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Yesterday, an automobile dealership sold exactly 15 vehicl

by varun289 » Sat Mar 02, 2013 6:20 am
Yesterday, an automobile dealership sold exactly 15 vehicles for a total of $225,000. Did at least one of the vehicles sell for more than $16,500?

(1) The median price for the 15 vehicles was $13,000.

(2) The range for the price of the 15 vehicles was $4,000.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked

Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient

EACH statement ALONE is sufficient to answer the question asked

Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed



IMO - D, a??
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Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Sat Mar 02, 2013 8:55 am
varun289 wrote:Yesterday, an automobile dealership sold exactly 15 vehicles for a total of $225,000. Did at least one of the vehicles sell for more than $16,500?

(1) The median price for the 15 vehicles was $13,000.

(2) The range for the price of the 15 vehicles was $4,000.
Target question: Did at least one of the vehicles sell for more than $16,500?

Given: Exactly 15 vehicles sold for a total of $225,000

Statement 1: The median price for the 15 vehicles was $13,000.
So, if we arrange the 15 prices in ascending order, they look like this:
_ _ _ _ _ _ _ 13 _ _ _ _ _ _ _ (let's use 13 to represent $13,000)
This tells us that 8 of the prices are less than or equal to 13.
Let's try to add some prices such none of them are greater than 16.5 (i.e., $16,500)
To do this, we'll need to minimize the range of the 15 values.
So, let's make all 7 prices to the left of the median 13 to get:
13 13 13 13 13 13 13 13 _ _ _ _ _ _ _
Since the sum of all 15 values must be 225, the remaining 7 values must add to 121
So, if we want to minimize the range of all 15 prices, we could make each value 121/7 to get:
13 13 13 13 13 13 13 13 121/7 121/7 121/7 121/7 121/7 121/7 121/7
Now recognize that 121/7 = 17.3 (approximately)
So, even when we try to minimize the range, we still get at least one price greater than 16.5

This tells us that it's impossible to have a median of 13 and have all of the prices less than 16.5.
In other words, it must be the case that at least one of the vehicles sold for more than $16,500

Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The range for the price of the 15 vehicles was $4,000.
There are several sets of values that meet this condition. Here are two:
Case a: the prices are {13,13,13,13,13,13,13,15,17,17,17,17,17,17,17} in which case at least one vehicle sold for more than $16,500
Case b: the prices are {12,12,12,13,16,16,16,16,16,16,16,16,16,16,16} in which case no vehicles sold for more than $16,500
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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