12
$\begingroup$

You have 2 movable weighing scales.
Their least count is 1 gram.
You have tried both to find out your weight, but they give you different answers.
You do not know whether both of them are broken, or only one of them is broken.

You know the problem of a weighing scale is:
If your actual weight is $A$, then it will show $A+C$.  ($C$ is a constant for that scale, which can be negative.)

Without fixing your scales, how do you determine your weight correctly?

Note:

  • If there is nothing on a scale, it shows no reading (so you cannot determine $C$ by "weighing nothing")
  • You do not know the constant errors ($C$) of your scales
  • You do not have any other things to check your weight
|improve this question
$\endgroup$
  • $\begingroup$ Good question by Jamal Senjaya.... $\endgroup$ – KSR Aug 24 '16 at 4:20
  • 1
    $\begingroup$ If the scale always shows a reading when there is "something" on the scale, one could press on the scale with a finger, varying pressure to determine what the minimum reading is. That reading should be C + 1 g. (Clearly this is not the intended solution, but the wording does not strictly prevent this interpretation. =) $\endgroup$ – Arkku Sep 19 '16 at 14:09
  • $\begingroup$ What does "broken" mean here? $\endgroup$ – Eric Sep 20 '16 at 17:15
15
$\begingroup$

  • Put one scale on top of the other one. Record the weight shown on the bottom scale.
  • Stand on the stack of scales. Record the weight shown on the bottom scale and subtract the previous weight. The constant error cancels out leaving your correct weight.

|improve this answer
$\endgroup$
  • $\begingroup$ I agree with f'' $\endgroup$ – KSR Aug 24 '16 at 4:18
  • $\begingroup$ This method is very useful if you use public weighting scale, but you do not know whether it shows right or wrong result. $\endgroup$ – Jamal Senjaya Aug 24 '16 at 4:53
  • 2
    $\begingroup$ "This method is very useful if you use public weighting scale" - not true. A far more likely form is $\hat A = c_1A + c_2$, ie the result is off by a constant multiplier (a stiffer spring, for instance) as well as an offset $\endgroup$ – Eric Aug 24 '16 at 7:49
  • 1
    $\begingroup$ @Eric Sounds like a nice followup puzzle! $\endgroup$ – David Zhang Aug 24 '16 at 7:56
  • 1
    $\begingroup$ @Eric: Agreed. My problem with the puzzle is the scale shows A + C for a weight A but 0, not C, for no weight. C is therefore either 0 or not a constant. (Obviously C is required to be a constant for the puzzle to work, but this is unrepresentative of any real-world scale.) $\endgroup$ – paolo Sep 20 '16 at 12:59
4
$\begingroup$

  • Check your weight on first scale with a result of A (your actual weight) + C1 (constant error of first scale)

  • Do the same for second scale with a result of A + C2 (constant error of second scale)

  • Stand on both scales (one leg on each) to get the result of A + C1 + C2

  • Your actual weight is the sum of the readings of each scale minus the reading from "both scales together" X + Y - Both = A+C1 + A+C2 - (A+C1+C2)

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.