Q. 424.1( 10 Votes )

Two years ago, a man's age was three times the square of his son's age. In three years’ time, his age will be four times his son's age. Find their present ages.

Answer :

Let son’s age 2 years ago be x years, Then

man’s age 2 years ago be 3x2 years


son’s present age = (x + 2) years


man’s present age = (3x2 + 2)years


In three years’ time :


son’s age = (x + 2 + 3) = (x + 5) years


man’s age = (3x2 + 2 + 3)years = (3x2 + 5) years


According to question


Man’s age = 4 son’s age


3x2 + 5 = 4(x + 5)


3x2 + 5 = 4x + 20


3x2 – 4x – 15 = 0


Using the splitting middle term – the middle term of the general equation is divided in two such values that:


Product = a.c


For the given equation a = 3 b = – 4 c = – 15


= 3. – 15 = – 45


And either of their sum or difference = b


= – 4


Thus the two terms are – 9 and 5


Difference = – 9 + 5 = – 4


Product = – 9.5 = – 45


3x2 – 9x + 5x – 15 = 0


3x(x – 3) + 5(x – 3) = 0


(x – 3) (3x + 5) = 0


(x – 3) = 0 or (3x + 5) = 0


x = 3 or x = – 5/3 (age cannot be negative)


x = 3


son’s present age = (3 + 2) = 5years


man’s present age = (3.32 + 2) = 29years


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