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Two pans of a balance are 46.3 cm apart. The fulcrum of the balance has been shifted 0.633 away from the center by a dishonest shopkeeper.

By what percentage is the true weight of the goods being marked up by the shopkeeper?
Answer in units of %

Respuesta :

distance of each pan from the center or fulcrum is given as

[tex]r = 23.15 cm[/tex]

now if dishonest shopkeeper shifted it by 0.633 cm from center

so distance on each side is given as

[tex]d_1 = 23.15 - 0.633 = 22.52 cm[/tex]

[tex]d_2 = 23.15 + 0.633 = 23.78 cm[/tex]

now the weight is balance as

[tex]W_1d_1 = W_2d_2[/tex]

[tex]W(22.52) = W_2(23.78)[/tex]

now we will have

[tex]W_2 = 0.95W[/tex]

now we can find the percentage change as

[tex]percentage = \frac{W - W_2}{W} \times 100[/tex]

[tex]percentage = 5%[/tex]

abiolataiwo2015

The percentage that is the true weight of the goods being marked up by the shopkeeper is 2.70%.

Percentage change:

Wd=W′d ′

W(46.3-0.633)=W'(46.3+0.633)

W45.667=W'46.933

W'=45.667/46.933

W'=0.9730W

Now let find the percentage change as:

W-W'÷W 100%

=W-0.9730W÷W 100%

Percentage change=100%-97.30%

Percentage change=2.70%

Inconclusion the percentage that is the true weight of the goods being marked up by the shopkeeper is 2.70%.

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

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