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.
Since the numbers are co-prime, they contain only 1 as the common factor.Also, the given two products have the middle number in common.So, middle number = H.C.F. of 551 and 1073 = 29;First number = 551/29 = 19; Third number = 1073/29 = 37.Required sum = (19 + 29 + 37) = 85.
EXAMIANS