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Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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27 Feb 2017, 07:17
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61% (01:42) correct 39% (01:55) wrong based on 199 sessions
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Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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27 Feb 2017, 08:02
St1: x^2  y^2 < 0 > x < y. Not sufficient as we do not know the signs of x and y.
St2: x + y > 0 x and y are positive > x > y or y > x or either x or y is positive > The positive value must be greater than the other value. Not Sufficient.
Combining St1 and St2, we know that y > x and x + y > 0. If y > x and x and y are positive, then y > x If y > x and one value is positive, then the value, which has a greater number, must be positive > y > x. In both the cases we get a definite yes answer for the question is x < y. Sufficient.
Answer: C



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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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27 Feb 2017, 08:32
LamaSalah wrote: Is x < y ?
(1) x^2  y^2 < 0 (2) x  y < 0 (1) \(x^2 < y^2 \iff x < y\) We have \(3^2 < 5^2 \implies 3 < 5\) However \(3^2 < (5)^2 \implies 3 >  5\). Hence, insufficient. (2) \(xy < 0 \implies x+y > 0\) If x=5, y=2 then \(x+y>0\) and \(x>y\) If x=2, y=5 then \(x+y>0\) and \(x<y\). Hence, insufficient. Combine (1) and (2): \((2) \implies x+y > 0\) \((1) \implies (xy)(x+y) < 0 \implies xy < 0 \implies x<y\). Sufficient The answer is C
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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28 Feb 2017, 07:20
x < y? stmt1: x^2  y^2 < 0 x < y here x or y or both can be postitive or negative. so we cannot decide if x < y or not. stmt2: xy < 0 x < y say x=2, y=1 then 2 < 1, so x>y. so answer to the given question is NO. say x=1, y=2 then 1 < 2, but here x < y. so answer to the given question is Yes. stmt1 + stmt2: x < y AND x < y then is x < y? x=1, y=2 satisfies both x < y AND x < y. in this case x < y (1<2). so answer is YES. x=1,y=2 satisfies both x < y AND x < y. in this case x < y (1<2). so answer is YES. Sufficient!
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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02 Mar 2017, 17:59
LamaSalah wrote: Is x < y ?
(1) x^2  y^2 < 0 (2) x  y < 0 Statement One Alone:x^2  y^2 < 0 We see that x^2 > y^2; however we still do not have sufficient information to determine whether x > y. For instance, if x = 2 and y = 1, then x is less than y. However, if x = 2 and y = 1, then x is greater than y. Statement one alone is not sufficient to answer the question. Statement Two Alone: x  y < 0 We can multiply the inequality in statement two by 1, remembering to reverse the inequality sign, and obtain: x + y > 0, or equivalently, 0 < x + y. However, we still cannot determine whether x > y. For instance if x = 2 and y = 1, then x is greater than y; however if x = 1 and y = 2, then x is less than y. Statement two alone is not sufficient to answer the question. Statements One and Two Together:Notice that x^2  y^2 in statement one is a difference of two squares: x^2  y^2 = (x + y)(x  y). So x^2  y^2 < 0 means (x + y)(x  y) < 0. From statement two, we know that x + y > 0. That means x  y < 0, since x + y is positive. Therefore, x  y must be negative in order for their product (x + y)(x  y) to be negative. Since x  y < 0, we have that x < y. Answer: C
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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21 Feb 2018, 13:40
broall wrote: LamaSalah wrote: Is x < y ?
(1) x^2  y^2 < 0 (2) x  y < 0 (1) \(x^2 < y^2 \iff x < y\) We have \(3^2 < 5^2 \implies 3 < 5\) However \(3^2 < (5)^2 \implies 3 >  5\). Hence, insufficient. (2) \(xy < 0 \implies x+y > 0\) If x=5, y=2 then \(x+y>0\) and \(x>y\) If x=2, y=5 then \(x+y>0\) and \(x<y\). Hence, insufficient. Combine (1) and (2): \((2) \implies x+y > 0\) \((1) \implies (xy)(x+y) < 0 \implies xy < 0 \implies x<y\). Sufficient The answer is C  Can we write the first statement as y<x<y since (x+y)(xy)<0, x falls in the range of y and y. This would make statement 1 alone sufficient to answer the question.



