If the average age of 4 children is 10 years old then at least one children is at least 10 years old.
I am unsure how to prove this problem with the approach of contradiction or contrapositive. I have done problems involving math but this one is different and I am lost on how to prove it.
On
Theorem: Let $a_i \in \mathbb{R}$ for $1 \leq i \leq n$. If $$\frac{1}{n} \sum_{i=1}^{n} a_i = k$$ then there exists $i$, $1 \leq i \leq n$, such that $a_i \geq k$.
Proof by contrapositive: Suppose $a_i < k$ for all $1 \leq i \leq n$. Then $$\sum_{i = 1}^{n} a_i < n \cdot k.$$ Divide by $n$ for the result.
Intuitively, this says that if you take the $\textit{average}$ of some finite list of numbers, there was at least one number on the list bigger than or equal to the average. You might notice this is slightly more general than your explicit question, but if you solve your question it generalizes to this proof immediately.
Now you know that if the average age of $\textit{any}$ number of children is $\textit{any number}$, at least one child was older than or equal to that average.
If all four are less than 10 years old then sum of ages is less than 40, so average is less than $40/4=10.$