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. 24 21 17 31 24 21 17 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 Find the least number which when divided by 20,25,35 and 40 leaves remainders 14,19,29 and 34 respectively. 1394 1494 1194 1294 1394 1494 1194 1294 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,(20-14) = 6,(25 – 19)=6,(35-29)=6 and (40-34)=6. Required number = (L.C.M. of 20,25,35,40) – 6 =1394.
Problems on H.C.F and L.C.M The Smallest number which when diminished by 7, is divisible by 12,16,18,21 and 28 is : 1015 1008 1032 1022 1015 1008 1032 1022 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M If the sum of two numbers is 55 and the H.C.F and L.C.M of these numbers are 5 and 120 respectively, then the sum of the reciprocal of the numbers is equal to: 120/11 55/601 11/120 601/55 120/11 55/601 11/120 601/55 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Three friends Raju , Ramesh and Sunil start running around a circular stadium and complete a single round in 24 s, 36 s and 40 s, respectively. After how many minutes will they meet against at the sta 8 minutes 6 minutes 5 minutes 7 minutes 8 minutes 6 minutes 5 minutes 7 minutes ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 24 = 3 × 2 × 2 × 2 = 3 × 2³ 36 = 3 × 3 × 2 × 2 = 3² × 2²and 40 = 2 × 2 × 2 × 5 = 5¹ × 23 LCM of 24, 36 and 40 = 3² × 2³ × 5 = 9 × 8 × 5 = 360Hence, they will meet again at the starting point after 360 s, i.e., 6 min
Problems on H.C.F and L.C.M The least number,which when divided by 48,60,72,108 and 140 leaves 38,50,62,98 and 130 as remainder respectively, is : 15110 15120 15210 11115 15110 15120 15210 11115 ANSWER DOWNLOAD EXAMIANS APP
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