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Five Years Hence, a Man’S Age Will Be Three Times the Sum of the Ages of His Son. Five Years Ago, the Man Was Seven Times as Old as His Son. Find Their Present Ages - Mathematics

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Question

Five years hence, a man’s age will be three times the sum of the ages of his son. Five years ago, the man was seven times as old as his son. Find their present ages

Solution

Let the present age of the man be x years and that of his son be y years.
After 5 years man’s age = x + 5
After 5 years ago son’s age = y + 5
As per the question
x + 5 = 3(y + 5)
⇒ x – 3y = 10                    ……………(i)
5 years ago man’s age = x – 5
5 years ago son’s age = y – 5
As per the question
x – 5 = 7(y – 5)
⇒ x – 7y = -30                           …….(ii)
Subtracting (ii) from (i), we have
4y = 40 ⇒ y = 10
Putting y = 10 in (i), we get
x – 3 × 10 = 10
⇒ x = 10 + 30 = 40

Hence, man’s present age = 40 years and son’s present age = 10 years.

 

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Chapter 3: Linear Equations in two variables - Exercises 4

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RS Aggarwal Mathematics [English] Class 10
Chapter 3 Linear Equations in two variables
Exercises 4 | Q 73

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