= 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 $u_2=-50\,\mathrm{cm}$.
$$
\frac1{v_2}=\frac1{30}-\frac1{-50}=\frac4{75}\,\mathrm{cm^{-1}},\qquad
\boxed{v_2=18.75\,\mathrm{cm}\text{ to the right of lens 2}.}
$$
Thus \b[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.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1-ia2010-lenses.png]
{title=Two converging lenses and a real image}
{description=Original paraxial ray construction with an arbitrary small object height. Two rays meet at the calculated inverted final image; dashed continuations show where lens 1 alone would have focused them.}
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