Question

ten years ago,the sum ages of two sons was one third of their father 's age .one son's 2 years older than the other and sum of their present ages is 14 years less than the father's present age ,find the present ages of all

Answer 1

Ten years ago
Let one son's age = x
Then other son's age = x+2
Let father's age = y
x+x+2 = y/3
2x+2 =y/3
2x-y/3 = - 2
6x-y = - 6 - - - - - (i)
x = (y-6)/6
At present
one son's age = x+10
other son's age = x+12
father's age = y+1 0
x+10+x+12 = y+10-14
2x+22 = y-4
2x-y = - 26 - - - - - (ii)
2{(y-6)/6} - y = - 26
(y-6)/3 - y = - 26
(-2y-6)/3 = - 26
-2y-6 = - 78
-2y = - 72
Y = 36
x = (y-6)/6 = (36-6)/6 = 5
Hence 10 years ago
One son's age = 5
other son's age = 7
father's age = 36

& present age
One son's age = 15
other son's age = 17
father's age = 46