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26 May 2017, 04:43
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A basketball team plays 10 games, and it scores 12 points on average. Has this team scored at least 40 points in one game?
(1) The team scored 55 points in the first 5 games.
(2) The score of each game is at least 9 points.
(1) The team scored 55 points in the first 5 games.
(2) The score of each game is at least 9 points.
26 May 2017, 04:43
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Official Solution:
If you look at the original condition, there are 10 variables (since you have to know the each score of 10 games) and 1 equation (since the average score of 10 games is 12). In order to match the number of variables to the number of equations, there must be 9 equations. Therefore, E is most likely to be the answer. By solving con 1) and con 2), the total sum of the scores is \(10 * 12 points = 120 points\), and since you scored 55points in 5 games,
(9pts) (11pts) (11pts) (12pts) (12pts) (a pts) (b pts) (c pts) (d pts) (40 pts) can be substituted.
Then, the last 5 scores have to be \(120 - 55 = 65 pts\), so it is impossible to get \(a + b + c + d = 65 - 40 = 25\). This is because it has to be \(12 \le a \le b \le c \le d\). Thus, no is the answer (CMT 1: no is also an answer), hence it is sufficient. The answer is C. However, this is an (hidden) integer question, one of the key questions, so you can apply "CMT 4(A: if you get C too easily, consider A or B)".
In the case of con 1),
(9pts) (11pts) (11pts) (12 pts) (12 pts) (1pt) (1pt) (1pt) (1pt) (22pt) (40pts) yes but,
(9pts) (11pts) (11pts) (12pts) (12pts) (1pt) (1pt) (21pts) (22pts) (20pts) no, hence, is not sufficient.
In the case of con 2),
Even if (9pts) (9pts) (9pts) (9pts) (9pts) (9pts) (9pts) (9pts) (9pts) (40pts) is possible, since \(9 * 9 + 40 = 121pts \ne 120pts\), the team could not have scored 40 points in any of the games. Hence no, it is sufficient (CMT 1: no is also an answer). The answer is B.
Answer: B
If you look at the original condition, there are 10 variables (since you have to know the each score of 10 games) and 1 equation (since the average score of 10 games is 12). In order to match the number of variables to the number of equations, there must be 9 equations. Therefore, E is most likely to be the answer. By solving con 1) and con 2), the total sum of the scores is \(10 * 12 points = 120 points\), and since you scored 55points in 5 games,
(9pts) (11pts) (11pts) (12pts) (12pts) (a pts) (b pts) (c pts) (d pts) (40 pts) can be substituted.
Then, the last 5 scores have to be \(120 - 55 = 65 pts\), so it is impossible to get \(a + b + c + d = 65 - 40 = 25\). This is because it has to be \(12 \le a \le b \le c \le d\). Thus, no is the answer (CMT 1: no is also an answer), hence it is sufficient. The answer is C. However, this is an (hidden) integer question, one of the key questions, so you can apply "CMT 4(A: if you get C too easily, consider A or B)".
In the case of con 1),
(9pts) (11pts) (11pts) (12 pts) (12 pts) (1pt) (1pt) (1pt) (1pt) (22pt) (40pts) yes but,
(9pts) (11pts) (11pts) (12pts) (12pts) (1pt) (1pt) (21pts) (22pts) (20pts) no, hence, is not sufficient.
In the case of con 2),
Even if (9pts) (9pts) (9pts) (9pts) (9pts) (9pts) (9pts) (9pts) (9pts) (40pts) is possible, since \(9 * 9 + 40 = 121pts \ne 120pts\), the team could not have scored 40 points in any of the games. Hence no, it is sufficient (CMT 1: no is also an answer). The answer is B.
Answer: B
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