Question

10 year ago the sum of ages of two sons was one third of the age of father . one son is two year older son. sum of their present age is 14years less than the age of father , find their present age.​

Answer 1

Answer:

let present ages of father be x and older son be y and younger son be z

now ten years ago

age was father x-10

older son y-10

younger son z-10

now a t q   y-10+z-10=1/3[x-10]

y+z-20 = 1/3[x-10]

3y+3z-60+10=x mark it one

now y=z+2 given

now y+z=x-14

here x=y+z+14

now in eq 1

y+z+14= 3y+3z-50

14+50=3y-y+3z-z

64=2y+2z

64/2=y+z

y+z=32 and y=Z+2

2z=32-2

z=30/2

z=15 so y= 15+2=17

for father x= y+z+14

32+14=46

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