Find the number of copper spheres in a box A box contains a mixture of small copper spheres and small lead spheres. The total volume of both metals is measured by the displacement of water to be $\pu{420 cm^3}$ and the total mass is $\pu{5.4 kg}$. What percentage of the spheres are copper?


I can't get the right answer. I did this:

Given: 

$m_{\text{total}} = 5.4\;\mathrm{kg} =5400\;\mathrm{g}$

$V=420\;\mathrm{cm^3}$

$\rho_{\ce{Cu}} = 8.96\;\mathrm{g} ~\mathrm{cm^-3}$

$V\times{\mathrm{\rho_{\ce{Cu}}}} = 3763.2\;\mathrm{g}$

$\frac{m_{\ce{Cu}}}{m_{\text{total}}}=69.9\%$

Can someone tell me how to approach this problem?
Here is your formulation :

$m_{\ce{Cu}} + m_{\ce{Pb}} = m_{\text{tot}}$

$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}V_{\ce{Pb}} = m_{\text{tot}}$

$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(V_{\text{tot}}-V_{\ce{Cu}}) = m_{\text{tot}}$

$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(420-V_{\ce{Cu}}) = 5400$

$V_{\ce{Cu}}(\rho_{\ce{Cu}}-\rho_{\ce{Pb}}) + 420\rho_{\ce{Pb}}= 5400$

Put values for $\rho_{\mathrm{Cu}}$ and $\rho_{\mathrm{Pb}}$ in the equation, and rest is simple arithmetic.