Question

iii. A father is 3 times as old as his son. In 10 years time, his age will be double his son's age. Find the present age​

Answer 1

At present:

Age of son = x years ,

Father's age = 3x years

After 10 years:

Age of son = ( x + 10 ) years ,

Father's age = ( 3x + 10 ) years

/* According to the problem given */

$Age \: of \: father = 2 \times (age \:of \:son )$

$\implies 3x + 10 = 2( x + 10 )$

$\implies 3x + 10 = 2x + 20$

$\implies 3x - 2x = 20 - 10$

$\implies x = 10$

Therefore.,

At present:

Age of son = x years = 10 years ,

Father's age = 3x years = 3 × 10 = 30 years .

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Answer 2

Question:

• A father is 3 times as old as his son. In 10 years time, his age will be double his son's age. Find the present age.

Given:

• Father age's 3 times old to his son 10 years time, his age will double to son's

To find:

• Their present ages?

Solution:

❍ Let's Consider, the present age of son be n years and his father's age be 3n years.

$\begin{gathered} { \color{navy}{\maltese} } \: \underline{\boldsymbol{ \sf{ \pmb{According\: to \:the\: Question :}}}}\\\end{gathered}$

• In 10 years, the age of father will be double of his son's age.

• 10 years later their ages will be: (n + 10) and (3n + 10).

⠀

$\begin{gathered}:\implies\sf Father_{\;(age)} = 2(Son_{\:(age)}) \\\\\\\end{gathered} \\ \begin{gathered}:\implies\sf (3n + 10) = 2(n + 10)\\\\\\\end{gathered} \\ \begin{gathered}:\implies\sf 3n + 10 = 2n + 20\\\\ \\ \end{gathered} \\ \begin{gathered}:\implies\sf 3n - 2n = 20 - 10\\\\\\\end{gathered} \\ \begin{gathered}:\implies \underline{\boxed{\frak{ n = 10}}} \pink\bigstar\\\\\end{gathered}$

Therefore,

• Son's age, n = 10
• Father's age, 3n = 3(10) = 30

$\\\underline{\textsf{Hence, their present ages are \textbf{10 years} \: and \textbf{30 years}}}.$

тнαηк үσυ

||TheUnknownStar||