Question

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 %

Answer 1

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

$r = 23.15 cm$

now if dishonest shopkeeper shifted it by 0.633 cm from center

so distance on each side is given as

$d_1 = 23.15 - 0.633 = 22.52 cm$

$d_2 = 23.15 + 0.633 = 23.78 cm$

now the weight is balance as

$W_1d_1 = W_2d_2$

$W(22.52) = W_2(23.78)$

now we will have

$W_2 = 0.95W$

now we can find the percentage change as

$percentage = \frac{W - W_2}{W} \times 100$

$percentage = 5%$