I KINDAAAA get it, but No I dont! Could someone please explain the logic. I understand why '1' is good, also, i understand algebraic approach..somehow not sure about the 'logical' approach
Is the number x positive?
(1) On the number line, 0 is closer to x - 1 than to x.
(2) On the number line, 0 is closer to x than to x + 1.
Solution is A
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synchrodestiny
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|x| = the distance between 0 and x.Is the number x positive?
(1) On the number line, 0 is closer to x - 1 than to x.
(2) On the number line, 0 is closer to x than to x + 1.
|x-y| = the distance between 0 and x-y.
|x+y| = the distance between 0 and x+y.
Statement 1: On the number line, 0 is closer to x - 1 than to x.
|x-1| < |x|.
Since there is absolute value notation on each side, we can square the inequality.
(x-1)² < x²
x² - 2x + 1 < x²
-2x < -1
x > 1/2.
Thus, x must be positive.
SUFFICIENT.
Statement 2: On the number line, 0 is closer to x than to x + 1.
|x| < |x+1|.
Since there is absolute value notation on each side, we can square the inequality.
x² < (x+1)²
x² < x² + 2x + 1
-2x < 1
x > -1/2.
Thus, x could be negative or positive.
INSUFFICIENT.
The correct answer is A.
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Brent@GMATPrepNow
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Another approach is the sketch the cases on a number line.Is the number x positive?
(1) On the number line, 0 is closer to x - 1 than to x.
(2) On the number line, 0 is closer to x than to x + 1.
Target question: Is x positive?
Statement 1: On the number line, 0 is closer to x - 1 than to x.
First, recognize that x-1 will always be to the left of x.
Second, recognize that there are 3 possible ways to place x-1 and x with relation to zero.
If zero is closer to x-1 than to x, then we can rule out case #2, leaving us with cases #1 and #3 as possible scenarios.
If case #1 is true, we can see that x must be positive
If case #3 is true, we can see that x must be positive
Since both possible cases yield the same answer to the target question, we can answer the target question with certainty.
So, statement 1 is SUFFICIENT
Statement 2: On the number line, 0 is closer to x than to x + 1.
Recognize that x+1 will always be to the right of x.
Also recognize that there are 3 possible ways to place x and x+1 with relation to zero.
If zero is closer to x than to x+1, then we can rule out case #2, leaving us with cases #1 and #3 as possible scenarios.
If case #1 is true, we can see that x is negative
If case #3 is true, we can see that x is positive
Since the two possible cases yield different answers to the target question, we cannot answer the target question with certainty.
So, statement 2 is NOT SUFFICIENT
Answer = A
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GMATinsight
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Question : Is X > 0?synchrodestiny wrote:I KINDAAAA get it, but No I dont! Could someone please explain the logic. I understand why '1' is good, also, i understand algebraic approach..somehow not sure about the 'logical' approach
Is the number x positive?
(1) On the number line, 0 is closer to x - 1 than to x.
(2) On the number line, 0 is closer to x than to x + 1.
Solution is A
Statement 1) On the number line, 0 is closer to x - 1 than to x
Which is TRUE ONLY if X is greater than 0.5
i.e. X can have any value greater than 0.5
SUFFICIENT
Statement 2) On the number line, 0 is closer to x than to x + 1
Which is TRUE ONLY if X is greater than -0.5
i.e. X can have any value greater than -0.5 which may be Positive or Negative
INSUFFICIENT
Answer: Option A
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Alternate approach:synchrodestiny wrote: Is the number x positive?
(1) On the number line, 0 is closer to x - 1 than to x.
(2) On the number line, 0 is closer to x than to x + 1.
|a| = the distance between a and 0.
|a-b| = the distance between a and b.
|a+b| = |a-(-b)| = the distance between a and -b.
Statement 1: On the number line, 0 is closer to x - 1 than to x.
Translated into math:
|x-1| < |x|.
Rephrased:
The distance between x and 1 is less than the distance between x and 0.
For x to be closer to 1 than to 0, x must be TO THE RIGHT OF 1/2 on the number line.
..........0..........1/2..........1..........
Since x must be within the red portion shown above, x>0.
SUFFICIENT.
Statement 2: On the number line, 0 is closer to x than to x + 1.
Translated into math:
|x| < |x+1|.
Rephrased:
The distance between x and 0 is less than the distance between x and -1.
For x to be closer to 0 than to -1, x must be TO THE RIGHT OF -1/2 on the number line.
..........-1..........-1/2..........0..........
Since x must be within the red portion shown above, it is possible that x≤0 or that x>0.
INSUFFICIENT.
The correct answer is A.
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Is the number x positive?synchrodestiny wrote:I KINDAAAA get it, but No I dont! Could someone please explain the logic. I understand why '1' is good, also, i understand algebraic approach..somehow not sure about the 'logical' approach
Is the number x positive?
(1) On the number line, 0 is closer to x - 1 than to x.
(2) On the number line, 0 is closer to x than to x + 1.
Solution is A
(1) On the number line, 0 is closer to x - 1 than to x.
Hence اx - 1ا is less than اxا, no non-positive value of x can satisfy this inequality, hence x is positive and (1) is sufficient.
(2) On the number line, 0 is closer to x than to x + 1.
Hence اxا is less than اx + 1ا, which holds true for x = 0 and for certain negative values of x too, not sure if x is positive, hence (2) is [spoiler]insufficient.
Pick (A)[/spoiler].
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Brent@GMATPrepNow
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Another approach is the sketch the cases on a number line.synchrodestiny wrote:Is the number x positive?
(1) On the number line, 0 is closer to x - 1 than to x.
(2) On the number line, 0 is closer to x than to x + 1.
Solution is A
Target question: Is x positive?
Statement 1: On the number line, 0 is closer to x - 1 than to x.
First, recognize that x-1 will always be to the left of x.
Second, recognize that there are 3 possible ways to place x-1 and x with relation to zero.
If zero is closer to x-1 than to x, then we can rule out case #2, leaving us with cases #1 and #3 as possible scenarios.
If case #1 is true, we can see that x must be positive
If case #3 is true, we can see that x must be positive
Since both possible cases yield the same answer to the target question, we can answer the target question with certainty.
So, statement 1 is SUFFICIENT
Statement 2: On the number line, 0 is closer to x than to x + 1.
Recognize that x+1 will always be to the right of x.
Also recognize that there are 3 possible ways to place x and x+1 with relation to zero.
If zero is closer to x than to x+1, then we can rule out case #2, leaving us with cases #1 and #3 as possible scenarios.
If case #1 is true, we can see that x is negative
If case #3 is true, we can see that x is positive
Since the two possible cases yield different answers to the target question, we cannot answer the target question with certainty.
So, statement 2 is NOT SUFFICIENT
Answer = A
Cheers,
Brent
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