Question

6.
Two years ago, Dilip was three times as old as his son and two years
hence, twice his age will be equal to five times that of his son. Find
their present ages.​

Answer 1

Given :-

Dilip's age 2 years ago = 3 × Age of his son

Dilip's present age = (3 × Age of his son) + 2

To Find :-

Present age of Dilip.

Present age of his son.

Solution :-

Let the age of son 2 years ago be x and then, Dilip's age would be 3x

Assume that,

• x = Age of son after 2 years
• 3x = Age of Dilip after 2 years
• x + 2 = Present age of son
• 3x + 2 = Present age of Dilip

After two years,

Son's age = $\sf x+2+2 =x+4$

Dilip's age = $\sf 3x+2+2=3x+4$

According to the question,

$\sf 5(x+4)=2(3x+4)$

$\sf 5x+20=6x+8$

$\sf 5x-6x=8-20$

$\sf -x=-12$

$\sf x=12$

Hence, the value of x is 12

Finding their present ages,

Present age of his son = $\sf x+2=12+2=14$

Present age of son = 14 years

Present age of Dilip = $\sf 3x+2=( 3\times 12)+2$

Present age of Dilip = $\sf 36+2=38$ years

Therefore, the present ages of Dilip and his son are 38 years and 14 years respectively.

Answer 2

Given :-

• Dilip's age 2 years ago = 3 × Age of his son

• Dilip's present age = (3 × Age of his son) + 2

To Find :-

• Present age of Dilip.

• Present age of his son.

Solution :-

Let the age of son 2 years ago be x

then, Dilip's age would be 3x.

Let ,

x + 2 = Present age of son

3x + 2 = Present age of Dilip

After two years,

• Son's age = $\sf x+2+2 =x+4$

• Dilip's age =$\sf 3x+2+2=3x+4$

Now,

$\sf 5(x+4)=2(3x+4)\\\\\\</p><p></p><p>\sf 5x+20=6x+8\\\\\\</p><p></p><p>\sf 5x-6x=8-20\\\\\\</p><p></p><p>\sf -x=-12\\\\\\</p><p></p><p>\sf x=12$

Hence, the value of x is 12

Now, Present ages :-

• Present age of his son = $\sf x+2=12+2=14$

• Present age of Dilip =$\sf 3x+2=( 3\times 12)+2$

=38 years

$\therefore \underline{\pink{ \: The\: present \: ages \: of \:Dilip\: and\: his \:son \:are \:38 \:years \:and \:14 years\: respectively}}.$

________________________________