Solving the Neighbors' Age Puzzles: A Mathematical Adventure

In today’s article, we solve a mathematical puzzle involving the ages of four neighbors: Jack, Sharon, Alex, and Terence. This puzzle requires us to use algebraic equations to determine the ages of each individual and ultimately identify the youngest among them. Let's dive into the details to unravel the mystery.

Defining Variables

This puzzle introduces us to four neighbors with the following ages:

- Jack: (mathsf{J}) - Sharon: (mathsf{S}) - Alex: (mathsf{A}) - Terence: (mathsf{T} 8)

Setting Up Equations

From the information given, we can set up the following equations:

Jack is half as old as Sharon: (mathsf{J} frac{1}{2} mathsf{S}) Sharon is three years older than Alex: (mathsf{S} mathsf{A} 3) The sum of Alex’s and Sharon’s ages equals 17: (mathsf{A} mathsf{S} 17) Terence is eight: (mathsf{T} 8)

Substituting and Solving

Let's substitute the value of ( mathsf{S} ) from the second equation (S A 3) into the third equation (A S 17):

(mathsf{A} (mathsf{A} 3) 17)

(mathsf{2A} 3 17)

(mathsf{2A} 14)

(mathsf{A} 7)

Now, we can find {S} by substituting ( mathsf{A} ) back into the second equation:

(mathsf{S} mathsf{A} 3 7 3 10)

Next, we can find {J} using the first equation:

(mathsf{J} frac{1}{2} mathsf{S} frac{1}{2} times 10 5)

So, the ages are:

Jack: 5 years Alex: 7 years Sharon: 10 years Terence: 8 years

Conclusion

Comparing the ages, we see that:

Jack: 5 years Alex: 7 years Sharon: 10 years Terence: 8 years

Therefore, Jack is the youngest at 5 years old.