runzun wrote:This table lists enrollment in an afterschool program by activity. There are 30 total students enrolled in the entire program. Students may participate in one, two, or three activities. How many students participate in all three activities?

(1) 21 students only participate in one activity.

(2) 6 students participate in both basketball and math.


OA is A.. but i am not able to understand the logic..
plz help

"This table lists" is this attached! am unable to view this attachment!

If we add up the participants in each activity, we have a total of 42 students. This discrepancy is accounted for by the fact that some students may participate in multiple activities.Since only 30 are enrolled, this means 12 places must be accounted for.

(1) If 21 students only participate in one activity, then 9 participate in multiple activities. However, we must account for 12 spaces. This means that 3 students are triple counted (12 - 9), and 6 are double counted. Therefore, 3 students participate in all three activities. To solve this using algebra, set D = the number of students in 2 activities and T = the number of students in all 3:

D + T = 9

21 + 2D + 3T = 42 {This accounts for the double and triple counting of students in multiple activities}
2D + 3T = 21

Combining the two equations, we get:
2(9 - T) + 3T = 21
18 - 2T+ 3T = 21
T = 3
SUFFICIENT

(2) The limitation of this statement is that it does not tell us how many of those 6 students, if any, also wrestle. It is possible that all 6 students participate in all activities. This would account for the table. Alternately, perhaps these 6 students only participate in these two activities, and another 6 students participate in a different pair of activities. We cannot reach a definitive conclusion.
INSUFFICIENT.

The credited response is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

this is the explanation but I am unable to understand..
plz help

Umm.. Where is the table??