Question

After five years, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer 1

Hint: Assume Jacob’s and his son’s present age to be unknown quantities (in years). Now use the two given conditions given in the question to form two simultaneous linear equations. Solve these two equations either by the method of elimination or by the method of substitution. Hence, the present ages of Jacob and his son are obtained.

Complete step-by-step answer:
Let us assume the present age of Jacob be x years and the present age of Jacob’s son be y years.
Now, it is given that 5 years from now, Jacob’s age will be three times of his son. Five years hence, Jacob’s age will be (x + 5) years and that of his son will be (y + 5) years. This condition can be formulated as:
$\begin{align}
  & \text{x + 5 = 3}\left( \text{y + 5} \right) \\
 & \Rightarrow \ \text{x + 5 = 3y + 15} \\
 & \therefore \text{ x }-\text{ 3y = 10 }....\text{(i)} \\
\end{align}$
Again, it is given that 5 years ago, Jacob’s age was 7 times of that of his son. Five years ago, Jacob’s age will be (x – 5) years and that of his son will be (y – 5) years. This condition can be formulated as:
$\begin{align}
  & \text{x }-\text{ 5 = 7}\left( \text{y }-\text{ 5} \right) \\
 & \Rightarrow \ \text{x }-\text{ 5 = 7y }-\text{ 35} \\
 & \therefore \text{ x }-\text{ 7y = }-3\text{0 }....\text{(ii)} \\
\end{align}$
Thus, we can find the values of x and y, using the method of substitution.
Now, in the equation (i), we express the variable x in terms of y as follows,
$\text{x = 3y + 10 }....\text{(iii)}$
Putting this value of x in the equation, we get,
$\begin{align}
  & 3\text{y + 10 }-\text{ 7y = }-30 \\
 & \Rightarrow \text{ 10 + 30 = 7y }-\text{ 3y} \\
 & \Rightarrow \text{ 4y = 40} \\
 & \therefore \text{ y = }\frac{40}{4}\text{ = 10} \\
\end{align}$
Thus, the present age of Jacob’s son is 10 years.
Putting the value of y = 10 in the equation (iii), we get,
$\begin{align}
  & \therefore \text{ x = $3 \times 10$ + 10} \\
 & \text{ = 40} \\
\end{align}$
Thus, Jacob’s present age is 30 years and his son’s present age is 10 years.

Note: The two simultaneous linear equations can alternatively be solved by the method of elimination. If we subtract equation (ii) from equation (i), then we get,
$\begin{align}
  & \left( \text{x }-\text{ 3y} \right)\text{ }-\text{ }\left( \text{x }-\text{ 7y} \right)\text{ = 10 + 30} \\
 & \Rightarrow \text{ 4y = 40} \\
 & \therefore \text{ y = 10} \\
\end{align}$
Putting the value of y = 10 in equation (i), we get x = 30.
Thus, by this method also, we get the same answers for Jacob and his son’s present ages.