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Re: M10-10 [#permalink]
Bunuel wrote:
Official Solution:


Notice that the range of ANY set is more than or equal to zero.

(1) All elements of set \(S\) are negative. The mean of a set with all negative elements is certainly negative so less than its range (which as discussed is always non-negative). Sufficient.

(2) The median of set \(S\) is negative. So, there is at least one negative term is the set. Now, consider two cases:

A. If all elements in set \(S\) are negative then we have the same scenario as above so \(Range \gt Mean\);

B. If not all elements in set \(S\) are negative then \(Range=Largest-Smallest\), which will mean that \(\text{Range} \gt \text{Largest Element}\) (that's because the smallest element in set \(S\) is negative. For example consider the following set {-1, -1, 2}: the range of that set is \(Range=2-(-1)=3 \gt 2)\). For the same reason the mean will be less than the largest element, so \(Range \gt Largest \gt Mean\).

So, in any case \(Range \gt Mean\). Sufficient.


Answer: D


What if the set has only one number Set: {x}. Can we rule out this option saying that range is not defined for set consisting of one number? Or range is 0 for this set?
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Re: M10-10 [#permalink]
Expert Reply
popov wrote:
Bunuel wrote:
Official Solution:


Notice that the range of ANY set is more than or equal to zero.

(1) All elements of set \(S\) are negative. The mean of a set with all negative elements is certainly negative so less than its range (which as discussed is always non-negative). Sufficient.

(2) The median of set \(S\) is negative. So, there is at least one negative term is the set. Now, consider two cases:

A. If all elements in set \(S\) are negative then we have the same scenario as above so \(Range \gt Mean\);

B. If not all elements in set \(S\) are negative then \(Range=Largest-Smallest\), which will mean that \(\text{Range} \gt \text{Largest Element}\) (that's because the smallest element in set \(S\) is negative. For example consider the following set {-1, -1, 2}: the range of that set is \(Range=2-(-1)=3 \gt 2)\). For the same reason the mean will be less than the largest element, so \(Range \gt Largest \gt Mean\).

So, in any case \(Range \gt Mean\). Sufficient.


Answer: D


What if the set has only one number Set: {x}. Can we rule out this option saying that range is not defined for set consisting of one number? Or range is 0 for this set?


The range of a single element set is 0.
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Re: M10-10 [#permalink]
I think this is a high-quality question and I agree with explanation. is range always greater than mean and median... if not, in which case?
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Re: M10-10 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M10-10 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M10-10 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M10-10 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M10-10 [#permalink]
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