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This is another problem from  American puzzle-maker Sam Lloyd..........

 

Five little girls taken in couples weigh 129 pounds, 125 pounds, 124 pounds, 123 pounds, 122 pounds, 121 pounds, 120 pound, 118 pounds, 116 pounds and 114 pounds on a weighing machine. 

What was the weight of each one of the five little girls if taken separately?

 

 

cool cool cool

CPhill  Sep 23, 2017
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7+0 Answers

 #1
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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 !!.

Guest Sep 24, 2017
edited by Guest  Sep 24, 2017
 #2
avatar+90581 
+2

Challenge accepted     laugh

 

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

 

score score  -121.2 
129D+E7.8 
125C+E3.8D=C+4
124?2.8 
123?1.8 
122?0.8 
121?-0.2 
120?-1.2 
118?-3.2 
116A+C-5.2C=B+2
114A+B-7.2 
sum 0 

 

ABCDE
ABB+2C+4=B+6E

 

A+B = -7.2  So   A= -7.2 -B

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

 

ABCDE
-7.2 - BBB+2C+4=B+61.8 - 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

 

 ABCDE
distance from mean-7.2 - -2.6 = -4.4-2.6-2.6+2=-0.6-2.6+6 =3.41.8 - -2.6= 4.4
Girls weight60.6-4.4=56 60.6-2.6=5860.6-0.6=6060.6+3.4=6460.6+4.4=65

 

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

 

Is that how you did it Chris?

Melody  Sep 24, 2017
 #3
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+2

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

Guest Sep 24, 2017
edited by Melody  Sep 24, 2017
 #4
avatar+90581 
+1

Thanks CPhill and thank you guest!  

Melody  Sep 24, 2017
 #5
avatar+729 
+1

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:

GingerAle  Sep 24, 2017
 #6
avatar+90581 
0

I think guest has done something similar to me only more simply but I have not looked at it closely.

 

My answer was mathematically precise.  There are no other possible answers. 

Melody  Sep 24, 2017
 #7
avatar+76974 
+2

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 + dc + 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 heaiest 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

 

 

 

cool cool cool

CPhill  Sep 24, 2017

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