Question

5 year ago , FAther age was 7 times the age of his son . 5 year later ,the father age will be 3 time the age of his son . find their present ages


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Answer 1

Answer:-

Let son's age be x and father's age be y.

Given:

5 years ago father was 7 times the age of his son.

• Son's age before 5 years = (x - 5) years

• Father's age before 5 years = (y - 5) years.

So,

(y - 5) = 7(x - 5)

⟶ y = 7x - 35 + 5

⟶ y = 7x - 30 -- equation (1)

And,

After 5 years father's age will be 3 times the age of the son.

• Son's age after 5 years = (x + 5) years

• father's age after 5 years = (y + 5) years.

So,

⟶ y + 5 = 3(x + 5)

Substitute the value of y from equation(1)

⟶ 7x - 30 + 5 = 3x + 15

⟶ 7x - 3x = 15 + 30 - 5

⟶ 4x = 40

⟶ x = 40/4

⟶ x = 10 years

Substitute the value of x in equation (1).

⟶ y = 7(10) - 30

⟶ y = 70 - 30

⟶ y = 40 years

∴

• Son's present age = x = 10 years.

• Father's present age = y = 40 years.

Answer 2

Step-by-step explanation:

★Question★ :-

5 years ago , Father's age was 7 times the age of his son. 5 years later, the Father's age will be 3 time the age of his son. Find their present ages.

★Answer★ :-

Let the father present age =x years and son present age =y years

According to the first condition

5 years later father will be (x+5) years old and son will be (y+5) years old.

⇒x+5=3(y+5)

⇒x+5=3y+15⇒x=3y+10years (equation 1)

According to the second condition

5 years ago father was (x−5) years old and son was (y−5) years old.

⇒x−5=7(y−5)

⇒x−5=7y−35

⇒x−7y=−30years (equation 2)

Put the value of x from eq1

⇒3y+10−7y=−30⇒−4y=−40⇒y=10

Put y=10 in (equation 1)

⇒x=3×10+10=40

Hence, Father's present age =40 years

Son's present age =10 years