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What is the minimum number of RECTANGULAR shipping boxes [#permalink]

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17 Apr 2013, 01:18

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C

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E

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Hi

I was going through Kaplan book DS question. I just wanted someone to verify/comment on what i m thinking. the question is:

What is the minimum number of RECTANGULAR shipping boxes Company L will need in order to ship 120 rectangular packages that are rectangular solids, all of which have exactly the same dimensions?

(1) The dimensions of the packages are 3 inches in length, 4 inches in depth, and 6 inches in height. (2) Each box is a cube of length one foot

Well certainly we can use both the statements to find out the answer easily, which is stated as the answer in the book too.. Just that what made me wonder was since we know the net volume of packages which is 3x4x6x120.. can we not equate this with volume of a BOX (i know dimensions are not given) whose dimensions we think upon to give the least number of boxes??

Let me make myself clear: 3x4x6x120= X (volume of box).. we need to minimize X or we can say maximise volume of box..

can the volume not be maximised by putting values based on assumptions ( which we tend to do in DS).. coz if we do so.. dimensions of the box can be 12x6x120 or 24x3x120 or etc etc.. each giving the same volume and making the number of rectangular boxes as ONE.. which would be least.. HENCE ''A'' SHOULD BE SUFFICIENT?

Please tell me where I am going wrong!! Thanks

p.s. i usually get embarrassed when ppl ask me to search for a topic before posting one , I swear i did this time.. and I believe the similar posts have slightly different doubt with a slightly moulded question..

Re: What is the minimum number of RECTANGULAR shipping boxes [#permalink]

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17 Apr 2013, 03:17

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I'm not sure I understood what you meant. By I can try to explain my view.

Given statement 1, you say that the volume of the boxes is \(120*3*4*6\) CORRECT. But if we don't know the dimension of the box, we cannot say how many of those we need. let me give you a simple example: \(AreaBox= 100000000\) (huge number) => one box is enough \(AreaBox = 20\) => more than one is necessary. A is not Sufficient

Let me know if this helps
_________________

It is beyond a doubt that all our knowledge that begins with experience.

I was going through Kaplan book DS question. I just wanted someone to verify/comment on what i m thinking. the question is:

What is the minimum number of RECTANGULAR shipping boxes Company L will need in order to ship 120 rectangular packages that are rectangular solids, all of which have exactly the same dimensions?

(1) The dimensions of the packages are 3 inches in length, 4 inches in depth, and 6 inches in height. (2) Each box is a cube of length one foot

Well certainly we can use both the statements to find out the answer easily, which is stated as the answer in the book too.. Just that what made me wonder was since we know the net volume of packages which is 3x4x6x120.. can we not equate this with volume of a BOX (i know dimensions are not given) whose dimensions we think upon to give the least number of boxes??

Let me make myself clear: 3x4x6x120= X (volume of box).. we need to minimize X or we can say maximise volume of box..

can the volume not be maximised by putting values based on assumptions ( which we tend to do in DS).. coz if we do so.. dimensions of the box can be 12x6x120 or 24x3x120 or etc etc.. each giving the same volume and making the number of rectangular boxes as ONE.. which would be least.. HENCE ''A'' SHOULD BE SUFFICIENT?

Please tell me where I am going wrong!! Thanks

p.s. i usually get embarrassed when ppl ask me to search for a topic before posting one , I swear i did this time.. and I believe the similar posts have slightly different doubt with a slightly moulded question..

Dear Swarman, I believe Zarrolou gave a fine response, but I am going to add my 2¢ as well.

The prompt says "What is the minimum number of RECTANGULAR shipping boxes Company L will need in order to ship 120 rectangular packages that are rectangular solids, all of which have exactly the same dimensions?' We do not know how many packages will fit in each box. Pay very close attention to the wording --- "packages" vs. "boxes" ---- both are rectangular, and we are putting the former inside the latter. We assume that the packages are smaller than the boxes, but we have no way of knowing how small. We also don't know, from the prompt, whether we will be able to pack the packages into the boxes neatly with no space left over.

As it happens, if we take both statement #1 & statement #2, we see that we can pack the packages neatly --- 4 packages wide, 3 packages deep, 2 packages high, or 24 packages total in each box. Without knowing the size of the box, we could not have calculated this, so we would have no way to know the size of the box.

Re: What is the minimum number of RECTANGULAR shipping boxes [#permalink]

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18 Apr 2013, 10:48

Thank you so much guys for the reply.. I certainly agree to what you say. Actually this question is of DS, which always makes me more sceptical towards its answers.. My approach simply was that the minimum no of packages can be one only if we can get net volume of the boxes to be as the product of 3 possible dimensions, which it certainly does in this case as the net volume of boxes an have several possible dimensions of the package which would allow us to use ONLY ONE package. is this thinking wrong ??

was i able to put across what the doubt in my mind is?

P.S.-i know it makes me sound like an insane.. had i read this question before starting with my prep i would have had the same approach for choosing C as well..probably i have been thinking in excess..

Re: What is the minimum number of RECTANGULAR shipping boxes [#permalink]

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18 Apr 2013, 11:06

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Ahahah, you are not thinking too much! The correct approach to solve DS is: look at the question => manipulate the question [ try to make it more friendly ie: x^2+1>0? becomes is x<-1 or x>1?] read the statement => manipulate the statements [to see if they help]

You have to be aware of the limits of the "manipulating" phase: you CANNOT add info or assume anything. I know this sounds obvious, but this is what happened here.

swarman wrote:

Thank you so much guys for the reply.. I certainly agree to what you say. My approach simply was that the minimum no of packages can be one only if we can get net volume of the boxes to be as the product of 3 possible dimensions, which it certainly does in this case as the net volume of boxes an have several possible dimensions of the package which would allow us to use ONLY ONE package. is this thinking wrong ??

This is way too much! Of course we can COULD use one package, given your conditions above, but "common sense" tells me that if I don't know how big a box is, I cannot know how many of those I need.

Stupid ( I know ) but we have to keep "common sense" with us to beat the GMAT!

Hope this helps
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: What is the minimum number of RECTANGULAR shipping boxes [#permalink]

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05 Nov 2015, 00:15

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Re: What is the minimum number of RECTANGULAR shipping boxes [#permalink]

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19 Nov 2015, 03:07

In option B, only length of the box is mentioned. It is possible that the width and height of the box are less than those of the rectangular package. Please explain why that case isn't taken into consideration. As per the info available, the answer must be E. Why isn't it so?

In option B, only length of the box is mentioned. It is possible that the width and height of the box are less than those of the rectangular package. Please explain why that case isn't taken into consideration. As per the info available, the answer must be E. Why isn't it so?

Dear sravanthi30, I'm happy to respond. In math, the details matter. Here's the exact wording of statement 2:

(2) Each box is a cube of length one foot

The word "cube" has a technical meaning in mathematics. It is not simply any random rectangular solid. A cube is a rectangular solid in which length and width and height are all equal. Thus, if the length of the cube is 1 ft, then it absolutely must be true that the height is 1 ft and the width is 1 ft. Part of success in math is reading carefully.

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