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Three boxes of supplies have an average (arithmetic mean) [#permalink]

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05 Dec 2010, 14:17

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Three boxes of supplies have an average(arithmetic mean) weight of 7 kilograms and a median weight of 9kg. What is the max possible weight, in kg, of the lightest box?

Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9kg. What is the max possible weight, in kg, of the lightest box?

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As the median value of 3 (odd) # of boxes is 9 then weights of the boxes in ascending order are: {a, 9, b}. Also as the mean equals to 7 then a+9+b=3*7=21;

Now, general rule for such kind of problems: to maximize one quantity, minimize the others; to minimize one quantity, maximize the others.

So we need to maximize the weight of the lightest box, so we want to maximize \(a\) --> to maximize \(a\) we should minimize \(b\) --> min value of \(b\) is 9 (it cannot be less than the median value), so we'll have \(a_{max}+9+9=21\) --> \(a_{max}=3\).

Thanks a lot Bunuel. Your reasoning and approach to Quant problems is the best. Too bad I didn't study well enough in Avg/Medians/SD. MGMAT isn't as elaborate in their treatment. But I promptly went through GMATClub topics on Avg/Medians/SD and solved some problems, without spending too much time as you rightly highlighted

Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

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05 Mar 2013, 08:22

we can't use 11 in this question as maximum value can you please exaplin this part. Thanks. I assumed 11 as the max value and got the answer as 1.
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Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

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05 Mar 2013, 15:22

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Hello Fozzy,

Hopefully I can help you with this one. The question asks us for the maximum weight of the lightest box.

Let us take an example here. Suppose three of your friends got a total of 30 marks in an exam. Now, what would be the maximum possible mark you got? Well, you could calculate the maximum possible mark you got if both of your friends scored 0 in the test(I pity the poor friends!). This would mean that you would score about 30 marks in the test. Any other arrangement would make them score more and consecutively, you would have to score less , right? Does this make sense?

Similarly, if you need to find the maximum possible weight of the lightest box, you would have to minimize the weight of the bigger boxes. Now, we know that one box weighs 9 lb for sure. What is the minimum weight that a box heavier than that must weigh so that it can appear at the end when arranged in ascending order based on weight. Well, the answer is 9kg.

If you consider the weight of the heaviest box to be 11, you would be minimizing the weight of the lightest box.

For example, let x be the lightest box and y the heaviest box.

x+9+y=21 implies, x+y=12. For x to be largest, y=9. x=3. If y=11, then x=1 which is lighter than the maximum possible weight.

Hope this clears your doubt! Let me know if I can help you further.

fozzzy wrote:

we can't use 11 in this question as maximum value can you please exaplin this part. Thanks. I assumed 11 as the max value and got the answer as 1.

Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

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15 Jun 2014, 09:15

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Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

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18 Jan 2016, 03:45

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Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

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01 Jun 2016, 02:26

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Three boxes Average=7 kg So total weight=21kg

Now various combinations are available 1,9,11 2,9,11 3,9,9 but we cannot go beyond 3,9,9 to 4,9,8 As numbers right to median must be equal or greater than the median value.

Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

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06 Dec 2016, 11:41

Very helpful in understanding!!

Ashishsteag wrote:

Three boxes Average=7 kg So total weight=21kg

Now various combinations are available 1,9,11 2,9,11 3,9,9 but we cannot go beyond 3,9,9 to 4,9,8 As numbers right to median must be equal or greater than the median value.

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