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

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Senior Manager
Joined: 12 Oct 2008
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Each of the 30 boxes in a certain shipment weighs either 10 [#permalink]

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09 Oct 2009, 23:25
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64% (02:02) correct 36% (02:06) wrong based on 795 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

[Reveal] Spoiler:
OA D IMO B
[Reveal] Spoiler: OA
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09 Oct 2009, 23:43
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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

[Reveal] Spoiler:
OA D IMO B

x = 20 lb box:

20x + 10(30-x) = 18(30)
x = (540 - 300)/10
x = 24

If k = removed 20 lb box:

20(24-k)+ 10(6) = 14(6+24-k)
480 - 20k + 60 = 420 - 14k
540 - 420 = 6k
540 - 420 = 6k
k = 120/6 = 20
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Joined: 02 Sep 2009
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10 Oct 2009, 00:07
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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

[Reveal] Spoiler:
OA D IMO B

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 10-pound boxes --> $$\frac{10x + 20*(30-x)}{30}=18$$ --> $$x=6$$;

We have $$6$$ 10-pound boxes and $$24$$ 20-pound boxes.

Let $$y$$ be number of 20-pound boxes that should be removed so that average to become 14 pounds --> $$\frac{10*6+20*(24-y)}{30-y}=14$$ --> $$y=20$$.

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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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13 Dec 2010, 02:47
aalriy wrote:
dont you have to reduce the average weight of 30 booxes to 14 pounds?

Yes, the average weight of the 30 boxes is to be reduced to 14 pounds by removing some of the 20-pound boxes.

We found that there were 6 10-pound boxes and 24 20-pound boxes.

So, if $$y$$ is the number of 20-pound boxes that should be removed so that average to become 14 pounds --> $$\frac{10*6+20*(24-y)}{30-y}=14$$, (note that as we are reducing the # of 20-pound boxes (24) by y then the total # of boxes (30) also will be reduced by y) --> $$y=20$$.

Hope it's clear.
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13 Dec 2010, 04:43
Bunuel wrote:
So, if $$y$$ is the number of 20-pound boxes that should be removed so that average to become 14 pounds --> $$\frac{10*6+20*(24-y)}{30-y}=14$$, (note that as we are reducing the # of 20-pound 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 $$y$$ 10-pound boxes. Does this have to be specifically mentioned in the question?
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Joined: 03 Sep 2006
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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

$$(10*x+20*y-20*?)/(30-?)=14$$

$$(540-20*?)= (420-14*?)$$

$$? = (120/6) = 20$$
Math Expert
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13 Dec 2010, 05:53
aalriy wrote:
Bunuel wrote:
So, if $$y$$ is the number of 20-pound boxes that should be removed so that average to become 14 pounds --> $$\frac{10*6+20*(24-y)}{30-y}=14$$, (note that as we are reducing the # of 20-pound 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 $$y$$ 10-pound 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 20-pound boxes, how many 20-pound boxes must be removed?" So stem mentions that the reduction in average weight should be made by removing some of the 20-pound boxes.

Hope it's clear.
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13 Dec 2010, 22:02
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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

[Reveal] Spoiler:
OA D IMO B

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/tough-ds-105651.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 20-pound boxes

Now 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 20-pound boxes must be removed?
There were 24 of these. Now there are only 4. 24 - 4 = 20 should be removed.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Status: mba here i come! Joined: 07 Aug 2011 Posts: 243 Re: Boxes [#permalink] ### Show Tags 21 Aug 2011, 10:59 6 This post received KUDOS 2 This post was BOOKMARKED total current weight = 30*18 = 540 let's assume that x number of 20lbs boxes will be removed $$\frac{540-20x}{30-x} = 14$$ x = 20 ............. D _________________ press +1 Kudos to appreciate posts Manager Joined: 02 Nov 2009 Posts: 131 Re: Boxes [#permalink] ### Show Tags 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 Intern Joined: 24 May 2013 Posts: 19 Location: United Kingdom WE: Project Management (Real Estate) Re: Each of the 30 boxes in a certain shipment weighs either 10 [#permalink] ### Show Tags 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 30-n =14 therefore 10*6 + 20(24-n)=14( 30-n) after solving the above equation we have n=20 Ans is D _________________ Correct me If I'm wrong !! looking for valuable inputs Manager Joined: 24 Nov 2012 Posts: 173 Concentration: Sustainability, Entrepreneurship GMAT 1: 770 Q50 V44 WE: Business Development (Internet and New Media) Re: Boxes [#permalink] ### Show Tags 31 May 2013, 13:55 1 This post was BOOKMARKED MBAhereIcome wrote: total current weight = 30*18 = 540 let's assume that x number of 20lbs boxes will be removed $$\frac{540-20x}{30-x} = 14$$ x = 20 ............. D Faster to use this logic _________________ You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprep-com/ - This is worth its weight in gold Economist GMAT Test - 730, Q50, V41 Aug 9th, 2013 Manhattan GMAT Test - 670, Q45, V36 Aug 11th, 2013 Manhattan GMAT Test - 680, Q47, V36 Aug 17th, 2013 GmatPrep CAT 1 - 770, Q50, V44 Aug 24th, 2013 Manhattan GMAT Test - 690, Q45, V39 Aug 30th, 2013 Manhattan GMAT Test - 710, Q48, V39 Sep 13th, 2013 GmatPrep CAT 2 - 740, Q49, V41 Oct 6th, 2013 GMAT - 770, Q50, V44, Oct 7th, 2013 My Debrief - http://gmatclub.com/forum/from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542 Manager Joined: 20 Jun 2012 Posts: 93 Location: United States Concentration: Finance, Operations GMAT 1: 710 Q51 V25 Re: Each of the 30 boxes in a certain shipment weighs either 10 [#permalink] ### Show Tags 10 Jun 2013, 09:24 2 This post received KUDOS 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 20-pound boxes, how many 20-pound boxes must be removed? A. 4 B. 6 C. 10 D. 20 E. 24 [Reveal] Spoiler: OA D IMO B Easy question: suppose 10 pound boxes = x 20 pound boxes = (30-x) Given in question >> [x*10+(30-x)*20]/30=18 => [x*10+(30-x)*20] = 540 ----eq. 1 .. Leave it like this. Now we have to remove some 20 Pound boxes let 20 pound boxes removed = a new equation will be >> [x*10+(30-x-a)*20]/(30-a)=14 >> x*10+(30-x)*20 - 20a = 14(30-a) 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 _________________ Forget Kudos ... be an altruist Current Student Joined: 19 Nov 2014 Posts: 3 Location: United States GMAT 1: 600 Q41 V31 GPA: 3.88 WE: Project Management (Pharmaceuticals and Biotech) Re: Each of the 30 boxes in a certain shipment weighs either 10 [#permalink] ### Show Tags 19 May 2015, 13:25 My solution is as below: 10x+ (30-x)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 24-4 = 20 boxes EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 11489 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: Each of the 30 boxes in a certain shipment weighs either 10 [#permalink] ### Show Tags 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 20-pound 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 10-pound boxes 24 20-pound boxes Since we want to remove some number of the 20-pound boxes and get an average weight of 14 pounds, we know that the number of 10-pound boxes MUST BE greater than the number of 20-pound boxes (this is a Weighted Average Number Property). Since we have 6 10-pound boxes, we need to remove at least 19 20-pound boxes. We can certainly TEST answer D first OR we can note that answer E (24) would eliminate ALL of the 20-pound 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: [Reveal] Spoiler: D GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: Each of the 30 boxes in a certain shipment weighs either 10   [#permalink] 11 Apr 2018, 20:08
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