Each of the 30 boxes in a certain shipment with either 10 : PS Archive
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# Each of the 30 boxes in a certain shipment with either 10

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Manager
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Each of the 30 boxes in a certain shipment with either 10 [#permalink]

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04 Oct 2005, 07:01
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Each of the 30 boxes in a certain shipment with either 10 pounds or 20 pounds, and the average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?
A. 4
B. 6
C. 10
D. 20
E. 24
Manager
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04 Oct 2005, 07:49
I don't know if I'm doing this right but I will give it a shot!

a= #of 10 pound boxes
b= # 20 pound boxes
x= # of 20 pound boxes that need to be taken away

(10a+20b)/30=18
a+2b=54
a=54-2b

and a+b=30

54-2b+b=30
24=b
therfore 6=a

hence [10(6)+20(24-x)]/30-x=14
(540-20x)=420-14x
120=6x
x=20

So I will pick D.
Manager
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04 Oct 2005, 07:58
I came to the same result. But you don't really need to include the variable for the 10pound Boxes. I calculated the sum of the averages (18*30) and then set this into relation of the new desired average, whereas X represents the number of 20pound boxes to be removed:

(540-20X)/(30-X)=14

solve it for X and you get 20.
Manager
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04 Oct 2005, 08:14
Thanks for the explanations. I got it now. Answer is D.
SVP
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04 Oct 2005, 20:46
finde here: http://www.gmatclub.com/phpbb/viewtopic ... ht=#135655

btw, tell me the source?????????????
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05 Oct 2005, 00:40
Let number of 20 pounds boxes be x, then number of 10 pound boxes = 30-x

Average weight = 18 = (20x + (30-x)10)/30

540 = 20x + 300 - 10x
240 = 10x
x = 24

So there are 24 20-pounds boxes and 6 10-pound boxes.

We need to reduce number of 20-pounds boxes by y boxes.

Average weight = 14 = (24-y)20 + 60/30-y
420-14y = 480 - 20y + 60
6y = 120
y = 20

Ans: D
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05 Oct 2005, 01:24
huskers wrote:
Each of the 30 boxes in a certain shipment with either 10 pounds or 20 pounds, and the average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?
A. 4
B. 6
C. 10
D. 20
E. 24

Let x be the number of smaller boxes,
y be the number of bigger boxes.
x + y = 30 -> (1)
(10X+20Y)/(X+Y) = 18 -> (2)

Solving equn 1 and 2 gives x = 6, and y = 24.

Let 'z' be the number after some number of bigger boxes are removed.
Then (10x + 20z)/(x+z) = 14, we found that x = 6, substituting we get
z = 4,

Number of boxes to be removed, Y - Z = 24 - 4 hence 20 boxes to be removed.
Re: PS - shipment   [#permalink] 05 Oct 2005, 01:24
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