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kevincan
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GRE 1: Q170 V170
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game over
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I think its D

For mean to be greater than 2.5. Sum must be greater than 1690.

St1: If median is 3.5 then minimum mean will be when
1---337 times
2---0 times
3---1 time
4---338 times
Sum = 4 * 338 + 337 + 3 = 1692 : SUFF

St2: If 4 is most common (i.e mode) then minimum mean will be when
1---169 times
2---169 times
3---168 times
4---170 times
Sum = 170 *4 + 169 * 3 + 169 * 2 + 168 = 1691 : SUFF
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game over
I think A is the correct answer.

Since the median is 3.5 we know that 676/2 pages must contain 4 spelling mistakes and at least one page must contain 3 spelling mistakes. Since every other page must contain at least one spelling mistake, the mean must be above 2.5.

(2) is not sufficient.
For example 226 = number of pages with 4 spelling mistakes,
225 = number of pages with 2 spelling mistakes = number of pages with
1 spelling mistake => mean mean=4.

[In my first example. the number of pages wasn't an integer]


Good work! You were successful because you figured that "sacrificing" the 3's would make the average the lowest, even if it meant more pages with 4 mistakes. A great solution would have been using (n-1)/3 (n-1)/3 and (n+2)/3 as the number of pages with 1,2 and 4 mistakes.



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