mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
Results for the 'Bales of Hay' Challenge Exercise
- Thread starter mmm4444bot
- Start date
D
Deleted member 4993
Guest
Since none of the pairs weigh the same, all the bales must weigh different.
lets assume (the bales are named a,b,c,d &e) a>b>c>d>e
then
a+b+c+d+e = 271
a+b = 136
d+e = 82
c = 53
Then
A + C = 125 ---> A = 72
A+B = 136 ---> B = 64
C+E = 88 ---> E = 35
D+E = 82 ---> D = 47
TADA.....
lets assume (the bales are named a,b,c,d &e) a>b>c>d>e
then
a+b+c+d+e = 271
a+b = 136
d+e = 82
c = 53
Then
A + C = 125 ---> A = 72
A+B = 136 ---> B = 64
C+E = 88 ---> E = 35
D+E = 82 ---> D = 47
TADA.....
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
Subhotosh Khan said:
D
Deleted member 4993
Guest
Re:
Yes - sort of - sum of all the weights = 4 * (a+b+c+d+e)mmm4444bot said:
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
Subhotosh Khan said:
Sort of how?
My "breakthrough" was realizing that twice the average bale weight equals the average of the paired weights.
\(\displaystyle 2 \cdot \left( \frac{a+b+c+d+e}{5} \right) \; = \; \left( \frac{\text{sum of paired weights}}{10} \right)\)
I used the same logic as Subhotosh. In the ten combinations, each bale appears 4 times. Add the sums of all the combinations, then divide it by 4. After getting my 5 equations, I just stuck them in a matrix and let the calculator do the dirty work.
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
wjm11 said:
Excellent!
If anybody else has yet another way to arrive at five equations, then please post it.
This thread is for posting results from the challenge exercise posted HERE.
Cheers,
~ Mark
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.
Alternatively:
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.
Cheers,
~ Mark
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.
Alternatively:
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.