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M28-58

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Re: M28-58  [#permalink]

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New post 30 Dec 2017, 10:09
sandysilva wrote:
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 x-2>0 i.e x>2
The equation becomes
x-2 < 2-y
x < 4-y
Now according the above inequality and the condition X>2 there is no value of x or y which satisfies the 2<x<4-y

What am i missing here ?

Case 2 x-2<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 <2-y
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 \(y-2 \leq 0\) 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 \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.

(2) \(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \leq 0\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), 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


The problem with that solution is that it does not consider the case when x = 2. When you open the modules for |x - 2| you should consider the case x <= 2 and x > 2 OR x < 2 and x >= 2, so you should include = sign for 2 either in the first case or in second. If you do, then you'll find that x = 2 and y = 1 are possible.

Hope it's clear.
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Re: M28-58  [#permalink]

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New post 18 Sep 2018, 12:24
Superb Question! Tests concepts very well.
If one is familiar with the concept, it can be solved under 2 minutes.
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Re: M28-58 &nbs [#permalink] 18 Sep 2018, 12:24

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