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26 May 2017, 04:41
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If \(0 < z < y < x < w\), is \(z < 4\)?
(1) \(\frac{1}{w} > \frac{1}{4}\)
(2) \(\frac{1}{w} + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1\)
(1) \(\frac{1}{w} > \frac{1}{4}\)
(2) \(\frac{1}{w} + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1\)
26 May 2017, 04:41
Kudos
Bookmarks
Official Solution:
If you look at the original condition, there are 4 variables \((w, x, y, z)\) and 1 equation \((0 < z < y < x < w)\). In order to match the number of variables to the number of equations, there must be 3 more equations. Therefore, E is most likely to be the answer. By solving con 1) and con 2),
In the case of con 1), \(\frac{1}{w} > \frac{1}{4}\) then \(w < 4\), and from \(0 < z < y < x < w < 4\), you get \(z < 4\), hence yes, it is sufficient. However, if you look at con 2), it is quite difficult. What you need in this ind of situation is "CMT 4(B: if you get A or B too easily consider D)". Con 2) is always yes, hence it is sufficient. That is because from \(\frac{1}{12} + \frac{1}{6} + \frac{1}{4} + \frac{1}{2} = 1\), you get \(2 < 4\), hence yes. However, in order for it to be no, it has to be \(z = 4 < 4\). From \(\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1\), this does not satisfy \(0 < z < y < x < w\), hence it is impossible. For any possible \(z\), \(z > 4\), so \(z\) that satisfies \(0 < z < y < x < w\) does not exist, hence yes, it is sufficient. The answer is D.
Answer: D
If you look at the original condition, there are 4 variables \((w, x, y, z)\) and 1 equation \((0 < z < y < x < w)\). In order to match the number of variables to the number of equations, there must be 3 more equations. Therefore, E is most likely to be the answer. By solving con 1) and con 2),
In the case of con 1), \(\frac{1}{w} > \frac{1}{4}\) then \(w < 4\), and from \(0 < z < y < x < w < 4\), you get \(z < 4\), hence yes, it is sufficient. However, if you look at con 2), it is quite difficult. What you need in this ind of situation is "CMT 4(B: if you get A or B too easily consider D)". Con 2) is always yes, hence it is sufficient. That is because from \(\frac{1}{12} + \frac{1}{6} + \frac{1}{4} + \frac{1}{2} = 1\), you get \(2 < 4\), hence yes. However, in order for it to be no, it has to be \(z = 4 < 4\). From \(\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1\), this does not satisfy \(0 < z < y < x < w\), hence it is impossible. For any possible \(z\), \(z > 4\), so \(z\) that satisfies \(0 < z < y < x < w\) does not exist, hence yes, it is sufficient. The answer is D.
Answer: D
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