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22 Jun 2017, 03:15
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Is |a-b|-(a-b)>0?
1) a<b
2) a>0 and b>0
==> If you modify the original condition and the question, what satisfies |a-b|>a-b? must be a-b<0.
Therefore, the answer is A.
Answer: A
1) a<b
2) a>0 and b>0
==> If you modify the original condition and the question, what satisfies |a-b|>a-b? must be a-b<0.
Therefore, the answer is A.
Answer: A
23 Jun 2017, 01:07
Kudos
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Is y^4<x^2y^2?
1) x<y
2) y<0
==> If you modify the original condition and the question, if y≠0 from y^4<x^2y^2?, y^4-x^2y^2<0?, y^2(y^2-x^2)<0?, you get y^2-x^2<0?. There are 2 variables, and in order to match the number of variables to the number of equations, there must be 2 equations as well. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) & con 2), from x<y<0, you get x^2-y^2>0, hence yes, it is always sufficient.
Therefore, the answer is C.
Answer: C
1) x<y
2) y<0
==> If you modify the original condition and the question, if y≠0 from y^4<x^2y^2?, y^4-x^2y^2<0?, y^2(y^2-x^2)<0?, you get y^2-x^2<0?. There are 2 variables, and in order to match the number of variables to the number of equations, there must be 2 equations as well. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) & con 2), from x<y<0, you get x^2-y^2>0, hence yes, it is always sufficient.
Therefore, the answer is C.
Answer: C
25 Jun 2017, 18:12
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If x≠0, is x^2<|x|?
1) x<1
2) x>-1
==> If you modify the original condition and the question, from x^2<|x|?, |x|^2<|x|?, when you divide both sides by |x|, (Since |x|>0, the inequality sign doesn’t change after division) you get |x|<1?, and then -1<x<1?.
The answer is C.
Answer: C
1) x<1
2) x>-1
==> If you modify the original condition and the question, from x^2<|x|?, |x|^2<|x|?, when you divide both sides by |x|, (Since |x|>0, the inequality sign doesn’t change after division) you get |x|<1?, and then -1<x<1?.
The answer is C.
Answer: C
