Problems on H.C.F and L.C.M 252 can be expressed as a product of primes as: 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 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 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Clearly, 252 = 2 x 2 x 3 x 3 x 7.
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: 111 107 185 101 111 107 185 101 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 In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are? 820,360 700,400 820,369 800,500 820,360 700,400 820,369 800,500 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width. Find the least number of square tiles of equal size required to cover the entire floor of the room? 117 96 124 110 117 96 124 110 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the lowest common multiple of 24, 36 and 40. 240 360 120 480 240 360 120 480 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
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 6 4 3 5 6 4 3 5 ANSWER DOWNLOAD EXAMIANS APP
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