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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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21 Feb 2018, 21:17
manimadhuri wrote: broall wrote: LamaSalah wrote: Is x < y ?
(1) x^2  y^2 < 0 (2) x  y < 0 (1) \(x^2 < y^2 \iff x < y\) We have \(3^2 < 5^2 \implies 3 < 5\) However \(3^2 < (5)^2 \implies 3 >  5\). Hence, insufficient. (2) \(xy < 0 \implies x+y > 0\) If x=5, y=2 then \(x+y>0\) and \(x>y\) If x=2, y=5 then \(x+y>0\) and \(x<y\). Hence, insufficient. Combine (1) and (2): \((2) \implies x+y > 0\) \((1) \implies (xy)(x+y) < 0 \implies xy < 0 \implies x<y\). Sufficient The answer is C  Can we write the first statement as y<x<y since (x+y)(xy)<0, x falls in the range of y and y. This would make statement 1 alone sufficient to answer the question. x^2  y^2 < 0; x^2 < y^2; x < y. This means that y is further from 0 than x is. Which is not enough to get whether x < y. Basically we can have the following cases: yx0 y0x x0y 0xy
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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22 Feb 2018, 23:50
LamaSalah wrote: Is x < y ?
(1) x^2  y^2 < 0 (2) x  y < 0 i will also go with c as by using both the statement we can conclude whether x and y are positive and negative and then we can say x<y.



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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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18 Aug 2019, 17:38
Hi ScottTargetTestPrep and Bunuel, thanks for your explanations. What values could we plug into statement 2 to make it agree with the constraints? I didn't think to multiply by 1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plugin approach
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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18 Aug 2019, 21:33
dcummins wrote: Hi ScottTargetTestPrep and Bunuel, thanks for your explanations. What values could we plug into statement 2 to make it agree with the constraints? I didn't think to multiply by 1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plugin approach For (2) x = 1 and y = 2 gives an YES answer, while x = 2 and y = 1 gives a NO answer.
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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18 Aug 2019, 21:36
Bunuel wrote: dcummins wrote: Hi ScottTargetTestPrep and Bunuel, thanks for your explanations. What values could we plug into statement 2 to make it agree with the constraints? I didn't think to multiply by 1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plugin approach For (2) x = 1 and y = 2 gives an YES answer, while x = 2 and y = 1 gives a NO answer. Damn. I realise how silly this mistake was. I didn't write "" in front of the x i.e. xy so was wondering how the hell you could get xy<0
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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21 Aug 2019, 09:40
dcummins wrote: Hi ScottTargetTestPrep and Bunuel, thanks for your explanations. What values could we plug into statement 2 to make it agree with the constraints? I didn't think to multiply by 1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plugin approach I actually addressed your question in my solution: Statement Two Alone: x  y < 0 We can multiply the inequality in statement two by 1, remembering to reverse the inequality sign, and obtain: x + y > 0, or equivalently, 0 < x + y. However, we still cannot determine whether x > y. For instance if x = 2 and y = 1, then x is greater than y; however if x = 1 and y = 2, then x is less than y. Statement two alone is not sufficient to answer the question.
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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21 Aug 2019, 09:53
ScottTargetTestPrep wrote: dcummins wrote: Hi ScottTargetTestPrep and Bunuel, thanks for your explanations. What values could we plug into statement 2 to make it agree with the constraints? I didn't think to multiply by 1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plugin approach I actually addressed your question in my solution: Statement Two Alone: x  y < 0 We can multiply the inequality in statement two by 1, remembering to reverse the inequality sign, and obtain: x + y > 0, or equivalently, 0 < x + y. However, we still cannot determine whether x > y. For instance if x = 2 and y = 1, then x is greater than y; however if x = 1 and y = 2, then x is less than y. Statement two alone is not sufficient to answer the question. @ScottTargetTestPrep@Bunuel Please evaluate the attached problem, the work done on this same question. I tried not to do it by putting values.
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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21 Aug 2019, 16:28
ScottTargetTestPrep wrote: dcummins wrote: Hi ScottTargetTestPrep and Bunuel, thanks for your explanations. What values could we plug into statement 2 to make it agree with the constraints? I didn't think to multiply by 1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plugin approach I actually addressed your question in my solution: Statement Two Alone: x  y < 0 We can multiply the inequality in statement two by 1, remembering to reverse the inequality sign, and obtain: x + y > 0, or equivalently, 0 < x + y. However, we still cannot determine whether x > y. For instance if x = 2 and y = 1, then x is greater than y; however if x = 1 and y = 2, then x is less than y. Statement two alone is not sufficient to answer the question. Thanks Scott, I must have overlooked this.
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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22 Aug 2019, 17:29
dcummins wrote: ScottTargetTestPrep wrote: dcummins wrote: Hi ScottTargetTestPrep and Bunuel, thanks for your explanations. What values could we plug into statement 2 to make it agree with the constraints? I didn't think to multiply by 1, making it easier to eliminate statement 2, so I'm trying to see what I missed in my plugin approach I actually addressed your question in my solution: Statement Two Alone: x  y < 0 We can multiply the inequality in statement two by 1, remembering to reverse the inequality sign, and obtain: x + y > 0, or equivalently, 0 < x + y. However, we still cannot determine whether x > y. For instance if x = 2 and y = 1, then x is greater than y; however if x = 1 and y = 2, then x is less than y. Statement two alone is not sufficient to answer the question. Thanks Scott, I must have overlooked this. My pleasure.
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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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