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4J The LCM of two numbers is 85 and their product is 1020. Their HCF will be a) 16 b)27 c) 12 d)22 5. The LCM and HCF of two numbers is 4284 and 32 respectively. If one of the number is 204 the other is a) 672 b) 576 c) 676 d) 572 The largest four digit number divisible by 48, 60 and 64 a) 7200 b) 9600 c) 8400 d) 1000 1. HCF of 420, 315 and 462 a) 24 b) 28 c) 27 d) 21 8. The smallest number exactly divisible by 3, 4, 6 and 8 a) 26 b) 24 c) 25 d) 28 9. The smallest three digit number exactly divisible by 12, 18 and 24 a) 720 b) 144 c) 180 d)224 10. The largest three digit number when divided by 6, 9, 12 leaves raminder 1 as in each case is a) 887 h) 987 973 d)

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Two numbers are in the ratio 8:15. Their HCF is 4. The numbers are a)32and60 b)l6and3O c) 80 and 150d) 64 and 120 The largest number that will divide 226 and 272 leaving 1 and 2 as remainders respectivelyis a) 36 b) 45 c) 55 d) 59 Which of the following is the greatest common divisor of 1170102 and a)8 b)4 c)6 d)3 A number when 3 is added to it become divisible by 36, 45, 50. Then smallest such number is a) 987 b) 798 c) 986 d) 897 The least perfect square which is completely divisible by 10, 20. 30 and 40 is a) 4800 b) 3600 ed by 6, 9 and 12 leaves 3 as c) 4400 d) 2500 remainder in each case is 16. The greatest number that will a) 575 b) 795 c) 975 d) 525 divide 366, 513 and 324 leaving 19. The greatest number that will the same remainder in each case divide 33, 64 and 80 leaving 3, 4 a) 21 b) 18 c) 27 d) 42 and 5 as remainders respective17. The sum of two numbers is 216 ly is and their IICF is 27. Then numa) 10 b) 20 c) 15 d) 22 bers are 20. The HCF of three numbers is a) 60 &90 b) 81&135 12. If the three numbers are in c) 64& 128 d) 30 84 the ratio 1:2:3, then the nurn1 8 • The largest three digit number bers are which when successively IV1(1 a) 14,28,42 b) 12,24,36 c) 15, 30, 45 d) 24, 48, 72

Find the greatest number that will exactly divide x,y,z Find the greatest number that 2 will divide x,y,z leaving remainders a,b,c respectively Find the least number which is exactly divisible by x,y,z Find the least number which 4 when divided by x,y,z leaves remainders a,b,c respectively

- Find the least number which 5 when divided by x,y,z leaves the same remainder 'r'in each case Find the greatest number that 6 will divide x,y,z leaving the same remainder in each case

- Find the n-digit greatest number which when divided by x,y,z (a) leaves no remainder (b) leaves remainder Pin each case

- Find the n-digit smallest number which when divided by x,y,z 8 (a) leaves no remainder (b) leaves remainder P'in each case

left according to the maximum decimal places among the given numbers LCM of decimal numbers —*Find the LCM of given numbers without decimal —Put the decimal point (in the LCM) from right to left at the place equal to the maximum decimal places among the given numbers HCF of fractions

-HCF of numbers x LCM of numbers is equal to the product of numbers (this rule is applicable only in the case of two numbers)

LCM of 10, 20, 30, 40 HCF of 105 and 462 is 21 is 2x2x5x3x2 required HCF 21 but the number is a perfect 8. b) LCM of 3, 4, 6, 8 24 square number required nurnber=2x2x5x5x3x3x2x2=3600 9. b) 16.a LCM of 12, 18, 24 is 72 difference between 513 and 366 The required three digit number is 147 and the difference which is smallest as well as between 513 and 324 is 189. divisible by 12, 18, 24 is 72x2 :. HCF of 147 189 is 21 144 The required number is 21 10. c) 17.b LCM of 6, 9, 12 is 36 let the numbers bex &y On dividing largest three digit x+y=216=27x8 number 999 by 36 we get 27 as Since HCF of two numbers is 27 remainder hence common factor of these 999-27=972 ANSWERS HCF of x,y,z two numbers is 27. Therefore I. c) required number = 972 + 1 = 973 these numbers will be of form 72=2x2x2x3x3 11.a 27x1, 27x2, 27x3...etc let the numbers be 8x & 15x 18= 2x3x3 HCF of (x—a), (y—b), (z—c) According to the condition in LCM of &x & lE=120x Common factors = 2x3x3 the problem sum of these num1st Number x 2nd number = HCF = 18 bers is 216 the numbers will be HCFx LCM 2. a) 274=81 and 27x5= 135 8xxl5x = 4x120x LCM = 2x2x2x3x5 = 120 LCM of x,y,z 18 c 3. b) 120x2 = 4x12th The LCM of 6, 9, 12 is 36 and Numbers are 8x4=32 and 1ICFof5 6 7 8 o (x—a)=(x—b)=(z—c)=k largest three digit number is 15x4=60 LCM of 6 7 8 9 10 = _L Number=[(LCM of(x,y,z)1—k 2520 999 HCF of 36 & 999 is 27 The 12. b largest three digit number corn226-1=225, 272-2=270 4. c) pletely divisible by 6, 9, 12 is HCF of 225 & 270 is 45 LCM xHCF = Product of numbers [LCM of (x,y,z)]+r 999-27 = 972 required number is 45 85xHCF = 1020 Since 3 is left as remainder in 13.c HCF = 1020 = 12 85 1170=2x5x3x3x13 each case required largest three digit number 972+3=975 102=2x3x17 a) HCF of (x—y), (y—z), (z—x) Greatest common divisor=24=6 19 C jst number x2WI number = 33-3=30, 64-4=60, 80-5=75 14.d LCM x HCF LCM of x,y,z=k HCF of 30, 60, 75 is 15 The smallest number will be 204 xx = 4284 x32 divide the n digit number by k, X = 4284x32 = 672 required largest number is 15 three less than the LCM of 36, we get remainder 'R' 204 20b 45,50 (a) 'n'digit number—R Let the numbers be x, 2 x, 3x LCM of 36, 45, 50 is 900 Second number = 672 (b) (n-digit greatest number-R) +P HCF of three numbers is 12, required number is 900-3=897 b) LCM of 48, 60, 64 = 960 therefore 12 is the largest corn15.b LCM of x,y=k ,z largest four digit No - 9999 mon factor of these three numdivide the n digit number by k, On dividing 9999 by 960 we get 399 bers. Thus x=12 the numbers as remainder we get remainder 'R' are 12, 24, 36 (a) 'n'digit number+(K—R) required number =9999 - 399 9600 (b) (n-digit smallest number d) HCF of 315&420 is 105 Grab

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: 1. l'he HCF of 72 and 18 is 9 b) 72 c) 18 d) 36 2.e LCM of 6, 8,10 and 12 120b)60c)240dl30 and 3. 7h)e HCF of --, a)

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