Bunuel wrote:
A certain high school with a total enrollment of 900 students held a science fair for three days last week. How many of the students enrolled in the high school attended the science fair on all three days?
(1) Of the students enrolled in the school, 30 percent attended the science fair on two or more days.
(2) Of the students enrolled in the school, 10 percent of those that attended the science fair on at least one day attended on all three days.
The trick here is that we don't know whether all of 900 students attended on at least one day. Meaning that there can be some students who didn't take part in the fair.
So we have the following groups:
A. Attended on only one day;
B. Attended on two days exactly;
C. Attended on all three days;
D. Did not attend at all.
And the sum of these groups is A+B+C+D=900. We want to determine the value of C.
(1) Of the students enrolled in the school, 30 percent attended the science fair on two or more days.
So, 900*0.3=270 attended 2 or more days:
Not sufficient.
(2) Of the students enrolled in the school, 10 percent of those that attended the science fair on at least one day attended on all three days.
10%*(A + B + C) = C;
A + B = 9C.
Not sufficient.
(1)+(2) B+C=270, A+B=9C and A+B+C+D=900. Three linear equations four unknowns. Not sufficient.
Answer: E.
But does "at least on one day" not imply that all 3 cases are possible: on one day, two days, and/or three days? Meaning, there is even higher uncertainty to answer this question. Equally, in S2, how can we consider A+B+C for at least one day? If we use a Venn Diagram, it could be all 3 cases, correct? Or did you just not go too much into detail as even if we consider one case, we are left with no sufficient answer?