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# 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. - ICSE Class 10 - Mathematics

#### Question

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.

#### Solution

Let the age of son 2 years ago be x years.
Then, father’s age 2 years ago = 3x2 years
Present age of son = (x + 2) years
Present age of father = (3x2 + 2) years
3 years hence:
Son’s age = (x + 2 + 3) years = (x + 5) years
Father’s age = (3x2 + 2 + 3) years = (3x2 + 5) years
From the given information,
3x2 + 5 = 4(x + 5)
3x2 – 4x – 15 = 0
3x2 – 9x + 5x – 15 = 0
3x(x – 3) + 5(x – 3) = 0
(x – 3) (3x + 5) = 0
x = 3,
Since, age cannot be negative. So, x = 3.
Present age of son = (x + 2) years = 5 years
Present age of father = (3x2 + 2) years = 29 years

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#### APPEARS IN

Selina Solution for Selina ICSE Concise Mathematics for Class 10 (2018-2019) (2017 to Current)
Chapter 6: Solving (simple) Problems (Based on Quadratic Equations)
Ex.6E | Q: 12

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Solution 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. Concept: Quadratic Equations.
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