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anonymous
 one year ago
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anonymous
 one year ago
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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Interesting question, give me some minutes to look into it. Are you sure that's all there is to it ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's quite simple really.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let b15 be the 5 books in question and v110 be the values of each 10 weightings (I'm too lazy to write them down  v1 = 108, v2=112,v3=113 and so forth).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let's write a system of equations now: b1+b2=v1 b1+b3=v2 b1+b4=v3 b1+b5=v4 b2+b3=v5 b2+b4=v6 b2+b5=v7 b3+b4=v8 b3+b5=v9 b4+b5=v10

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Scratch that in the definition, b15 represents the weight of each of those 5 books. Regardless, now we have a system with 10 equations and 10 unknown variables which should be solved easily.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.05C2=5!/(2! * (52)!)=5!/(2! *3!)=10 yes, just checking that there are ten possible combos:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Easiest way to do it is to take 3 equations out of that bunch that use exactly 3 different variables. So let's say, these ones: b1+b2=v1 b1+b3=v2 b2+b3=v5 See ? 3 equations that only use 3 different variables. The system is solving itself now. Add the first two together: 2b1+b2+b3=v1+v2 And replace b2+b3 from the third equation 2b1+v5=v1+v2 And now you have b1=(v1+v2v5)/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, didn't know that it was supposed to be a system of equations problem! I was almost prepared to do it manually) I think I kind of get the jist of it now. Thank you so much!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And from then onward, you can get b2 out of the first equation, and then b3 out of the second one and so forth.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, that's the beauty of mathematics! I didn't know either but once you start writing down the question in mathematical language things get so much clearer. The beauty of it is that there may very well be another way of solving it that we haven't figured yet. For the time being, this will do though.
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