Question 885747: Ten years ago sum of ages of two sons was one third of their father.one son is 2yrs older than the other and sum of their present ages is 14years less than their fathers presenr age.find their present age?
Answer by momentman(1)   (Show Source):
You can put this solution on YOUR website! Let the ages of the sons be x and y
x-Older son
y-Younger son
Let the fathers age be z
10 years ago their ages would have been
(x-10), (y-10) and (z-10)
10 years ago,The ages of two sons equals 1/3 of their fathers
(x-10)+(y-10)=1/3*(z-10)
Simplify and get...3x+3y-z=50--------------------(1)
Since one son is 2 years older, x=y+2----------------(2)
Currently the sum of the children's ages is the Father's less 14 meaning
x+y = z-14-------------------------------------------------(3)
if you substitute eqn 2 in 3 you get
z=2y+16
Now go back to equation 1 and make sure its all in terms of y
3(y+2)+3y-(2y+16) = 50
4y=60
y=15
therefore x=2+15 =17
z=2*15+16 = 46
Their present ages are 15,17,46
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