 Solution

For two equal lenses with a virtual intermediate object, use positive distances for real objects on a lens's incident side and real images on its outgoing side. The first lens gives $1/v_1=1/f-1/u_1=1/30-1/60$, hence $v_1=60\mathrm{cm}$. At the second lens these rays are already converging toward a point $50\mathrm{cm}$ to its right: its object distance is therefore $u_2=-50\mathrm{cm}$.Thus the final image is real, $28.75\mathrm{cm}$ to the right of lens 1. Its transverse magnification is $(-60/60)(-18.75/[-50])=-3/8$: it is inverted. The intermediate focus is not a real object placed before lens 2.