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For five numbers, the median is the 3rd number when ordered. The current list contains 2, 5, 7, 9, and xx. To keep the median at 7, 7 must stay in the middle position; thus xx must be ≥9\ge 9 or ≤5\le 5. Now impose the mean condition: mean 7 implies total sum =5×7=35= 5 \times 7 = 35. The sum of known four numbers is 2+5+7+9=232 + 5 + 7 + 9 = 23. Therefore x=35−23=12x = 35 - 23 = 12. Check ordering: with x=12x=12, the sorted set is 2,5,7,9,122, 5, 7, 9, 12; the median remains 7 and the mean is (2+5+7+9+12)/5=35/5=7(2+5+7+9+12)/5 = 35/5 = 7. Option C is correct. A common error is to choose a value that maintains the median but breaks the mean requirement, or vice versa. Another pitfall is to forget that with an odd count the median is the middle element, not the average of the two middle ones (that only applies to even counts). Always verify both constraints before finalizing the answer.