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Logic Puzzle # 38 Logic Problems Help



Logic Problem Solution:
The Skating Elvises

Paul (clue 1), Wayne (4), and Art (6) finished 3rd-5th in some order in the figure skating judging. If Paul in clue 1 had placed 3rd with the Toronto Elvis 2nd and the "Heartbreak Hotel" skater 1st, the Baltimore Elvis in clue 4 would have to be Paul, which he isn't (11), or the skater to "Heartbreak Hotel," which he isn't (9). So, Paul was 4th or 5th. If Art in clue 6 had placed 3rd with Burnsides 2nd and the Elvis from Glasgow 1st, since the Baltimore skater isn't Burnsides (11), the Baltimore Elvis would have been 3rd followed by the skater to "Jailhouse Rock" 4th and Wayne 5th (4). Paul then would have skated to "Jailhouse Rock"--no (8). So, Art wasn't 3rd in the event; Wayne was. By clue 4, therefore, the Baltimore Elvis skated to 1st place and the performer to "Jailhouse Rock" was 2nd. Since the Baltimore skater's song choice wasn't "Heartbreak Hotel" (9), in clue 1, Wayne in 3rd place performed to "Heartbreak Hotel," with the Toronto Elvis then 4th and Paul 5th in the "Elvises on Ice" competition. Then Art in clue 6 placed 4th, with the Glasgow skater 2nd and Burnsides 3rd. Wayne Burnsides is from Memphis and Paul from Paris (10). By clue 2, Stan won the figure skating and Gold finished 2nd. Gold is Lou. The one who skated to "Can't Help Falling in Love" isn't Art or Paul from Paris (3) and is Stan. By clue 7, Art performed to the Presley hit "The Wonder of You," and Paul is Soleil. Paul chose "Return to Sender" for his background song. Art is Cadillac and Stan Parker (5). The 1st-5th place finishers in "Elvises on Ice" are

  • 1st - Stan Parker, Baltimore, "Can't Help Falling in Love"
  • 2nd - Lou Gold, Glasgow, "Jailhouse Rock"
  • 3rd - Wayne Burnsides, Memphis, "Heartbreak Hotel"
  • 4th - Art Cadillac, Toronto, "The Wonder of You"
  • 5th - Paul Soleil, Paris, "Return to Sender"



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