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16 Sep 2014, 01:07
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If \([x]\) denotes the largest integer smaller than \(x\), is \([x] > [-x]\)?
(1) \(x = [x] + 1\)
(2) \(x + 1 > 0\)
(1) \(x = [x] + 1\)
(2) \(x + 1 > 0\)
16 Sep 2014, 01:07
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Official Solution:
If \([x]\) denotes the largest integer smaller than \(x\), is \([x] > [-x]\)?
We are given a function, [ ], which rounds DOWN a number to the nearest integer. For example:
\([2.7] = 2\), because 2 is the largest integer less than 2.7;
\([3] = 2\), because 2 is the largest integer less than 3;
\([-1.7] = -2\), because -2 is the largest integer less than -1.7.
As a result, for \([x] > [-x]\) to hold true, \(x\) must be greater than 0. This is because if \(x < 0\), \([x]\) will be the greatest integer less than a negative value, which is negative, while \([-x]\) will be the greatest integer less than a positive value, which is positive (or 0 if \(-1 \leq x < 0\)). If \(x = 0\), both \([x]\) and \([-x]\) will be -1, making them equal to each other. However, when \(x > 0\), \([x]\) will be the greatest integer less than a positive value, which is positive (or 0 if \(0 < x \leq 1\)), while \([-x]\) will be the greatest integer less than a negative value, which is negative. Therefore, the question essentially asks whether \(x > 0\).
(1) \(x = [x] + 1\).
The given statement is true for all integers. For example, if x = 3, then \([3] = 2\) (the largest integer smaller than 3), so 3 = 2 + 1. Therefore, this statement simply implies that \(x\) is an integer, which is not sufficient to determine whether \(x > 0\). For instance, if \(x = 1\), then \([x] = [1] = 0\) and \([-x] = [-1] = -2\), resulting in a YES answer for \([x] > [-x]\). However, if \(x = -1\), then \([x] = [-1] = -2\) and \([-x] = [1] = 0\), leading to a NO answer for \([x] > [-x]\). Not sufficient.
(2) \(x + 1 > 0\).
This statement implies that \(x > -1\), which is clearly insufficient to answer the question.
(1)+(2) Combining the statements, we can conclude that \(x\) is an integer greater than -1: 0, 1, 2, 3, and so on. If \(x\) is a positive integer, the answer will be YES. However, if \(x = 0\), then \([x] = [0] = 0\) and \([-x] = [0] = 0\), and the answer will be NO. Not sufficient.
Answer: E
If \([x]\) denotes the largest integer smaller than \(x\), is \([x] > [-x]\)?
We are given a function, [ ], which rounds DOWN a number to the nearest integer. For example:
\([2.7] = 2\), because 2 is the largest integer less than 2.7;
\([3] = 2\), because 2 is the largest integer less than 3;
\([-1.7] = -2\), because -2 is the largest integer less than -1.7.
As a result, for \([x] > [-x]\) to hold true, \(x\) must be greater than 0. This is because if \(x < 0\), \([x]\) will be the greatest integer less than a negative value, which is negative, while \([-x]\) will be the greatest integer less than a positive value, which is positive (or 0 if \(-1 \leq x < 0\)). If \(x = 0\), both \([x]\) and \([-x]\) will be -1, making them equal to each other. However, when \(x > 0\), \([x]\) will be the greatest integer less than a positive value, which is positive (or 0 if \(0 < x \leq 1\)), while \([-x]\) will be the greatest integer less than a negative value, which is negative. Therefore, the question essentially asks whether \(x > 0\).
(1) \(x = [x] + 1\).
The given statement is true for all integers. For example, if x = 3, then \([3] = 2\) (the largest integer smaller than 3), so 3 = 2 + 1. Therefore, this statement simply implies that \(x\) is an integer, which is not sufficient to determine whether \(x > 0\). For instance, if \(x = 1\), then \([x] = [1] = 0\) and \([-x] = [-1] = -2\), resulting in a YES answer for \([x] > [-x]\). However, if \(x = -1\), then \([x] = [-1] = -2\) and \([-x] = [1] = 0\), leading to a NO answer for \([x] > [-x]\). Not sufficient.
(2) \(x + 1 > 0\).
This statement implies that \(x > -1\), which is clearly insufficient to answer the question.
(1)+(2) Combining the statements, we can conclude that \(x\) is an integer greater than -1: 0, 1, 2, 3, and so on. If \(x\) is a positive integer, the answer will be YES. However, if \(x = 0\), then \([x] = [0] = 0\) and \([-x] = [0] = 0\), and the answer will be NO. Not sufficient.
Answer: E
01 Nov 2015, 00:42
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Hi Bunuel,
I have only one doubt.. when we combined two statements.. (1) & (2)..
We understood that "x is an integer" and "x>-1"
So my Q is when x = 0 then [x] = -1; [-x] = 1; Hence [x]<[-x] and for all other cases [x]>[-x]..
Hence insufficient to draw a conclusion, so option E is correct.
But as per you when x is 0 ([0] = [0]).. it is not as per the context of question which says (largest integer smaller than x)
Am I right? let me know.. thanks
I have only one doubt.. when we combined two statements.. (1) & (2)..
We understood that "x is an integer" and "x>-1"
So my Q is when x = 0 then [x] = -1; [-x] = 1; Hence [x]<[-x] and for all other cases [x]>[-x]..
Hence insufficient to draw a conclusion, so option E is correct.
But as per you when x is 0 ([0] = [0]).. it is not as per the context of question which says (largest integer smaller than x)
Am I right? let me know.. thanks
