Question

Five years hence, a father's age will be three times the age of his son. Five years ago, a father was seven times as old as his son. Find their present ages
A. Father’s age = 80 years, Son’s age = 25 years
B. Father’s age = 60 years, Son’s age = 30 years
C.Father’s age = 75 years, Son’s age = 42 years
D. Father’s age = 40 years, Son’s age = 10 years

Answer 1

Hint- In order to find out the present age of man and his son first we will assume the present age of man and son as variable then according to the given statement we will form two equations with two variables and after solving it we will get the answer.

Complete step-by-step solution -
Let the present age of the man be $x$ years
 And his son’s be $y$ years
After 5 years man’s age =$x + 5$
After 5 years ago son’s age = $y + 5$
According to the question
$
   \Rightarrow x + 5 = 3(y + 5) \\
  x - 3y = 10..............(1) \\
$
5 years ago, man’s age = $x - 5$
5 years ago, son’s age=$y - 5$
As per the question
$
   \Rightarrow x - 5 = 7(y - 5) \\
   \Rightarrow x - 7y = - 30.............(2) \\
$
Subtracting (2) from (1), we have
$
   \Rightarrow 4y = 40 \\
   \Rightarrow y = 10 \\
$
Put the value of $y$ in Eq (1)
$
   \Rightarrow x - 3y = 10 \\
   \Rightarrow x - 3 \times 10 = 10 \\
   \Rightarrow x = 10 + 30 \\
  x = 40 \\
$
Hence, man’s present age = 40 years
And son’s present age = 10 years.

Note- In order to solve problems related to time and ages students must start the problem by considering the present age of the people as some unknown variable and must find the age in past and future in terms of the same unknown variable. If the students consider the age in the past or future as unknown, the problem becomes more difficult.