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

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Manager
Joined: 04 Jan 2008
Posts: 117
Each of the 30 boxes in a certain shipment weighs either 10 [#permalink]

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06 Sep 2008, 00:28
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Question Stats:

0% (00:00) correct 100% (02:17) wrong based on 1 sessions

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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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Director
Joined: 23 Sep 2007
Posts: 767

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06 Sep 2008, 00:59
dancinggeometry 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

D

let x = number of 10lbs boxes
let y = number of 20lbs boxes

10x + 20y = 18(x+y)
y/x = 4/1 or 24/6

so 6 10lbs boxes, 24 20lbs boxes

new average = 14
6(10) + y(20) = 14(6 + y)
y = 4

so 24 - 4 = 20
VP
Joined: 05 Jul 2008
Posts: 1367

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06 Sep 2008, 11:46
1st eq is

10 X + 20 Y = 30 X 18

2nd eq is

10 X + 20 (y-k) = (30-k) X 14

use (1) in (2)

540 -20k = 420 -14k

6k= 120 means k =20
Intern
Joined: 03 Sep 2008
Posts: 22

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06 Sep 2008, 15:42
As much as possible you want to avoid algebra when doing GMAT problems.

You can think of this as a weighted average problem with 2/10 of the weight on the value 10 and 8/10 on 20 so there must be 6 of the 10 lbs boxes and 24 of the 20 lbs boxes. Since 14 is closer to 10 than 20, after removing the 20 lbs boxes there must be more of the 10 lbs boxes than the 20 lbs boxes. Answer D is the only answer that makes sense.
Manager
Joined: 04 Jan 2008
Posts: 117

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06 Sep 2008, 23:29
OA is D. Bin 2 problem.
Retired Moderator
Joined: 18 Jul 2008
Posts: 920

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07 Sep 2008, 11:40
I don't quite understand how you get 2/10 of the weight vs 8/10 on the weight.

Can you explain more closely? Thanks.

lsmv479 wrote:
As much as possible you want to avoid algebra when doing GMAT problems.

You can think of this as a weighted average problem with 2/10 of the weight on the value 10 and 8/10 on 20 so there must be 6 of the 10 lbs boxes and 24 of the 20 lbs boxes. Since 14 is closer to 10 than 20, after removing the 20 lbs boxes there must be more of the 10 lbs boxes than the 20 lbs boxes. Answer D is the only answer that makes sense.
SVP
Joined: 07 Nov 2007
Posts: 1756
Location: New York

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07 Sep 2008, 13:11
1
KUDOS
dancinggeometry 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

Fastest way:
Total weight = 30*18=540
say n= no.of 20 lb boxes to be removed

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

Weighted average method: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

twently 20 pound boxes need to be removed.
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Current Student
Joined: 06 Oct 2008
Posts: 38

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23 Oct 2008, 21:45
Hey x2suresh,

Your fastest way is so stupidly easy. I love it! thanks!
I can't believe I didn't think of it like that! I did it the long method and solved it in 4 mins (wasn't happy with my time)

Sumi
VP
Joined: 30 Jun 2008
Posts: 1018

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23 Oct 2008, 22:34
x2suresh wrote:
dancinggeometry 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

Weighted average method: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

twently 20 pound boxes need to be removed.

Can someone explain the parts in red ?? :stupid

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Re: Zumit PS 002   [#permalink] 23 Oct 2008, 22:34
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