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The median of n consecutive odd integers is 30. If the fifth

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by BTGmoderatorLU » Sun May 12, 2019 3:04 am

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Source: e-GMAT

The median of n consecutive odd integers is 30. If the fifth term is 33, then which of the following can be the last term of the sequence?

A. 19
B. 21
C. 25
D. 33
E. 41

The OA is A
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by Ian Stewart » Tue May 14, 2019 8:01 am
This question makes no sense - they are misusing terminology. If you talk about a list of "consecutive" odd integers, say, then the second thing in the list follows the first thing, in sequence - the second term must be larger than the first term. But here they intend the sequence of "consecutive" odd integers to be decreasing from one term to the next. They're not using the word "consecutive" correctly.

If instead we adopt the interpretation they intend, that we have a sequence of odd integers that looks like n, n-2, n-4, and so on, then we know the fifth term, n-8, is equal to 33, so the first term is 41. Since the median, 30, is halfway between the smallest and largest things in any equally spaced list, the smallest term is thus 19.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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