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Bunuel
In a set of data, the average (arithmetic mean) equals the median. Which of the following must be true?

I. The set consists of evenly spaced numbers.
II. The set consists of an odd number of terms.
III. The set has no mode.

A. I only
B. I and II
C. II and III
D. I, II, and III
E. None of the above

We need to find whether the following 'must be true':

I. The set consists of evenly spaced numbers.
Is it necessary? Can we not have a set in which mean = median but the set is not evenly spaced?
Such as 1, 4, 4, 5, 6
Here mean = median = 4, but the set is not evenly spaced.
So this condition is not necessary.

II. The set consists of an odd number of terms.
Is this condition necessary?
3, 4, 4, 5 has an even number of terms but mean = median = 4.
So this condition is also not necessary.

III. The set has no mode.
The example above 3, 4, 4, 5 has a mode but mean = median = 4.
So this condition is also not necessary.

Answer (E)

Responding to a pm:

Quote:

Just to confirm, I think logic in this question is inversely true.
Meaning, if a set contains evenly spaced numbers, and have odd number of terms then the mean and median are same. is this always right?

In an evenly spaced set, mean will be equal to median. It doesn't matter whether the number of elements is even or odd.
1, 3, 5
Mean = median = 3

1, 3, 5, 7
Mean = median = 4
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Bunuel
In a set of data, the average (arithmetic mean) equals the median. Which of the following must be true?

I. The set consists of evenly spaced numbers.
II. The set consists of an odd number of terms.
III. The set has no mode.

A. I only
B. I and II
C. II and III
D. I, II, and III
E. None of the above

Let’s analyze each Roman numeral with examples to determine whether it must be true.

I. The set consists of evenly spaced numbers.


This does not have to be true. For example, let’s say the set consists of the following numbers: 1, 2, 4, 6, 7. The mean and median are both 4, but this is not a set of evenly spaced numbers (notice that 2 is one more than 1, but 4 is two more than 2).

II. The set consists of an odd number of terms.

This does not have to be true. For example, let’s say the set consists of the following numbers: 1, 1. The mean and median are both 1, but this set has 2 terms.

III. The set has no mode.

This does not have to be true. For example, let’s say the set consists of the following numbers: 1, 2, 2, 3. The mean and median are both 2, which is also the mode of the set.

Answer: E
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Since, mean=median, the frequency distribution graph should be symmetric and we can think of something like 1,1 2,2,3,3 for this particular question.

This could be an easier way to find combinations, may be.
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