[ARCHIVED THREAD] - Give me my dollar. Math problem. (Page 1 of 2)
Posted: 9/10/2008 11:12:46 AM EDT
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Math is for suckers. Three guys walk into a hotel and are told that for one room (with three beds) it is 30 dollars. Each guy coughs up 10 bucks and go up to their room and watch football. The hotel manager realizes he overcharged since the room was only 25 bucks. So he takes the five bucks, realizes he can't split it between three guys equally, and gives each guy back 1 dollar and keeps 2 dollars. So each guy has now paid a total of 9 dollars and the manager kept 2... 3 guys times 9 dollars = 27 dollars plus the remaining 2 dollars = 29 dollars! If you started with 30 dollars, where did the last dollar go? Show work please. |
Leave the bellboy out. How did the three guys pay 27 dollars???? 
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K, the classic version involves a bellboy. I see now that you left that part out. In this case, the manager has $25 + $2, which equals $27, plus the $3 the men have equals $30. All dollars are accounted for. |
There's your answer 
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You're not thinking about trading in your truck, are you? |
$30-$5=$25 $5 remaining gets divided ($1 X 3men =$3) and $2 for the manager. $25 for the room +$3 (returned amount) =$28 $28 +$2 (manager hold back) $30 What was the question again? |
I am sure she is an avid NEA supporter, and strongly against merit pay as "unfair." |
 
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As the missing dollar approaches the event horizon, it's velocity approaches the speed of light and it's time, as seen from outside, will appear to stop. It will appear stationary. It will also be incredibly red shifted, and will essentially appear invisible. 9x3, +2, + the dollar stuck at the event horizon = 30. Moral of the story - Don't stay at the Hadron bed and breakfast. |
You have got to be kidding me? They EACH paid 9 dollars, and received 1 dollar EACH back. That is 9 dollars times 3 - or 27 dollars, not 28. That is two dollars more than the cost of the room, which explains why the manager still has... wait for it... TWO DOLLARS EXTRA! |
Interestingly, my most memorable "stupid math teacher" moment was 5th grade as well - it was in fiundiung the area of a shape that was essentially two rectangles. I remember even making a 5th grader verison of a proof to show her why she was wrong. I also had a college professor who had no knowledge of basic trigonometry - and was amazed when I showed him how you could calculate the area of a triangular field knowing only the length of two sides and the included angle. |