Question 1

The HCF of two numbers is 18. If the numbers are 54 and 90, verify this is correct. What is their LCM?

Accepted Answer

Step 1: Verify HCF(54, 90) = 18. 54 = 2 × 3³, 90 = 2 × 3² × 5. HCF = 2 × 3² = 18 ✓. Step 2: Use HCF × LCM = 54 × 90. Step 3: 18 × LCM = 4860. Step 4: LCM = 4860 ÷ 18 = 270.

Question 2

Three numbers are in the ratio 2:3:4. Their HCF is 5. Find the largest number.

Accepted Answer

Step 1: Let the numbers be 2k, 3k, and 4k where k is a constant. Step 2: HCF(2k, 3k, 4k) = k × HCF(2, 3, 4) = k × 1 = k. Step 3: Given HCF = 5, so k = 5. Step 4: The numbers are 10, 15, and 20. Step 5: Largest number = 20.

Question 3

The HCF of two numbers is 18. If the numbers are 54 and 90, verify which statement is correct.

Accepted Answer

Step 1: Find HCF of 54 and 90 using prime factorization. 54 = 2 × 3³, 90 = 2 × 3² × 5. Step 2: HCF = 2 × 3² = 2 × 9 = 18. Step 3: The given HCF value of 18 is correct. Therefore, the answer is 18 is the correct HCF.

Question 4

What is 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, the answer is 144.

Question 5

Find the LCM of 24, 36, and 48.

Accepted Answer

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

Question 6

Three bells ring at intervals of 18 minutes, 24 minutes, and 30 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 18, 24, and 30. Step 2: Prime factorization: 18 = 2 × 3², 24 = 2³ × 3, 30 = 2 × 3 × 5. Step 3: LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360 minutes. Step 4: Convert to hours: 360 ÷ 60 = 6 hours. Step 5: 9:00 AM + 6 hours = 3:00 PM. Therefore, the answer is 3:00 PM.

Question 7

Two numbers have an HCF of 18 and an LCM of 540. How many pairs of such numbers are possible?

Accepted Answer

Step 1: Let the two numbers be 18m and 18n, where HCF(m,n) = 1 (m and n are coprime). Step 2: LCM of 18m and 18n = 18 × m × n = 540. Step 3: Therefore, m × n = 540 ÷ 18 = 30. Step 4: Find all pairs of coprime factors of 30: 30 = 2 × 3 × 5. Step 5: Coprime factor pairs of 30 are: (1,30), (2,15), (3,10), (5,6). Step 6: All four pairs are coprime. Therefore, there are 4 pairs of numbers possible.

Question 8

The HCF of two numbers is 16. If the numbers are in the ratio 3:5, what is their LCM?

Accepted Answer

Step 1: Let the two numbers be 3x and 5x, where x is a positive integer. Step 2: Since HCF = 16, we have x = 16. Step 3: The two numbers are 3(16) = 48 and 5(16) = 80. Step 4: Use HCF × LCM = Product of numbers: 16 × LCM = 48 × 80 = 3840. Step 5: LCM = 3840 ÷ 16 = 240. Therefore, the answer is 240.

Question 9

Three bells ring at intervals of 18 minutes, 24 minutes, and 36 minutes respectively. If they ring together at 10:00 AM, at what time will they ring together again?

Accepted Answer

Step 1: Find LCM of 18, 24, and 36. Step 2: Prime factorization: 18 = 2 × 3², 24 = 2³ × 3, 36 = 2² × 3². Step 3: LCM = 2³ × 3² = 8 × 9 = 72 minutes. Step 4: 72 minutes = 1 hour 12 minutes. Step 5: 10:00 AM + 1 hour 12 minutes = 11:12 AM.

Question 10

A rectangular hall has dimensions 48 m × 36 m. It is to be paved with square tiles of the largest possible size such that no tile is cut. What is the side length of each tile?

Accepted Answer

Step 1: The largest square tile that fits without cutting is determined by the HCF of the two dimensions. Step 2: Find HCF(48, 36). Step 3: 48 = 2⁴ × 3, 36 = 2² × 3². Step 4: HCF = 2² × 3 = 12. Step 5: Side length of each tile = 12 m.

Question 11

Two numbers have an HCF of 18. Their LCM is 216. How many pairs of numbers satisfy these conditions?

