Question

By what percentage is a true weight of the girls being marked up by the shop keeper?

Answer 1

ANSWER

3. 5.93432

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,

$\begin{gathered} \tau_1=\tau_2 \\ F_1d_1=F_2d_2_{} \end{gathered}$

The relationship between the forces is,

$F_2=\frac{d_1}{d_2}F_1$

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,

$\frac{F_2-F_1}{F_1}\cdot100\%=\frac{1.0563432F_1-F_1}{F_1}\cdot100\%=0.0593432\cdot100\%=5.93432\%$

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