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Re: A perfect square is an integer whose square root is an integ [#permalink]
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Quantity A: (1^2 + 2^2+ 3^2+.....+100^2)/100.
Now, we know, the sum of the first n square numbers is basically, n(n+1)(2n+1)/6. So quantity A is= 100*101*201/600= 3383.5 (Use GRE Calculator)

Quantity B: It is the midpoint of the ordered list and that is an average of 50^2 and 51^2. which is 2550.5

So, Quantity A> Quantity B
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Re: A perfect square is an integer whose square root is an integ [#permalink]
We can solve this quickly, I think so in an increasing series with increasing gaps (as we move forward in the series) the average becomes greater than the median of the series.
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Re: A perfect square is an integer whose square root is an integ [#permalink]
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Aaka wrote:
We can solve this quickly, I think so in an increasing series with increasing gaps (as we move forward in the series) the average becomes greater than the median of the series.


I think the fastest way proposed by raf above is the most convenient.

After all is nasty question this one..

On the GRE you cannot pretend to solve ALWAYS a question in 30 seconds, sometimes you do need more. The most important thing is to be on the right path.

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Re: A perfect square is an integer whose square root is an integ [#permalink]
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Re: A perfect square is an integer whose square root is an integ [#permalink]
I solved in 30 seconds cause I just reasoned that the numbers get super big very quickly as we go from 1, 4,9, 25, 36 ,49, 64, etc.

And from distribution/skews, we know that outliers on the right make the mean>median because the average is pulled up by all those super big numbers. Is that the correct way to approach? Or did I just get lucky?
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Re: A perfect square is an integer whose square root is an integ [#permalink]
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