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?
L.C.M. of 6, 9, 15 and 18 is 90.Let required number be 90k + 4, which is multiple of 7.Least value of k for which (90k + 4) is divisible by 7 is k = 4.Required number = (90 x 4) + 4 = 364.
The number students in each room = HCF of 70,98 and 126 =14 In each room , maximum 14 students can be seated. Total number students = 70 + 98 +126 =294 Number of rooms required = 294/14 =21
EXAMIANS