Problems on H.C.F and L.C.M In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are? 820,369 700,400 820,360 800,500 820,369 700,400 820,360 800,500 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the HCF of 54, 288, 360 16 19 18 17 16 19 18 17 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 Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together. 17 24 21 31 17 24 21 31 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.Now 60/2 = 30Adding one bell at the starting it will 30+1 = 31
Problems on H.C.F and L.C.M The H.C.F. of two numbers is 11 and their L.C.M. is 693 . If one of the numbers is 77, find the other . 95 99 88 77 95 99 88 77 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The least number which when increased by 5 is divisible by each one of 24, 32, 36 and 54, is 427 4320 859 869 427 4320 859 869 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. 2 6 4 8 2 6 4 8 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43) = H.C.F. of 48, 92 and 140 = 4.
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