Question

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

Answer 1

Answer:

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

Step-by-step explanation: