Bunuel wrote:
Last year, in a certain housing development, the average (arithmetic mean) price of 20 new houses was $160,000. Did more than 9 of the 20 houses have prices that were less than the average price last year?
(1) Last year the greatest price of one of the 20 houses was $219,000.
(2) Last year the median of the prices of the 20 houses was $150,000.
The total price of all 20 house was 160,000 x 20 = $3,200,000. We want to determine whether there are more than 9 houses with prices less than the average price of $160,000.
Statement One Alone:
Last year the greatest price of one of the 20 houses was $219,000.
That means the total price of the remaining 19 houses is $3,200,000 - 219,000 = $2,981,000. It’s possible that more than 9 houses were priced less than $160,000. For example, say 14 houses were priced $150,000 each for a total of $2,100,000. Then the last 5 houses could have been priced (2,981,000 - 2,100,000)/5 = 881,000/5 = $176,200 each. However, it’s also possible that no more than 9 houses were priced less than $160,000. For example, say 9 houses were priced $150,000 each for a total of $1,350,000. Then the last 10 houses could have been priced (2,981,000 - 1,350,000)/10 = 1,631,000/10 = $163,100 each.
Statement one alone is not sufficient.
Statement Two Alone:
Last year the median of the prices of the 20 houses was $150,000.
That means the average price of the 10th and 11th houses (assuming the prices of the houses are listed in ascending order) was $150,000. It is possible that the prices of these two houses were $150,000. If that is the case, then there will be at least 11 houses with prices less than the average price of $160,000. It is also possible that the price of the 10th house is less than $150,000 and the price of the 11th house is more than $150,000. If that is the case, we still have at least 10 houses with prices less than the average price of $160,000. Therefore, in either case, we have more than 9 houses with prices less than the average price.
Statement two alone is sufficient.
Answer: B
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