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Manager
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n/2 [#permalink] New post 21 Aug 2006, 11:24
...

Last edited by lan583 on 11 Feb 2007, 07:54, edited 1 time in total.
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 [#permalink] New post 21 Aug 2006, 11:39
A.

1) [(n/3)*74.5 + (2n/3)*70]/n. n can be eliminated, hence we can have the answer.

2) Insufficient.
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 [#permalink] New post 21 Aug 2006, 14:36
A

St1: (n/3) * 74.5 + (2n/3) * 70 = n * A, where A is average of all
So A can be calculated.

St2: We don't know total numer of people.
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Re: GMATPrep [#permalink] New post 21 Aug 2006, 19:38
lan583 wrote:
...


(1) (Sum of height of n/3 ppl) / (n/3) = 6ft 2.5 in
(Sum of height of n/3 ppl) = n*6ft2.5in/3 --------(a)

(Sum of height of 2n/3 ppl) / (2n/3) = 5ft 10 in
(Sum of height of 2n/3 ppl) = 2n*5ft10in/3-----(b)

From (a) & (b)
sum heights of n ppl= (Sum of height of n/3 ppl)+(Sum of height of 2n/3 ppl)

Hence we can cal avg

AD

(2) Insuff No information abt n

Hence A

Heman
Re: GMATPrep   [#permalink] 21 Aug 2006, 19:38
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