Question

the ratio of father and son is 7:2 .After 5 years their ratio becomes 8:3.find their present ages

Answer 1

$let \: the \: age \: of \: father = 7x \: years \\ \: \: \: \: \: \: \: \: \: \: the \: age \: of \: son \: \: \: = 2x \: years \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: after \: 5 \: years - - \\ \\ now \: \: age \: of \:father = (7x + 5) \: years \\ age \: of \: son = (2x + 5) \: years \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: acc. \: to \: question - - \\ \\ = > \frac{7x + 5}{2x + 5} = \frac{8}{3} \\ \\ = >3( 7x + 5) = 8(2x + 5) \\ = > 21x + 15 = 16x + 40 \\ = > 21x - 16x = 40 - 15 \\ = > 5x = 25 \\ \\ = > x = \frac{25}{5} \\ \\ = > x = 5 \\ \\ present \: age \: of \: father = 7(5) = 35 \: years \\ present \: age \: of \: son = 2(5) = 10 \: years \\ \\ \\ please \: mark \: it \: as \: brainliest.$

Answer 2

$\frac{fathers \: age}{sons \: age} = \frac{7x}{2x}$
After 5 years,

$\frac{fathers \: age}{sons \: age} = \frac{7x + 5}{2x + 5} = \frac{8y}{3y}$
$\frac{fathers \: age}{sons \: age} = \frac{7x + 5}{2x + 5} = \frac{8y}{3y} \\ \\ \frac{7x + 5}{2x + 5} = \frac{8y}{3y} \\ \\ 21xy + 15y = 16xy + 40y \\ \\ 5xy =40y - 15y \\ 5xy = 25y \\ x = \frac{25y}{5y} = 5$


Therefore,
The present age of father = 7x = 7(5)=35 years.
The present age of son = 2x = 2(5)= 10 years.