A lens forms 3 times magnified image of an object on screen. When the distance between the screen and the lens is increased by 10 cm and object is also moved to get 5 times magnified image again on the screen. Then

For first case
Image distance = 3 times object distance.
$$\begin{gathered} \therefore \frac{1}{\upsilon }-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{3\upsilon }-\frac{1}{(-u)}=\frac{1}{f} \\ \Rightarrow \frac{1}{3\upsilon }+\frac{1}{u}=\frac{1}{f} \dots \dots \dots \left(i\right) \end{gathered}$$

When distance between screen and lens is increased by 10 cm

$$\begin{gathered} \upsilon =3u+10 \\ \left|u^{\mathbf{|}}\right|=\frac{\upsilon }{5} \\ \therefore \frac{1}{\upsilon }-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{3u+10}-\frac{1}{u^{\mathbf{|}}}=\frac{1}{f} \\ \Rightarrow \frac{1}{3u+10}+\frac{5}{3u+10}=\frac{1}{f} \end{gathered}$$

From equations (i) and (ii)

$$\begin{gathered} \therefore u=\frac{20}{3}cm \\ \therefore f = 5 \end{gathered}$$