Question

Two pans of a balance are 22.4 cm apart.
The fulcrum of the balance has been shifted
1.01 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 %. Don't round answer.

Answer 1

$$Wd=W'd'$$

$$W(22.4-1.01)=W'(22.4+1.01)$$

$$W'=0.9137W$$

Now we can find the percentage change as

$$\frac{W-W'}{W}100\%=\frac{W-0.9137W}{W}100\%=8.63\%$$