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