Problems on H.C.F and L.C.M Find the HCF of 54, 288, 360 16 17 19 18 16 17 19 18 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 will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ? 420 540 630 770 420 540 630 770 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP L.C.M. of 12, 18, 21, 30 2 | 12 - 18 - 21 - 30 ---------------------------- = 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15 ---------------------------- Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5 = 630.
Problems on H.C.F and L.C.M Find the least number which when divided by 6, 7, 8, 9 and 10 leaves 1, 2, 3, 4 and 5 as remainders respectively, but when divided by 19 leaves no remainder? 5016 5054 5035 5073 5016 5054 5035 5073 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The traffic light at three different road crossing change after every 48 sec, 72 sec. and 108 sec. respectively . If they all change simultaneously at 8 : 20 : 00 hours, then at what time will they ag 5 : 27 : 12 hrs. 8 : 27 : 12 hrs. 6 : 27 : 12 hrs. 7 : 27 : 12 hrs. 5 : 27 : 12 hrs. 8 : 27 : 12 hrs. 6 : 27 : 12 hrs. 7 : 27 : 12 hrs. ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The maximum number of student amoung 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 : 1911 910 1001 91 1911 910 1001 91 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the H.C.F. of 513, 1134 and 1215. 27 58 63 42 27 58 63 42 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 1134 ) 1215 ( 1 1134 --------- 81 ) 1134 ( 14 81 --------- 324 324 --------- x ---------- H.C.F. of 1134 and 1215 is 81.So, Required H.C.F. = H.C.F. of 513 and 81. ______81 ) 513 ( 6 __486____ 27) 81 ( 3 81 0H.C.F. of given numbers = 27.
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