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Each of the 30 boxes in a certain shipment weighs either 10 [#permalink]
09 Oct 2009, 22:25

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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?

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?

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?

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) --> \(y=20\).

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) --> \(y=20\).

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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?

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) --> \(y=20\).

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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.

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?

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. _________________

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

Re: Each of the 30 boxes in a certain shipment weighs either 10 [#permalink]
10 Jun 2013, 08:24

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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?

=> [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 _________________

Re: Each of the 30 boxes in a certain shipment weighs either 10 [#permalink]
16 Aug 2014, 11:07

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