Five women out shopping come upon a pay weight scale (50 cents) in the store. Amongst them, they have only enough coins to activate the machine but once. The youngest woman in the group (A Smartkit regular 
) asserts that by getting on the machine in couples, and then exchanging places, one at a time, she could figure out the correct weight of all of five women.
In couples, they weighted 129 kilograms, 125 kilograms, 124 kilograms, 123 kilograms, 122 kilograms, 121 kilograms, 120 kilograms, 118 kilograms, 116 kilograms, and 114 kilograms.
What do you think? Has our young bright regular played 1 game too many of SmileyPuzzle?! Or can this actually be done?
as always, answers submitted below in the comment section will be unmasked in 2 days. Thanks!
Okay
Assuming the order they get on is like this
(Lady 1=A, Lady 2=B ect.)
A+B=129
A+C=125
A+D=124
A+E=123
B+C=122
B+D=121
B+E=120
C+D=118
C+E=116
D+E=114
I figured A=C+7, by figuring A+B=129 and B+C=122(So A must be 7 more than C)
So That Means if A+C=125 than C+C+7=125, so C=59
Once you get C=59 you can use substitution for the rest
A+C=125, 125-59=66
A=66, C=59
A+B=129, 129-66=64
A=66, B=63, C=59
C+D=118, 118-59=59?
But that would mean C=D, but A+C=/=A+D
This is where I got stuck. I’m sure you can do it, but the question also confused me because it didn’t explain the order well.
Best I got is, no it’s not possible, but I know it is so =\
This problem would be *A LOT* easier to solve if they had a better idea.
2 women get on the machine, and 1 switches out each time (the same woman!). Thus leaving 1 woman of the 5 standing on the machine the entire time.
At the end, before stepping off, simply record her actual weight, and then subtract that from all the coupled weighings!
Although it IS possible. I’m not sure if there is only one solution, but the following weights work:
(from lightest to heaviest)
A = 56
B = 58
C = 60
D = 64
E = 65
I started with the lowest and highest weights:
A + B = 114
D + E = 129
Given that A != B (not required for the problem, so there are most definitely other solutions?) 114/2 = 57 –> +/-1 = 56 and 58
D+E must be the biggest 2 numbers possible so 65 and 64.
after adding up 56,58,64 and 65 to find out what totals I had, I simply subtracted 65 from the next biggest number (E + C) to get 60. the final weight.
Their weights: 65 kilos, 64 kilos, 60 kilos, 58 kilos, 56 kilos. So, yeah, possible.
Total weighing = 1212kg
Divide by 4 = 303kg = total weight of the 5 women
five women from heaviest: a, b, c, d, e
a + b + c + d + e = 303
a + b = 129 (Two heaviest)
d + e = 114 (Two lightest)
Therefore c = 60
a + c = 125 (Heaviest with 3rd heaviest)
Therefore a = 65, b = 129 – 65 = 64
e + c = 116 (Lightest with 3rd lightest)
Therefore e = 56, d = 114 -56 = 58
The women’s weights are: 65, 64, 60, 58, and 56kg.
For the sake of order. I labeled each of the 5 women a-e with a being the heaviest.
The largest number had to be the sum of the two heaviest women: a+b=129
The smallest number had to be the sum of the two lightest women: d+e=114
The second largest number had to be the sum of the heaviest and the third heaviest woman: a+c=125
The second smallest number had to be the sum of the lightest and the third lightest woman: e+c=116
This left me with 4 equations and 5 unknowns.
To be honest, I took an educated guess for my 5th equation. I knew that the weights of the women needed to be fairly close in order to get the run of numbers 120-125kg, so I used the smallest sum, 114, and divided it by two. That gave me 57 so I assumed that the lightest weight was 56=e. After solving this series, I listed out all the possible sums and got the 10 numbers that were given in the problem. I’ll be very interested to see if anyone managed to use more logic than an educated guess to determine that last equation.
