Question

A man is 5 times as old as his son.he will be three times as old a his after 10 years.find their present ages​

Answer 1

Given:-

• At present a man is 5 times as old as his son

• After 10 years he will be three times as old as his son

Find:-

• Their present ages

Solution:-

• Let, Age of a man's be 'x' yrs
• Let, Age of a man's be 'x' yrsand age of his son's be 'y' yrs

$\large{\blue{\textsf{According To Question:-}}}$

A men is 5 times the age of his son

$\sf \implies x = 5y\:(Equation\:1)$

Now, 

After 10yrs His age will be three times the age of his son.

$\sf \implies x + 10= 3(y + 10)\: (Equation\:1)$

$\underline{\sf Substituting\:equation \:1\:in\:equation\:2:-}$

$\begin{gathered} \sf \dashrightarrow x + 10= 3(y + 10) \\ \\ \end{gathered}$

$\begin{gathered} \sf \dashrightarrow (5y) + 10= 3(y + 10) \\ \\ \end{gathered}$

$\begin{gathered} \sf \dashrightarrow 5y + 10= 3y + 30 \\ \\ \end{gathered}$

$\begin{gathered} \sf \dashrightarrow 5y - 3y= 30 - 10\\ \\ \end{gathered}$

$\begin{gathered} \sf \dashrightarrow 2y= 20\\ \\ \end{gathered}$

$\begin{gathered} \sf \dashrightarrow y= \dfrac{20}{2}\\ \\ \end{gathered}$

$\begin{gathered} \sf \dashrightarrow y= 10yrs\\ \\ \end{gathered}$

$\underline{\sf Substituting\:value\:of\:y\:in\:equation\:1:-}$

$\begin{gathered} \sf \implies x = 5y \\ \\ \end{gathered}$

$\begin{gathered} \sf \implies x = 5(10) \\ \\ \end{gathered}$

$\begin{gathered} \sf \implies x = 50yrs\\ \\ \end{gathered}$

$\sf \red{Therefore:-}$

$\underline{\boxed{\therefore \textsf{Age of a man, x = 50yrs}}}$

$\underline{\boxed{\therefore \textsf{Age of his son, y = 10yrs}}}$

Answer 2

Method (by two variables)

Let the age of the man be x and age of his son be y.

A/q , Man is 5 times old his son.

• x=5y ---(1)

Also after 10 years the man will be 3 times older than his son

• (x+10)=3(y+10)

Now Substituting the value of 1 in above equation, we get

⇒5y + 10 = 3y +30

⇒2y = 20

⇒ y=10

Now , x=5y ⇒x= 5×10=50 years .

Thus, the age of the man is 50 years and age of his son is 10 years.