Question

10 year ago, a fathes was 12 time as old as his
Son and lo, year here, he will be twice
as old as his son will be then find their
present age!​

Answer 1

Correct question :-

→ Ten years ago, a father was 12 times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages.

Answer :-

･ Father's present age = 34

･ Son's present age = 12

Solution :-

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

Case - I :

10 years ago,

Father's age = x - 10

Son's age = y - 10

According to the given condition,

Father was 12 times as old as his son.

⇒x - 10 = 12(y - 10)

⇒x - 10 = 12y - 120

⇒x = 12y - 120 + 10

⇒x = 12y - 110 .... eq. i.)

Case - II :

10 years hence,

Father's age = x + 10

Son's age = y + 10

According to the given condition,

Father will be twice as old as his son

⇒x + 10 = 2(y + 10)

⇒x + 10 = 2y + 20

⇒x = 2y + 20 - 10

⇒x = 2y + 10

Substituting the value of x from eq.i.)

⇒12y - 110 = 2y + 10

⇒12y - 2y = 10 + 110

⇒10y = 120

$\tt{ => y = \dfrac{120}{10}}$

⇒y = 12

⇒Son's present age = 12

Now, finding the present age of father by substituting the value of y in eq.i.)

⇒x = 12y - 110

⇒x = 12 × 12 - 110

⇒x = 144 - 110

⇒x = 34

⇒Father's present age = 34

More questions to solve :-

1. Suppose there are two monuments. Monument A is thrice as old as

monument B. After 12 years, monument A will be twice as old as monument B.

How many years have passed since the construction of each monument?

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2. The denominator of a fraction is greater than it's numerator by 12. If the numerator is decrease by 2 and the denominator is increased by 7 the new fraction is equivalent with 1/2. Find the fraction

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