Finding the Present Ages of Father and Son

Video Explanation

Question

Ten years ago, a father was twelve times as old as his son. Ten years hence, he will be twice as old as his son will be then. Find their present ages.

Solution

Step 1: Let the Variables

Let present age of father = \(x\) years

Let present age of son = \(y\) years

Step 2: Form the Equations

Ten years ago:

\[ x – 10 = 12(y – 10) \]

\[ x – 10 = 12y – 120 \]

\[ x – 12y = -110 \quad (1) \]

Ten years hence:

\[ x + 10 = 2(y + 10) \]

\[ x + 10 = 2y + 20 \]

\[ x – 2y = 10 \quad (2) \]

Step 3: Solve by Elimination Method

Subtract equation (2) from equation (1):

\[ (x – 12y) – (x – 2y) = -110 – 10 \]

\[ x – 12y – x + 2y = -120 \]

\[ -10y = -120 \]

\[ y = 12 \]

Step 4: Find the Value of x

Substitute \(y = 12\) in equation (2):

\[ x – 2(12) = 10 \]

\[ x – 24 = 10 \]

\[ x = 34 \]

Conclusion

Present age of father:

\[ \boxed{34 \text{ years}} \]

Present age of son:

\[ \boxed{12 \text{ years}} \]