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26 May 2017, 04:43
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If the median of 3 positive integers is 6, is the average (arithmetic mean) of the integers greater than 6?
(1) The smallest number of the integers is 1
(2) The largest number of the integers is 12
(1) The smallest number of the integers is 1
(2) The largest number of the integers is 12
26 May 2017, 04:45
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Official Solution:
You need preliminary knowledge in order to solve this question. In general, if the question contains "greater than", you must get the least value. This is because all other numbers are "greater than" the least value.
In the original condition, there are 2 variables \((a,6,b)\). In order to match the number of variables to the number of equations, there must be 2 equations. Therefore, C is most likely to be the answer. By solving con 1) & con 2), you need to get the least value of the average, which is \(\frac{1+6+12}{3} = \frac{19}{3}\). Since it is always \(\frac{19}{3} > 6\), it is yes, hence sufficient. However, this is also a statistics question, one of the key questions, so you should apply "CMT 4(A: if you get C too easily, consider A or B)". In the case of con 1), when the least of the average is \(\frac{1+6+6}{3} = \frac{13}{3} < 6\), so no, and from \(\frac{1+6+20}{3} = 9 > 6\), it is yes, hence not sufficient.
In the case of con 2), if you also get the least value of the average, from \(\frac{1+6+12}{3} = \frac{19}{3} > 6\), you get yes, hence sufficient. Therefore, the answer is B.
Answer: B
You need preliminary knowledge in order to solve this question. In general, if the question contains "greater than", you must get the least value. This is because all other numbers are "greater than" the least value.
In the original condition, there are 2 variables \((a,6,b)\). In order to match the number of variables to the number of equations, there must be 2 equations. Therefore, C is most likely to be the answer. By solving con 1) & con 2), you need to get the least value of the average, which is \(\frac{1+6+12}{3} = \frac{19}{3}\). Since it is always \(\frac{19}{3} > 6\), it is yes, hence sufficient. However, this is also a statistics question, one of the key questions, so you should apply "CMT 4(A: if you get C too easily, consider A or B)". In the case of con 1), when the least of the average is \(\frac{1+6+6}{3} = \frac{13}{3} < 6\), so no, and from \(\frac{1+6+20}{3} = 9 > 6\), it is yes, hence not sufficient.
In the case of con 2), if you also get the least value of the average, from \(\frac{1+6+12}{3} = \frac{19}{3} > 6\), you get yes, hence sufficient. Therefore, the answer is B.
Answer: B
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