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Logic Puzzle # 58 Logic Problems Help

Logic Problem Solution:
Dime Darts

Three boys are mentioned in both clues 4 and 5, so either there is one or more guests common to the two clues or all six are named between the two. From the introduction, all six boys got at least one dime in the bull's-eye, and no two boys scored the same number of coins at Dime Darts. By clue 4, Lars won 30 cents more than the Potter boy, who won twice as much as Harry; and by clue 5, the Jansen boy won 20 cents more than David, who took home twice as many dimes as the Ciccarelli boy. The possible commonality between the clues is that 1) Harry is Jansen, 2) that Lars is Ciccarelli, 3) that Lars is Jansen, or 4) that David is Potter and Harry Ciccarelli (since no two won the same amount, David can't be Potter and Harry and Ciccarelli be different boys or vice versa). Trying the first possibility, that Harry is Jansen, if the Ciccarelli boy had won only one dime, David would have scored two, Harry Jansen four (5), the Potter boy eight, and Lars 11 for $1.10 in prize money--no, since the most won was $1.00 (clue 1). Similarly, if Lars were the Ciccarelli boy and Harry won only one dime, the Potter boy would have garnered two, Lars Ciccarelli would have won 50 cents (4), David would have taken home $1.00, and the Jansen boy would have scored 12 dimes--again contradicting clue 1. Since the most any guest won at Dime Darts was $1.00, by clue 4, Lars couldn't have won the top prize or Harry would have tallied 35 cents at the game--not possible in tossing 10-cent pieces. So, if Lars were the Jansen boy, the most he could have won would be 90 cents, with the Potter boy taking home 60 and Harry 30 (4), David 70 and the Ciccarelli boy 35 (5). Given that this total of $2.85 is the most the five boys could have won if Lars were Jansen, clue 1 conflicts, since the five boys who didn't win Big scored $3.00 among them. So, Lars isn't the Jansen boy. As from above, Lars couldn't have won the $1.00 top amount (1) or Harry would have totalled 35 cents (4). If David were Potter and Harry Ciccarelli, then the most dimes Harry could have put in the bull's-eye would be 3. If Harry Ciccarelli had won 20 cents, then by clues 4 and 5, David Potter would have taken home 40 cents, Lars would have hit for 70, and the Jansen boy would have won 60 cents. Summing, the four would have won $1.90 among them, leaving $1.00 for the game winner (1)--but $1.10 for the sixth guest and a contradiction (1). If Harry Ciccarelli had put only one dime into the bull's-eye, the above totals would be even less ($1.20) for the four in clues 4 and 5 and the sixth boy would have won $1.80. If Harry Ciccarelli had won 30 cents, then by clues 4 and 5, David Potter would have taken home 60 cents, Lars would have won 90 cents, and the Jansen boy would have hit for 80 cents. Since these sums total $2.60, by clue 1, the other two guests would have won $1.00 and 40 cents. By clue 2, Mario hit the bull's-eye with one more dime than the Freeman boy. Mario would have to be the $1.00 scorer and Lars Freeman--no (6). So, Harry and David aren't Ciccarelli and Potter. All six boys are named between clues 4 and 5, and either Lars or the Jansen boy then won Big by hitting 10 Dime Darts bull's-eyes (1). Again, Lars didn't win the $1.00 or Harry would have won 35 cents (4). The Jansen boy won $1.00, David won 80 cents, and the Ciccarelli boy won 40 (5)-a total of $2.20, leaving $1.80 won (1) among the three in clue 4. Letting Harry's winnings equal X, the Potter boy won 2X and Lars 2X + .30--so that 5X + .30 = 1.80 and X would be .30. Harry won 30 cents, the Potter boy took home 60 cents, and Lars scored 90. Since Lars isn't the Freeman boy (6), by clue 2, Mario is the Ciccarelli boy and Harry Freeman. Adam is Jansen and Ricky Potter (7). David is Benes and Lars Steiner (3). The six boys won at Dime Darts as follows:

  • Adam Jansen -- $1.00 & Big
  • Lars Steiner -- .90
  • David Benes -- .80
  • Ricky Potter -- .60
  • Mario Ciccarelli -- .40
  • Harry Freeman -- .30 Copyright © 2000-2007 by Calvin J. Hamilton & Randall L. Whipkey. All rights reserved. Privacy Statement.