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A has 300 bags (comprising some 100 lb and some 200 lb) with

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A has 300 bags (comprising some 100 lb and some 200 lb) with [#permalink] New post 18 May 2005, 05:32
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A has 300 bags (comprising some 100 lb and some 200 lb) with an average of 180 lb. how many bags should A remove to get 140 lb as average weight of the bags.
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 [#permalink] New post 18 May 2005, 07:23
(100x+200y)/(x+y)=180
finally you'll end up with y=4x
knowing that x+y=300, you find that x=60 and y=240

Even with X=0 you can see that the average is already > 140 so we can deduce that we will certainly have to move some 200 lbs bags.

let's call V the number of 200 lbs bags we would need if we keep the 60*100lbs :
200V+60(100)/V+60 = 140 ; 200V+6000=140V+840 ; 6V=240
V=40

As we know Y=240, we can see that we need to remove 200 bags to reach 40 bags (240-200=40)

Answer should be 200 bags
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Re: PS: Bags [#permalink] New post 18 May 2005, 14:28
HIMALAYA wrote:
A has 300 bags (comprising some 100 lb and some 200 lb) with an average of 180 lb. how many bags should A remove to get 140 lb as average weight of the bags.


Is it a selective removal of 200lb bags? (I mean this is obvious, but just to confirm- If you remove bags randomly, then the average would tend to 150 lbs, and would not approach 140).

If n bags of 200lbs each are removed,

180x300 - 200n / (300-n) = 140
Solving for n, 54000 - 200 n = 42000 - 140n => 12000 = 60n => n = 200
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Re: PS: Bags   [#permalink] 18 May 2005, 14:28
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A has 300 bags (comprising some 100 lb and some 200 lb) with

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