Logic Problem Solution: Last-Minute Lester at the Beach By clue 1, Last-Minute Lester got a room at the Sand Dollar Motel the night before staying in Full Moon Bay; and by clue 5, the family stayed in Full Moon Bay the night before staying in a motel whose room cost twice as much as the one in Full Moon Bay. By clue 3, the Beachcomber Motel stay was the night before the stay in the room that cost $50 less than the Beachcomber did and two nights before the overnight in Cape June. So, there must be some overlap between the two sets of three night stays: the Beachcomber can't be in Full Moon Bay, since clues 3 and 5 would conflict, so the two possibilities are that Last-Minute Lester found a room at the Beachcomber Motel for Sunday night and then at the Sand Dollar in Cape June for Tuesday, or that Lester found the room at the Sand Dollar on Sunday and at the Beachcomber on Tuesday. Trying the first arrangement, by clue 6, Wednesday's room cost $100 more than Monday's did, so if we let X equal Monday's room cost, Wednesday's in Full Moon Bay would be X + 100 and Sunday's at the Beachcomber Motel would be X + 50 (3). Thursday's cost would be 2X + 200 (5). By clue 2, Lester found a room in Atlantic Palisades earlier in the week than when the family stayed at the Dolphin Inn, which cost $100 less than the lodging in Atlantic Palisades. The family then couldn't have stayed at the Dolphin Inn on Monday, since the Sunday stay cost only $50 more than Monday's; nor could the family have stayed at the Dolphin Inn on Wednesday or Thursday and in Atlantic Palisades on Sunday or Monday, since both nights later in the week would have cost more than either Sunday or Monday night. Therefore, Lester found a room at the Sand Dollar on Sunday night, in Full Moon Bay on Monday, at the Beachcomber Motel on Tuesday for twice as much as the room in Full Moon Bay, on Wednesday for $50 less than what he paid at the Beachcomber, and on Thursday night in Cape June. Letting Monday's room cost equal X, Tuesday's would then be 2X (5) and Wednesday's X + 100 (6). By clue 3, Wednesday's room cost $50 less than Tuesday's, so X + 100 = 2X - 50. Solving, X = 150. So, Monday's room cost $150, Tuesday's was $300, and Wednesday's was $250. By clue 2, if the Dolphin Inn were Monday's stay, the Sand Dollar room would have cost $250, the same as Wednesday's stop--no (7). If the Dolphin Inn were Wednesday's lodging, the Atlantic Palisades room would have cost $350 (2) and, by clue 7, the lodging in Cape June would have cost $250--the same as that in Full Moon Bay--and again no (7). Lester found a room at the Dolphin Inn on Thursday night. If the Ebb Tide Hotel were Wednesday's lodging, the room in Snug Harbor would have cost $350 (4) on Monday. But then, by clue 7, the Dolphin Inn room would have cost $250, the same as the Ebb Tide and a conflict with clue 7. So, Lester found a room at the Ebb Tide Hotel on Monday and at the Isla del Sol Resort on Wednesday. By clues 7 and 4, the Isla del Sol is in Snug Harbor. By clue 7, the remaining two nights cost a total of $600. If Atlantic Palisades were Sunday's stop, then between that lodging and the Dolphin Inn the $600 would have been split $350/$250 (2), the latter total being the same as the Costa del Sol room and a contradiction with clue 7. So, Lester found the room in Atlantic Palisades on Tuesday night for $300 and paid $200 at the Dolphin Inn. By elimination, the family was in Ocean City Sunday night and paid $400 for the room there (7). Last-Minute Lester found rooms at the beach as follows:
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