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# 3 boxes of supplies have an average (arith.mean) weight of 7

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VP
Joined: 22 Nov 2007
Posts: 1104
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3 boxes of supplies have an average (arith.mean) weight of 7 [#permalink]  26 Dec 2007, 07:12
3 boxes of supplies have an average (arith.mean) weight of 7 kilograms and a median weight of 9 kilograms. what is the maximum possible weight, in kilograms, of the lightest box?

A. 1
B. 2
C. 3
D. 4
E. 5
CEO
Joined: 17 Nov 2007
Posts: 3574
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 414

Kudos [?]: 2245 [0], given: 359

Expert's post
C

{a,9,b} a - box with minimum weight, b - box with maximum weight

av=7=a+b+9/3
a=12-b

max(a)=max(12-b)=max(12-9)=3
VP
Joined: 22 Nov 2007
Posts: 1104
Followers: 8

Kudos [?]: 215 [0], given: 0

walker wrote:
C

{a,9,b} a - box with minimum weight, b - box with maximum weight

av=7=a+b+9/3
a=12-b

max(a)=max(12-b)=max(12-9)=3

let's say: a+b+c=21. the median is 9, so b=9, then a+c=12..so 1+11=12
2+10=12, 3+9=12....4+8=12 but it is not valid for the median to be 9...
SVP
Joined: 28 Dec 2005
Posts: 1579
Followers: 2

Kudos [?]: 91 [0], given: 2

also got C

in order for the lightest to be the heaviest it can, we make the heaviest box the same as the median, ie. x+9+9+=21 ... so x is 3
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