bonanova Posted February 3, 2009 Report Share Posted February 3, 2009 (edited) Every day at noon, a Greyhound tour bus leaves New York for San Francisco, and another [Greyhound] bus leaves SF for NY. The cross-country trip takes 168 hours [7 days and 7 nights] and follows Interstate 80 from end to end [George Washington Bridge to Oakland Bay Bridge.] How many Greyhounds [buses] will the bus leaving New York today encounter on its trip to San Francisco? Edited February 3, 2009 by bonanova added [clarifying material] Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 3, 2009 Report Share Posted February 3, 2009 Every day at noon, a Greyhound tour bus leaves New York for San Francisco, and another bus leaves SF for NY. The cross-country trip takes 168 hours [7 days and 7 nights] and follows Interstate 80 from end to end [George Washington Bridge to Oakland Bay Bridge.] How many Greyhounds will the bus leaving New York today encounter on its trip to San Francisco? Assuming all buses are Greyhounds: 7, or 8 if you include the bus that's leaving as you arrive in San Francisco. You would meet your first bus halfway, leaving 84 hours left to drive. Every 12 hours the bus would meet another bus. The last (eighth) bus would be pulling out as you pulled in. However, since you specified that Greyhounds go from NY to SF at noon but did not specify that the buses traveling from SF to NY at noon are Greyhounds, I can only assume that because you specified "How many GREYHOUNDS will the bus leaving NY encounter?" that this is a trick question and there is not enough information (The buses coming the opposite direction may or may not be greyhounds). Quote Link to comment Share on other sites More sharing options...
0 HoustonHokie Posted February 3, 2009 Report Share Posted February 3, 2009 If the westbound bus leaves NY on day 0, it will encounter every eastbound bus that left SF between days -7 to 7 inclusive, or 15 buses. That assumes that you count the two encounters in the bus stations in NY & SF on days 0 & 7, respectively. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 3, 2009 Author Report Share Posted February 3, 2009 Assuming all buses are Greyhounds: 7, or 8 if you include the bus that's leaving as you arrive in San Francisco. You would meet your first bus halfway, leaving 84 hours left to drive. Every 12 hours the bus would meet another bus. The last (eighth) bus would be pulling out as you pulled in. However, since you specified that Greyhounds go from NY to SF at noon but did not specify that the buses traveling from SF to NY at noon are Greyhounds, I can only assume that because you specified "How many GREYHOUNDS will the bus leaving NY encounter?" that this is a trick question and there is not enough information (The buses coming the opposite direction may or may not be greyhounds). You're correct about the wording of the OP, but no trick intended. With all the correct assumptions, tho, you're missing a few encounters. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 3, 2009 Author Report Share Posted February 3, 2009 If the westbound bus leaves NY on day 0, it will encounter every eastbound bus that left SF between days -7 to 7 inclusive, or 15 buses. That assumes that you count the two encounters in the bus stations in NY & SF on days 0 & 7, respectively. Yup. Quote Link to comment Share on other sites More sharing options...
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bonanova
Every day at noon, a Greyhound tour bus leaves New York for San Francisco, and another [Greyhound] bus leaves SF for NY.
The cross-country trip takes 168 hours [7 days and 7 nights] and follows Interstate 80 from end to end
[George Washington Bridge to Oakland Bay Bridge.]
How many Greyhounds [buses] will the bus leaving New York today encounter on its trip to San Francisco?
Edited by bonanovaadded [clarifying material]
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