Two pans of a balance are 68 cm apart.
The fulcrum of the balance has been shifted 0.897 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 %.
To solve this problem, we can use the concept of lever arms in a balance. The distance between the pans of the balance is given as 68 cm.
Let's assume that the true weight of the goods on one side of the balance is W kg. The lever arm for this weight is the distance from the center to the fulcrum, which is 0.897 cm.
Since the balance is in equilibrium, the sum of the moments on both sides of the balance should be zero. The moment is calculated by multiplying the weight by the lever arm. Therefore, we can write:
W * 0.897 = (100 – P) * W * (68 - 0.897),
where P is the percentage markup by the shopkeeper.
Simplifying the equation, we get:
0.897 = 67.103 * (100 – P),
Dividing both sides by 67.103, we have:
0.01338 = 100 – P,
Rearranging the equation to solve for P, we find:
P = 100 – 0.01338,
Calculating P, we get:
P ≈ 99.98662.
Therefore, the true weight of the goods is being marked up by approximately 99.98662%.