Bunuel wrote:

An artist's portfolio consisting of 1,000 photos is divided into 20 subjects. After an extensive photo shoot, two more subjects are then added to the portfolio. Is the average (arithmetic mean) number of photos in each subject greater than 55?

(1) Each of the new subjects has fewer than 117 photos.

(2) Each of the new subjects has more than 71 photos.

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTIONCorrect Answer: E

From the question stem, the average of the current portfolio is 50 photos (1000/20). Statement (1) indicates that each of the two new subjects has less than 117 photos. At most, each of the two new subjects would contain 116 photos. This would bring the new average to a maximum 56 photos (1232/22). While this is the maximum, there could always be fewer photos. This allows for both greater than 55 photos on average and fewer than 55 photos on average. Since this statement yields both "yes" and "no" as a response, statement (1) is not sufficient. Statement (2 tells us that each new subject has more than 71 photos. At least, each of the two new subjects would contain 72 photos. This means that the new average must be at least 52 photos (1144/22). Of course, this also allows for both greater than 55 and fewer than 55 photos on average, so statement (2) is not sufficient. Taking the two statements together, we know that the new average is between 52 and 56; however this is still not sufficient to say whether the average is greater than 55 or not. The two statements together are not sufficient and the answer is accordingly E. Tip: A quicker way to evaluate the statements is to maintain the average of 50 first, and then see whether there are enough additional photos to push the average to 55. For statement (1), subtract 50 photos from the maximum possible of 116. That 50 subtracted is enough to maintain the average; so at this point, we have an average of 50 with an additional 66 photos per subject. Since we have 2 new subjects, we have 132 additional photos (66 * 2) over 22 subjects to try and push the average over 55. If 132/22 yields a number larger than 5, then the total average will be greater than 55. This technique requires much easier calculating than dividing 1232 by 22. Repeat the same procedure for statement (2).

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