Exactly 50\% of the X-rays emitted from a Coolidge tube are able to...

The maximum energy of X-ray photons emitted from a Coolidge tube is given by E_{\text\max} = eV, where $V$ is the potential difference between the target and the filament.

When the potential difference $V$ is increased, the energy of the emitted X-rays increases. These higher energy X-rays are known as hard X-rays and have a greater penetrating power.

The intensity of transmitted X-rays is given by $I = I_{0} e^{-\mu x}$, where $\mu$ is the absorption coefficient. As the energy of X-rays increases, the absorption coefficient $\mu$ of the material decreases.

Because $\mu$ decreases, the transmitted fraction $\dfrac{I}{I_{0}}$ increases.

Since initially $50\%$ of the X-rays were transmitted, the new fraction of X-rays penetrating the foil will be $> 50\%$.