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M02-25

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Re: M02-25  [#permalink]

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New post 28 May 2017, 07:34
Bunuel
I 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. x-2>y-2 = 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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New post 28 May 2017, 07:54
kerin wrote:
Bunuel
I 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. x-2>y-2 = 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: M02-25  [#permalink]

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New post 27 Jun 2017, 19:50
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<x-2<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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New post 27 Jun 2017, 22:08
kreel11 wrote:
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<x-2<1 and hence 1<x<3 so x=2 which is prime.

Kindly confirm if the above approach for statement 1 is correct.

______________
Yes, that's correct.
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Re: M02-25  [#permalink]

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New post 25 Dec 2017, 08:37
Hi Bunuel

Please let me know if my approach makes sense

Given: x, y -> +ve
x -> prime?

(1) |x-2| < 2-y
Opening up the Absolute Value:
when +ve
x-2 < 2-y
x+y < 4
OR when -ve
-x+2 < 2-y
-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+y-3 = |1-y|
when +ve
x+y-3 = 1-y
x+2y = 4
Testing Cases:
x = 2, y = 1: Only option that works

OR when -ve
x+y-3 = -1+y
x=2 -> prime

Sufficient

Answer: D

Thanks
Prathamesh
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New post 25 Dec 2017, 10:26
ppnimkar wrote:
Hi Bunuel

Please let me know if my approach makes sense

Given: x, y -> +ve
x -> prime?

(1) |x-2| < 2-y
Opening up the Absolute Value:
when +ve
x-2 < 2-y
x+y < 4
OR when -ve
-x+2 < 2-y
-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+y-3 = |1-y|
when +ve
x+y-3 = 1-y
x+2y = 4
Testing Cases:
x = 2, y = 1: Only option that works

OR when -ve
x+y-3 = -1+y
x=2 -> prime

Sufficient

Answer: D

Thanks
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: M02-25  [#permalink]

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New post 25 Dec 2017, 20:42
Bunuel wrote:
ppnimkar wrote:
Hi Bunuel

Please let me know if my approach makes sense

Given: x, y -> +ve
x -> prime?

(1) |x-2| < 2-y
Opening up the Absolute Value:
when +ve
x-2 < 2-y
x+y < 4
OR when -ve
-x+2 < 2-y
-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+y-3 = |1-y|
when +ve
x+y-3 = 1-y
x+2y = 4
Testing Cases:
x = 2, y = 1: Only option that works

OR when -ve
x+y-3 = -1+y
x=2 -> prime

Sufficient

Answer: D

Thanks
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 Bunuel

Thanks 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 & Limiting
-2 < 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
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Re: M02-25  [#permalink]

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New post 26 Jan 2018, 07:30
1
1
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 \(y-2 \le{0}\) 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 \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.

(2)\(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \le {0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.


Answer: D


Bunuel
(if \(y\) is more than or equal to 2, then \(y-2 \le{0}\) and it cannot be greater than \(|x - 2|\))

Shouldn't the highlighted part be actually \(2-y \le{0}\) and not \(y-2 \le{0}\)?
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New post 26 Jan 2018, 08:03
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 \(y-2 \le{0}\) 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 \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.

(2)\(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \le {0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.


Answer: D


Bunuel
(if \(y\) is more than or equal to 2, then \(y-2 \le{0}\) and it cannot be greater than \(|x - 2|\))

Shouldn't the highlighted part be actually \(2-y \le{0}\) and not \(y-2 \le{0}\)?


Yes. Edited the typo. Thank you.
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New post 17 Jun 2018, 11:29
I think this is a high-quality question and I agree with explanation.
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New post 12 Sep 2018, 15:13
I think this is a high-quality question and I agree with explanation.
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New post 19 Nov 2018, 23:55
I think this is a high-quality question and I agree with explanation.
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New post 23 Dec 2018, 09:20
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.

Posted from my mobile device
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New post 23 Dec 2018, 11:07
saurabhrampal wrote:
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.

Posted from my mobile device


This is a hard question and its discussion goes on previous two pages. Please re-read it and if the question is still no clear please ask specific question.
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Re: M02-25 &nbs [#permalink] 23 Dec 2018, 11:07

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