On a final note, in pondering the way to solve the problem, I realized that had the young lady been really smart, she would have had them step on the scale in this order. A, AB, B, BC, C, CD, D, DE, E. That way, the scale would never have hit 0, and all of their individual weights would be known without any calculations required.
Absolutely. Label their weights A, B, C, D, and E, with A being the heaviest.
All ten weights add up to 1212, so
(A+B)+(A+C)+ … +(C+E)+(D+E)=1212, 4(A+B+C+D+E)=1212, A+B+C+D+E=303
The heaviest weight must be A+B, and the lightest D+E, so
(A+B)+C+(D+E)=303, 129+C+114=303, C=60
The second heaviest weight must be A+C, and the second lightest C+E, so
A+C=125, A+60=125, A=65
C+E=116, 60+E=116, E=56
and finally
A+B=129, 65+B=129, B=64
D+E=114, D+56=114, D=58
Their weights are 65, 64, 60, 58, and 56 kg.
I’m not sure if this is a unique solution, as i only did a bit of trial and error, but i did find a solution, with the weights (in kg) being 56, 58, 60, 64, and 65
I labeled them, from smallest to largest, as x1, x2, x3, x4, x5
set
x2=114-x1
x3=116-x1
x4=129-x5
if you plot the 10 points, xi+xj, you’ll see that you always have 114, 116, and 129
after changing the values of x1 and x5, it’s easy to see that the other points move around in predictable patterns, and it didn’t take long to get a solution after that
Are the weights in that order or is it jumbled? That is, the first 4 values include are the weights of the first person + x or could any 4 values be assigned to the first person + x?
Hi Migrated- I don’t believe the weights have to be in that order…
The easiest way is to label which 2 weighed the weights! Then you need only 5 weighs.
However, this way/weigh makes it much more interesting! Pun intended.
The heaviest weigh was made by the 2 most full size gals. The lightest weigh was made by the 2 skinniest gals. The second to the heaviest weigh and the second to the lightest weigh must have included the lady in the middle, with heaviest and lightest on the scales with her as well. A little playing around with the numbers and you see…
Heaviest lady is 65 kg, followed by 64, 60, 58.
The lightest lady is 56 kg.
Where they really will get in trouble is if they try to use the same fortune and lottery numbers!
That is a puzzle of another day I suppose…
Think it is
65
64
60
58
56
56,58,60,64,65
@Migrated
Hint: Your assumption of the smallest 4 numbers being X + the other 4 people, is correct.
By “weight” I mean the weight of one person, and by “weighing” I mean the measurement from one application of the scale.
The weights are 65, 64, 60, 58 and 56.
I called the five weights A, B, C, D and E with:
A > B > C > D > E.
It is clear that the heaviest weighing must be A + B, and the second heaviest is A + C, and similarly D + E is the lightest and C + E the second lightest. So:
A + B = 129,
A + C = 125,
C + E = 116,
D + E = 114.
From this, it can be seen that B is 4 more pounds than C, and therefore 6 pounds more than D, and A is 9 pounds more than E. From this we know:
A + D = 123,
B + E = 120.
There are only 4 weighings left unknown, which are:
A + E, B + C, B + D, and C + D, which have to be equal to in some order: 124, 122, 121, and 118.
But, since we know B = C + 4 = D + 6, we know that:
B + C = B + D + 2 (Add B – 4 to the last two expressions), and
B + C = C + D + 6 (Add C to first and third expressions).
The only number out of the four that we have to pick such that there is another two less, and another 6 less is 124 (122 and 118 respectively.
Therefore:
B + C = 124,
B + D = 122,
C + D = 118, and therefore
A + E = 121.
Now that we know the identity of all ten weighings, then it is a simple matter to use the relations we already know (A – E = 9, B – C = 4, B – D = 6)
to figure out the 5 weights. These are:
A = 65, B = 64, C = 60, D = 58, and E = 56.
