Posted by Zephine 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  Steve, 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
(a5)+(b5)+(c5)+(d5) = 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