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 4 8 6 2 4 8 6 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.
Problems on H.C.F and L.C.M Find the greatest number that will divide 172, 205 and 304 so as to leave the same remainder in each case? 33 99 None of these 66 33 99 None of these 66 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The greatest number that will divide 172,205 and 304 leaving the same remainder in each case is HCF ((205-172), (304-205), (304-172)) = HCF of 33, 99, and 132 33=31x11 99=3 x 3 x 11 132=3 x 2 x 2 x11 Required number =3 x 11=33
Problems on H.C.F and L.C.M The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is: 40 20 10 30 40 20 10 30 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let the numbers be 2x and 3x.Then, their L.C.M. = 6x.So, 6x = 48 or x = 8The numbers are 16 and 24.Hence, required sum = (16 + 24) = 40.
Problems on H.C.F and L.C.M The smallest fraction, which each of 6/7, 5/14, 10/21 will divide exactly is: 60/147 50/294 30/7 30/98 60/147 50/294 30/7 30/98 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Let the leas number of six digits. which when divided by 4, 6, 10 and 15 leaves in each case the same remainder of 2 be N. The sum of the digits in N is 4 6 5 3 4 6 5 3 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M 252 can be expressed as a product of primes as: 2 x 2 x 2 x 3 x 7 2 x 3 x 3 x 3 x 7 3 x 3 x 3 x 3 x 7 2 x 2 x 3 x 3 x 7 2 x 2 x 2 x 3 x 7 2 x 3 x 3 x 3 x 7 3 x 3 x 3 x 3 x 7 2 x 2 x 3 x 3 x 7 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Clearly, 252 = 2 x 2 x 3 x 3 x 7.
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