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# One Year Ago, a Man Was 8 Times as Old as His Son. Now His Age is Equal to the Square of His Son’S Age. Find Their Present Ages. - ICSE Class 10 - Mathematics

#### Question

One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.

#### Solution

Let the present age of the son be x years.
∴ Present age of man = x2 years
One year ago,
Son’s age = (x – 1) years
Man’s age = (x2 – 1) years
It is given that one year ago; a man was 8 times as old as his son.
∴ (x2 – 1) = 8(x – 1)
x2 – 8x – 1 + 8 = 0
x2 – 8x + 7 = 0
(x – 7) (x – 1) = 0
x = 7, 1
If x = 1, then x2 = 1, which is not possible as father’s age cannot be equal to son’s age.
So, x = 7.
Present age of son = x years = 7 years
Present age of man = x2 years = 49 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.6D | Q: 5

#### Video TutorialsVIEW ALL [5]

Solution One Year Ago, a Man Was 8 Times as Old as His Son. Now His Age is Equal to the Square of His Son’S Age. Find Their Present Ages. Concept: Quadratic Equations.
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