ziyuen wrote:
There are a total 10 movie theaters. Minimum number of tickets that each movie theaters sold was 80 tickets. Is total number of average tickets sold more than 90 tickets?
1) Number of movie theaters that sold at least 100 tickets were more than half of the total number of movie theaters.
2) In 4 movie theaters of the total number of movie theaters, number of each tickets sold were more than 105.
OFFICIAL EXPLANATION
FROM
MathRevolutionIf you modify the original condition and the question, the average number of tickets sold>90? And this becomes the total number of tickets sold>90(10)=900? Then, there are 10 variables (10 movie theaters) and 1 equation (minimum of 80 tickets were sold), and in order to match the number of variables to the number of equations, there must be 9 more equations. Therefore, E is most likely to be the answer. By solving con 1) and con 2) together,
If you assume that the number of movie theaters that sold at least 100 tickets as 6 and from the 6 movie theaters, 4 of them sold 106 tickets, from the total number of tickets sold=100(6)+106(4)=1,024>900, it is always yes, hence it is sufficient. The answer is C. However, this is an integer question, one of the key questions, so you apply CMT 4 (A: if you get C too easily, consider A and B) and CMT 4 (B: if you get A or B too easily, consider D).
For con 1), even if you assume the number of theaters that sold at least 100 tickets as 5, since the rest of the 5 movie theaters sold at least 80 tickets, so the total number of tickets sold becomes 100(5)+80(5)=900. Since it said that it is half greater than the total number of movie theaters, which is 10, so it is always greater than 900, hence yes, it is sufficient.
For con 2), if you assume that from the total of 10 movie theaters, 4 movie theaters sold 105 tickets and the 6 movie theaters sold 80 tickets, then the total number of tickets sold is 105(4)+80(6)=900, and since it said that the 4 movie theaters sold more than 105 tickets each, so it is always greater than 900, hence it is yes and sufficient. Therefore, the answer is D.
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