LCM and HCF if mathematics of csat and asc.pdf
- 1. LCM AND HCF A F E E F A S L A H . T . P 1
- 2. PYQ What is the greatest length x such that 31 2 m and 8 3 4 m are integral multiples of x? (2020) a) 11 2 b) 11 3 c) 11 4 d) 13 4 Afeef Aslah .T.P 2
- 3. INTRODUCTION TO LCM AND HCF LCM (Least Common Multiple) • It is the smallest number that is a multiple of two or more numbers. HCF (Highest Common Factor) • Also known as GCD (Greatest Common Divisor) • It is the largest number that divides two or more numbers without leaving a remainder. Afeef Aslah .T.P 3
- 4. METHODS TO FIND LCM 1. Prime Factorization Find the prime factors of each number and take the highest power of each prime number. Example: LCM of 12 and 15: 12 = 22 × 3, 15 = 3 × 5. LCM = 22 × 3 × 5 = 60. 2. Division Method Divide the numbers by their common prime factors until the quotient is 1 or there are no common prime factors. Example: LCM of 12 and 15. 3 12, 15 4, 5 LCM = 3 × 4 × 5 = 60 Afeef Aslah .T.P 4
- 5. PRACTICE QUESTION Find LCM of 18, 36, and 108 Afeef Aslah .T.P 5
- 6. LCM Find LCM of 14, 18 and 24. a) 120 b) 126 c) 252 d) 504 Afeef Aslah .T.P 6
- 7. LCM Answer: (d) Afeef Aslah .T.P 7
- 8. METHODS TO FIND HCF 1. Prime Factorization Find the prime factors of each number and take the lowest power of each common prime number. Example: HCF of 12 and 15: 12 = 22 × 3, 15 = 3 × 5. HCF = 3. 2. Repetitive Division Subtract the smaller number from the larger one repeatedly until you reach 0. The last non- zero remainder is the HCF. Example: HCF of 12 and 15: 1 12 15 12 4 3 12 12 0 Last non-zero remainder is 3. HCF = 3 Afeef Aslah .T.P 8
- 9. PRACTICE QUESTION Find HCF of 18, 36, and 108 Afeef Aslah .T.P 9
- 10. Find HCF of 34 and 85 a) 13 b) 17 c) 19 d) 21 Afeef Aslah .T.P 10 PRACTICE QUESTION
- 11. Answer: (b) Afeef Aslah .T.P 11 PRACTICE QUESTION
- 12. Find HCF of 48, 72 and 108. a) 8 b) 12 c) 16 d) 24 Afeef Aslah .T.P 12 PRACTICE QUESTION
- 13. Answer: (b) Afeef Aslah .T.P 13 PRACTICE QUESTION
- 14. RELATIONSHIP BETWEEN LCM AND HCF • For any two numbers, the product of their LCM and HCF is equal to the product of the numbers. • Example: For 12 and 15, HCF = 3, LCM = 60, HCF × LCM = 3 × 60 = 180. 12 × 15 = 180. Afeef Aslah .T.P 14
- 15. • If one number is a multiple of the other, their: Least Common Multiple (LCM) is the larger number, and Highest Common Factor (HCF) is the smaller number. • For example, if the numbers are 4 and 12: LCM: The LCM of 4 and 12 is 12 HCF: The HCF of 4 and 12 is 4. RELATIONSHIP BETWEEN LCM AND HCF Afeef Aslah .T.P 15
- 16. LCM AND HCF OF FRACTIONS LCM of Fractions LCM = LCM of numerators HCF of denominators Example: LCM of 5/8 and 15/4 HCF of Fractions HCF = HCF of numerators LCM of denominators Example: HCF of 5/8 and 15/4 Afeef Aslah .T.P 16
- 17. PYQ What is the least four digit number when divided by 3, 4, 5 and 6 and leaves a remainder 2 in each case (UPSC 2020) a) 1012 b) 1022 c) 1122 d) 1222 Afeef Aslah .T.P 17
- 18. PYQ Answer: (b) Afeef Aslah .T.P 18
- 19. PRACTICE QUESTION Find the largest three digit number which when divided by any of the numbers 6, 12, 15 and 20 leaves a remainder 5 in each case. a) 935 b) 955 c) 965 d) 995 Afeef Aslah .T.P 19
- 20. Answer: (c) PRACTICE QUESTION Afeef Aslah .T.P 20
- 21. In a school every student is assigned a unique identification number. A student is a football player if and only if the identification number is divisible by 4, whereas a student is a cricketer if and only if the identification number is divisible by 6. If every number from 1 to 100 is assigned to a student, then how many of them play both cricket and football. (UPSC 2019) a) 4 b) 8 c) 10 d) 12 PYQ Afeef Aslah .T.P 21
- 22. The numbers should be divisible by both 4 and 6. i.e., COMMON MULTIPLEs of 4 and 6. LCM (4 and 6) = 12. So, multiples of 12 will play both cricket and football. 100/12 = 8.33.. i.e., between 1 to 100, there are 8 multiples of 12. Answer: (b) PYQ Afeef Aslah .T.P 22
- 23. PRACTICE QUESTION Anjali, Athira and Amritha has 300, 400 and 320 chocolates with them respectively. Each of them decided to divide it among a group of students equally. When Anjali, Athira and Amritha divided their chocolates among the students, they were left with 20, 15 and 5 chocolates respectively. Then what was the number of students in the group? a) 35 b) 37 c) 39 d) 40 Afeef Aslah .T.P 23
