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Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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27 Feb 2017, 06:17
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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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02 Mar 2017, 16:59
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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27 Feb 2017, 07:02
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, 07:32
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, 06:20
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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21 Feb 2018, 12:40
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, 20:17
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, 22:50
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, 16:38
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, 20:33
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, 20:36
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, 08:40
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, 08:53
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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IMG_20190821_222100.jpg [ 1.66 MiB  Viewed 3633 times ]



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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, 15:28
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, 16:29
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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26 Oct 2019, 21:30
Detail Solution 
1) x2 y2 < 0 >(xy)(x+y) <0, > either xy>0 and x+y<0 or xy<0 and x+y>0 > hen x>y or x<y both possibleNot Sufficient  A,D rejected 2)xy<0 > x+y>0 Not ufficientB rejected.
combining 1+2 > x+y>0 hence xy<0 >x<y ...sufficient option C is correct.
you can also solve by putting random ve +ve values.



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Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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18 Nov 2019, 04:00
Official explanation
Explanation: (1) The values 2 and 3 will give us a YES answer to the question stem, but the values 2 and 3 will give us a response of NO. Another way to approach this is to realize that you can factor the left hand side of the inequality to get (x  y)(x + y) < 0. This means that either the first term, (x  y), is negative or the second term, (x + y), is negative, but not both. If the first term is negative, then YES is answer to the question. However, if the second term is negative, the answer could be YES or NO. Accordingly, this statement is insufficient.
(2) You can simplify this by multiplying both sides by 1, which means you need to flip the direction of the inequality. You'll now have x + y > 0. This, however, doesn't tell us which value is larger, and this statement is insufficient.
Together, we know that (x  y)(x + y) < 0 and x + y > 0. This means that x  y < 0 . Remember, for the product of two values to be negative, one of them and only one of them must be negative. Because we know that (x + y) must be positive, then (x  y) < 0. Simply adding to each side of that inequality, we find that x < y, and know that the statements together are sufficient. Accordingly, the answer is C.




Re: Is x < y ? (1) x^2  y^2 < 0 (2) x  y < 0
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