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26 May 2017, 04:41
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If \(m\) and \(n\) are positive decimals smaller than 1, is \(m + n > 1\)?
(1) All decimal digits of \(m\) and \(n\) are greater than 5
(2) \(mn > \frac{1}{2}\)
(1) All decimal digits of \(m\) and \(n\) are greater than 5
(2) \(mn > \frac{1}{2}\)
26 May 2017, 04:41
Kudos
Bookmarks
Official Solution:
If you look at the original condition, there are 2 variables \((m, n)\) and 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) and con 2),
In the case of con 1), it is yes, too easily. This is because all the decimals of both m and n are greater than 5. Even if it is the least number \(m = n = 0.6\), you get \(m + n = 0.6 + 0.6 = 1.2 > 1\), hence always yes, and it is sufficient. (When the question is "greater than", you should get the minimum value. This is because every other number is "greater than" minimum value.
In the case of con 2), it is difficult. What you need in this case is "CMT 4(B: if you get A or B too easily, consider D)". You can apply it and you get yes, hence it is sufficient because \(m + n \ge 2\sqrt{mn} > 2\sqrt{\frac{1}{2}} = 2(\frac{1}{\sqrt{2}}) = (\sqrt{2})^{2}(\frac{1}{\sqrt{2}}) = \sqrt{2} > 1\), then \(m + n > 1\). Therefore, the condition is sufficient. The answer is D.
Answer: D
If you look at the original condition, there are 2 variables \((m, n)\) and 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) and con 2),
In the case of con 1), it is yes, too easily. This is because all the decimals of both m and n are greater than 5. Even if it is the least number \(m = n = 0.6\), you get \(m + n = 0.6 + 0.6 = 1.2 > 1\), hence always yes, and it is sufficient. (When the question is "greater than", you should get the minimum value. This is because every other number is "greater than" minimum value.
In the case of con 2), it is difficult. What you need in this case is "CMT 4(B: if you get A or B too easily, consider D)". You can apply it and you get yes, hence it is sufficient because \(m + n \ge 2\sqrt{mn} > 2\sqrt{\frac{1}{2}} = 2(\frac{1}{\sqrt{2}}) = (\sqrt{2})^{2}(\frac{1}{\sqrt{2}}) = \sqrt{2} > 1\), then \(m + n > 1\). Therefore, the condition is sufficient. The answer is D.
Answer: D
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