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Re M2858 [#permalink]
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16 Sep 2014, 00:44
Official Solution: (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 \(0 \lt 2  y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \leq 0\) 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 \(1y \leq 0\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient. Answer: D
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Re: M2858 [#permalink]
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21 Jul 2015, 18:10
Bunuel wrote: Official Solution:
(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 \(0 \lt 2  y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \leq 0\) 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 \(1y \leq 0\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D Bunuel, Why in (1) you made X2 = 0? I didn't understand why LHS should be the least value. BR,



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Re: M2858 [#permalink]
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22 Jul 2015, 00:12
lucassandrade wrote: Bunuel wrote: Official Solution:
(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 \(0 \lt 2  y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \leq 0\) 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 \(1y \leq 0\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D Bunuel, Why in (1) you made X2 = 0? I didn't understand why LHS should be the least value. BR, Have you read the highlighted part? In \(x  2 \lt 2  y\), the left hand side of the inequality (x  2) is an absolute value. An absolute value of a number cannot be negative, so the left hand side of the inequality (x  2) is not negative, which means that the right hand side of the inequality, which is greater than the left hand side, also cannot be negative. This gives \(0 \lt 2  y\).
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Re: M2858 [#permalink]
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22 Jul 2015, 10:40
Bunuel wrote: lucassandrade wrote: Bunuel wrote: Official Solution:
(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 \(0 \lt 2  y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \leq 0\) 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 \(1y \leq 0\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D Bunuel, Why in (1) you made X2 = 0? I didn't understand why LHS should be the least value. BR, Have you read the highlighted part? In \(x  2 \lt 2  y\), the left hand side of the inequality (x  2) is an absolute value. An absolute value of a number cannot be negative, so the left hand side of the inequality (x  2) is not negative, which means that the right hand side of the inequality, which is greater than the left hand side, also cannot be negative. This gives \(0 \lt 2  y\). I still don't getting it. I know that absolute value cant be negative, but why you made the LHS equal 0? Couldn't it be 1 (if x=1)?



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Re: M2858 [#permalink]
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22 Jul 2015, 10:55
lucassandrade wrote: Bunuel wrote: lucassandrade wrote: Bunuel,
Why in (1) you made X2 = 0? I didn't understand why LHS should be the least value.
BR,
Have you read the highlighted part? In \(x  2 \lt 2  y\), the left hand side of the inequality (x  2) is an absolute value. An absolute value of a number cannot be negative, so the left hand side of the inequality (x  2) is not negative, which means that the right hand side of the inequality, which is greater than the left hand side, also cannot be negative. This gives \(0 \lt 2  y\). I still don't getting it. I know that absolute value cant be negative, but why you made the LHS equal 0? Couldn't it be 1 (if x=1)? Where did I make something equal to 0? It's a simple logic LHS: is >= 0, because it's an absolute value. Therefore RHS must also be >= 0.
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Re M2858 [#permalink]
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16 Oct 2015, 07:17
I think this is a highquality question and I agree with explanation. This question is repeated.



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Re M2858 [#permalink]
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29 Jun 2016, 03:58
1
This post received KUDOS
I think this is a highquality question and I don't agree with the explanation. I didn't quite understand how we can prove that y is prime from statement 2.
Statement 1 is completely clear.



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Re: M2858 [#permalink]
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30 Jun 2016, 07:11
TDK82 wrote: I think this is a highquality question and I don't agree with the explanation. I didn't quite understand how we can prove that y is prime from statement 2.
Statement 1 is completely clear. Here is a logic for (2): We know from the stem that y is a positive integer. Since y is a positive integer (1, 2, 3, ...), then 1  y = 1  positive = nonpositive, which measn that 1y=(1y). Therefore, for (2) we'd have \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient. Hope it's clear.
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Re: M2858 [#permalink]
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30 Jun 2016, 18:59
In the practice test question, you have asked whether y is prime as well.



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Re M2858 [#permalink]
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27 Jul 2016, 09:59
I think this is a highquality question and I don't agree with the explanation. in option 2, 1−y≤0. . In your explanation you have considered only the situation in which 1y<0 and arrived at the answer that statement 2 is also sufficient to prove that x =2. However if 1y = 0 then the equation would become x+y = 3 leaving the possibility that x or y could be 2 / 1 or vice versa.



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Re: M2858 [#permalink]
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27 Jul 2016, 10:10



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27 Jul 2016, 10:18
thanks Bunuel sorry! that is a classic example of where i could fail on the real GMAT! Such a blasphemous oversight! Apologies as well!



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Re: M2858 [#permalink]
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28 Jul 2016, 06:24
Bunuel wrote: Official Solution:
(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 \(0 \lt 2  y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \leq 0\) 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 \(1y \leq 0\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D Hi Bunuel, Can you please check and criticize my way solution below??? I used critical point method: Statement (2): x+y−3= 1−y y<1: is not viable as the prompt states that y is positive integer. Eliminate this possibility. y≥1: x+y+3=  1+y. then always x=2=prime. Sufficient Statement (1) x−2 < 2−y X<2: x+2< 2y ……. –x< y……… x>y .....As per prompt above, only x=1 so y =0 . This solution is not viable as y must be positive integer. Eliminate this possibility X≥2: x2<2y………x+y <4….the only combination that satisfies the condition is when x=2 & y=1 So sufficient. Answer: D Thanks



