silverprince wrote:
Hi
Could anyone please advise of an efficient way of testing statement 1? I found that it took me awfully long to come up with the bookend scenarios: 1 house priced less than average and 19 houses priced less than average.
EMPOWERgmatRichC could you please advise?
Thanks
Hi silverprince,
We're told that last year, in a certain housing development, the AVERAGE (arithmetic mean) price of 20 new houses was $160,000. We're asked if MORE than 9 of the 20 houses have prices that were LESS than the average price last year. This question can be solved with a bit of Arithmetic and TESTing VALUES (although you don't have to do much math to get to the solution).
To start, since the average price of the 20 houses was $160,000, we know that the SUM of those 20 prices is (20)($160,000) = $3,200,000. We're ultimately looking to see how we can distribute that total sum to see if more than 9.... OR exactly 9 or less than 9 of the houses are below the $160,000 threshold. I want to reiterate that you DO NOT need to be doing lots of math - as long as you understand how the 20 values "relate" to the $160,000 average.
(1) Last year the greatest price of one of the 20 houses was $219,000.
With Fact 1, we know that the HIGHEST PRICE was $59,000 more than the average... but what about the other 19 prices? To bring the average down to $160,000.....
-We could 'distribute' that $59,000 overage equally to the remaining 19 houses and make ALL 19 below average in price... and the answer to the question is YES.
-We could 'distribute' that $59,000 overage equally to 8 of the 19 houses and make those 8 below average in price...and keep the other 11 AT the exact average price... and the answer to the question is NO.
Fact 1 is INSUFFICIENT
(2) Last year the median of the prices of the 20 houses was $150,000.
With Fact 2, we're dealing with the MEDIAN of the group. If we put the prices of the 20 houses in order from least to greatest, the median would be the AVERAGE of the two "middle prices" (re: the 10th and 11th prices). Since the median price is BELOW the AVERAGE price, we know that 1 OR both of the 10th and 11th prices are BELOW the average.
For example,
-IF the two prices were $150,000 and $150,000, then they are BOTH below the average
-IF the two prices were $130,000 and $170,000, then ONE price is below the average and one is above the average
-IF the two prices were $140,000 and $160,000, then ONE price is below the average and one IS the average
In ALL THREE situations, since we're dealing with the MEDIAN, we know that there are 9 other prices that are equal to (or below) the lower price of that pair (remember, we had to put the numbers in order from least to greatest before we got the median). Thus, we KNOW that those 9 prices are below the average AND we have 1 or 2 additional prices that are below the average (depending on the prices that you chose for the two 'median' houses). Thus, there will ALWAYS be either 10 or 11 houses that are below the average... so the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT
Final Answer:
GMAT assassins aren't born, they're made,
Rich