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?

CPhill
Sep 23, 2017

#1**+2 **

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**+2 **

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**

score | score -121.2 | ||

129 | D+E | 7.8 | |

125 | C+E | 3.8 | D=C+4 |

124 | ? | 2.8 | |

123 | ? | 1.8 | |

122 | ? | 0.8 | |

121 | ? | -0.2 | |

120 | ? | -1.2 | |

118 | ? | -3.2 | |

116 | A+C | -5.2 | C=B+2 |

114 | A+B | -7.2 | |

sum | 0 |

A | B | C | D | E |

A | B | B+2 | C+4=B+6 | E |

A+B = -7.2 So A= -7.2 -B

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

A | B | C | D | E |

-7.2 - B | B | B+2 | C+4=B+6 | 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

A | B | C | D | E | |

distance from mean | -7.2 - -2.6 = -4.4 | -2.6 | -2.6+2=-0.6 | -2.6+6 =3.4 | 1.8 - -2.6= 4.4 |

Girls weight | 60.6-4.4=56 | 60.6-2.6=58 | 60.6-0.6=60 | 60.6+3.4=64 | 60.6+4.4=65 |

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

Is that how you did it Chris?

Melody
Sep 24, 2017

#3**+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

#5**+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

#7**+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

CPhill
Sep 24, 2017