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Re: M0225 [#permalink]
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28 May 2017, 08:34
BunuelI was taught that when handling inequalities like x−2<2−y, we need to consider 2 equations as follows. 1. x−2<2−y = x+y<4 2. x2>y2 = x>y And the relationship between (1) and (2) is "OR". Is that the wrong approach to handle inequalities and how does the above relates to your sample answer? Thanks!



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28 May 2017, 08:54
kerin wrote: BunuelI was taught that when handling inequalities like x−2<2−y, we need to consider 2 equations as follows. 1. x−2<2−y = x+y<4 2. x2>y2 = x>y And the relationship between (1) and (2) is "OR". Is that the wrong approach to handle inequalities and how does the above relates to your sample answer? Thanks! 1. What contraindications do you see between what you say and the solution? 2. There are several different ways to solve absolute value questions and this question requires some out of the box reasoning. Having said that, if you notice, the reasoning there is based on the basic properties of absolute values. For (1) we used that absolute value cannot be negative and for (2) we used that if a <= 0, then a = a. Hope it helps.
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Re: M0225 [#permalink]
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27 Jun 2017, 20:50
Hi Bunuel, So from statement 1 what I understand is that x−2<2−y will only happen if x being positive will be equal to 2. Since it is absolute value and will be positive so for RHS Y cannot be >2 since then RHS<LHS (ve value). so y has to be only 1 (positive integer) and hence x has to be 2 as no other value will make it less than RHS (1). Also alternatively, now since x−2 < 1 (value with y=1), 1<x2<1 and hence 1<x<3 so x=2 which is prime. Kindly confirm if the above approach for statement 1 is correct.
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27 Jun 2017, 23:08



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Re: M0225 [#permalink]
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25 Dec 2017, 09:37
Hi BunuelPlease let me know if my approach makes sense Given: x, y > +ve x > prime? (1) x2 < 2y Opening up the Absolute Value: when +vex2 < 2y x+y < 4 OR when vex+2 < 2y x<y x>y Test Cases: x=2 y=1 for x+y < 4 or = 3 [Given x, y > 0 & x>y, so total has to be 3] and so x is prime Sufficient(2) x+y3 = 1y when +vex+y3 = 1y x+2y = 4 Testing Cases: x = 2, y = 1: Only option that works OR when vex+y3 = 1+y x=2 > prime SufficientAnswer: DThanks Prathamesh



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Re: M0225 [#permalink]
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25 Dec 2017, 11:26
ppnimkar wrote: Hi BunuelPlease let me know if my approach makes sense Given: x, y > +ve x > prime? (1) x2 < 2y Opening up the Absolute Value: when +vex2 < 2y x+y < 4 OR when vex+2 < 2y x<y x>y Test Cases: x=2 y=1 for x+y < 4 or = 3 [Given x, y > 0 & x>y, so total has to be 3] and so x is prime Sufficient(2) x+y3 = 1y when +vex+y3 = 1y x+2y = 4 Testing Cases: x = 2, y = 1: Only option that works OR when vex+y3 = 1+y x=2 > prime SufficientAnswer: DThanks Prathamesh I guess when you test values for (1), you consider x + y < 4 and x > y as simultaneous equations, and that is why you test only x = 2 and y = 1, whereas x + y < 4 and x > y are two different cases (true in certain conditions).
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Re: M0225 [#permalink]
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25 Dec 2017, 21:42
Bunuel wrote: ppnimkar wrote: Hi BunuelPlease let me know if my approach makes sense Given: x, y > +ve x > prime? (1) x2 < 2y Opening up the Absolute Value: when +vex2 < 2y x+y < 4 OR when vex+2 < 2y x<y x>y Test Cases: x=2 y=1 for x+y < 4 or = 3 [Given x, y > 0 & x>y, so total has to be 3] and so x is prime Sufficient(2) x+y3 = 1y when +vex+y3 = 1y x+2y = 4 Testing Cases: x = 2, y = 1: Only option that works OR when vex+y3 = 1+y x=2 > prime SufficientAnswer: DThanks Prathamesh I guess when you test values for (1), you consider x + y < 4 and x > y as simultaneous equations, and that is why you test only x = 2 and y = 1, whereas x + y < 4 and x > y are two different cases (true in certain conditions). Hi BunuelThanks for your response I thought if it's the same parent absolute equation, you can combine & limit them eg. x+y < 2 x+y < 2 OR x+y > 2 Combining & Limiting2 < x+y < 2 Please can you provide examples in which these two parts need to be considered as two different equations where the simultaneous method will not work? Thanks Prathamesh Merry Christmas



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Re: M0225 [#permalink]
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04 Jan 2018, 09:11
I think this is a highquality question and I agree with explanation. hey, the solution is D yet the answer shows that it is B. please check. thanks!



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04 Jan 2018, 09:16



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Re: M0225 [#permalink]
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26 Jan 2018, 08:30
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Bunuel wrote: Official Solution:
If x and y are positive integers, is x a prime number?
(1)\(x  2 \lt 2  y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2  y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \le{0}\) and it cannot be greater than \(x  2\)). Next, since given that \(y\) is a positive integer, then \(y=1\). So, we have that: \(x  2 \lt 1\), which implies that \(1 \lt x2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient. (2)\(x + y  3 = 1y\). Since \(y\) is a positive integer, then \(1y \le {0}\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D Bunuel (if \(y\) is more than or equal to 2, then \(y2 \le{0}\) and it cannot be greater than \(x  2\)) Shouldn't the highlighted part be actually \(2y \le{0}\) and not \(y2 \le{0}\)?



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Re: M0225 [#permalink]
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26 Jan 2018, 09:03
sushforgmat wrote: Bunuel wrote: Official Solution:
If x and y are positive integers, is x a prime number?
(1)\(x  2 \lt 2  y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2  y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \le{0}\) and it cannot be greater than \(x  2\)). Next, since given that \(y\) is a positive integer, then \(y=1\). So, we have that: \(x  2 \lt 1\), which implies that \(1 \lt x2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient. (2)\(x + y  3 = 1y\). Since \(y\) is a positive integer, then \(1y \le {0}\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D Bunuel (if \(y\) is more than or equal to 2, then \(y2 \le{0}\) and it cannot be greater than \(x  2\)) Shouldn't the highlighted part be actually \(2y \le{0}\) and not \(y2 \le{0}\)? Yes. Edited the typo. Thank you.
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