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26 May 2017, 04:41
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If \(a \le b \le c \le d\), is \(a \ge 20\)?
(1) \(a + b + c + d = 170\)
(2) \(5a \ge 2d\)
(1) \(a + b + c + d = 170\)
(2) \(5a \ge 2d\)
26 May 2017, 04:41
Kudos
Bookmarks
Official Solution:
In the original condition, there are 4 variables \((a, b, c, d)\) and 1 equation \((a \le b \le c \le d)\). In order to match the number of variables to the number of equations, there should be 3 more equations, so E is most likely to be the answer. By solving con 1) &con 2), you get \(a = 20\), \(b = c = d = 50\), hence yes, it is sufficient and the answer becomes C. In other words, if \(a = 19\), then \(5a = 5 * 19 \ge 2*47\). If so, you get \(a + b + c + d = 19 + 47 + 47 + 47 < 170\) and \(a + b + c + d\) cannot be \(170\), so it always becomes yes. Therefore, the answer is C.
Answer: C
In the original condition, there are 4 variables \((a, b, c, d)\) and 1 equation \((a \le b \le c \le d)\). In order to match the number of variables to the number of equations, there should be 3 more equations, so E is most likely to be the answer. By solving con 1) &con 2), you get \(a = 20\), \(b = c = d = 50\), hence yes, it is sufficient and the answer becomes C. In other words, if \(a = 19\), then \(5a = 5 * 19 \ge 2*47\). If so, you get \(a + b + c + d = 19 + 47 + 47 + 47 < 170\) and \(a + b + c + d\) cannot be \(170\), so it always becomes yes. Therefore, the answer is C.
Answer: C
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