ANSWER
3. 5.93432
Step-by-step explanation
The pans are 67.6cm apart. If the fulcrum was centered, the distance from each pan to the fulcrum would be,
![\frac{67.6\operatorname{cm}}{2}=33.8\operatorname{cm}]()
Now, the fulcrum was moved 0.974cm towards one of the pans,
The new distances are,
![\begin{gathered} d_1=33.8cm+0.974\operatorname{cm}=34.774cm \\ d_2=33.8cm-0.974\operatorname{cm}=32.826cm \end{gathered}]()
When the shopkeeper weighs the products, the balance is in equilibrium. This means that the torque each weight produces must be equal,

The relationship between the forces is,

Replace with the distances,
![\begin{gathered} F_2=\frac{34.774\operatorname{cm}}{32.826\operatorname{cm}}\cdot F_1 \\ \\ F_2=1.0593432F_1 \end{gathered}]()
Now we have to find the percent change,

Hence, the true weight of the goods is being marked up by 5.93432%.