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Three boxes have an average weight of 7kg and a median weigh [#permalink]
 22 Aug 2010, 21:19
Question Stats:
60% (01:31) correct
40% (00:41) wrong based on 31 sessions
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 [#permalink]
22 Aug 2010, 21:29
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 [#permalink]
22 Aug 2010, 21:30
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 [#permalink]
22 Aug 2010, 22:48
@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 [#permalink]
23 Aug 2010, 05:56
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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 [#permalink]
23 Aug 2010, 10:51
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 [#permalink]
23 Aug 2010, 12:28
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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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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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 3-term 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 600-700 level question as this is easily solvable in ~1 minute for most people. Thoughts? Thanks! -Ron
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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 [#permalink]
09 May 2013, 12:14
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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Re: Three boxes have an average weight of 7kg and a median weigh
[#permalink]
09 May 2013, 12:14
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