(C) Standard deviation is a measure of the dispersion of the numbers on a list from the list’s mean, or average. When the teacher adds 10 points to every score, the scores and their average are all shifted upward by 10, but the spread of the test scores from their mean remains unchanged. Therefore, the standard deviation of the new set is the same as that of the original set, and (C) is correct. To compute an actual standard deviation will not be required on the GRE. The complicated formula involves subtracting every number on the list from the mean, then averaging the squares of these differences, and finally taking the square root.

(D) We know that 0 < w < x < y < 100, but we don’t know how far apart the variables w, x, and y are spread between 0 and 100. For example, if w = 1, x = 2, and y = 3, then Quantity (A), 100 − y, equals 97, while Quantity (B), y − w, equals only 2. But if w = 10, x = 20, and y = 99, then Quantity (A) equals only 1, while Quantity (B) equals 89. Therefore, there is not enough information to determine, and (D) is correct.

($2,940) Since the four shares together constitute 100% of the total, Heir D inherits 100% minus the sum of the other three heirs’ percentages. Then his share is 10% of the total inheritance, because 100 − (35 + 30 + 25) = 100 − 90 = 10. If Heir D’s $840 inheritance is 10% of the total, then the total is 10 × 840 = $8400. Heir A received 35% of the total, which is .35 × 8400 = $2,940.

(A) At Store D, the left-hand column, Beverages, is higher than for any other store. For Store D, the ratio of Prepared Food to Packaged Food is 15% to 40%, or 15:40. Divide both numbers in the ratio by 5, the greatest common factor of 15 and 40, to obtain the reduced ratio 3:8, choice (A).

(C) In order for one category to account for at least as much revenue as the other two categories combined, the one category must comprise at least half, or 50%, of the total revenue. Only Stores A and B have a column that reaches the 50% level, so these two stores meet the condition described, choice (C).

(C) If each of the four stores earned exactly the same amount of total revenue, we might as well suppose that this number was $100—an unrealistic figure—but one that makes computing percents easy. Then the percentage of revenue from beverages for the four stores—30%, 25%, 40%, and 45%—would correspond precisely to the dollar figures $30, $25, $40, and $45. So the revenue from beverages would be 30 + 25 + 40 + 45 = $140 out of 4 × 100 = $400 total revenue. Therefore beverages would account for (140/400 = 0.35=35%) of total revenue, choice (C).

(A, B) Let b = the marked price of the book. The sales tax is 9% of the book’s marked price, so tax = .09b. Then the cost of the book plus tax is given by 1b + .09b, which equals 1.09b. Since John received change for his $20 bill, and since the change was less than $5, he must have spent between $15 and $20. Then $15 < 1.09b < $20. Divide all three parts of this inequality by 1.09 to obtain $13.76 < b < $18.35. This result implies both (A) and (B), because b is certainly greater than $13.50 but less than $18.50. Choice (C) is not true, because if the marked price were $18, for example, then the sales tax would have been .09 × $18 = $1.62, which is greater than $1.50, while the total price remains low enough, $19.62, to leave John change, as required.