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mean median

by buoyant » Thu Dec 25, 2014 7:36 pm
For a certain set of n numbers, where n > 2, is the average
(arithmetic mean) equal to the median?
(1) The n numbers are positive, consecutive even
integers.
(2) The average of the n numbers is equal to the
average of the largest and smallest numbers
in the set.

[spoiler]OA:A[/spoiler]
[spoiler]
why is statement 2 insufficient?
i tried different combinations of numbers and got mean=median:
e.g. 3,5,5,7 median=5, mean=5 mean=median . even though the set is not evenly spaced
3,4,6,7 median=5, mean=5 mean=median . also,set is not evenly spaced
i don't find a case to disprove statement 2[/spoiler]
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Source: — Data Sufficiency |

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by [email protected] » Thu Dec 25, 2014 8:07 pm
Hi buoyant,

Your idea to TEST VALUES is perfect for this scenario. Fact 2 gives us a tricky scenario to deal with because you instinctively want to focus on positive numbers. The key to finding the "exception" that you're looking for is to think about numbers that are NOT positive and/or duplicate values.

I used (-1,-1, 3, 4, 5).

Average of biggest and smallest = (-1 + 5)/2 = 4/2 = 2
Average of all 5 terms = (12 - 2)/5 = 10/5 = 2
Median = 3

Fact 2 is INSUFFICIENT

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by Brent@GMATPrepNow » Sat Dec 27, 2014 10:43 am
buoyant wrote:For a certain set of n numbers, where n > 2, is the average (arithmetic mean) equal to the median?

(1) The n numbers are positive, consecutive even integers.
(2) The average of the n numbers is equal to the average of the largest and smallest numbers in the set.
Target question: Is the average (arithmetic mean) EQUAL to the median?

Statement 1: The n numbers are positive, consecutive even integers.
There's a nice rule that says, "In a set where the numbers are equally spaced, the mean will equal the median."
For example, in each of the following sets, the mean and median are equal:
{7, 9, 11, 13, 15}
{-1, 4, 9, 14}
{3, 4, 5, 6}

Statement 1 tells us that the numbers are consecutive even integers, which means they are equally spaced.
As such, we can be certain that the mean and median are equal.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The average of the n numbers is equal to the average of the largest and smallest numbers in the set.
This statement doesn't FEEL sufficient, so I'm going to try testing some different values.

Aside: For more on this idea of testing values when a statement doesn't feel sufficient, you can read my article: https://www.beatthegmat.com/mba/2013/10/ ... -in-values

There are several different sets that satisfy this condition. Here are two:
Case a: the numbers are {1, 2, 3}. Here, the mean is equal to the average of the biggest and smallest numbers. In this case the median and mean ARE equal
Case b: the numbers are {-3, -3, 1, 2, 3}. Here, the mean is equal to the average of the biggest and smallest numbers. In this case the median and mean are NOT equal
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

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by GMATGuruNY » Sat Dec 27, 2014 4:09 pm
An alternate way to evaluate statement 2:
(2) The average of the n numbers is equal to the
average of the largest and smallest numbers
in the set.
Let n=5, implying a set of 5 numbers.
Let the 5 numbers, in ascending order, be a, b, c, d and e.

Since the average of the 5 numbers is equal to the average of a and e, we get:
(a+b+c+d+e)/5 = (a+e)/2
2(a+b+c+d+e) = 5(a+e)
2(b+c+d) + 2(a+e) = 5(a+e)
2(b+c+d) = 3(a+e)
(a+e)/(b+c+d) = 2/3.

Let a+e = 8 and b+c+d = 12, so that (a+e)/(b+c+d) = 8/12 = 2/3.
Average of the 5 numbers = (a+b+c+d+e)/5 = (8+12)/5 = 4.
Let a=0 and e=8, so that a+e = 8.

Case 1: The numbers are evenly spaced
Let b=2, c=4, and d=6, so that b+c+d = 12.
Resulting set:
0, 2, 4, 6, 8.
In this case, the average of the 5 numbers (4) is equal to the median (also 4).

Case 2: The numbers are NOT evenly spaced
Let b=3, c=3, and d=6, so that b+c+d = 3+3+6 = 12.
Resulting set:
0, 3, 3, 6, 8.
In this case, the average of the 5 numbers (4) is NOT equal to the median (3).

Thus, statement 2 is INSUFFICIENT.
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