Question

please help me out !! ​

Answer 1

Answer:

Father's age =40 years

Son's age = 10 years

Answer 2

$\huge\sf\red{Given\::}$

• Five years ago,a man was seven times as old as his son.
• Five years hence, the father will be three times as old as his son.

$\huge\sf\purple{To\:find\::}$

• Present ages of father and son

$\huge\sf\pink{Solution:-}$

$\sf Let,$

• $\sf Present\: age\: of\: father\: =\: x\: years$
• $\sf Present\: age\: of\: son\: =\: y\: years$

$\sf 5\; years\: ago,$

• $\sf Age \:of \:father\: =\: (x-5)\: years$
• $\sf Age \:of \:son\: = \:(y-5) \:years$

$\leadsto\sf{x-5=7(y-5)}$

$\leadsto\sf{x-5=7y-35}$

$\leadsto\sf{x=7y-35+5}$

$\leadsto\sf{x=7y-30............(1)}$

$\sf After \:5\: years,$

• $\sf Age \:of \:father\: =\: (x+5)\: years$
• $\sf Age \:of \:son\: = \:(y+5) \:years$

$\leadsto\sf{x+5=3(y+5)}$

$\leadsto\sf{x+5=3y+15}$

$\leadsto\sf{7y-30+5=3y+15\:(\:put\:x=7y-30\: from\:eq(1)\:)}$

$\leadsto\sf{7y-3y=15+30-5}$

$\leadsto\sf{4y=40}$

$\leadsto\sf{y=10}$

$\hookrightarrow\:\sf\underline\red{Present\: age\: of\: son\: =\: 10 \:years.}$

$\sf Now ,\: substitute\: y\: value\: in\: eq \:(1)$

$\leadsto\sf{x=7y-30}$

$\leadsto\sf{x=7\times\:10-30}$

$\leadsto\sf{x=70-30}$

$\leadsto\sf{x=40}$

$\hookrightarrow\:\sf\underline\red{Present \:age\: of\: father\: =\: 40\: years.}$