Problems on H.C.F and L.C.M Find the least number which when divided by 6,7,8,9, and 12 leave the same remainder 1 each case 404 505 707 606 404 505 707 606 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 3 | 6 - 7 - 8 - 9 - 12----------------------------- 4 | 2 - 7 - 8 - 3 - 4 --------------------------- 2 | 2 - 7 - 2 - 3 - 1 -------------------------- | 1 - 7 - 1 - 3 - 1 L.C.M = 3 X 2 X 2 X 7 X 2 X 3 = 504.Hence required number = (504 +1) = 505
Problems on H.C.F and L.C.M What is the smallest number which when diminished by 8, is divisible by 9, 6, 12 and 18? 45 44 50 60 45 44 50 60 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The least number which when divided by 5,6,7 And 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is : 3363 1683 2523 1677 3363 1683 2523 1677 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M A, B and C start running around a circular stadium and complete one round in 27 s, 9 s and 36 s, respectively. In how much time will they meet again at the starting point? 4 minute 48 seconds 2 minute 48 seconds 1 minute 48 seconds 3 minute 48 seconds 4 minute 48 seconds 2 minute 48 seconds 1 minute 48 seconds 3 minute 48 seconds ANSWER EXPLANATION DOWNLOAD EXAMIANS APP LCM of 27, 9 and 36 = 108 So they will meet again at the starting point after 108 sec. i.e., 1 min 48 sec.
Problems on H.C.F and L.C.M The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: 107 185 111 101 107 185 111 101 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let the numbers be 37a and 37b.Then, 37a x 37b = 4107 ab = 3.Now, co-primes with product 3 are (1, 3).So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.
Problems on H.C.F and L.C.M The least number which is a perfect square and is divisible by each of the numbers 16,20, and 24, is : 1600 11400 6400 3600 1600 11400 6400 3600 ANSWER DOWNLOAD EXAMIANS APP
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