the age of the father 8years ago was 5times that of his son 8years hence age 8years more than twice the age of his son then the present age of father
Answers
Answered by
1
Let 8years ago age of son = x years
Present age of son= (x+8 )
let 8 years ago age of father=5x years
present age of father= (5x+8)
let 8 years hence age of son = x+8+8=(x+4)
" " " " " " father=5x+8+8=5x+4
Present age of son= (x+8 )
let 8 years ago age of father=5x years
present age of father= (5x+8)
let 8 years hence age of son = x+8+8=(x+4)
" " " " " " father=5x+8+8=5x+4
Answered by
0
let the age of son will be = x
father age before 8 year =5x
after 8. year ,
father age twice = 5x+8
son =x+8
atq
5x+8=2x
5x+8 =2x+8
5x-2x=8-8
3=o
x=3
father age before 8 year =5x
after 8. year ,
father age twice = 5x+8
son =x+8
atq
5x+8=2x
5x+8 =2x+8
5x-2x=8-8
3=o
x=3
Similar questions