In Set T, the average (arithmetic mean) equals the median. Which of th

Question

In Set T, the average (arithmetic mean) equals the median. Which of the following must be true?

I. Set T consists of evenly spaced numbers.
II. Set T consists of an odd number of terms.
III. Set T has no mode.
IV. None of the above.

A. I only
B. I and II
C. II and III
D. I, II, and III
E. IV only

Kudos for a correct  solution .

gmat6nplus1 · Accepted Answer

I. consider T (1, 2, 3) median=mean=2. Consider T (5, 9, 2, 19, 10) median=mean=9
II. Consider T (1, 2, 3) median=mean=2. Consider T (1, 1, 1, 1) median=mean=1
III. Consider T (1, 2, 3) mean=median=2 and there is no mode. Consider T (1, 1, 1) median=mean=1 and mode is 1.

Answer E.

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION

Correct Answer: E

None of the proposed statements must be true; to prove this, you might consider a hypothetical: Set T could be  , in which the average and median are both 2. This set proves I, II, and III wrong. (Note: This set has two modes, 1 and 3)

Alchemist14 · Answer

This is a very good 'trap question'.

Consider Set A= { 0,0,0,0 } Mean=Median=Mode=0- Eliminates 1,2 and 3 OR
Consider Set B= {1,2,2,3 } Mean=Median=Mode=2- Eliminates 1,2 and 3

If you think that only in 'consecutive integers' the average is equal to the median you might fall for 'D'.

Ans: E

stonecold · Answer

This is an Excellent Question.
Here is my take on this one =>

Firstly, 
 
Also =>

Now We are told that a Set T is such that its mean = median
We are asked which of the statements must be true.

Statement 1-->
T is evenly Spaced

Consider two sets => 5,6,7,8,9,10,11
Mean = Median=8
Now Removing 7 and 9 wont affect the mean as 8-7 will be balanced by 8-9 and the sum of deviations around 8 will still be zero.
Hence 5,6,8,10,11 => Mean = 8 and Median = 8
But the above set isn't every spaced.

Hence this statement is not always true.
Statement 2=>

Consider a set => 7,7,7,7
Here => Mean=Median=7
And the number of terms is even.
Hence this statement also isn't always true.

Statement 3-->
Consider a set => 7,7,7,7
Here Mode = 7
Hence Set T will can have a mode=> The statement isn't always true.
Statement 4--> None of the above => TRUE.

Hence E

ashikaverma13 · Answer

So, in conclusion we can takeway from this question that for an evenly spaced set mean=median does not necessarily mean that the vice versa is true. Just because for a set mean = median does not render the set as an evenly spaced one.

Bunuel · Answer

_______________
Yes, that's true.

ScottTargetTestPrep · Answer

Let’s analyze each Roman numeral statement.

I. Set T consists of evenly spaced numbers.

This is not necessarily true. For example, if set T = {1, 2, 4, 4, 9}, we see that average = median, but T is not an evenly spaced set.

II. Set T consists of an odd number of terms.

This is not necessarily true. For example, if set T = {3, 3}, we see that average = median, but T has an even number of terms.

III. Set T has no mode.

This is not necessarily true. For example, if we use the same set T as in Roman numeral I, we see that average = median, but T does have a mode.

Answer: E

bumpbot · Answer

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In Set T, the average (arithmetic mean) equals the median. Which of th

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In Set T, the average (arithmetic mean) equals the median. Which of th

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