contestada

Two pans of a balance are 53 cm apart.
The fulcrum of the balance has been shifted
0.784 cm away from the center by a dishonest
shopkeeper.
By what percentage is the true weight of the
goods being marked up by the shopkeeper?
Assume the balance has negligible mass.
Answer in units of %.

Respuesta :

The percentage of the true weight is 3%.

Percentage change

To calculate the percentage of mass indicated by the balance, use the following expression:

                                                   [tex]W'd' = Wd[/tex]

Applying the values ​​given by the statement we have:

                                      [tex]W(53-0.784) = W'(53+0.784)[/tex]

                                           [tex]52.21 W = 53.78W'[/tex]

                                              [tex]W' = 0.97W[/tex]

Finally, just find the percentage:

                                                    [tex]\frac{W-W'}{W} \times 100[/tex]

                                                  [tex]\frac{W-0.97W}{W} \times 100[/tex]

                                                         = 3%

So, the percentage that is the true weight of the goods being marked up by the shopkeeper is 3.00%.        

Learn more about percentage change in: brainly.com/question/809966

Otras preguntas