$(a)\;x=15\;cm \\ (b)\;x=10\;cm \\ (c)\;x=20\;cm \\ (d)\;x=-25\;cm $

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Linear magnification $m= \large\frac{-2 \;cm}{1\;cm}$$=-2$

ie image is real and inverted $|v|=2 |u|$

Let $|u|=x$ then $|v|=2x$

Now $|u|+|v|=50-(-40)=90$

$\therefore x+2x=90\;or\;x=30\;cm$

Hence distance of object from the lens is $30\;cm$ and of object is $60\;cm$.

$\therefore$ lens should be located at $x=-10\;cm$ as shown.

Hence b is the correct answer.

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