Let Amy's, Ben's, and Chris's ages be $a$, $b$, and $c$, respectively. We have the equations \begin{align*} \tag{1}
\frac{a+b+c}{3}=6 \Rightarrow a+b+c&=18 \\ \tag{2}
c-4&=a\\ \tag{3}
b+4&=\frac{3}{5}(a+4)
\end{align*} From Equation (3), we have $b=\frac{3}{5}(a+4)-4$. We substitute Equation (2) into Equation (3) to eliminate $a$, to get $b=\frac{3}{5}(c)-4$. Substituting this last equation and Equation (2) into Equation (1) to eliminate $a$ and $b$, we have  \[[c-4]+[\frac{3}{5}(c)-4]+c=18\] Solving for $c$, we find that $c=10$. Thus, Chris's age is $\boxed{10}$.
Question:
The average of Amy's, Ben's, and Chris's ages is 6. Four years ago,  Chris was the same age as Amy is now. In four years, Ben's age will be $\frac{3}{5}$ of Amy's age at that time. How many years old is Chris now?
Answer:
Let's think step by step.

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10
The average of Amy's, Ben's, and Chris's ages is 6. Four years ago,  Chris was the same age as Amy is now. In four years, Ben's age will be $\frac{3}{5}$ of Amy's age at that time. How many years old is Chris now?
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