Acme Posted February 12, 2014 Share Posted February 12, 2014 Hope this is new here; didn't see it scanning the category.5 pails of water are weighed two at a time in each possible pairing producing the following list. What is the weight, in pounds, of each of the 5 pails of water? 110 lbs.112 lbs.113 lbs.114 lbs.115 lbs.116 lbs.117 lbs.118 lbs.120 lbs.121 lbs. Link to comment Share on other sites More sharing options...
imatfaal Posted February 13, 2014 Share Posted February 13, 2014 54,56,58,59,62 I really hope that's right method below you can set up a system of 5 simultaneous equations with 5 unknowns - which is solvable weights from x1->x5 getting heavier x1+x2 = 110 x4+x5 = 121 x3 = x4-1 x3 = x2+2 4(x1+x2+x3+x4+x5) = 1156 2 Link to comment Share on other sites More sharing options...
michel123456 Posted February 13, 2014 Share Posted February 13, 2014 (edited) 54,56,58,59,62 I really hope that's right method below you can set up a system of 5 simultaneous equations with 5 unknowns - which is solvable weights from x1->x5 getting heavier x1+x2 = 110 x4+x5 = 121 x3 = x4-1 x3 = x2+2 4(x1+x2+x3+x4+x5) = 1156 I can find x3 but where do x3 = x4-1 come from? --------------- (edit) otherwise I get the same result. Edited February 13, 2014 by michel123456 Link to comment Share on other sites More sharing options...
Acme Posted February 13, 2014 Author Share Posted February 13, 2014 54,56,58,59,62 I really hope that's right method below you can set up a system of 5 simultaneous equations with 5 unknowns - which is solvable weights from x1->x5 getting heavier x1+x2 = 110 x4+x5 = 121 x3 = x4-1 x3 = x2+2 4(x1+x2+x3+x4+x5) = 1156 Correct! Implicit in your approach is that one has to realize the lightest pair-weight must be from the lightest 2 pails and the heaviest pair-weight from the heaviest 2 pails. Nice work and thanks for playing. Link to comment Share on other sites More sharing options...
michel123456 Posted February 13, 2014 Share Posted February 13, 2014 Correct! Implicit in your approach is that one has to realize the lightest pair-weight must be from the lightest 2 pails and the heaviest pair-weight from the heaviest 2 pails. Nice work and thanks for playing. Yes, that helps to find x3 (I made an approach with abcde, finding c) But i still miss imatfaal's next steps.(though he seems to get it right). ? x3 = x4-1 ? ? x3 = x2+2 ? Link to comment Share on other sites More sharing options...
Acme Posted February 14, 2014 Author Share Posted February 14, 2014 Yes, that helps to find x3 (I made an approach with abcde, finding c) But i still miss imatfaal's next steps.(though he seems to get it right). ? x3 = x4-1 ? ? x3 = x2+2 ? I better let imatfaal make his own clarification as I took a slightly different approach. So what I did was with C found to be 58 I then reasoned since A & B are the smallest two and D & E the largest two, then C is the middle weight. So any pairing to give 121 has to have both terms greater than 58. There are only 2 possibilities: 60+61 & 59+62. But, if you let E=61, then E + C = 119 and that weight's not on the list, so the answer is D=59 and E=62. Similarly I found A & B. Hope that's helpful; thanks for playing. Link to comment Share on other sites More sharing options...
Mayflow Posted March 31, 2014 Share Posted March 31, 2014 Hope this is new here; didn't see it scanning the category. 5 pails of water are weighed two at a time in each possible pairing producing the following list. What is the weight, in pounds, of each of the 5 pails of water? 110 lbs. 112 lbs. 113 lbs. 114 lbs. 115 lbs. 116 lbs. 117 lbs. 118 lbs. 120 lbs. 121 lbs. It doesn't seem to be a very good test. Do all the pails weigh the same, and also what is the weight of the various pails when empty? Also how can 5 be evenly divided by 2? Link to comment Share on other sites More sharing options...
Acme Posted March 31, 2014 Author Share Posted March 31, 2014 It doesn't seem to be a very good test. Do all the pails weigh the same, and also what is the weight of the various pails when empty? Also how can 5 be evenly divided by 2? Yes; all pails weigh the same. In effect the weight of the pails is not germane to the solution, whether they are full or empty. Substitute 5 rocks if you like. As to 'how can 5 be evenly divided by 2?' question, there are no fractions to figure out in the problem. [still, 5 does divide evenly by 2. 5/2=2.5] Link to comment Share on other sites More sharing options...
imatfaal Posted April 1, 2014 Share Posted April 1, 2014 It doesn't seem to be a very good test. Do all the pails weigh the same, and also what is the weight of the various pails when empty? Also how can 5 be evenly divided by 2? 5 pails of water are weighed two at a time in each possible pairing producing the following list. It is a perfectly good test 5 pails named a,b,c,d, and e weigh a+b, a+c, a+d, a+e, b+c, b+d, b+e, c+d, c+e, d+e Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now