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M28-58

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M28-58 [#permalink]

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Re M28-58 [#permalink]

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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
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Re: M28-58 [#permalink]

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New post 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 \(y-2 \leq 0\) 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


Bunuel,

Why in (1) you made X-2 = 0? I didn't understand why LHS should be the least value.

BR,
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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 \(y-2 \leq 0\) 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


Bunuel,

Why in (1) you made X-2 = 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: M28-58 [#permalink]

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New post 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 \(y-2 \leq 0\) 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


Bunuel,

Why in (1) you made X-2 = 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: M28-58 [#permalink]

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New post 22 Jul 2015, 10:55
lucassandrade wrote:
Bunuel wrote:
lucassandrade wrote:

Bunuel,

Why in (1) you made X-2 = 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 M28-58 [#permalink]

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New post 16 Oct 2015, 07:17
I think this is a high-quality question and I agree with explanation. This question is repeated.
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Re M28-58 [#permalink]

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I think this is a high-quality 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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New post 30 Jun 2016, 07:11
TDK82 wrote:
I think this is a high-quality 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 = non-positive, which measn that |1-y|=-(1-y). Therefore, for (2) we'd have \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.

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

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New post 30 Jun 2016, 18:59
In the practice test question, you have asked whether y is prime as well.
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Re M28-58 [#permalink]

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New post 27 Jul 2016, 09:59
I think this is a high-quality 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 1-y<0 and arrived at the answer that statement 2 is also sufficient to prove that x =2. However if 1-y = 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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I think this is a high-quality 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 1-y<0 and arrived at the answer that statement 2 is also sufficient to prove that x =2. However if 1-y = 0 then the equation would become x+y = 3 leaving the possibility that x or y could be 2 / 1 or vice versa.


Dear Senthil7, EVERY question in quant tests is correct.

As for your doubt: if 1 - y = 0, then y = 1. So, x = 2.
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Re: M28-58 [#permalink]

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New post 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: M28-58 [#permalink]

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New post 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 \(y-2 \leq 0\) 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



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< 2-y ……. –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: x-2<2-y………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: M28-58 [#permalink]

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New post 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

|x-2|<2-y

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 = |1-y|

again, y>0 and integer, so choose y=1, y=2. y=3 y=4

y=1 ; x+1-3=0 --- x=2 --- Good Case
y=2 ; x+2-3=1 ----- x=2 --- Good case
y=3 ; x+3-3=2 --- X=2 --- GooD Case
y=4 ; x+4-3=3 --- x=2 -- Good Case


Therefore x=2, x is prime and this is Sufficient

Therefore overall D is answer
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New post 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 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
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M28-58 [#permalink]

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New post 06 Feb 2017, 04:00
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


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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M28-58 [#permalink]

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New post 04 Dec 2017, 10:06
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



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 y-2< 0?

(if \(y\) is more than or equal to 2, then \(y-2 \leq 0\) and it cannot be greater than \(|x - 2|\))
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New post 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 \(y-2 \leq 0\) 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



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 y-2< 0?

(if \(y\) is more than or equal to 2, then \(y-2 \leq 0\) 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: M28-58 [#permalink]

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New post 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 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
Re: M28-58   [#permalink] 30 Dec 2017, 07:36

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