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# 3 boxes of supplies have an average (arithmetic mean) weight

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Director
Joined: 11 Sep 2006
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3 boxes of supplies have an average (arithmetic mean) weight [#permalink]  02 Nov 2006, 19:08
3 boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?

1
2
3
4
5
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VP
Joined: 25 Jun 2006
Posts: 1173
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Kudos [?]: 79 [0], given: 0

C 3 for me.

plug in and solve.
Manager
Joined: 17 Dec 2004
Posts: 72
Followers: 2

Kudos [?]: 7 [0], given: 0

I get B, 2.

Mean = 7, so total weight of all 3 boxes is 21.

Median = 9, so minimum possible weight for heaviest box is 10.

10 + 9 = 19

21 - 19 = 2

2 is the maximum possible weight of the lightest box.
Director
Joined: 18 Jul 2006
Posts: 531
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halahpeno wrote:
I get B, 2.

Mean = 7, so total weight of all 3 boxes is 21.

Median = 9, so minimum possible weight for heaviest box is 10.

10 + 9 = 19

21 - 19 = 2

2 is the maximum possible weight of the lightest box.

Why it can't be 3, 9, 9?
In this case, max possible weight is 3.
Manager
Joined: 03 Jul 2005
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Location: City
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I would go with 3 (c). There is nothing in the question that states that all the weights are different. You could have 3 9 9 and the median is 9.
Senior Manager
Joined: 24 Oct 2006
Posts: 339
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Since it says max possible weight, and the mean is 9 for 3,9,9 I go with C
Senior Manager
Joined: 05 Oct 2006
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Re: 3 boxes... [#permalink]  03 Nov 2006, 21:11
choice c

what's the oa?
Senior Manager
Joined: 30 Aug 2006
Posts: 374
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yea C

(x+y+z)/3 = 7
(x+y+z) = 21

y must be 9 if median is 9, so :

x + 9 + z = 21
x + z = 12

to maximise x, z = 9, therefore x = 3
Director
Joined: 11 Sep 2006
Posts: 514
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OA was 3 (C) - thanks, all.
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