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Each of the 30 boxes in a certain shipment weighs either 10
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09 Oct 2009, 23:25
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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 20pound boxes, how many 20pound boxes must be removed? A. 4 B. 6 C. 10 D. 20 E. 24
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Re: Boxes
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10 Oct 2009, 00:07
reply2spg 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 20pound boxes, how many 20pound boxes must be removed? A. 4 B. 6 C. 10 D. 20 E. 24 This is a weighted average question: \(Weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\). Let \(x\) be number of 10pound boxes > \(\frac{10x + 20*(30x)}{30}=18\) > \(x=6\); We have \(6\) 10pound boxes and \(24\) 20pound boxes. Let \(y\) be number of 20pound boxes that should be removed so that average to become 14 pounds > \(\frac{10*6+20*(24y)}{30y}=14\) > \(y=20\). Answer: D.
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Re: Boxes
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21 Aug 2011, 10:59
total current weight = 30*18 = 540 let's assume that x number of 20lbs boxes will be removed \(\frac{54020x}{30x} = 14\) x = 20 ............. D
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Re: Boxes
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09 Oct 2009, 23:43
reply2spg 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 20pound boxes, how many 20pound boxes must be removed? A. 4 B. 6 C. 10 D. 20 E. 24 x = 20 lb box: 20x + 10(30x) = 18(30) x = (540  300)/10 x = 24 If k = removed 20 lb box: 20(24k)+ 10(6) = 14(6+24k) 480  20k + 60 = 420  14k 540  420 = 6k 540  420 = 6k k = 120/6 = 20
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Re: Boxes
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13 Dec 2010, 00:28
dont you have to reduce the average weight of 30 booxes to 14 pounds?



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Re: Boxes
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13 Dec 2010, 04:43
Bunuel wrote: So, if \(y\) is the number of 20pound boxes that should be removed so that average to become 14 pounds > \(\frac{10*6+20*(24y)}{30y}=14\), (note that as we are reducing the # of 20pound boxes (24) by y then the total # of boxes (30) also will be reduced by y> is this implied or an assumption) > \(y=20\).
. See my comment in Red Italics. What if we make an attempt to keep total number of boxes 30 by adding \(y\) 10pound boxes. Does this have to be specifically mentioned in the question?



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Re: Boxes
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13 Dec 2010, 05:15
x boxes of 10pounds y boxes of 20pounds x+y = 30
\((10*x+20*y)/30=18\)
\(10x+20y=540\)
Let ? be the number of 20 pound boxes which should be removed in order to get the average weight of 14pounds
\((10*x+20*y20*?)/(30?)=14\)
\((54020*?)= (42014*?)\)
\(? = (120/6) = 20\)



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Re: Boxes
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13 Dec 2010, 05:53
aalriy wrote: Bunuel wrote: So, if \(y\) is the number of 20pound boxes that should be removed so that average to become 14 pounds > \(\frac{10*6+20*(24y)}{30y}=14\), (note that as we are reducing the # of 20pound boxes (24) by y then the total # of boxes (30) also will be reduced by y> is this implied or an assumption) > \(y=20\).
. See my comment in Red Italics. What if we make an attempt to keep total number of boxes 30 by adding \(y\) 10pound boxes. Does this have to be specifically mentioned in the question? You should read the stem more carefully: "If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20pound boxes, how many 20pound boxes must be removed?" So stem mentions that the reduction in average weight should be made by removing some of the 20pound boxes. Hope it's clear.
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Re: Boxes
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13 Dec 2010, 22:02
reply2spg 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 20pound boxes, how many 20pound boxes must be removed? A. 4 B. 6 C. 10 D. 20 E. 24 Long question stem. Best approach here would be to analyze one sentence at a time. 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 average of 10 and 20 is 18, the ratio of no. of 10 pound boxes: no of 20 pound boxes is 1:4 (If this concept is not clear, check out http://gmatclub.com/forum/toughds105651.html#p828579) So out of 30 boxes, 6 are 10 pound boxes and 24 are 20 pound boxes. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20pound boxesNow average of 10 and 20 pound boxes is to be 14. So ratio of no. of 10 pound boxes:no. of 20 pound boxes = 3:2 The number of 10 pound boxes remain the same so we still have 6 of them. To get a ratio of 3:2, the no of 20 pound boxes must be 4. how many 20pound boxes must be removed?There were 24 of these. Now there are only 4. 24  4 = 20 should be removed.
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Re: Boxes
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10 Feb 2012, 06:35
N1+ N2 = 30
10(N1) + 20(N1) _____________ =18 N1+ N2
 > solve 10(N1) + 20(N2) = 18(N1) + 18(N2)
So n1 =6 n2 = 24
Now N1 will not change ONLY n2 will change so
reduce the average to 14
10(6) + 20(N2) = 14(6) + 14(N2)
N2 =4
So it has to reduce by 20



