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

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

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New post 05 Aug 2008, 09:53
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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New post 05 Aug 2008, 09:59
D
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Re: Equation question [#permalink]

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New post 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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New post 05 Aug 2008, 11:04
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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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Re: Equation question [#permalink]

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New post 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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Re: Equation question [#permalink]

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New post 05 Aug 2008, 17:52
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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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Re: Equation question [#permalink]

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New post 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!

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Re: Equation question [#permalink]

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New post 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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Re: Equation question   [#permalink] 06 Aug 2008, 01:43
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Each of the 30 boxes in a certain shipment weighs either 10

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