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by GMATGuruNY » Thu Jul 28, 2016 3:32 am
Mo2men wrote:If x and y are positive integers, is x a prime number?

(1) |x−2|< 2−y

(2) x+y−3=|1−y|

Statement 1: |x−2|< 2−y
Since the absolute value on the left side cannot be less than 0, the right side must be GREATER THAN 0:
2-y > 0
2 > y
y < 2.
Since the prompt constrains x and y to positive integers, y = 1.

Substituting y=1 into |x-2| < 2-y, we get:
|x-2| < 1.
For the left side to be less than 1, the expression inside the absolute value must be equal to 0:
x-2 = 0
x = 2.
SUFFICIENT.

Statement 2: x+y−3=|1−y|
Case 1: y=1
Substituting y=1 into x+y−3=|1−y|, we get:
x+1-3 = |1-1|
x-2 = 0
x=2.

Case 2: y=2
Substituting y=2 into x+y−3=|1−y|, we get:
x+2-3 = |1-2|
x-1 = 1
x=2.

Case 3: y=100
Substituting y=100 into x+y−3=|1−y|, we get:
x+97 = |1-100|
x+97 = 99
x=2.

In every case, x=2.
SUFFICIENT.

The correct answer is D.
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by Mo2men » Thu Jul 28, 2016 6:25 am
GMATGuruNY wrote:
Statement 1: |x−2|< 2−y
Since the absolute value on the left side cannot be less than 0, the right side must be GREATER THAN 0:
2-y > 0
2 > y
y < 2.
Since the prompt constrains x and y to positive integers, y = 1.

Substituting y=1 into |x-2| < 2-y, we get:
|x-2| < 1.
For the left side to be less than 1, the expression inside the absolute value must be equal to 0:
x-2 = 0
x = 2.
SUFFICIENT.

Statement 2: x+y−3=|1−y|
Case 1: y=1
Substituting y=1 into x+y−3=|1−y|, we get:
x+1-3 = |1-1|
x-2 = 0
x=2.

Case 2: y=2
Substituting y=2 into x+y−3=|1−y|, we get:
x+2-3 = |1-2|
x-1 = 1
x=2.

Case 3: y=100
Substituting y=100 into x+y−3=|1−y|, we get:
x+97 = |1-100|
x+97 = 99
x=2.

In every case, x=2.
SUFFICIENT.

The correct answer is D.
Hi Mitch,

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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by GMATGuruNY » Thu Jul 28, 2016 6:54 am
Mo2men wrote:Hi Mitch,

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
Looks good!
Nice work.
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by Mo2men » Mon Oct 24, 2016 7:40 pm
GMATGuruNY wrote:
Mo2men wrote:Hi Mitch,

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
Looks good!
Nice work.
Dear Mitch,

I have a general question about critical points using the problem above as an example.

In statement 1, when we determine the critical point to be 2, where is the equal sign should be placed? is it X<=2 or X≥2? can I use in both? Do we have any restrcition to use equal sign with either < or > ??

Whatever you answer is, is it applicable for absolute values for absolute equation questions and inequality questions?

thanks in advance for your support
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by GMATGuruNY » Tue Oct 25, 2016 11:55 am
Mo2men wrote:Dear Mitch,

I have a general question about critical points using the problem above as an example.

In statement 1, when we determine the critical point to be 2, where is the equal sign should be placed? is it X<=2 or X≥2? can I use in both? Do we have any restrcition to use equal sign with either < or > ??

Whatever you answer is, is it applicable for absolute values for absolute equation questions and inequality questions?

thanks in advance for your support
If x≥2, then |x-2| = x-2.
If x<2, then |x-2| = -x+2.
Apply the equals sign only to the inequality in blue.
Last edited by GMATGuruNY on Tue Oct 25, 2016 12:37 pm, edited 1 time in total.
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by Mo2men » Tue Oct 25, 2016 12:22 pm
GMATGuruNY wrote:
Mo2men wrote:Dear Mitch,

I have a general question about critical points using the problem above as an example.

In statement 1, when we determine the critical point to be 2, where is the equal sign should be placed? is it X<=2 or X≥2? can I use in both? Do we have any restrcition to use equal sign with either < or > ??

Whatever you answer is, is it applicable for absolute values for absolute equation questions and inequality questions?

thanks in advance for your support
If x≥2, then |x-2| = x-2.
If x<2, then |x-2| = -x+2.
The equals sign is applied only to the inequality in blue.
What about Statement 1? Is the following correct?

If y<1, then |1-y| = 1-y
If y≥1, then |1-y| = y-1

or
If y<=1, then |1-y| = 1-y
If y>1, then |1-y| = y-1


Thanks for your help
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by GMATGuruNY » Tue Oct 25, 2016 12:49 pm
Mo2men wrote:What about Statement 1? Is the following correct?

If y<1, then |1-y| = 1-y
If y≥1, then |1-y| = y-1

or
If y<=1, then |1-y| = 1-y
If y>1, then |1-y| = y-1


Thanks for your help
To avoid confusion, change the signs for the expression inside the absolute value only if necessary.
Given expression: |1-y|.
The signs for 1-y do not need to change if y≤1.
Thus:
|1-y| = 1-y for all values of y such that y≤1.
The signs for 1-y must change only if y>1.
Thus:
|1-y| = -1+y for all values of y such that y>1.
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by Mo2men » Wed Oct 26, 2016 12:20 am
GMATGuruNY wrote:

Statement 2: x+y−3=|1−y|
Case 1: y=1
Substituting y=1 into x+y−3=|1−y|, we get:
x+1-3 = |1-1|
x-2 = 0
x=2.

Case 2: y=2
Substituting y=2 into x+y−3=|1−y|, we get:
x+2-3 = |1-2|
x-1 = 1
x=2.

Case 3: y=100
Substituting y=100 into x+y−3=|1−y|, we get:
x+97 = |1-100|
x+97 = 99
x=2.

In every case, x=2.
SUFFICIENT.
Dear Mitch,

I tried different way to solve statement 2 but It did not led to any to correct solution.

x+y−3=|1−y|

RHS has NON NEGATIVE value.
So x+y-3≥0

x+y≥3

Based on the solution above, there are many values such as x=2, y=1 or even x=10, y=1.

My solution contradicts that x is always 2.

Where I did go wrong?

Thanks
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by GMATGuruNY » Wed Oct 26, 2016 3:27 am
Mo2men wrote:
Dear Mitch,

I tried different way to solve statement 2 but It did not led to any to correct solution.

x+y−3=|1−y|

RHS has NON NEGATIVE value.
So x+y-3≥0

x+y≥3

Based on the solution above, there are many values such as x=2, y=1 or even x=10, y=1.

My solution contradicts that x is always 2.

Where I did go wrong?

Thanks
The inequality in red does not fully satisfy the constraint that x+y−3=|1−y|.
While it is true that x+y-3≥0, the equation in blue must still be satisfied.

If y≤1, then |1-y| = 1-y.
Substituting |1-y| = 1-y into x+y−3=|1−y|, we get:
x+y-3 = 1-y.
In this case -- since y≤1, and the prompt indicates that y is a positive integer -- the only valid option for y is y=1.
Substituting y=1 into x+y-3 = 1-y, we get:
x+1-3 = 1-1
x-2 = 0
x=2.

If y>1, then |1-y| = -1+y.
Substituting |1-y| = -1+y into x+y−3=|1−y|, we get:
x+y-3 = -1+y
x-3 = -1
x=2.

Thus, in every case, x=2.
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