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Jeopardy (Posted on 2003-06-14) Difficulty: 4 of 5
When the long-time popular TV program "Jeopardy!" filmed a show in the Summerset area recently, host Alex Trebek quizzed three local contestants, including one from Peyton Park, as they vied for cash in the game's three rounds.
During the action, the contestants, including Jay, "questioned answers" against each other during the "Jeopardy!" and "Double Jeopardy!" rounds; each then wagered part or all of the amount he or she had won on a "Final Jeopardy!" question to determine the final sums and game winner.

From the clues below, determine each contestant's full name, home town, and score before and after the final round on the perennial quiz favorite:
  1. During "Final Jeopardy!", Ben doubled the money he had won through "Double Jeopardy!" and McNabb added $2,000 in winnings, but the contestant from Summerset missed the final answer and lost 1/3 of the winnings through the first two rounds.

  2. The winner, who isn't Hardy, had $3,000 more than the second-place contestant.

  3. Smith and the contestant from Thoreau Falls both missed the same $2,000 question in the category "Crossword Words" near the end of "Double Jeopardy!".

  4. Stefanie finished in third place with a final total of $4,000.

  5. After the "Double Jeopardy!" round and going into "Final Jeopardy!", the three contestants had amassed $13,000 among them.

(puzzle originally from www.allstarpuzzles.com)

See The Solution Submitted by DJ    
Rating: 3.8333 (12 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): solution | Comment 6 of 12 |
(In reply to re(2): solution by DJ)

When you say "Even knowing the result, however, I cannot find any [algebraically correct] reasoning that would lead to the solution you previously found, although it works and fits all the clues," I assume you mean to avoid the enumeration of cases. That is, the six possible permutations of which place finisher 1-3 could be matched against the three Final Jeopardy outcomes: doubling, adding 2000 and losing 1/3.

Since there are only six, each could be handled separately, and the solution originally presented was the result of assuming the 1st place doubled; the 2nd place lost 1/3 and the 3rd place added 2000.

That resulted in, 3rd place adding 2000 to 2000 to get the 4000 she was said to have ended up with, and if x is the amount the 2nd place finisher had going into double jeopardy, he had 2x/3 after Final, and as Stefanie went into Final with 2000, 11000 was left between 1st and 2nd, meaning 1st went in with 11000-x, and since he doubled, came out with 22000-2x. As the final difference between first and second was 3000, then
22000-2x - 2x/3 = 3000
66000-6x -2x = 9000
8x = 57000
x=7125

This is the algebraically correct reasoning. Unfortunately I had applied incorrect numbers to the other valid case, 1st place doubled, second place gained 2000 and third place lost 1/3, thereby making me think the case was invalid.

So, yes, if you are referring to using algebra to avoid breaking the problem down into six cases, I don't see a way. But what a present above is certainly valid algebra for the case you are wondering about.

  Posted by Charlie on 2003-06-15 16:33:05

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