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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 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 20-pound boxes, how many 20-pound boxes must be removed?

A. 4
B. 6
C. 10
D. 20
E. 24

OA D IMO B
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Re: Boxes  [#permalink]

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New post 10 Oct 2009, 00:07
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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

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 --> \(\frac{10*6+20*(24-y)}{30-y}=14\) --> \(y=20\).

Answer: D.
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Re: Boxes  [#permalink]

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New post 21 Aug 2011, 10:59
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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
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Re: Boxes  [#permalink]

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New post 09 Oct 2009, 23:43
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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

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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Re: Boxes  [#permalink]

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New post 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  [#permalink]

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New post 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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Re: Boxes  [#permalink]

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New post 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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Re: Boxes  [#permalink]

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New post 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

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

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

\(? = (120/6) = 20\)
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Re: Boxes  [#permalink]

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New post 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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Re: Boxes  [#permalink]

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New post 13 Dec 2010, 22:02
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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

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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Re: Boxes  [#permalink]

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New post 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  [#permalink]

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New post 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
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Re: Boxes  [#permalink]

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New post 31 May 2013, 13:55
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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
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Re: Each of the 30 boxes in a certain shipment weighs either 10  [#permalink]

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New post 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 20-pound boxes, how many 20-pound boxes must be removed?

A. 4
B. 6
C. 10
D. 20
E. 24

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

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

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New post 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:

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

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Re: Each of the 30 boxes in a certain shipment weighs either 10   [#permalink] 20 May 2019, 20:12
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