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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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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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