Solved Examples

Example 1 (Prime factorization method): Find HCF and LCM of 24 and 36. - 24 = 2³ × 3¹ - 36 = 2² × 3² - HCF = 2^min(3,2) × 3^min(1,2) = 2² × 3¹ = 12 - LCM = 2^max(3,2) × 3^max(1,2) = 2³ × 3² = 72 - Check: HCF × LCM = 12 × 72 = 864 = 24 × 36 ✓

Example 2 (Division method): Find the HCF of 4,296 and 5,220 using Euclid’s algorithm. - 5220 = 4296 × 1 + 924 - 4296 = 924 × 4 + 600 - 924 = 600 × 1 + 324 - 600 = 324 × 1 + 276 - 324 = 276 × 1 + 48 - 276 = 48 × 5 + 36 - 48 = 36 × 1 + 12 - 36 = 12 × 3 + 0 → last non-zero divisor is 12 Answer: HCF = 12

Example 3: The HCF and LCM of two numbers are 12 and 336 respectively. If one number is 48, find the other. - Product of numbers = HCF × LCM = 12 × 336 = 4032 - Other number = 4032 ÷ 48 = 84

Example 4 (Fractions): Find the LCM and HCF of 4/9, 8/21, and 10/63. - Numerators: 4, 8, 10 → LCM = 40; HCF = 2 - Denominators: 9, 21, 63 → HCF = 3; LCM = 63 - LCM of fractions = LCM(numerators)/HCF(denominators) = 40/3 - HCF of fractions = HCF(numerators)/LCM(denominators) = 2/63

Example 5 (Classic word problem): Three bells ring at intervals of 12, 15, and 18 minutes respectively. If they ring together at 9:00 AM, at what time will they next ring together? - This is an LCM problem — the bells will ring together again after a time interval equal to LCM(12, 15, 18). - 12 = 2²×3, 15 = 3×5, 18 = 2×3². LCM = 2²×3²×5 = 180 minutes = 3 hours. - Next time together = 9:00 AM + 3 hours = 12:00 PM (noon)

Example 6 (Classic word problem): Find the greatest length of tape that can exactly measure lengths of 7m, 3m 85cm, and 12m 95cm. - This is an HCF problem — convert everything to the same unit (cm): 700 cm, 385 cm, 1295 cm. - 700 = 2²×5²×7; 385 = 5×7×11; 1295 = 5×7×37 - HCF = 5×7 = 35 cm Answer: 35 cm (Rule of thumb: “greatest measure that exactly divides several given quantities” always signals HCF; “smallest quantity exactly divisible by / smallest time when events coincide” always signals LCM.)

Example 7 (Three-number HCF word problem): Three pieces of timber measuring 42 m, 49 m, and 63 m have to be cut into planks of the same length, with no wastage. Find the greatest possible length of each plank. - This is an HCF problem: HCF(42, 49, 63). - 42 = 2 × 3 × 7; 49 = 7²; 63 = 3² × 7 - The only prime common to all three is 7 (2 is missing from 49 and 63; 3 is missing from 49). HCF = 7^min(1,2,1) = 7¹ = 7 Answer: 7 m

Example 8 (Three-number LCM): Find the LCM of 8, 12, and 20. - 8 = 2³; 12 = 2² × 3; 20 = 2² × 5 - LCM = 2^max(3,2,2) × 3^max(0,1,0) × 5^max(0,0,1) = 2³ × 3 × 5 = 8 × 3 × 5 = 120 Answer: 120 (Check: 120 ÷ 8 = 15, 120 ÷ 12 = 10, 120 ÷ 20 = 6 — all exact, and note the HCF×LCM=product identity is NOT used here since it does not extend to three numbers.)