Wait...Find the Weight?
On a trip to the corner drugstore, you meet five girls who are from your school. They have just weighed themselves on a coinoperated scale but since they had only one quarter, they could not weigh each person individually.
They weighed themselves in the following way: Two of the girls got on the scale and noted their combined weight. Then one girl got off and another girl got on. They did this until they had the following readings: 183, 186, 187, 190, 191, 192, 193, 194, 196, 200.
Knowing that you are a mathematics whiz, they ask if you can tell them what their individual weights are?
Source: B.S. (former student and now a teacher in WA)
Hint: Can you set up some equations? What role does combinatorics play in this situation?
Solution Commentary: B.S. offers this solution commentary: "There are ten weights so we know that each of the 10 possible pairs of girls has been weighed. Label the weights of the five girls from lighest to heaviest as A, B, C, D, and E.
We know the lighest combination must be A+B and that equals 183. We also know the heaviest combination is D+E and equals 200. Now look at the pairings AB, AC, AD, AE, BC, BD, BE, CD, CE, DE, and it can be seen that each girl has been weighed exactly four times, so the total of all ten readings (1,912) divided by 4 equals 478 which is the combined weight of the five girls. Therefore, A+B+C+D+E = 478. Also, remembering A+B = 183 and D+E = 200, so A+B+D+E = 383. Then C must equal 95. A+C is the second lightest weight so A+C = 186. Then A must equal 91. A+B = 183 so B equals 92. We also know that the second heaviest weight C+E = 196 so E must equal 101 and D+E = 200 so D equals 99. Therefore the weights of the five girls are 91, 92, 95, 99, 101.