There are no five weights that can produce these ten resultants. Following bobq’s advice and assuming that the next lightest pairing results in 121kg and than solving for the lighter of the two you determine that a=59.5kg (a+b=121kg, b=a+2kg, a+a+2=121kg, 2a=119kg, a=59.5kg). Based on this you can than solve for x (x+a=114, x+59.5=114, x=54.5). Therefore the individual weights would be x=54.5kg, a=59.5kg, b=61.5kg, c=63.5kg, and d=65.5kg. However, these five women can not have a combined weight of 122kg or 124kg, and b+d=127kg. This discrepancy in combined weights (like you will find in all other pairings) is only isolated to this example’s provided resultants. If weighted properly in pairs, with all ten resultants correct, it would be very simple to determine each woman’s weight.
It’s not in order. I tried it, and there was a conflict when I compared 1+4 and 1+5 with 4+5. The difference in weights of the first two former combos was only 1 pound, while the total weight of 4+5 was 114.
If there is a way to determine the orders of the combinations, it can be done. Anyone have any ideas besides trial and error?
Ok, I got it. Thanks for the hint, bobg.
Ok, I didn’t get it. I don’t think it can be done without knowing the order of the weigh-ins.
The highest combined weight is for the two heaviest ladies.
The lowest combined weight is for the two lightest ladies.
The total weight of 1212 kg is for ten pairs, 20 ladies, so it is easy to work out the total weight of the five. Subtracting the top and bottom weights gives the weight of the middle lady, then the second highest combined total must be for the heaviest and the middle ladies. Knowing the middle you now know the heaviest and following through you get all five as follows:
65, 64, 60, 58, 56 kg.
@migrated
Actually, I meant to say that your assupmtion was possible, but it is entirely possible to guess at least 1 set of weights that fits the pair sums.
Just take the smallest sum for the lighter 2 people and the biggest sum for the heaviest 2 people. work backward from there to guess the middle weight.
Not that this is the best way, but it worked for me eventually. I’ll give the broad steps to my solution.
With the 5 weights a,b,c,d,e, say that a>b>c>d>e. (No two weights can be equal to eachother, if you think about it). With that, you can identify 4 of the pairs exactly (a+b, a+c, c+e, and d+e).
@bobg, don’t think you are right. If the c, d, and e are well below b, then c+d will be less than b+e…
So, what I did is get a+e in terms of c, and d+c in terms of e, and b+c in terms of a.
Then… I looked at the two possibilities for b+c — its either the third highest weight (if b+c>a+d) or the fourth (if b+c<a+d). I carried those two possibilities for a while and *eventually* it resolved.
I think it cannot be done especially exchanging places one at the time. Explanation:
the weights: 129,125,124,123,122,121,120,118,116,114
all the weights are different in couples, this means for me, that you have to find the unique combination of 2 in 5 persons:
lets say the GIRLS are: A,B,C,D,E
the ONLY UNIQUE combinations are:
AB,AC,AD,AE,BC,BD,BE,CD,CE,DE from this and that the weights listed are all different you get that all of the ladies have to have unique weights, so lets assume this order, It will work for any order I think:
A>B>C>D>E
THE LONGEST DECEDENT CHAIN THAT I COULD FIND IS THIS:
AB->AC->BC->BD->CD->CE->DE ,until this is ok and work just fine, THE NEXT COULD BE: BE or AE BUT THIS CONTRADICT THE INITIAL ORDER, THAT SHOULD HAPPEN WITH ANY INTIAL ORDER, cool.
Let’s call the girls a,b,c,d and x.
