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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. - Mathematics

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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 Concise Mathematics for Class 10 ICSE (2020 (Latest))
Chapter 6: Solving (simple) Problems (Based on Quadratic Equations)
Exercise 6(D) | Q: 5 | Page no. 78
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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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