SOLUTION: Ten years ago, the sum of the ages of two sons was one third of their father’s age. One son is
two years older than the other and sum of their present ages is 14 years less than t
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Question 778562: Ten years ago, the sum of the ages of two sons was one third of their father’s age. One son is
two 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 by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Ten years ago, the sum of the ages of two sons was (1/3) of their father’s age.
One son is two 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
--------
Equations:
s1-10 + s2-10 = (1/3)f
s1 = s2+2
s1+s2 = f-14
---------------------
Simplify::
s1+s2-20 = (1/3)f
Sustitute for s1+s2
f-14 + (1/3)f
3f-42 = f
2f = 42
father's age is 21 years
----
s1+s2 = 21-14 = 7
---
Substitute for s1 to get:
s2+2 + s2 = 7
2s2 = 5
s2 = 2 1/2 years
Then s1 = 4 1/2 years
=====================
Cheers,
Stan H.
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