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Given pairs of weights find individual values

Puzzling Asked on May 20, 2021

The problem is as follows:

A kid has five marbles. These marbles have different weights and the
child weighs them in pairs in all possible ways. He records the
weights in his notebook. These are the results: 10g, 12g, 13g, 14g,
15g, 16g, 17g, 18g, 20g and 21g. Using this information, what is the
weight of the lightest marble?

The choices given in my book are as follows:

  1. 4g
  2. 3g
  3. 2g
  4. 5g

For reference I found this problem in my collection of puzzles book Reason and Logic. From the style I believe it is an adaptation from the contents found in Martin Gardner’s 50’s book on Recreational Puzzles.

I’m having trouble with accounting for the weights being paired.

So far the only thing I could noticed is that if I were in that situation I would label the marbles as:

A B C D E

All the combinations without repetitions (which I’m assuming is the intended meaning) would be:

AB, AC, AD, AE, BC, BD, BE, CD, CE, DE

which indicates the 10 pairs given in the problem. But that’s it. I don’t know if this can be used to get an answer.

Does an easier way to make some equations or get an answer exist?

Another way would be to build a set of 10 equations with 10 unknowns. But I don’t think that would be the intended method of solution. Even if such set is made, which would correspond to which weight?

Can someone help me here? Does a way to simplify this situation exist?

Please only give detailed, step-by-step solutions. No matter how I look at this question, I get tangled with equations.

4 Answers

Here is my brute-force method for those who are interested in brute-force solutions (only takes two seconds to return the output)!

Output:

Answered by risky mysteries on May 20, 2021

Using the corrected weights:

Answered by Bubbler on May 20, 2021

I say the minimum weight is 4.

Since we have five different weights and we take two at a time, then if we apply the formula of combinatorics with no repetitions, the number of combinations is ten. Since the question assigns the weights the letters A,B,C,D,E we obtain the following ten combinations.

AB  BC   CD   DE

AC  BD   CE 

AD  BE

AE

Let AB be the minimum, AB=10, and DE=21 the maximum. If we set A=4 then, according to the given facts, we have

AB 10-4=6 so B=6

AC 12-4=8 so C=8

AD 13-4=9 so D=9

DE 21-9=12 SO E=12

So from the weights 4, 6, 8, 9, 12 we can obtain all ten combinations.

Answered by Vassilis Parassidis on May 20, 2021

Answered by Retudin on May 20, 2021

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