Acme Posted February 12, 2014 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.
imatfaal Posted February 13, 2014 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
michel123456 Posted February 13, 2014 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
Acme Posted February 13, 2014 Author 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.
michel123456 Posted February 13, 2014 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 ?
Acme Posted February 14, 2014 Author 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.
Mayflow Posted March 31, 2014 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?
Acme Posted March 31, 2014 Author 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]
imatfaal Posted April 1, 2014 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
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