Problems on H.C.F and L.C.M Find the HCF of 54, 288, 360 16 17 18 19 16 17 18 19 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Lets solve this question by factorization method.18=2×3²,288=2×2×2×2×2×3²,360=2³×3²×5So HCF will be minimum term present in all three, i.e.2×3²=18
Problems on H.C.F and L.C.M What is the lowest common multiple of 12, 36 and 40? 360 380 240 260 360 380 240 260 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M What is the greatest number which divides 29, 32, 38 and leaves same remainder in each case? 5 3 4 2 5 3 4 2 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is: 432 268 1015 689 432 268 1015 689 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Required number = (L.C.M. of 12,16, 18, 21, 28) + 7= 1008 + 7= 1015
Problems on H.C.F and L.C.M Six belts commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes. how many times do they toll together ? 4 16 15 10 4 16 15 10 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M In a school, the number of students in Sections A , B and C of 10th class are 70,98 and 126 respectively. Due to overload of student in each class , the administration wants to increase the number roo 14 21 17 28 14 21 17 28 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The number students in each room = HCF of 70,98 and 126 =14 In each room , maximum 14 students can be seated. Total number students = 70 + 98 +126 =294 Number of rooms required = 294/14 =21
EXAMIANS