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Ten years ago,the sum of the ages of 2 sons was one third of their father's age. One son is 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


Solution

Ten years ago,the sum of the ages of 2 sons was one third of their father's age. One son is 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

let the present age of the father = x
& present age of two sons be y+2 and y
by question ten yrs ago,
 [(y+2-10) + (y-10)]  = (x-10)/3 
2y-18 = (x-10)/3
6y-54=x-10
6y-44= x

 At present ( y+2)+y] = x-14
2y+16 = x

thus we find  6y-44 = 2y+16
or 4y = 60
y = 15 
y+2 = 15+2=17
x = 2y+16=2*15+16= 46

father's age=46,
two son's ages are 15 & 17  Ans.


 

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