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26 May 2017, 04:42
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If \(a, b, c\), and d are different positive odd numbers, what is the median of the numbers?
(1) \(a + b + c + d=16\)
(2) The product of the 3 numbers of \(a, b, c,\) and \(d\) is 15
(1) \(a + b + c + d=16\)
(2) The product of the 3 numbers of \(a, b, c,\) and \(d\) is 15
26 May 2017, 04:43
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Official Solution:
If you look at the original condition, there are 4 variables \((a, b, c, d)\) and in order to match the number of variables to the number of equations, there must be 4 equations. Therefore, E is most likely to be the answer. By solving con 1) and con 2),
4 integers always have to be 1, 3, 5, and 7. Hence, \(median = \frac{3+5}{2} = 4\), hence it is unique and sufficient. The answer is C. However, this is an integer question, one of the key questions, so you can apply "CMT 4(A: if you get C too easily, consider A or B, B: if you get A or B too easily, consider D)".
In the case of con 1), since a, b, c, d are each different odd number, only \((a, b, c, d) = (1, 3, 5, 7)\) is possible. \(Median = \frac{3+5}{2} = 4\), hence it is unique and sufficient.
In the case of con 2), if \(abc=15, a, b, c\) are each different odd number, so only \((a, b, c) = (1, 3, 5)\) is possible. If so, regardless of the value of d, \(d \ne 1,3,5,\) so d has to be an odd number greater than 5. Therefore, from \(1, 3, 5, d,\) you get \(median = \frac{3+5}{2} = 4\), hence it is unique and sufficient.
The important thing here is "CMT 4(B: if you get A or B too easily consider D)". In other words, in the case of con 1), it is easy, but in the case of con 2), it is very difficult. However, con 2) is also sufficient, so D is the answer. In order to solve 5051 level question, you must approach it with logics based on CMT.
Answer: D
If you look at the original condition, there are 4 variables \((a, b, c, d)\) and in order to match the number of variables to the number of equations, there must be 4 equations. Therefore, E is most likely to be the answer. By solving con 1) and con 2),
4 integers always have to be 1, 3, 5, and 7. Hence, \(median = \frac{3+5}{2} = 4\), hence it is unique and sufficient. The answer is C. However, this is an integer question, one of the key questions, so you can apply "CMT 4(A: if you get C too easily, consider A or B, B: if you get A or B too easily, consider D)".
In the case of con 1), since a, b, c, d are each different odd number, only \((a, b, c, d) = (1, 3, 5, 7)\) is possible. \(Median = \frac{3+5}{2} = 4\), hence it is unique and sufficient.
In the case of con 2), if \(abc=15, a, b, c\) are each different odd number, so only \((a, b, c) = (1, 3, 5)\) is possible. If so, regardless of the value of d, \(d \ne 1,3,5,\) so d has to be an odd number greater than 5. Therefore, from \(1, 3, 5, d,\) you get \(median = \frac{3+5}{2} = 4\), hence it is unique and sufficient.
The important thing here is "CMT 4(B: if you get A or B too easily consider D)". In other words, in the case of con 1), it is easy, but in the case of con 2), it is very difficult. However, con 2) is also sufficient, so D is the answer. In order to solve 5051 level question, you must approach it with logics based on CMT.
Answer: D
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