Question 512900: The present ages of 4 brothers are consecutive multiples of three. 5 years ago the sum of their afes was 46. Find their ages now.
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! The present ages of 4 brothers are consecutive multiples of three. 5 years ago the sum of their afes was 46. Find their ages now.
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Let n = the age of the youngest brother
Since the ages of the other brothers are consecutive multiples of 3, their ages are:
n+3, n+6, and n+9
5 years ago, each brother was 5 years younger, so we have to subtract 5 from each of their ages:
n + n+3 + n+6 + n+9 - 4*5 = 46
Solve for n:
4n - 2 = 46
4n = 48
n = 12
So their present ages are: 12, 15, 18, and 21 years old
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