A father's age is three times more than his son's age. After 8 years, the father's age will be 2.5 times his son's age, and after 7 years, how many times will the father's age be his son's age?

Explanation

Let, the son's age = $x$ years. $\therefore$ Father's age = $x + 3x = 4x$ years Son's age after 8 years = $x + 8$ Father's age after 8 years = $4x + 8$ According to the question $$4x + 8 = 2.5 \times (x + 8)$$ $$4x + 8 = 2.5x + 20$$ $$4x - 2.5x = 20 - 8$$ $$1.5x = 12$$ $$x = \frac{12}{1.5}$$ $$x = 8 years$$ Son's age after 7 years = $x + 7 = 8 + 7 = 15$ years Father's age after 7 years = $4x + 7 = 4 \times 8 + 7 = 39$ years $\therefore$ Required times = $\frac{39}{15}$