 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Faulty Scales (Posted on 2004-05-23) Good scales have equal arms (arms are the the things that connect the actual scale to the center), but in one grocery stall, the arms of the scale are not equal. Pending replacement, the manager wonders if he can give correct weight this way:

"I'll balance a 1-pound weight on the left with sugar on the right, and then I'll balance the 1-pound weight on the right with some more sugar on the left, and the sugar will add up to exactly 2 pounds."

Will it? What are other (assuming that the above works, it may not) ways of weighing 2 pounds of sugar, if you also have a lead shot with you to help weigh? (Note and hint: The lead shot has an unknown weight. You can make it whatever weight you choose. Remember, the arms aren't equal, and you need 2 pounds of sugar.)

 See The Solution Submitted by Victor Zapana Rating: 4.2500 (4 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Error factor | Comment 3 of 8 | Continuing from "First part", the grocer makes 2 weighings:
the 2 weighings result in
w1=a,  and w2=1/a  where the long side is L, the short side is aL
let e be the error
w1 + w2 = 2 + e
a + 1/a = 2 + e
a - 2 + 1/a = e
(a^2 - 2a + 1)/a = e
(a-1)^2 / a = e

So the error can be quantified if "a" is known.  If W pounds is the goal, rather than 2 pounds, the the grocer's method will overestimate by:
W*(a-1)^2 / 2a

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By the way, I have a question:
For the second part of the problem, are we allowed to do any callibration?  For example, do we have 2 different weights that are known to be 1 pound?  If so, then we could find the proper balance point by putting a 1 pound weight on each arm.  The one reading heavier is the longer arm.  We could shorten the effective length of the longer arm either by bending it or by moving the weight closer to the fulcrum.

 Posted by Larry on 2004-05-23 10:33:30 Please log in:

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