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.

 

 

 

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

 

 

 

 

 

 

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, 201411 yr by michel123456

 

 

 

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.

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 ?

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.

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?

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]

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