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(B) Let the present ages of the three persons be $4x$,$7x$,and $9x$ years respectively.Eight years ago,each person's age was $8$ years less than their

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$16, 28, 36$

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(B) Let the present ages of the three persons be $4x$,$7x$,and $9x$ years respectively.Eight years ago,each person's age was $8$ years less than their present age.Therefore,the sum of their ages $8$ years ago was $(4x - 8) + (7x - 8) + (9x - 8) = 56$.Simplifying the equation: $20x - 24 = 56$.$20x = 56 + 24 = 80$.$x = 80 / 20 = 4$.Now,calculating their present ages:First person: $4x = 4 	imes 4 = 16$ years.Second person: $7x = 7 	imes 4 = 28$ years.Third person: $9x = 9 	imes 4 = 36$ years.Thus,their present ages are $16, 28, 36$ years.

The present ages of three persons are in the ratio $4: 7: 9$. Eight years ago,the sum of their ages was $56$. Find their present ages (in $years$).

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