algebra

The present ages in years of four cousins are consecutive multiples of 3. Five years ago the sum of their ages was 46. What is their ages now?

asked by Zephine
  1. if their ages are a,b,c,d, then

    a+b+c+d = 3n + 3n+3 + 3n+6 + 3n+9 = 12n + 18
    (a-5)+(b-5)+(c-5)+(d-5) = 46
    a+b+c+d = 66

    so, 12n+18 = 66
    n=4

    so, the ages are 12,15,18,21
    5 years ago, they were 7,10,13,16 and added up to 46

    posted by Steve

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