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Re: Each of the 30 boxes in a certain shipment weighs either 10
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29 May 2013, 05:13
This is how I solved X+Y =30 from condition given 10X+20Y = 540 therefore X=6 and Y = 24 Condition 2 X= 6 Y changes by "n" and avg weight of 30n =14 therefore 10*6 + 20(24n)=14( 30n) after solving the above equation we have n=20 Ans is D
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Re: Boxes
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31 May 2013, 13:55
MBAhereIcome wrote: total current weight = 30*18 = 540 let's assume that x number of 20lbs boxes will be removed
\(\frac{54020x}{30x} = 14\)
x = 20 ............. D Faster to use this logic
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Re: Each of the 30 boxes in a certain shipment weighs either 10
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10 Jun 2013, 09:24
reply2spg 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 20pound boxes, how many 20pound boxes must be removed? A. 4 B. 6 C. 10 D. 20 E. 24 Easy question: suppose 10 pound boxes = x 20 pound boxes = (30x) Given in question >> [x*10+(30x)*20]/30=18 => [x*10+(30x)*20] = 540 eq. 1 .. Leave it like this. Now we have to remove some 20 Pound boxeslet 20 pound boxes removed = a new equation will be >> [x*10+(30xa)*20]/(30a)=14 >> x*10+(30x)*20  20a = 14(30a) From eq 1 >> 540  20a = 420 14a solve for a, and you'll get a=20 .. I made it look quite big, but I solved it under 2 minutes using this in first try on such question
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Re: Each of the 30 boxes in a certain shipment weighs either 10
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19 May 2015, 13:25
My solution is as below:
10x+ (30x)20 = 540 x=6 (no. of 10 pound boxes) hence 24 is no. of 20 pound boxes
If y is the new no. of 20 pound boxes required to make the average 14 then,
(10*6+20y) /(6+y) = 14
y=4
hence the no. of 20 pound boxes to be removed is 244 = 20 boxes



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Re: Each of the 30 boxes in a certain shipment weighs either 10
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11 Apr 2018, 20:08
Hi All, This question can be solved in a number of different ways  including by TESTing THE ANSWERS. There's a Number Property rule that can actually make that solution a bit faster. We're told that the average weight is 18 pounds, which makes the total weight 540 pounds (18 pounds x 30 boxes = 540 pounds) Since the box weights are 10 or 20 pounds each, you can figure out the number of each rather quickly. If there were 30 20pound boxes, then the total weight would be 600 pounds. Since the total weight is 540 pounds, that means that 6 of the boxes weigh 10 pounds instead of 20 pounds. So, there are: 6 10pound boxes 24 20pound boxes Since we want to remove some number of the 20pound boxes and get an average weight of 14 pounds, we know that the number of 10pound boxes MUST BE greater than the number of 20pound boxes (this is a Weighted Average Number Property). Since we have 6 10pound boxes, we need to remove at least 19 20pound boxes. We can certainly TEST answer D first OR we can note that answer E (24) would eliminate ALL of the 20pound boxes, which would make the average of the remaining boxes 10 pounds, which is too low (and would prove that answer D was correct). Final Answer: GMAT assassins aren't born, they're made, Rich
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