Question

ten years ago a father was 12 times as old as his son and ten years hence he will be twice as oldas his son will be then. find their present ages?
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Answer 1

Solution :-

Let :-

Present age of son = x

Present age of father = y

10 years ago :-

Age of son = x - 10

Age of father = y - 10

10 years hence :-

Age of son = x + 10

Age of father = y + 10

According to the first condition :-

$\longrightarrow \sf 12(x-10) = y - 10 \\\\\longrightarrow 12x-120 = y - 10 \\\\\longrightarrow 12x - y = 110 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ....................(1)$

According to the second condition :-

$\longrightarrow \sf 2(x+10) = y + 10 \\\\\longrightarrow 2x+20 = y +10 \\\\\longrightarrow 2x-y = - 10 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ....................(2)$

Equation (1) - Equation (2) :-

$\longrightarrow \sf 10x = 120 \\\\\longrightarrow \bf x = 12$

Substitute the value of x in eq (2) :-

$\longrightarrow \sf 2\times 12 -y = -10 \\\\\longrightarrow 24 -y = -10 \\\\\longrightarrow -y = -34 \\\\\longrightarrow \bf y = 34$

Present age of son = 12 years

Present age of father = 34 years

Answer 2

Answer:

Given :-

• Tens years ago father age was 12 times as old as his son age at that time and ten years hence he will be twice as old as his son.

To Find :-

• What is their present ages.

Solution :-

Let, the age of son before 10 years be x

And, the age of the father before 10 years be 12x

10 years hence he will be twice as old as his son.

According to the question,

⇒ 2(x + 10 + 10) = (12x + 10 + 10)

⇒ 2x + 20 + 20 = 12x + 10 + 10

⇒ 2x - 12x = 10 + 10 - 20 - 20

⇒ 10x = 20

⇒ x = $\sf\dfrac{\cancel{20}}{\cancel{10}}$

➠ x = 2 years

Hence, the required ages are :

❒ Present age of son = x + 10 = 2 + 10 = 12 years

❒ Present age of his father = 12x + 10 = 12(2) + 10 = 24 + 10 = 34 years

$\therefore$ The age of son is 12 years and his father is 34 years .

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