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# Each of the 30 boxes in a certain shipment weighs either 10

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Joined: 22 Jul 2008
Posts: 42
Each of the 30 boxes in a certain shipment weighs either 10  [#permalink]

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05 Aug 2008, 09:53
Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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

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Re: Equation question  [#permalink]

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05 Aug 2008, 09:59
D
7-t37573
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Joined: 07 Nov 2007
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Re: Equation question  [#permalink]

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05 Aug 2008, 10:24
rnemani wrote:
Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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

10 (x/x+y) + 20 (y/x+y) = 18
should be same as

0 (x/x+y) + 10 (y/x+y) = 8 --> 10*y/30 =8
--> y=24
x=6

-- for 14 weighted average
0 (x/x+y)+10 (y/6+y) = 4 --> 10y = 24+6y
--> y=4

20 pound boxes need to be removed.

D
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Re: Equation question  [#permalink]

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05 Aug 2008, 11:04
1
Sum/30 = 18
s = 540

Let the number of boxes that need to be removed = n

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

n = 20

IMO D
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Joined: 07 Nov 2007
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Location: New York
Re: Equation question  [#permalink]

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05 Aug 2008, 11:06
x97agarwal wrote:
Sum/30 = 18
s = 540

Let the number of boxes that need to be removed = n

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

n = 20

IMO D

You are right this is fastest way.

+1 for you
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VP
Joined: 17 Jun 2008
Posts: 1279
Re: Equation question  [#permalink]

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05 Aug 2008, 17:52
1
rnemani wrote:
Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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

IMO D
14(30-x)=(18*30)-20x where x=number of 20 pound boxes to be removed
solving x=20
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Joined: 04 Sep 2007
Posts: 205
Re: Equation question  [#permalink]

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05 Aug 2008, 19:38
x97agarwal wrote:
Sum/30 = 18
s = 540

Let the number of boxes that need to be removed = n

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

n = 20

IMO D

Great work, Agarwal. Deserves +1!
Director
Joined: 27 May 2008
Posts: 530
Re: Equation question  [#permalink]

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06 Aug 2008, 01:43
rnemani wrote:
Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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

befor removing if there are x 10 lb boxes
10x + 20(30-x) = 18*30 = 540
x = 6

10 lb boxes = 6
20 lb boxes = 24

after removing some 20lb boxes, the average moves closer to 10 than 20 (its 14) means there'll be more 10 lb boxes than 20 lb boxes.. left 20lb boxes should be less than 6.

obviously we can not remove all 2o lb boxes (24) .. option E is out

only option D fits the requirement...

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This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: Equation question &nbs [#permalink] 06 Aug 2008, 01:43
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