- 24. Therefore the number of students in the group will be factor of 280, 385 and 315. Hence, we need to find out the HCF here. HCF (280, 315, 385) = 35 Answer: (a) PRACTICE QUESTION Person Chocolates with them Left with Chocolates distributed Anjali 300 20 300-20 = 280 Athira 400 15 400-15 = 385 Amritha 320 5 320-5 = 315 Afeef Aslah .T.P 24
- 25. PRACTICE QUESTION Mathew, Alan and Bobby start at same time, same point and in same direction to run around a circular ground. Mathew complete the round in 150 seconds, Alan in 300 seconds and Bobby in 250 seconds. Find after what time they will meet again? a) 15 min b) 20 min c) 25 min d) 30 min Afeef Aslah .T.P 25
- 26. PRACTICE QUESTION When will they meet? When they complete a round in the SAME TIME. SAME TIME means Common Multiple. Times, 250 s, 300 s, 150 s. Their LCM will be the NEXT COMMON TIME. LCM of 250, 300 and 150 Afeef Aslah .T.P 26
- 27. PRACTICE QUESTION Answer: (c) Afeef Aslah .T.P 27
- 28. PRACTICE QUESTION A bell rings every 6 minutes. A second bell rings every 8 minutes. A third bell rings every 9 minutes. If all the three bells ring at same time at 8 o’ clock in the morning, after how much minutes will they all ring together? a) 48 b) 72 c) 96 d) 144 Afeef Aslah .T.P 28
- 29. PRACTICE QUESTION Answer: (b) Afeef Aslah .T.P 29
- 30. There are three traffic signals. Each signal changes colour from green to red and then from red to green. The first signal takes 25 seconds, the second signal takes 39 seconds and the third signal takes 60 seconds to change the colour from green to red. The durations for green and red are the same. At 2:00 pm, they together turn green. At what time will they change to green next, simultaneously? (UPSC 2023) a) 4:00 pm b) 4:10 pm c) 4:20 pm d) 4:30 pm PYQ Afeef Aslah .T.P 30
- 31. 1st traffic signal – green to red + red to green = 25+25 = 50 2nd traffic signal – green to red + red to green = 39+39 = 78 3rd traffic signal – green to red + red to green = 60+60 = 120 LCM of 50, 78 and 120 = 7800 seconds = 7800/60 minutes = 130 minutes 2:00 + 130 minutes = 4:10 Answer: (b) PYQ Afeef Aslah .T.P 31
- 32. PRACTICE QUESTION A milk man has three different kinds of milk 32 litres, 36 litres and 40 litres. Find the minimum number of equal size containers required to store all the milk without mixing. a) 26 b) 27 c) 28 d) 29 Afeef Aslah .T.P 32
- 33. PRACTICE QUESTION Use HCF Afeef Aslah .T.P 33
- 34. PRACTICE QUESTION HCF of 32, 36, 40 = 4,. That means the size of container should be 4L. Total number of container = (32+36+40)/4 = 27 Answer: (b) Afeef Aslah .T.P 34
- 35. PRACTICE QUESTION A person has to completely put each of the three liquids 403 litres of petrol, 465 litres of diesel and 496 litres of engine oil in bottles of equal size without mixing of any of the above three types of liquids such that each bottle is completely filled. What is the least possible number of bottles required ? a) 34 b) 38 c) 44 d) 46 Afeef Aslah .T.P 35
- 36. Answer: (c) PRACTICE QUESTION Afeef Aslah .T.P 36
- 37. PYQ What is the greatest length x such that 31 2 m and 83 4 m are integral multiples of x? (UPSC 2020) a) 11 2 b) 11 3 c) 11 4 d) 13 4 Afeef Aslah .T.P 37
- 38. PYQ Answer: (d) Afeef Aslah .T.P 38
- 39. PYQ A person X wants to distribute some pens among six children A, B, C, D, E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F. What is the minimum number of pens X should buy so that the number of pens each gets is an even number (UPSC 2022) a) 147 b) 150 c) 294 d) 300 Afeef Aslah .T.P 39
- 40. PYQ • A = 2b A = 3c A = 4d A = 5e A = 6f • Now, we want to find the minimum number of pens X should buy so that each child gets an even number of pens. • To do this, we can find the least common multiple (LCM) of 2, 3, 4, 5, and 6. Afeef Aslah .T.P 40
- 41. PYQ LCM(2, 3, 4, 5, 6) = 60 So, A must receive at least 60 pens to satisfy the conditions. A = 60 B = A/2 = 60/2 = 30 C = A/3 = 60/3 = 20 D = A/4 = 60/4 = 15 E = A/5 = 60/5 = 12 F = A/6 = 60/6 = 10 Multiplying by 2 A = 120, B = 60, C = 40, D = 30, E = 24, F = 20 Total pens = 120 + 60 + 40 + 30 + 24 + 20 = 294 Answer: (c) Afeef Aslah .T.P 41