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Re: M2858 [#permalink]
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06 Sep 2016, 19:28
Hey guys
Here is what helped me solve the problem  I had trouble at first and trouble understanding the solution. So hopefully this helps someone.
So we know X>0 and Y>0
Lets focus on 1
x2<2y
with x>0 lets use x=1, x=2. x=3 as cases
x=1 ; 1 < 2  y (y >0 and integer) Not possible.  BAD CASE x=2 ; 0 < 2  y (y>) and interger, there fore y=2)  GOOD x=3 ; 1 < 2 y  BAD x=4 ; 2 < 2  y  BAD
Therefore for this case x must 2 and 2 is prime ..... SUFFICIENT
Prompt 2
x + y  3 = 1y
again, y>0 and integer, so choose y=1, y=2. y=3 y=4
y=1 ; x+13=0  x=2  Good Case y=2 ; x+23=1  x=2  Good case y=3 ; x+33=2  X=2  GooD Case y=4 ; x+43=3  x=2  Good Case
Therefore x=2, x is prime and this is Sufficient
Therefore overall D is answer



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Re: M2858 [#permalink]
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06 Feb 2017, 00:32
dear Bunuel, I got your solution, it really good ! how ever can you please see where i am going wrong in my approach for choice A. A. x−2<2−y i intent to open the mod and test the values of x and Y. Therefore 2 cases come up while opening the mod Case 1 x2>0 i.e x>2 The equation becomes x2 < 2y x < 4y Now according the above inequality and the condition X>2 there is no value of x or y which satisfies the 2<x<4y What am i missing here ? Case 2 x2<0 i.e x<2 if we analyze this condition carefully the only value that x can take here is 1 coz, 0<x<2 . However the equations becomes, x+2 <2y x>y Now this cannot be correct as y does not have any valid +integer value. 0<y<x<2 This is not a optimal approach, but i am keen to know gap do i have in my concepts ? thanks Bunuel wrote: Official Solution:
(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 \(0 \lt 2  y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \leq 0\) 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 \(1y \leq 0\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D



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grsm wrote: dear Bunuel, I got your solution, it really good ! how ever can you please see where i am going wrong in my approach for choice A.
A. x−2<2−y i intent to open the mod and test the values of x and Y. Therefore 2 cases come up while opening the mod
Case 1 x2>0 i.e x>2 The equation becomes x2 < 2y x < 4y Now according the above inequality and the condition X>2 there is no value of x or y which satisfies the 2<x<4y
What am i missing here ?
Case 2 x2<0 i.e x<2 if we analyze this condition carefully the only value that x can take here is 1 coz, 0<x<2 . However the equations becomes, x+2 <2y x>y Now this cannot be correct as y does not have any valid +integer value. 0<y<x<2
This is not a optimal approach, but i am keen to know gap do i have in my concepts ?
thanks
Hi grsm, You have missed equality sign in case1. If you put the equality sign in case1, x=2 and y =1 will satisfy the condition. In case2, there is no valid solution. Please note the definiton of modulus: \(x = \begin{cases} x, & \mbox{if } x \ge 0 \\ x, & \mbox{if } x < 0. \end{cases}\) Hope this helps.



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Bunuel wrote: Official Solution:
(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 \(0 \lt 2  y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \leq 0\) 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 \(1y \leq 0\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D Can someone explain me this. why we are checking if y more than or equal to 2 when we already know that y < 2. And even if y> 2 how can y2< 0? (if \(y\) is more than or equal to 2, then \(y2 \leq 0\) and it cannot be greater than \(x  2\))



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Re: M2858 [#permalink]
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04 Dec 2017, 10:10
unbroken wrote: Bunuel wrote: Official Solution:
(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 \(0 \lt 2  y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \leq 0\) 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 \(1y \leq 0\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D Can someone explain me this. why we are checking if y more than or equal to 2 when we already know that y < 2. And even if y> 2 how can y2< 0? (if \(y\) is more than or equal to 2, then \(y2 \leq 0\) and it cannot be greater than \(x  2\)) Explained above: In \(x  2 \lt 2  y\), the left hand side of the inequality (x  2) is an absolute value. An absolute value of a number cannot be negative, so the left hand side of the inequality (x  2) is not negative, which means that the right hand side of the inequality, which is greater than the left hand side, also cannot be negative. This gives \(0 \lt 2  y\).
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Re: M2858 [#permalink]
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30 Dec 2017, 07:36
grsm wrote: dear Bunuel, I got your solution, it really good ! how ever can you please see where i am going wrong in my approach for choice A. A. x−2<2−y i intent to open the mod and test the values of x and Y. Therefore 2 cases come up while opening the mod Case 1 x2>0 i.e x>2 The equation becomes x2 < 2y x < 4y Now according the above inequality and the condition X>2 there is no value of x or y which satisfies the 2<x<4y What am i missing here ? Case 2 x2<0 i.e x<2 if we analyze this condition carefully the only value that x can take here is 1 coz, 0<x<2 . However the equations becomes, x+2 <2y x>y Now this cannot be correct as y does not have any valid +integer value. 0<y<x<2 This is not a optimal approach, but i am keen to know gap do i have in my concepts ? thanks Bunuel wrote: Official Solution:
(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 \(0 \lt 2  y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \leq 0\) 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 \(1y \leq 0\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D I have the same query.Can any experts please address what GRSM said. I used the same approach. Thanks







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