26 Sep 03
Five people want to weigh themselves on a coin-operated scale. They want to pay only once. Two of the people climb onto the scale, drop in a coin, and record their combined weight. One person gets off, another person gets on, and they record their combined weight. This continues until the people have the following weights recorded: 188, 192, 196, 199, 203, 204, 207, 208, 212, and 219.
1.Show all work and determine what each person weighs.
2.Explane your answer in complete sentences.
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26 Sep 03
Originally posted by Lord BurnLabel the peoples' weights in order a, b, c, d and e where a is the lightest.
Five people want to weigh themselves on a coin-operated scale. They want to pay only once. Two of the people climb onto the scale, drop in a coin, and record their combined weight. One person gets off, another person gets on, and they record their combined weight. This continues until the people have the following weights recorded: 188, 192, 196, 1 ...[text shortened]... ll work and determine what each person weighs.
2.Explane your answer in complete sentences.
The sum of all the weighings is 2028
each person has been on the scale 4 times, ie 2028 = 4(a+b+c+d+e)
therefore a+b+c+d+e = 2028/4 = 507
188, the lightest total, must be the weight of the two lightest people, i.e 188 = a+b
219, the heaviest total, must the weight of the two heaviest people, i.e 219 = d+e
so a+b+d+e = 188+219 = 407
Because, from above, a+b+c+d+e =507, c must weigh 100
192, the second lightest total, must be the weight of the lightest person and the 3rd lightest person, i.e 192 = a+c, so a = 92
212, the second heaviest total, must be the weight of the heaviest person and the third heaviest person, i.e. 212 = c+e, so e = 112
Above we stated that a+b = 188, now we know a = 92 so b = 96
Also above we stated that d + e = 219, now we know e = 112, so d = 107
To summarise, the weights are 92, 96, 100, 112, 107 🙂