Assuming the smallest value must include the lightest person, I will call her x. I took the smallest values and combined x with the 4 other girls to get them. However, we had 4 uneven numbers, So I had to take the smallest uneven value for d, which is the heaviest girl. Considering that the last gap between the uneven 3rd and 4th values was of 4, I had to suppose that the gap between b and c was of 4. I came up with this:
x + a = 114
x + b = 116
x + c = 120
x + d = 121
With the gap of weight between a, b c and d revealed, it was now easy to assert the rest of the scales:
d + a = 123
d + b = 125
d + c = 129
a + b = 118
a + c = 122
c + b = 124
With that in hand, and knowing there was a gap of 2 kilograms between a and b, I divided 118 in a half and substracted/added 1 to make it to 58 and 60. The rest was easy, consisting in a serie of substractions to get the values. My final answers are:
x = 56
a = 58
b = 60
c = 64
d = 65
65,64,60,58,56 kg
This one was tough. Bizarette18, Aaornlau, Fuzzy, Bobo The Bear, Oneiric, Barryt, Jimmy Anders, Esalarc, Skyhawk, Michaelc, Bobg, Poopsicle, Blusummers13 got the correct answers. For a better understanding on how to solve, check out some of their good explanations.
Of course, if they had just kept track of who they were weighing, they could have figured it out in 5 weighings
mathematically is all good, congrats to the solvers however I understood that the women exchange places one at the time and that the order of the weightings was 129,125,124,123,122,121,120,118,116,114, assume what most of the solvers did: FROM A TO E DESCENDENT(A,B,C,D,E)
AB AC BC BD CD CE DE, thats the maximun descendent chain for different weights following the rules, so how can you record 10 descendent weights, maybe they were a little dizzy or am I the only dizzy in here, if someone can please make me come to my senses, cool
for this to be possible the order of the weightings should have been this:
129,125,116,114,123,118,124,122,120,121
AB, AC, CE, DE, AD, CD, BC, BD, BE, AE
If the order of the weightings was this it was possible aand all was good but if the order was the one presented in the problem no way jose, coolkit
I have been quite busy and just noted this nice puzzle.
The complete solution is as follows:
take the following order: A>B>C>D>E (they are different, otherwise at least two of the sums would be equal)
It follows that:
AB > AC > BC > BD > CD > CE > DE
AC > AD > BD
BD > BE > CE
AD > AE > BE
which implies that AB is the highest weight, and AC the next highest
and DE is the lowest weight, and CE the next lowest
The sums are (in order) 129, 125, 124, 123, 122, 121, 120, 118, 116, 114
Hence AB=129, AC=125, CE=116, DE=114
The remaining weights are 124, 123, 122, 121, 120, 118
AB-AC=B-C=129-125=4
then BD-CD=B-C=4
The possible (BD,CD) combinations are: (124,120) and (122,118)
CE-DE=C-D=116-114=2
then BC-BD=C-D=2
The only possible (BC,BD) combination is (124,122)
Hence BC=124 and BD=122
BD-CD=4 -> 122-CD=4 -> CD=118
The remaining weights are: 123, 121, 120 for AD, AE, BE
Since AD > AE > BE, then AD=123, AE=121, and BE=120
To recapitulate:
AB=129, AC=125, CE=116, DE=114
AD=123, AE=121, BE=120
BC=124, BD=122, CD=118
AB-AC=B-C=129-125=4 -> C=B-4
AC-CE=A-E=125-116=9 -> E=A-9
CE-DE=C-D=116-114=2 -> D=C-2
AE-BE=A-B=121-120=1 -> B=A-1
Hence:
C=B-4=A-1-4=A-5
D=C-2=A-5-2=A-7
A+B=129 -> A+A-1=129 -> A=65
B=A-1 -> B=64
C=A-5 -> C=60
D=A-7 -> D=58
E=A-9 -> E=56
Nice puzzle
I had posted about a week ago (after the comments were up) a complete solution to the problem. Too bad the post did not make it somehow. I am reluctant to rewrite it again as it had taken me a good half hour to do so.
I just discovered this site. Here is how I approached the problem.
total of all weighings – 1212kg
divided by 4 (each woman weighed 4 times) 303kg total weight of the 5 women
D + E (129) plus A + B (114)= 243
C = 60 kg
A + C = 116kg A = 56kg
A + B = 114kg B = 58kg
C + E = 125kg E = 65kg
D + E = 129kg D = 64kg