Question

4. A father is 7 times as old as his son. Two years ago the
father was 13 times as old as his son.what are their present
age
the number​

Answer 1

Given :

• A father is 7 times as old as his son.
• Two years ago the father was 13 times as old as his son.

To Find :

• Present Ages

Solution :

• Let The present age of Son be = x
• The Present age of his father = 7x

2 years ago,

• Son's age = (x - 2)
• Father's age = (7x - 2)

According to Question :

⟹ 7x - 2 = 13 (x - 2)

⟹ 7x - 2 = 13x - 26

⟹ 26 - 2 = 13x - 7x

⟹ 24 = 6x

⟹ x = 24/6

⟹ x = 4

Present Ages :

• Son's Age = 4 years
• Father's age = 4*7 = 28 years

Answer 2

$\large{\underline{\red{\textsf{\textbf{Given...}}}}}$

It's given that :

• A father is 7 times older than his son.
• Two years ago, father was 13 times as old as his son.

$\large\underline{\red{\textsf{\textbf{To find...}}}}$

We have to find the ir present ages.

$\large\underline{\red{\textsf{\textbf{Solution...}}}}$

Let, the son's age be x years and let the father's age be 7x years.

So, two years ago their ages are :

• Father's age = (7x-2) years
• Son's age = (x-2) years

According to the question, the father is 13 times as old as his son two years ago.

So, the equation goes this way :

$\red{☞}$ (7x-2) = 13 (x-2)

$\red{☞}$ 7x - 2 = 13x - 26

$\red{☞}$ 7x - 13x = -26 + 2

$\red{☞}$ - 6x = - 24

$\red{☞}$ x = - 24 / - 6

$\red{☞}$ x = 4

So, now substituting the value of x = 4 :

• Father's age = 7x = 28 years
• Son's age = x = 4 years

__________________________________

Therefore, the father's age is 28 years and the son's age is 4 years.