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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]

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02 Nov 2006, 20:08
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

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3
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5
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VP
Joined: 25 Jun 2006
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02 Nov 2006, 20:40
C 3 for me.

plug in and solve.

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Manager
Joined: 17 Dec 2004
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02 Nov 2006, 20:45
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.

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Director
Joined: 18 Jul 2006
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02 Nov 2006, 23:23
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.

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Manager
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03 Nov 2006, 00:32
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.

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Senior Manager
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03 Nov 2006, 21:19
Since it says max possible weight, and the mean is 9 for 3,9,9 I go with C

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Senior Manager
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03 Nov 2006, 22:11
choice c

what's the oa?

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Senior Manager
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03 Nov 2006, 22:12
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

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Director
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03 Nov 2006, 23:40
OA was 3 (C) - thanks, all.
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03 Nov 2006, 23:40
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# 3 boxes of supplies have an average (arithmetic mean) weight

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