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Outwitting the Weighing Machine - Solution
Puzzles -> Sam Loyd Puzzles
 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? |
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The 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.
Solution provided by "Chris" |
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