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KCC's Quizzes AQQ292 about weighing using 2-plates scale

1. First, the quote of the week: "Some people create their own storms and then get mad when it rains" - Unknown

                                             

 2. New challenge with AQQ292 about using a two-plates scale.

This is a kind proposal from a frequent participant: Herman Neufeld, Analog & Mixed Signals Specialist):

You are asked to design a two-dish scale that weighs amounts of a substance from 1 to 40 grams in integer amounts.

Hence your task is to determine:

  1. What is the minimum number counterweights you need?
  2. What are the weigh values of these counterweights?

These, in order to weigh this substance,

Good luck and try to be among the first ones!

P.S. Please forward such quizzes to friends and colleagues who want also to "relax" their brains...

Please share your answer to view other submitted answers
Parents
  • I hope that the weights can also be placed on a dish with a weighed object.
    The weights must have an odd value, so their sum can reach both odd and even numbers. The first two are definitely 1g and 3g. The first higher value above their sum is 5g, so this can be achieved by the combination 9g-3g-1g. And again the first highest unattainable for these 3 weights is the value 1g+3g+9g+1g = 14g, so the next weight value is these 14g and the sum of everything = 27g.

    The question is whether we can create all the combinations now. I don't have to examine even numbers and the same goes for the exact weight values.


    • 5g +3g+1g= 9g
    • 7g +3g = 9g+1g
    • 11g +1g = 9g+3g
    • 13g 9g+3g+1g
    • 15g +3g+9g = 27g
    • 17g +1g+9g = 27g
    • 19g +9g = 27g+1g
    • .
    • .
    • 39g = 27g+9g+3g

      All combinations are available. Thus we need 4 counterweights: 1g, 3g, 9g, 27g
Reply
  • I hope that the weights can also be placed on a dish with a weighed object.
    The weights must have an odd value, so their sum can reach both odd and even numbers. The first two are definitely 1g and 3g. The first higher value above their sum is 5g, so this can be achieved by the combination 9g-3g-1g. And again the first highest unattainable for these 3 weights is the value 1g+3g+9g+1g = 14g, so the next weight value is these 14g and the sum of everything = 27g.

    The question is whether we can create all the combinations now. I don't have to examine even numbers and the same goes for the exact weight values.


    • 5g +3g+1g= 9g
    • 7g +3g = 9g+1g
    • 11g +1g = 9g+3g
    • 13g 9g+3g+1g
    • 15g +3g+9g = 27g
    • 17g +1g+9g = 27g
    • 19g +9g = 27g+1g
    • .
    • .
    • 39g = 27g+9g+3g

      All combinations are available. Thus we need 4 counterweights: 1g, 3g, 9g, 27g
Children