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The Sum of Ages of a Man and His Son is 45 Years. Five Years Ago, the Product of Their Ages Was Four Times the Man'S Age at the Time. Find Their Present Ages. - Mathematics

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Question

The sum of ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man's age at the time. Find their present ages.

Solution

Let the present age of the man be x years

Then present age of his son is  (45 - x) years

Five years ago, man’s age = (x - 5) years

And his son’s age (45 - x - 5) = (40 - x) years

Then according to question,

(x - 5)(40 - x) = 4(x - 5)

40x - x2 + 5x - 200 = 4x - 20

-x2 + 45x - 200 = 4x - 20

-x2 + 45x - 200 - 4x + 20 = 0

-x2 + 41x - 180 = 0

x2 - 41x + 180 = 0

x2 - 36x - 5x + 180 = 0

x(x - 36) -5(x - 36) = 0

(x - 36)(x - 5) = 0

So, either 

x - 36 = 0

x = 36

Or

x - 5 = 0

x = 5

But, the father’s age never be 5 years

Therefore, when x = 36 then

45 - x = 45 - 36 = 9

Hence, man’s present age is36 years and his son’s age is 9 years.

  Is there an error in this question or solution?
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 RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 4: Quadratic Equations
Ex. 4.9 | Q: 2 | Page no. 61
 RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 4: Quadratic Equations
Ex. 4.9 | Q: 2 | Page no. 61
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The Sum of Ages of a Man and His Son is 45 Years. Five Years Ago, the Product of Their Ages Was Four Times the Man'S Age at the Time. Find Their Present Ages. Concept: Solutions of Quadratic Equations by Factorization.
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