Math, asked by shreeyas, 3 months ago

Two years ago, father 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.

Answers

Answered by Sheeman
3

Answer:

38 years

Step-by-step explanation:

Let us consider

Son’s age = X

Two years ago son’s age = X – 2

His Father’s age at that time = 3(X – 2)

Present age of Father = 3X – 6 + 2

The present age of Father = 3X – 4

Two years hence father’s age = 3X – 4 + 2

Two years hence father’s age = 3X – 2

Two years hence son’s age = X + 2

Given:

5 × (X + 2) = 2 × (3X – 2)

5X + 10 = 6X – 4

X = 14

Son’s present age X = 14 years.

Father’s present age = (3 × 14) – 4

Father’s present age = 38 years.

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