How Much Do They Weigh ????

Let us designate the girls as: a, b, c, d, e
Each girl is weighed 4 times in such manner:
a+b, a+c, a+d, a+e. Similarly for b, c, d, e 
So if  we add their total weights given in the puzzle, we get:
129+ 125+124+123..........+114 =1,212 Pounds.
Since this is the total of 4 weighing for 5 girls, we, therefore, have: 1212 / 4 =303 pounds. This is the combined weight of the 5 girls. Since 2 of them weigh 129 pounds, then 129/2 =64.5 pounds. Will round these 2 as 65 and 64 pounds.
Then the weight of the 3 remaining girls must be: 303 - 65 - 64 =174 pounds. And 174/3 =58 pounds. To get an average of 58 pounds for the 3 girls, we could distribute them as:
56, 58, 60 =174 pounds. So, the presumed age of the 5 girls could be: 56, 58, 60, 64, 65 pounds. This combination seems to balance ALL the paired weighings given in the question. That is:65+64=129. 65+60=125, 64+60=124.........etc. all the way down to 56+58=114.
This is not very rigorous mathematically, but it seems to work !!.

Challenge accepted     

Let the girls be a,b,c,d, and e   where a is the lightest and e is the heaviest.

If I add all those paired weithgts together I get  1212 pounds

Each girl is included 4 times so

4(a+b+c+d+e) = 1212 pounds

a+b+c+d+e     =   303

The average weigt of the children is  60.6 pounds

The average weight of each pair of girls is 121.2 pounds

So now it is important to know how far each pair  of girls deviate from the average

A+B = -7.2  So   A= -7.2 -B

D+E= 7.8   so     B+6+E = 7.8   so      E = 1.8 - B

These are all deviations so they have to add up to 0.

-7.2 - B +B +B+2 +B+6+1.8-B=0

2.6+B=0

B=-2.6

So the girls weigh  56, 58, 60, 64 and 65 pounds respectively.       

Is that how you did it Chris?

If any of you are interested in his riddles, there's a site full of his riddles!

Site: https://www.mathsisfun.com/puzzles/sam-loyd-puzzles-index.html

https://www.mathsisfun.com/puzzles/sam-loyd-puzzles-index.html

This is not very rigorous mathematically, but it seems to work !

No kidding?

This is not very rigorous etymologically, either.

So, the presumed age of the 5 girls could be:

Good job, Melody and guest   !!!!

Here's my solution :

Let  a > b > c > c > d > e       where these are the weights of the girls

And  each is weighed 4 times.....and their total weight  = 1212 lbs

So we have that    4 ( a + b + c + d + e)  = 1212  ⇒  a + b + c + d + e   = 303

And here are the equations that we are sure of

The two heaviest  = a + b  = 129  (1)

Since a > b, then a + c > b + c....so a + c is the second heaviest weight

a + c  =  125  (2)

The two lightest =  d + e  =  114   (3)

Since c > d, then c + e >  d + e  ......so c + e is the second lightest weight

c + e  =  116  (4)

Subtract  (2) from (1)  =  b - c  = 4  ⇒  b  = c + 4

Subtract  (4) from (3)  =  c - d  = 2  ⇒  c  =  d + 2

So this implies that   b =  d + 6

Now.....add (1) and (2)

2a + b + c  =  254   ⇒   b + c    = 154 - 2a  ⇒  (d + 6) + (d + 2)  = 154 - 2a  ⇒

2d + 8  = 154 - 2a    →  2a + 2d  = 246  ⇒  a + d  = 123  ⇒  a = 123 - d

Similarly.....add (3) + (4)  and we have that

c + d  + 2e  =  230  ⇒  (d + 2) + d =  230 - 2e ⇒  2d = 228 - 2e ⇒ d = 114 - e ⇒ e = 114 - d

So

a + b + c + d + e  = 303    and substituitng, we have

(123 - d) + (d + 6) + (d + 2)  + d + (114 - d)  = 303

d + 245  =  303

d = 58

So

a = 123 - d  =  65 lbs

b = d + 6 =  64 lbs

c = d + 2  =  60 lbs

d = 58 lbs

e = 114 - d = 56 lbs