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Tackle These Three Tests (Posted on 2005-08-07) Difficulty: 3 of 5
Arthur, Bert and Charles took three exams. In all exams, the first got X points, the second Y points, and the third Z points, X, Y and Z being positive integers such that X>Y>Z.

Summing the three results, Albert got 20 points; Bert, 10; and Charles, 9. Albert was 2nd in Algebra. Who was 2nd in Geometry?

See The Solution Submitted by Old Original Oskar!    
Rating: 2.5000 (8 votes)

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Solution Solution assuming that Albert is Arthur's real name... | Comment 2 of 7 |

The total of the three examinees on the three tests was 39, so the total for the three on each of the tests was 13.

The only sets of three different positive integers that add to 13 are (6,4,3), (6,5,2), (7,4,2), (7,5,1), (8,3,2), (8,4,1) and (9,3,1).

In order for Albert to score 20 points, there must be a score of at least 7 available. But neither of the sets that have a 7 would allow a total of 20, nor would the set that has a 9.  In fact, the only set that allows a total of 20 is (8,4,1).

With 8, 4 and 1 being the allowable scores, in order to get 9 with three of the numbers added together, Charles would have to get 4+4+1. Bert would have had to get 8+1+1 to make 10, and Albert would have 8+8+4.  So the only other person to have a second place finish (with a score of 4 on that test) would be Charles, who was, therefore, 2nd in Geometry.

  Posted by Charlie on 2005-08-07 03:37:50
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