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why do people say new years is a "fresh new start" when for most people it starts with a hangover ...and...nothing changes, there are still going to be the same problems as the day before..yeah, its a new year, but you say you'll change for the new year, why cant you do that any day? quit smoking, lose weight, just do it!! ...i think its just an excuse..new years eve is now mostly about getting wasted and partying...then "having a fresh new start"? thats messed up in my opinion...does anyone agree?...what are your thoughts on new years?...what are your traditions...?

2006-12-30 14:27:19 · 5 answers · asked by Anonymous in Society & Culture Holidays Other - Holidays

5 answers

It's just another day. But we measure time by years and so it is a chance to start over. Of course, you can start over any time you want. Not everyone drinks, you know.

2006-12-30 14:39:18 · answer #1 · answered by Beth T 5 · 2 0

New Year is a fresh start, since alot of people like to make changes in their lives and leave the past year behind them. Yes I do get drunk but I'm lucky that I don't suffer from hangovers. I don't see it as an excuse to make resolutions to change things and make me a better person. I see it an New Year and a New me!

2006-12-30 22:37:57 · answer #2 · answered by mrsrhowell 3 · 1 0

Speak for yourself. I don't get drunk and hungover, and it's not someting everyone has to do. If people choose to do that then that's on them. I personally have a blast on New Years with lots of friends, family, and loved ones. I believe in fresh starts, but it is up to each individual either to make the most out of life or waste it.

2006-12-30 23:14:53 · answer #3 · answered by Consuming Fire 7 · 1 1

New Year's Eve is for party animals.
New Year's day is for all who seek
a fresh start.
It's been this way since I can remember.

2006-12-30 22:33:38 · answer #4 · answered by kyle.keyes 6 · 1 0

I'd rather be partying this new year's eve in NYC instead of a dumpy antiparty small town.

2007-01-01 02:10:59 · answer #5 · answered by chuck h 5 · 1 0

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