First balance this scale by moving some or all of the weights from one (only one) of the piles from the left side to the right side at the point indicated with a question mark. Then calculate the weight on each side.
a. From which pile did you move the weights?
b. What is the weight on each side after balancing?
The degree that a given weight tends to pull down the left side is proportional to the product of the weight and its distance from the fulcrum. (You know this intuitively when playing on a see-saw with someone whose weight is different from yours -- the heavier person has to sit closer to the fulcrum so that the product of each person's weight and their distance from the fulcrum equals that of the other person.)
Note that "A" is 7 units from the fulcrum. "B" is 6 units, and "C" is 4 units. So the "A" weight is pushing down with 8*7 = 56 units of rotational force (properly called "torque"), the "B" weight is pushing down with 5*6=30 units of torque, and the "C" weight is pushing down with 15*4=60 units of torque.
On the right side, the 10-weight that's three units from the fulcrum is pushing down with 30 units of torque, and the 4-weight that's 7 units from the fulcrum is pushing down with 28 units of torque.
To see the effect of moving a weight, cover it with your hand, and look at the remaining torque on the left side (the top numbers). Then add four times the weight to the 58 known units of torque on the right side (because the question mark is four units from the fulcrum).
So if the "A" weight is moved, 90 units of torque will be left on the left, and it adds 8*4=32 units of torque to the 58 already on the right side, for a total of 90 units on the right side. So the scale will balance if A is moved in its entirety.
For completeness, let's see if the other weights will also balance the scale.
If "B" is moved, 116 units will remain on the left side, and 5*4=20 units will be added to the right, for a total of 78 units on the right side. Even if partial movement of 1, 2, 3, or 4 of the weights from B are allowed, the two sides are never even.
If "C" is moved, 86 units will remain on the left side, and 15*4=60 units will be added to the right, for a total of 118 units on the right side. But if only 11 of the 15 weights are moved from "B" to "?", then there will be 102 units of torque on each side.
So there are two answers...
The first answer:
The second answer:
Either answer should be counted as correct.
See the NEXT puzzle in the Academic Decathlon 2004 Logic Quiz
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