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August 23, 2014

Homework Help: maths

Posted by Emma on Friday, July 6, 2007 at 9:59am.

five bales of hay are weighted two at a time in all possible combinations. the weights, in kilogrames, are:

110,12,113,114,115,116,117,118,120 and 121

what is the weight, in kilograms, of the heaviest bale?

can someone please show me how to do this question?thx.

Something is missing. If there are four bales, two at a time, then there should be 12 sets of weights . There are ten. If there are three bales, 2 at a time, there should be six sets of weight. So ten weights does not make sense to me.

Suppose five bales of hay are weighed two at a time in all possible ways. The weights in pounds are 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121. How much does each bale weigh? >>


Allow me to lead you through the logic and see if you can get to the right answer yourself.
1--With 5 bales, a-b-c-d-e, being weighed 2 at a time in all possible combinations, leads to each bale being weighed ? times.
2--The sum of all the weighings is ____?
3--What then is the weight of the 5 individual bales?
4--Which bale can you determine the weight of right away, knowing the sum of the 5 bales and the sums of the lightest two, a + b, and the heaviest two, d + e?
5--Having this one bale weight, how can you proceed to derive the others?

If you did not have any success with these clues, the solution is below, way below.
























The easy solution--
One of the simpler solutions is as follows:
If you add the list of ten different weighings, you get a total of 1156.
Since each bale was weighed four times, 4A + 4B + 4C + 4D + 4E = 1156.
Dividing the 1156 total by four yields A+B+C+D+E=289.
Since A<B<C<D<E, A + B = 110 and D + E = 121.
Substituting into A+B+C+D+E=289 yields C=58.
The second lightest combined weight, A+C must be 112. Thus A=54. From here he rest is pretty self explanatory.










The longer solution--
Solution: Clearly each bale weighs a different amount or there would be less than 10 different weighings. Assign the weight of each bale to A through E such that A < B < C < D < E . Clearly, A + B = 110 and D + E = 121. Since A + B = 110 and A < B, the greatest A can be is 54 and the least B can be is 56. Similarly, since D +
E = 121 and D < E, the greatest D can be is 60 and the least E can be is 61. As C > B and < D, A + C = 112 and it follows that C = B + 2. Similarly, since D > C, A + D = 113 and it follows that D = C + 1. A boundry of our problem is therefore A = 54, B = 56, D = 60 and E = 61. Since D - B = 4 at this boundry, it follows that B could be
greater than 56 and/or D could be less than 60 , as long as we meet the two conditions derived earlier, that C = B + 2 and D = C + 1. If the boundry condition were the answer, then C = 58, D = 60 and E = 61. But C + E = 119, which is not one of the given totals and no sum gives us 120, which is a requirement. Therefore B is greater than 56 and/or D is less than 60. If B = 57, then C = 59, D = 60 and E = 61. But C + D = 119, which is not one of the given totals. Therefore B is not equal to 57. If B = 58, then C = 60, D = 61 and E = 60. But D must be less than E and/or both C + D and D + E add up to 121, which violates our given weighings. Thus, B is not equal to 58 nor can it be greater than 58. If D = 59, then C = 58, B = 56, A = 54 and E = 62. This solution satisfies all the given weighings, and is therefore a valid solution. What if D = 58 however? If D = 58, then E = 63, C = 57, B = 55 and A must equal 55, which violates our condition that A < B. Thus A = 54, B = 56, C = 58, D = 59 and E = 62. Not as elegantly straight forward but just another viewpoint.





!!I got it!!
thank you SOO much!!

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