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Interesting question. Looks like GMATPrep has infinite number of questions.
I. Lets pick the best case scenario:
8 houses have pirce equal to median price 130k
Their sum = 130k*8 = 1040k
Mean = 150k
Sum = 15*150k = 2250k
So rest of 7 houses should have average price = (2250k-1040k)/7 = 1210/7 = 172.xx k
Since the best case average of 7 houses is 172.xxk, atlease one of them should be > 165k
II. Not necessarily true. Consider the above example where 8 houses are 130k and 7 houses are 172.xxk, it is not true
Interesting question. Looks like GMATPrep has infinite number of questions.
I. Lets pick the best case scenario: 8 houses have pirce equal to median price 130k Their sum = 130k*8 = 1040k Mean = 150k Sum = 15*150k = 2250k So rest of 7 houses should have average price = (2250k-1040k)/7 = 1210/7 = 172.xx k Since the best case average of 7 houses is 172.xxk, atlease one of them should be > 165k
II. Not necessarily true. Consider the above example where 8 houses are 130k and 7 houses are 172.xxk, it is not true
Interesting question. Looks like GMATPrep has infinite number of questions.
I. Lets pick the best case scenario: 8 houses have pirce equal to median price 130k Their sum = 130k*8 = 1040k Mean = 150k Sum = 15*150k = 2250k So rest of 7 houses should have average price = (2250k-1040k)/7 = 1210/7 = 172.xx k Since the best case average of 7 houses is 172.xxk, atlease one of them should be > 165k
II. Not necessarily true. Consider the above example where 8 houses are 130k and 7 houses are 172.xxk, it is not true
III. Same example as above. Not necessarily true
Hence A.
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Girish you rock.
However, in exam pressure how should we approach?
Well I am no Guru to give advise 'coz I get stumped by such questions too.
Of course it is difficult to solve such problems exhaustively on the test. But you need to also remember that most GMAT questions are based on the same concepts. The testmakers don't invent new porblems everyday; They just can't
For this question you can start with III. rd condition and eliminate it by taking an example of 8 houses being 130k. Try to use the same condition for 2nd and first.
Also you must have seen several problems based on the same concept - Given an average & median, to get the least value of numbers greater than the Median, we need to consider the case where all the numbers smaller than median are equal to the median itself. Believe me this is tested over an over. I can't be more precise because of the confidentiality agreement with GMAC.
Another trick would be to think in terms of smaller number of values. Say pick only 3 numbers. IN MOST CASES what is true of 3 or 5 numbers will be true of 15 or 21 numbers.
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