Three lenses of focal lengths $+10 \mathrm{cm},-10 \mathrm{cm}$ and $+30 \mathrm{cm}$ are placed at a distance of $30 \mathrm{cm},35 \mathrm{cm}$ and $45 \mathrm{cm}$ respectively from an object. The distance between the object and the image formed is

Refer above diagram,

For lens $A$, $u=-30 \mathrm{cm}$, $f_B=+10 \mathrm{cm}$

The using lens formula, the image distance for lens $A$ is 

$\frac{1}{v_A}-\frac{1}{u_A}=\frac{1}{f_A}$

$\Rightarrow \frac{{\displaystyle 1}}{{\displaystyle v_A}}-\frac{{\displaystyle 1}}{{\displaystyle -30}}=\frac{{\displaystyle 1}}{{\displaystyle +10}}$

The image distance from lens $A$is given by

$\Rightarrow v_A=15 \mathrm{cm}$

For lens $B$, $u=(v_A-5)=(15-5)=10 \mathrm{cm}$ and $f_B=-10 \mathrm{cm}$

Now image distance for lens $B$ is 

$\frac{{\displaystyle 1}}{{\displaystyle v_B}}-\frac{{\displaystyle 1}}{{\displaystyle +10}}=\frac{{\displaystyle 1}}{{\displaystyle -10}}\Rightarrow v_B=\infty$

Similarly, for lens $C$

$u=\infty$ and $f_C=30 \mathrm{cm}$

$\frac{{\displaystyle 1}}{{\displaystyle v_c}}-\frac{{\displaystyle 1}}{{\displaystyle \infty }}=\frac{{\displaystyle 1}}{{\displaystyle 30}}\Rightarrow v_c=30 \mathrm{cm}$

Hence, total distance between object & image is 

$30+45=75 \mathrm{cm}$