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(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.
(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.
(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 \(0 \lt 2 - y\).
_________________
(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)?
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.
_________________
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.
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.
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.
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.
_________________
(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.
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
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.
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:
\(|x| = \begin{cases} x, & \mbox{if } x \ge 0 \\ -x, & \mbox{if } x < 0. \end{cases}\)
(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 \(|x - 2|\))
(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 \(0 \lt 2 - y\).
_________________
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.
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47% (00:38) correct 
