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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 06 Sep 2008, 00:28
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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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Re: Zumit PS 002 [#permalink]

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New post 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

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Re: Zumit PS 002 [#permalink]

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New post 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

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Re: Zumit PS 002 [#permalink]

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New post 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.

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Re: Zumit PS 002 [#permalink]

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New post 06 Sep 2008, 23:29
OA is D. Bin 2 problem.

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Re: Zumit PS 002 [#permalink]

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New post 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.

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Re: Zumit PS 002 [#permalink]

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New post 07 Sep 2008, 13:11
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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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Re: Zumit PS 002 [#permalink]

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New post 23 Oct 2008, 21:45
Hey x2suresh,

*hits himself on the forehead*

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

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Re: Zumit PS 002 [#permalink]

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

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