Question

21. Five
years ago aman was seven times as old as his son.
Five years the father will be three times as old as hig
son. Find their
ages.
hence,
present​

Answer 1

Answer: 40yrs

Step-by-step explanation:

Age of the son =x

Age of the father=7x

Present ages=x+5 and 7x+5

future age of the son=x+10

father=7x+10

7x+10=3(x+10)

x=5

7x=35

x+5=5+5=10

son age =10

father age =7x+5=7*5+5=40

Answer 2

Solution :

Let the present age of Son's be R years

Let the present age of Man's be M years

A/q

$\underbrace{\bf{5\:years\:ago\::}}}}$

The age of Son's will (R-5) years.

The age of Man's will (M-5) years.

$\longrightarrow\sf{7(R-5)=M-5}\\\\\longrightarrow\sf{7R-35=M-5}\\\\\longrightarrow\sf{7R-M=-5+35}\\\\\longrightarrow\sf{7R-M=30.......................(1)}$

$\underbrace{\bf{After\:5\:years\::}}}}$

The age of Son's will (R+5) years.

The age of Man's will (M+5) years.

$\longrightarrow\sf{3(R+5)=M+5}\\\\\longrightarrow\sf{3R+15=M+5}\\\\\longrightarrow\sf{3R-M=5-15}\\\\\longrightarrow\sf{3R-M=-10}\\\\\longrightarrow\sf{3R=-10+M}\\\\\longrightarrow\sf{R=-10+M/3......................(2)}$

∴Putting the value of R in equation (1),we get;

$\longrightarrow\sf{7\bigg(\dfrac{-10+M}{3} \bigg)-M=30}\\\\\\\longrightarrow\sf{\dfrac{-70+7M}{3} -M=30}\\\\\\\longrightarrow\sf{-70+7M-3M=90}\\\\\longrightarrow\sf{-70+4M=90}\\\\\longrightarrow\sf{4M=90+70}\\\\\longrightarrow\sf{4M=160}\\\\\longrightarrow\sf{M=\cancel{160/4}}\\\\\longrightarrow\bf{M=40\:years}$

∴Putting the value of M in equation (2),we get;

$\longrightarrow\sf{R=\dfrac{-10+40}{3} }\\\\\\\longrightarrow\sf{R=\cancel{\dfrac{30}{3} }}\\\\\longrightarrow\bf{R=10\:years}$

Thus;

$\boxed{\sf{The\:present\:age\:of\:Son's\:= R=\boxed{\bf{10\:years}}}}\\\boxed{\sf{The\:present\:age\:of\:Man's\:= M=\boxed{\bf{40\:years}}}}$