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Intern  Joined: 04 Aug 2010
Posts: 13
Three boxes have an average weight of 7kg and a median weigh  [#permalink]

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13 00:00

Difficulty:   25% (medium)

Question Stats: 69% (01:00) correct 31% (01:21) wrong based on 704 sessions

### HideShow timer Statistics 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
Math Expert V
Joined: 02 Sep 2009
Posts: 56303
Re: Box Weight word problem  [#permalink]

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3
3
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 $$a_{max}+9+9=21$$ --> $$a_{max}=3$$.

Hope it's clear.
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Re: Box Weight word problem  [#permalink]

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3
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]

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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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Naveenan Ramachandran
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Re: Box Weight word problem  [#permalink]

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@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.
Senior Manager  Joined: 18 Sep 2009
Posts: 256
Re: Box Weight word problem  [#permalink]

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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)
Intern  Joined: 15 Aug 2010
Posts: 17
Location: Mumbai
Schools: Class of 2008, IIM Ahmedabad
Re: Box Weight word problem  [#permalink]

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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
Veritas Prep GMAT Instructor Joined: 11 Dec 2012
Posts: 312

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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

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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Math Expert V
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Posts: 56303

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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

Merging similar topics.
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Re: Three boxes have an average weight of 7kg and a median weigh  [#permalink]

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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  [#permalink]

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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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GRE 1: Q169 V154 Re: Three boxes have an average weight of 7kg and a median weigh  [#permalink]

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Nice One.
Here is my approach to this one ->

Let the boxes be
w1
w2
w3

In increasing order

w1+w2+w3=21
w2=9

To maximises w1 -> w2=9 And w2=9

Hence w1=> 21-18=3

Hence C

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Re: Three boxes have an average weight of 7kg and a median weigh  [#permalink]

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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

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.

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Re: Three boxes have an average weight of 7kg and a median weigh  [#permalink]

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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 $$a_{max}+9+9=21$$ --> $$a_{max}=3$$.

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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"Do not watch clock; Do what it does. KEEP GOING." Re: Three boxes have an average weight of 7kg and a median weigh   [#permalink] 24 Nov 2018, 06:00
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