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Math Topic - HCF & LCM

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Helpful Notes for students who are preparing for competitive exams.

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    PBI PBI INSTITUTE 1. HCF & LCM HCF (Highest Common factor): HCF of two or more numbers is the largest number by which each given number is divisible without leaving any remainder. HCF of Fractions: If alb, c/d, e/f be the proper fraction, their HCF is equal to (HCF of Numerators) / LCM of Denominators LCM (Least Common Multiple): LCM of two or more given numbers is the least number which is exactly divisible by each of them. Eg. 6 is the LCM of 2 & 3. LCM of Fractions: If alb, c/d, e/f be the proper fraction, their LCM is equal to (LCM of Numerators) / HCF of Denominators HCF x LCM = I st Number x 2nd Number The smallest number which when diminished by 7, is divisible by 12, 16, 18, 21 and 28 is A. c. Solution 1008 1022 B. D. 1015 1032 Required numbers of + 7 = 1015. 2. The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275, then the other is A. c. Solution 279 308 B. D. 283 318 Other number 3. = / 275) = 308. The product of two numbers is 4107. If the H.C.F of those numbers is 37, then the
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    PBI greater number is PBI INSTITUTE A. c. Solution 101 109 B. D. Let the numbers be 37a and 37 b. Then, 37a = 4107 ab = 3. Now, co-primes with product 3 are (1, 3). So, the required numbers are (37 x 1, 37 x 3) i.e , (1 , 107 111 111) 4. The greatest possible length which can be used to measure exactly the length 7m, 3m 85cm, 12 m 95 cm is A. C. Solution 15 cm 35 cm B. D. 25 cm 42 cm Required length = H.C.F of 700 cm, 385 cm and 1295 cm 35 cm. 5. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. A. C. Solution 4 9 7 13 Required number - H.C.F of (91 - 43), (183 - 91) and (183 = H.C.F of 48, 92 and 140 D. - 43) 6. The G.C.D of 1.08 , 0.36 and 0.9 is
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    PBI PBI INSTITUTE A. c. Solution 0.03 0.18 B. D. 0.9 0.108 Given numbers are 1.08 , 0.36 and 0.90. H.C.F of 108 and 90 is 18. H.C.F of given numbers = 0.18. 7. Three numbers are in the ratio 1 A. 4, 8, 12 c. 10, 20, 30 Solution Let the required numbers be x, 2x and 3x. Then, their H.C.F = x. so x = 12. The numbers are 12, 24, 36. : 3 and their H.C.F is 12. The numbers are B. 5, 10,15 D. 12, 24, 36 8. A rectangular courty 3.78 metres long and 5.25 metres wide is to be paved exactly with square tiles, all of the same size. What is the largest size of the tile which could be used for the purpose ? A. 14 cms C. 42 cms Solution Largest size of the tile . H.C.F of 378cm and 525 cm = 21 cm. B. 21 cms D. None of these 9. The product of two numbers is 1320 and their H.C.F is 6. The L.C.M of the numbers is A. 220 c. 1326 Solution L.C.M 10. B. 1314 D. 7920 = product of numbers / H.C.F 1320/6 Product of two co-prime numbers is 117. Their L.C.M should be B. 117 D. cannot be calculated C. equal to their H.C.F Solution H.C.F of co-prime numbers is 1. so, L.C.M = 117/1 < = > 117. 11. The maximum numbers of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is A. 91 c. 1001 B. 910 D. 1911
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    PBI Solution Required number of students. PBI INSTITUTE 12. = H.C.F of 1001 and 910. < = > 91. 252 can be expressed as a product of prime as A. 2x2x3x3x7 c. 3x3x3x3x7 Solution Clearly, 252 x 2 x 3 x 3 B. 2x2x2x3x7 D. 2x3x3x3x7 13. Three different containers contain 496 litres, 403 litres and 713 litres of mixtures of milk and water respectively. What biggest measure can measure all the different quantities exactly? A. 1 litre C. 31 litre Solution Required measurement 14. The H.C.F of two numbers is 8. c. 60 D. 41 litre = H.C.F of (496, 403, 713) litres litres. Which one of the following can never be their L.C.M ? B. 12 D. 72 Solution H.C.F of two numbers divides their L.C.M exactly. Clearly, 8 is not a factor of 60. 15. The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 10, then their difference is A. 10 c. 70 Solution Let the numbers be x and (100 - x) . Then, x (100 - x) = 5 - + 2475 = o. (x -55) (x - 45) = o. x = 55 or x = 45. Therefore, the numbers are 45 and 55. Required difference = (55 - 45) = 10. B. 46 D. 90 16. Let N be the greatest numbers that will divide 43, 91 and 183 so as to leave the same remainder in each case.
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    PBI Solution Required number D. 13 = H.C.F of (91 - 43), (183 (183 - 43) = H.C.F of 48, 92 and 140 PBI INSTITUTE - 91) and

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