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Tom, Jane, and Sue each purchased a new house. The average

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trickygmat
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Tom, Jane, and Sue each purchased a new house. The average [#permalink] New post   10 Oct 2005, 22:03
Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was $120,000. What was the median price of the three houses?
(1) The price of Tom’s house was $110,000.
(2) The price of Jane’s house was $120,000.



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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink] New post   10 Oct 2005, 22:32
Ans is C

Both combined together would give median (120,000)
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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink] New post   10 Oct 2005, 22:32
We need both the statements to arrive at median.

Given only one price and average would only give us combined price of other two that can affect the median price.
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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink] New post   10 Oct 2005, 22:34
B

1. T=110K: J,S = 150,100 -> med T -or- J,S = 120,130-> med J
2. J=120K
=> T+S=120K,
Either T=K=S => med = 120 or T>J>S or T<J<S
In any case, median is 120K
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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink] New post   10 Oct 2005, 22:47
Oops. I fell for the trap. Thanks for correcting gsr.
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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink] New post   11 Oct 2005, 05:50
gsr wrote:
B

1. T=110K: J,S = 150,100 -> med T -or- J,S = 120,130-> med J
2. J=120K
=> T+S=120K,
Either T=K=S => med = 120 or T>J>S or T<J<S
In any case, median is 120K


Agree. B is sufficient to calculate the median. Good trap question.



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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink]
11 Oct 2005, 05:50
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