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28 May 2017, 23:51
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Is x^2>y^2?
1) |x|>y
2) x>|y|
==> In the original condition, there are 2 variables (x,y), and in order to match the number of variables to the number of equations, there must be 2 variables. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from x>|y|≥0, it is x>0, so from x>|y|, it is |x|>|y|, which becomes x^2>y^2, hence yes, it is sufficient. The answer is C. However, this is an absolute value question, one of the key questions, so you apply CMT 4 (A: if you get C too easily, consider A or B). For con 1), you get (x,y)=(2,1) yes (2,-3) no, hence not sufficient. For con 2), from x>|y|≥0, you get x>0, which becomes |x|>|y|, then x^2>y^2, it is always yes and sufficient. Therefore, the answer is B.
Answer: B
1) |x|>y
2) x>|y|
==> In the original condition, there are 2 variables (x,y), and in order to match the number of variables to the number of equations, there must be 2 variables. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from x>|y|≥0, it is x>0, so from x>|y|, it is |x|>|y|, which becomes x^2>y^2, hence yes, it is sufficient. The answer is C. However, this is an absolute value question, one of the key questions, so you apply CMT 4 (A: if you get C too easily, consider A or B). For con 1), you get (x,y)=(2,1) yes (2,-3) no, hence not sufficient. For con 2), from x>|y|≥0, you get x>0, which becomes |x|>|y|, then x^2>y^2, it is always yes and sufficient. Therefore, the answer is B.
Answer: B
01 Jun 2017, 18:00
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If m and n are positive integers, is mn a multiple of 9?
1) m+n is a multiple of 3
2) mn is a multiple of 3
==> In 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. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from the multiple of mn=3, it becomes a multiple of m or n=3. According to con 1), you always get a multiple of m=n=3, hence yes, it is sufficient.
The answer is C.
Answer: C
1) m+n is a multiple of 3
2) mn is a multiple of 3
==> In 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. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from the multiple of mn=3, it becomes a multiple of m or n=3. According to con 1), you always get a multiple of m=n=3, hence yes, it is sufficient.
The answer is C.
Answer: C
04 Jun 2017, 18:00
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We define the average (harmonic mean) as the reciprocal of the average (arithmetic mean) of reciprocals. What is the average (harmonic mean) of 2, 3, and 6?
A. 1/3
B. 1/2
C. 1
D. 2
E. 3
==> If you find the average of the reciprocals, you get \(\frac{++}{3}=13\) . Since it is the reciprocal of that, the answer is E.
Answer: E
A. 1/3
B. 1/2
C. 1
D. 2
E. 3
==> If you find the average of the reciprocals, you get \(\frac{++}{3}=13\) . Since it is the reciprocal of that, the answer is E.
Answer: E
