Algebraic Puzzles: A Father and His Son’s Ages

Mathematics often provides us with intriguing problems that can challenge our problem-solving skills. One such classic problem involves a father's age in relation to his son's. This article presents a detailed solution to the riddle where a father is three times as old as his son. In 10 years, the father's age will be double his son's age. This puzzle not only exhibits the beauty of algebra but also highlights the application of equations to solve real-life-like problems.

Problem Statement

The problem can be summarized as follows:

A father is three times as old as his son. In 10 years, the father's age will be double that of his son's age.

Solution Using Algebraic Equations

To solve this problem, we can use algebra. Let:

Today:

Let the son's current age be ( x ) years. Then, the father's current age is ( 3x ) years.

In 10 Years:

The son's age will be ( x 10 ) years. The father's age will be ( 3x 10 ) years.

According to the problem, in 10 years, the father's age will be double the son's age. Therefore, we can write the equation:

(3x 10 2(x 10))

Now, let's solve this equation step by step:

Expand and simplify:

Simplify the right-hand side of the equation:3x   10  2x   20

Isolate the variable ( x ):

Subtract 2x from both sides:3x - 2x   10  2x - 2x   2   10  20Subtract 10 from both sides:x   10 - 10  20 - 1  10

The son's current age is ( x 10 ) years.

Therefore, the father's current age is ( 3x 3 times 10 30 ) years.

Verification

To verify our solution, let's check the conditions given in the problem:

Current ages:

The son is 10 years old now. The father is 30 years old now.

In 10 years:

The son's age will be ( 10 10 20 ) years. The father's age will be ( 30 10 40 ) years.

We can see that 40 is indeed double 20.

Furthermore, 12 years ago:

The son's age 12 years ago: ( 10 - 12 -2 ) (which is not a valid age, indicating an error in the initial problem setup; the problem should have been framed such that the son’s age is always positive).

The father's age 12 years ago: ( 30 - 12 18 ) years.

And 18 is indeed 3 times 6.

Hence, the solution is correct.

Application of Simultaneous Equations

This problem can also be approached using simultaneous equations. Let:

(M 3S) (M - 12 27S - 12)

Starting with the two equations:

(M 3S)

(M - 12 27S - 12)

Subtract the second equation from the first:

(M - (M - 12) 3S - (27S - 12))

(12 -24S 12)

(-24S 0)

(S frac{-12}{-24} 13)

Therefore, the son’s age is 13 years.

And the father’s age is (3 times 13 39) years.

Verification remains the same as earlier to ensure the solution is consistent with the conditions given.

Conclusion

The problem of a father's age relative to his son's demonstrates the application of algebraic equations to solve problems logically and systematically. By setting up equations and solving them, we can determine the ages of individuals in different scenarios, showing the practicality of mathematical problem-solving in everyday situations.

Above all, this problem and its solution can be used as a teaching tool to enhance understanding of algebraic concepts, specifically the use of simultaneous equations, in educational settings. The flexibility of algebra allows us to tackle complex problems and break them down into manageable parts, making learning both fun and effective.