Question

9. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of his son. Find their present ages

Answer 1

Answer:

The present ages of Father and son are 42 Years and 10 Years respectively.

Step-by-step explanation:

Given :

• Two ears ago = father was five times as old as son
• Two years later = father age will be 8 more than three times the age of his son.

To Find :

• Find their present ages.

Solution :

✯ $\textbf{\small{\underline{Let the Ages two years ago as -}}}$

• Son as x
• Father as 5x

$\rule{300}{1.5}$

✯ $\textbf{\small{\underline{Present ages - }}}$

• Son --> x + 2
• Father --> 5x + 2

$\rule{300}{1.5}$

✯ $\textbf{\small{\underline{Ages after two years -}}}$

• Son --> x + (2 + 2)
• Father --> 5x + (2 + 2)

$\rule{300}{1.5}$

✯ $\textbf{\small{\underline{According to the Question -}}}$

Two years later, father age will be 8 more than three times the age of his son.

★ $\boxed{5x + (2+2)= 8 + 3(x + 4) }$

$\longrightarrow \: 5x + (2 + 2) = 8 + 3(x + 4)$

$\longrightarrow \: 5x + 4 = 8 + 3x + 12$

$\longrightarrow \: 5x + 4 = 3x + 20$

$\longrightarrow \: 5x - 3x= 20 - 4$

$\longrightarrow \:2x= 16$

$\longrightarrow \:x= \dfrac{16}{2}$

$\longrightarrow \:x = 8 \: .....\textbf{[ Son's age before 2 years ]}$

$\rule{300}{1.5}$

★ Value of (x + 2)

$\longrightarrow \: 8 + 2$

$\longrightarrow \: 10$

Son's present age = 10 Years

$\rule{300}{1.5}$

★ Value of 5x + 2

$\longrightarrow \: 5(8) + 2$

$\longrightarrow \: 40 + 2$

$\longrightarrow \: 42$

Father's present age = 42 years

$\therefore$ The present ages of Father and son are 42 Years and 10 Years respectively.

$\rule{300}{1.5}$

Verification :-

As mentioned in the Question, two years later, father's age will be 8 more than three times the age of his son.

✯ $\textbf{\small{\underline{Present Ages -}}}$

• Father = 42
• Son = 10

✯ $\textbf{\small{\underline{Ages after 2 years -}}}$

• Father = 42 + 2 --> 44
• Son = 10 + 2 --> 12

$\longrightarrow \: 44 = 8 + 3(12)$

$\longrightarrow \: 44 = 8 + 36$

$\longrightarrow \: 44 = 44$

$\therefore$ The present ages of Father and son are 42 Years and 10 Years respectively.

Answer 2

Answer:

Two years ago :

Suppose the age of son = X

and the age of father = 5x

The Present ages :

Son present age = X + 2

Father present age = 5x + 2

After two years :

Son age = X + (2 + 2)

Father age = 5x + (2 + 2)

Two years later , his father age will be more than three times the age of her son :

5x + (2 + 2) = 8 + 3 (X + 4)

5x + 4 = 8 + 3x + 12

5x + 4 = 3x + 20

5x - 3x = 20 - 4

2x = 16

X = 8

The age of son's after 2 years = x + 2

The value of X = 8

= 8 + 2

= 10

The age of father after 2 years = 5x + 2

= 5(8) + 2

= 40 + 2

= 42

The age of son's = 10 years .

The age of father = 42 years .