GMATPrep Q23

median is 130,000 let price of rest of 14 homes is X.

(130,000 + 14X)/15 = 150,000
(130,000 + 14X) = 2,250,000
14X=1,12,0000
X=80,000

So avg price of rest of the 14 homes is 80,000.
with the assumption that the three choices are given excluding the median priced home the choice would be III only.

What is the OA?

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jayhawk2001: Community Manager; Posts: 789; Joined: Sun Jan 28, 2007 3:51 pm; Location: Silicon valley, California; Thanked: 30 times; Followed by:1 members

by jayhawk2001 » Tue May 29, 2007 9:20 pm

Is it A i.e. I only.

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800GMAT: Master | Next Rank: 500 Posts; Posts: 174; Joined: Mon Mar 26, 2007 5:51 am; Thanked: 1 times

by 800GMAT » Tue May 29, 2007 9:59 pm

Hi jayhawk,
can u pls explain how did u arrive at A? Thnkx

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jayhawk2001: Community Manager; Posts: 789; Joined: Sun Jan 28, 2007 3:51 pm; Location: Silicon valley, California; Thanked: 30 times; Followed by:1 members

by jayhawk2001 » Wed May 30, 2007 12:17 pm

If we consider first 8 home-prices (sorted list) 130k, then we have a
solution for which

8*130k + 7*x = 15 * 150k
x ~ 172k

Since the above condition can exist, we can discard II and III.

To validate I, consider the least price for the max-priced home.
This can happen if all homes are the average (150k) but we know
that median is 130k.

So, least value for max-priced home = 130k * 8 + x*7 i.e. the last
7 homes are of same value. Even this condition shows x ~ 172k and
so there should be atleast 1 home priced > 165k.

Hence A