Question 1

Find the HCF of 48 and 64.

Accepted Answer

Step 1: Find prime factorization of 48 = 2⁴ × 3. Step 2: Find prime factorization of 64 = 2⁶. Step 3: HCF is the product of lowest powers of common prime factors = 2⁴ = 16. Therefore, the answer is 16.

Question 2

What is the LCM of 12 and 18?

Accepted Answer

Step 1: Prime factorization of 12 = 2² × 3. Step 2: Prime factorization of 18 = 2 × 3². Step 3: LCM is the product of highest powers of all prime factors = 2² × 3² = 4 × 9 = 36. Therefore, the answer is 36.

Question 3

What is the smallest number that is divisible by both 15 and 25?

Accepted Answer

Step 1: The smallest number divisible by both 15 and 25 is their LCM. Step 2: Prime factorization of 15 = 3 × 5. Step 3: Prime factorization of 25 = 5². Step 4: LCM = 3 × 5² = 3 × 25 = 75. Therefore, the answer is 75.

Question 4

Find the HCF of 36, 60, and 84.

Accepted Answer

Step 1: Prime factorization of 36 = 2² × 3². Step 2: Prime factorization of 60 = 2² × 3 × 5. Step 3: Prime factorization of 84 = 2² × 3 × 7. Step 4: HCF is the product of lowest powers of common prime factors = 2² × 3 = 4 × 3 = 12. Therefore, the answer is 12.

Question 5

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 be a common divisor of both dimensions. Step 2: Find HCF(48, 36). Step 3: 48 = 2⁴ × 3, 36 = 2² × 3². Step 4: HCF = 2² × 3 = 12 m. Step 5: The largest square tile has side 12 m.

Question 6

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 can be written as 15a and 15b where HCF(a,b) = 1 (coprime). Step 2: LCM(15a, 15b) = 15 × LCM(a,b) = 15 × a × b (since a and b are coprime). Step 3: 15 × a × b = 180, so a × b = 12. Step 4: Find coprime pairs (a,b) whose product is 12: (1,12), (3,4). Step 5: This gives number pairs (15,180) and (45,60). Step 6: Total pairs = 2.

Question 7

Two numbers are in the ratio 3:5. Their LCM is 180. Find the sum of the two numbers.

Accepted Answer

Step 1: Let the numbers be 3k and 5k. Step 2: Since HCF(3k, 5k) = k (as HCF(3,5)=1), we have LCM(3k, 5k) = 3 × 5 × k = 15k. Step 3: Given LCM = 180, so 15k = 180 → k = 12. Step 4: Numbers are 3(12) = 36 and 5(12) = 60. Step 5: Sum = 36 + 60 = 96.

Question 8

The product of two numbers is 2880 and their HCF is 12. If one number is 48, what is the LCM of these two numbers?

Accepted Answer

Step 1: Use HCF(a,b) × LCM(a,b) = a × b. Step 2: 12 × LCM = 2880. Step 3: LCM = 2880 ÷ 12 = 240. Step 4: Verify: If a=48, then b = 2880÷48 = 60. HCF(48,60) = 12 ✓ and LCM(48,60) = 240 ✓.

Question 9

Three bells ring at intervals of 24, 36, and 48 minutes respectively. They all ring together at 9:00 AM. At what time will they next ring together?

Accepted Answer

Step 1: Find LCM(24, 36, 48). Step 2: 24 = 2³ × 3, 36 = 2² × 3², 48 = 2⁴ × 3. Step 3: LCM = 2⁴ × 3² = 16 × 9 = 144 minutes. Step 4: 144 minutes = 2 hours 24 minutes. Step 5: 9:00 AM + 2 hours 24 minutes = 11:24 AM.

Question 10

The HCF of two numbers is 18. Their LCM is 1080. How many pairs of numbers satisfy this condition?

Accepted Answer

Step 1: If HCF = 18, both numbers can be written as 18m and 18n where HCF(m,n) = 1. Step 2: LCM = 18 × m × n = 1080, so m × n = 60. Step 3: Find coprime pairs (m,n) whose product is 60. Step 4: Factorize 60 = 2² × 3 × 5. Coprime pairs: (1,60), (3,20), (4,15), (5,12). Step 5: Count = 4 pairs.

SSC MTS 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