Neon is mixed with carbon dioxide. Treat carbon dioxide as rigid…

Step 1. The $C_v$ needed. $C_p = C_v + R$, so $\gamma = 1 + \dfrac{R}{C_v}$. We want $C_v$, so $C_v = \dfrac{R}{\gamma - 1} = \dfrac{R}{0.60} = \tfrac53 R = 13.9\ \text{J/(mol K)}$.

Step 2. The mixture. With $x$ moles of neon and $1$ mole of carbon dioxide, $C_v = \dfrac{x \cdot \tfrac32 R + 1 \cdot \tfrac52 R}{x + 1}$.

Step 3. Solve. $1.5x + 2.5 = \tfrac53 (x + 1)$. Multiply by 6: $9x + 15 = 10x + 10$, so $x = 5.0$.