Problems on H.C.F and L.C.M Three numbers are in the ratio 3:4:5 and their L.C.M is 2400. Their H.C.F is: 120 200 40 80 120 200 40 80 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is: 120 40 200 80 120 40 200 80 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let the numbers be 3x, 4x and 5x.Then, their L.C.M. = 60x.So, 60x = 2400 or x = 40. The numbers are (3 x 40), (4 x 40) and (5 x 40).Hence, required H.C.F. = 40.
Problems on H.C.F and L.C.M Find the HCF of 54, 288, 360 19 18 16 17 19 18 16 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 The L.C.M. of two numbers is 45 times their H.C.F. If one of the numbers is 125 and the sum of H.C.F. and L.C.M. is 1150, the other number is: 220 225 235 215 220 225 235 215 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The largest four – digit number which when divided by 4, 7 or 13 leaves a remainder of 3 in each case, is 9831 9893 9834 8739 9831 9893 9834 8739 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 7 minutes 8 minutes 6 minutes 5 minutes 7 minutes 8 minutes 6 minutes 5 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
EXAMIANS