# Outwitting the Weighing Machine Puzzle - Solution

### The Puzzle:

The five school children 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?

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

### Our Solution:

Let e > d > c > b > a be the girls weights.

NB no two girls have the same weights, otherwise there would

have been duplicate weighings.

e+d must be the heaviest and e+c the second heaviest weighing.

a+b must be the lightest and a+c the second lightest weighing.

It might be the a+d or b+c could be next, fortunately we don't need to know.

The sum of the weighings is 1212 and each girl is weighed four times

=> a+b+c+d+e = 1212/4 = 303

c = (a+b+c+d+e) - (a+b) - (d+e) = 303 - 114 - 129 = 60

a+c = 116 => a = 116 - 60 = 56

a+b = 114 => b = 114 - 56 = 58

c+e = 125 => e = 125 - 60 = 65

e+d = 129 => d = 129 - 65 = 64

So the girls weights are: 56, 58, 60, 64 and 65.

NB no two girls have the same weights, otherwise there would

have been duplicate weighings.

e+d must be the heaviest and e+c the second heaviest weighing.

a+b must be the lightest and a+c the second lightest weighing.

It might be the a+d or b+c could be next, fortunately we don't need to know.

The sum of the weighings is 1212 and each girl is weighed four times

=> a+b+c+d+e = 1212/4 = 303

c = (a+b+c+d+e) - (a+b) - (d+e) = 303 - 114 - 129 = 60

a+c = 116 => a = 116 - 60 = 56

a+b = 114 => b = 114 - 56 = 58

c+e = 125 => e = 125 - 60 = 65

e+d = 129 => d = 129 - 65 = 64

So the girls weights are: 56, 58, 60, 64 and 65.

*Solution provided by "Chris"*Puzzle Author: Loyd, Sam

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