor pairs of 60: (1,60), (3,20), (4,15), (5,12). Step 7: Verify each: HCF(1,60)=1✓, HCF(3,20)=1✓, HCF(4,15)=1✓, HCF(5,12)=1✓. Step 8: Total = 4 pairs.

Question 13

The HCF of two numbers is 16 and the sum of the two numbers is 144. If the numbers are in the ratio 3:5, find the larger number.

Accepted Answer

Step 1: Since HCF is 16, let the two numbers be 16m and 16n where HCF(m,n) = 1. Step 2: Given ratio is 3:5, so m:n = 3:5. Step 3: Let m = 3k and n = 5k (where k = 1 since HCF(m,n) = 1). Step 4: Numbers are 16(3) = 48 and 16(5) = 80. Step 5: Verify sum: 48 + 80 = 128 ≠ 144. Recalculate: 16m + 16n = 144 → m + n = 9. Step 6: With m:n = 3:5, we have 3x + 5x = 9 → 8x = 9 → x = 9/8. Step 7: Numbers are 16 × (27/8) = 54 and 16 × (45/8) = 90. Step 8: Larger number = 90.

Question 14

Two numbers have an HCF of 15 and an LCM of 180. How many pairs of such numbers exist?

Accepted Answer

Step 1: If HCF = 15, then both numbers are of the form 15m and 15n where HCF(m,n) = 1. Step 2: LCM(15m, 15n) = 15 × LCM(m,n) = 15 × m × n (since HCF(m,n) = 1). Step 3: 15 × m × n = 180 → m × n = 12. Step 4: Find coprime pairs (m,n) whose product is 12: (1,12), (3,4). Step 5: This gives number pairs: (15,180) and (45,60). Step 6: Count = 2 pairs.

Question 15

The HCF of two numbers is 16 and the sum of the two numbers is 144. If one number is 48, what is the other number?

Accepted Answer

Step 1: Since HCF is 16, both numbers are multiples of 16. Let the numbers be 16a and 16b where HCF(a,b) = 1. Step 2: First number = 48 = 16 × 3, so a = 3. Step 3: 16a + 16b = 144 → 16(3) + 16b = 144 → 48 + 16b = 144. Step 4: 16b = 96 → b = 6. Step 5: Other number = 16 × 6 = 96.

Question 16

Two numbers have an HCF of 15 and an LCM of 180. How many pairs of such numbers exist?

Accepted Answer

Step 1: If HCF = 15, both numbers are multiples of 15. Write them as 15a and 15b where HCF(a,b) = 1. Step 2: LCM(15a, 15b) = 15 × LCM(a,b) = 180 → LCM(a,b) = 12. Step 3: Find coprime pairs (a,b) whose LCM is 12. Step 4: 12 = 2² × 3. Coprime pairs: (1,12), (3,4). Step 5: Two pairs exist: (15,180) and (45,60).

Question 17

Four ropes of lengths 24 m, 36 m, 48 m, and 60 m are to be cut into equal pieces without any waste. What is the maximum length of each piece?

Accepted Answer

Step 1: The maximum length of each piece is the HCF of all four lengths. Step 2: Find HCF(24, 36, 48, 60). Step 3: 24 = 2³ × 3, 36 = 2² × 3², 48 = 2⁴ × 3, 60 = 2² × 3 × 5. Step 4: HCF = 2² × 3 = 4 × 3 = 12 m.

Question 18

Three bells ring at intervals of 8, 12, and 18 minutes respectively. If they all ring together at 9:00 AM, at what time will they ring together again?

Accepted Answer

Step 1: Find LCM of 8, 12, and 18. Step 2: 8 = 2³, 12 = 2² × 3, 18 = 2 × 3². Step 3: LCM = 2³ × 3² = 8 × 9 = 72 minutes. Step 4: 9:00 AM + 72 minutes = 10:12 AM.

Question 19

A rectangular hall has dimensions 48 m × 36 m. It needs to be tiled with square tiles of equal size such that no tile is cut. What is the side length of the largest square tile that can be used?

Accepted Answer

Step 1: The side of the largest square tile must divide both dimensions evenly. Step 2: Find HCF(48, 36). Step 3: 48 = 2⁴ × 3, 36 = 2² × 3². Step 4: HCF = 2² × 3 = 4 × 3 = 12 m. Step 5: The largest square tile has side 12 m.

Question 20

The HCF of two numbers is 18. If the numbers are in the ratio 2:3, find the larger number.

Accepted Answer

Step 1: Since HCF is 18, the two numbers can be written as 18a and 18b, where HCF(a,b) = 1. Step 2: Given ratio is 2:3, so a:b = 2:3. Step 3: Let a = 2k and b = 3k where HCF(2k, 3k) = 1, which means k = 1. Step 4: Numbers are 18×2 = 36 and 18×3 = 54. Step 5: Larger number = 54.

RRB NTPC LCM and HCF

LCM and HCF— Rules & Concept

Key Points to Remember

Exam-Specific Tips

LCM and HCF — Practice Questions

60-Second Revision — LCM and HCF