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Three boxes have an average weight of 7kg and a median weigh
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22 Aug 2010, 20:19
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Three boxes have an average weight of 7kg and a median weight of 9kg. What is the maximum weight of the smallest box? A. 1 B. 2 C. 3 D. 4 E. 5
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Re: Box Weight word problem
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23 Aug 2010, 04:56
kwhitejr wrote: Three boxes have an average weight of 7kg and a median weight of 9kg. What is the maximum weight of the smallest box?
A. 1 B. 2 C. 3 D. 4 E. 5 Three boxes have an average weight of 7kg > \(a+b+c=3*7=21\); Three boxes have a median weight of 9kg > median of a set with odd terms is middle term, hence \(b=9\); So we have a, 9, c. Question: \(a_{max}=?\) General rule for such kind of problems: to maximize one quantity, minimize the others; to minimize one quantity, maximize the others. So to maximize \(a\) we should minimize \(c\) > minimum value of \(c\) is 9 (\(c\), the third term, can not be less than median, the second term) > so \(a_{max}+9+9=21\) > \(a_{max}=3\). Answer: C. Hope it's clear.
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Re: Box Weight word problem
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22 Aug 2010, 20:29
kwhitejr wrote: Three boxes have an average weight of 7kg and a median weight of 9kg. What is the maximum weight of the smallest box?
A. 1 B. 2 C. 3 D. 4 E. 5 Given that the median is 9 that leaves us with the other two boxes with a combined weight of 12 kg. [Average * number = total  7 * 3 = 21. And total  median = sum of heaviest and smallest box = 12kg] Now 9 should be the median and the smallest box has to have the max weight. To leave the median as 9, the heaviest box should be at least 9 kg ... which gives us 3 kg for the smallest box. Hence the weight of the box should be 3, 9, 9. Hence answer is C.
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Re: Box Weight word problem
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22 Aug 2010, 20:30
Hi, Start with assuming that the smallest is a little less than 9 kg. Say, it is 8 kg. Now 8 + 9 + Largest = (3*3) = 21. This is a contradiction because 'Largest' becomes 4. Hence, reduce the 'smallest' to 7 kg. Now, 7+9+Largest = 21. This will also result in a contradiction. Continue doing this until the 'Largest' is indeed the 'Larger' than 9 kgs (which is the median). Hope this helps. Thanks.
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Re: Box Weight word problem
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22 Aug 2010, 21:48
@4gmatmumbai
Thanks for your suggestion. That is actually what I did originally to solve the question, but I came up with the max weight for the smallest box to be 2. Thus the three boxes would be 2, 9 and 10. However, if the boxes are 3, 9 and 9 then the median is still 9? This is what throws me off.



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Re: Box Weight word problem
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23 Aug 2010, 09:51
lets assume 3 boxes are X,Y,Z x+y+z= 21 Median = 9 ( middle number ) By back solving smallest maximum weight X Y(median) Z 1 9 11 2 9 10 3 9 9 4 9 8 ( it cant be the answer because Z value is below the Y (median value)) 5 9 7 (it cant be the answer because Z value is below the Y (median value))
so the smallest maximum weight is 9 (option C)



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Re: Box Weight word problem
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23 Aug 2010, 11:28
kwhitejr wrote: @4gmatmumbai
Thanks for your suggestion. That is actually what I did originally to solve the question, but I came up with the max weight for the smallest box to be 2. Thus the three boxes would be 2, 9 and 10. However, if the boxes are 3, 9 and 9 then the median is still 9? This is what throws me off. Hi kwhitejr, Definitely  the median of 3,9 and 9 is indeed the 2nd term > 9. Just pick the middle term after sorting them in ascending order  no matter what. Hope this helps. Thanks.
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14. Three boxes of supplies have an average (A.M.) weight of 7 kg and median of 9 kg. What is the maximum possible weight in kg of the lightest box? a)1 b)2 c)3 d)4 e)5



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Re: median
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07 May 2013, 07:58
Option C. 3Kg.
Since we need to maximize the minimum weight , let the weights of the 2 heaviest boxes be equal to the median = (kg.
So total weight of all boxes is 3*7 =21 and the combined weight of the 2 heaviest boxes = 9+9 = 18Kg.
so the maximum weight of the remaining box can be 3 KG



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Re: median
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07 May 2013, 09:32
mdbharadwaj wrote: Option C. 3Kg.
Since we need to maximize the minimum weight , let the weights of the 2 heaviest boxes be equal to the median = (kg.
So total weight of all boxes is 3*7 =21 and the combined weight of the 2 heaviest boxes = 9+9 = 18Kg.
so the maximum weight of the remaining box can be 3 KG This solution is correct, and the whole question hinges only on whether you have the correct definitions for median and average. If the average is 7 then the total weight must be 7x3=21 kg. The median of a 3term set is the middle term, so the middle term must be 9. The biggest term can be no smaller than 9, so at a minimum it's 9, leaving only 3 kg for the smallest entry. My question here is then about the difficulty level. I'm not sure I'd put this as a 600700 level question as this is easily solvable in ~1 minute for most people. Thoughts? Thanks! Ron
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Re: median
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07 May 2013, 09:55
rochak22 wrote: 14. Three boxes of supplies have an average (A.M.) weight of 7 kg and median of 9 kg. What is the maximum possible weight in kg of the lightest box? a)1 b)2 c)3 d)4 e)5 I also agree with md .... I got the answer as C..........
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Re: Three boxes have an average weight of 7kg and a median weigh
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09 May 2013, 11:14
3 Boxes: S, M, L M=9 Therefore L≥M, L≥9
Average= (S+M+L)/3 = 7 Sum= S+M+L = 21 S+L = 12
S must be 3 since L must be at least 9 and we are looking for the maximum S.
Must be C



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Three boxes have an average weight of 7kg and a median weigh
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24 Sep 2016, 14:52
C.
x1 + x2 + x3 = 21 > plug in x2 = 9. We want to minimize x3 in order to maximize x1, so we'll keep x3 @ 9
x2 + x3 = 18 > x1 = 3



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Re: Three boxes have an average weight of 7kg and a median weigh
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02 Jul 2018, 08:49
kwhitejr wrote: Three boxes have an average weight of 7kg and a median weight of 9kg. What is the maximum weight of the smallest box?
A. 1 B. 2 C. 3 D. 4 E. 5 The sum of the weights of the boxes is 3 x 7 = 21 kg, and the median is 9 kg. To make the smallest box as heavy as possible, we can make the largest box also 9 kg. So the heaviest possible weight of the smallest box is 21  18 = 3 kg. Answer: C
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Re: Three boxes have an average weight of 7kg and a median weigh
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24 Nov 2018, 05:00
Bunuel wrote: kwhitejr wrote: Three boxes have an average weight of 7kg and a median weight of 9kg. What is the maximum weight of the smallest box?
A. 1 B. 2 C. 3 D. 4 E. 5 Three boxes have an average weight of 7kg > \(a+b+c=3*7=21\); Three boxes have a median weight of 9kg > median of a set with odd terms is middle term, hence \(b=9\); So we have a, 9, c. Question: \(a_{max}=?\) General rule for such kind of problems: to maximize one quantity, minimize the others; to minimize one quantity, maximize the others. So to maximize \(a\) we should minimize \(c\) > minimum value of \(c\) is 9 (\(c\), the third term, can not be less than median, the second term) > so \(a_{max}+9+9=21\) > \(a_{max}=3\). Answer: C. Hope it's clear. Bunuel, If it is mentioned that all 3 boxes have different weights, then answer will be 2(option B), as we have to take c as 10 and not equal to median that is 9.
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