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Posted by on Thursday, November 17, 2011 at 8:17pm.

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?

  • algebra - , Friday, November 18, 2011 at 12:23pm

    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

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