Accepted Answer

Step 1: If HCF = 18, both numbers can be written as 18a and 18b where HCF(a, b) = 1 (coprime). Step 2: LCM(18a, 18b) = 18 × LCM(a, b) = 18 × a × b = 216 (since a and b are coprime). Step 3: a × b = 216 ÷ 18 = 12. Step 4: Find pairs (a, b) where a × b = 12 and HCF(a, b) = 1. Step 5: Coprime factor pairs of 12: (1,12), (3,4). Step 6: Two pairs exist: (18, 216) and (54, 72). Step 7: If order matters, 4 pairs; if unordered, 2 pairs. Standard interpretation: 2 unordered pairs.

Question 12

Three numbers are in the ratio 2:3:4. Their LCM is 144. Find the sum of the three numbers.

Accepted Answer

Step 1: Let the three numbers be 2k, 3k, and 4k. Step 2: Find LCM(2k, 3k, 4k). Step 3: LCM = k × LCM(2, 3, 4) = k × 12. Step 4: Given LCM = 144, so k × 12 = 144, thus k = 12. Step 5: Numbers are 24, 36, and 48. Step 6: Sum = 24 + 36 + 48 = 108.

Question 13

The LCM of two numbers is 180 and their HCF is 6. If the sum of the two numbers is 66, find the difference between them.

Accepted Answer

Step 1: Let the numbers be 6a and 6b where HCF(a,b) = 1. Step 2: LCM(6a, 6b) = 6 × LCM(a,b) = 180, so LCM(a,b) = 30. Step 3: Since HCF(a,b) = 1 and LCM(a,b) = 30, we have a × b = 30. Step 4: From sum: 6a + 6b = 66, so a + b = 11. Step 5: Solve: a and b are roots of t² - 11t + 30 = 0, giving (t-5)(t-6) = 0, so a=5, b=6. Step 6: Numbers are 30 and 36. Difference = 36 - 30 = 6.

Question 14

Four bells ring at intervals of 6, 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(6, 8, 12, 18). Step 2: Prime factorizations: 6=2×3, 8=2³, 12=2²×3, 18=2×3². Step 3: LCM = 2³ × 3² = 8 × 9 = 72 minutes. Step 4: 72 minutes = 1 hour 12 minutes. Step 5: 9:00 AM + 1 hour 12 minutes = 10:12 AM. Therefore, they ring together at 10:12 AM.

Question 15

Two numbers have an HCF of 15 and an LCM of 1260. How many such pairs of numbers exist?

Accepted Answer

Step 1: Let the numbers be 15a and 15b where HCF(a,b) = 1. Step 2: LCM(15a, 15b) = 15 × LCM(a,b) = 1260, so LCM(a,b) = 84. Step 3: Since HCF(a,b) = 1, we have a × b = 84. Step 4: Find coprime factor pairs of 84: 84 = 2² × 3 × 7. Step 5: Coprime pairs (a,b) where a×b=84: (1,84), (3,28), (4,21), (7,12). Step 6: Each pair gives a unique unordered pair of numbers. Therefore, 4 pairs exist.

Question 16

The LCM of two numbers is 12 times their HCF. If the sum of the LCM and HCF is 156, find the HCF.

Accepted Answer

Step 1: Let HCF = h and LCM = L. Given: L = 12h and L + h = 156. Step 2: Substitute L = 12h into the sum equation: 12h + h = 156. Step 3: 13h = 156, so h = 12. Step 4: Therefore, LCM = 12 × 12 = 144. Step 5: Verify: 144 + 12 = 156 ✓. The HCF is 12.

Question 17

Three numbers are in the ratio 2:3:4. The HCF of these three numbers is 8. What is their LCM?

Accepted Answer

Step 1: Let the numbers be 2k, 3k, and 4k where k is a constant. Step 2: HCF(2k, 3k, 4k) = k × HCF(2,3,4) = k × 1 = k. Step 3: Since HCF = 8, we have k = 8. Step 4: Numbers are 16, 24, 32. Step 5: LCM(16, 24, 32) = 96. Therefore, the LCM is 96.

Question 18

Three bells ring at intervals of 18, 24, and 30 minutes respectively. If they all ring together at 9:00 AM, at what time will they ring together again for the second time?

Accepted Answer

Step 1: Find the LCM of 18, 24, and 30 to determine when all three ring together. Step 2: Prime factorise: 18 = 2 × 3², 24 = 2³ × 3, 30 = 2 × 3 × 5. Step 3: LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360 minutes. Step 4: Convert 360 minutes to hours: 360 ÷ 60 = 6 hours. Step 5: They ring together every 6 hours. First time together: 9:00 AM. Second time together: 9:00 AM + 6 hours = 3:00 PM. Therefore, they ring together for the second time at 3:00 PM.

SSC CHSL 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