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15 Sep 2020, 10:38
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (02:11) correct
63%
(02:14)
wrong
based on 194
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Not Attempted Yet
We need to find if
\(x^2\) < x-|y|?
rearranging we get
|y|<x-\(x^2\) ?
Statement 1: y>x
we certainly know that if y>x then for any value of x
|y| > x - \(x^2\)
Hence Statement 1 is sufficient
Statement 2: x<0
|y|< x - \(x^2\)
We know that |y| will be a positive number hence
0< x - \(x^2\)
Taking x common we get
x(1-x)>0
So the conditions are x>0 and 1-x>0 which implies x<1
hence the range of values is between 0<x<1. For values of x between 0 and 1
but since it's give x<0
Statement 2 is sufficient to say that |y| > x-\(x^2\)
Hence both the statements alone are sufficient
In both the cases |y| is not less than x - \(x^2\). Hence D
Kudos if it helps
\(x^2\) < x-|y|?
rearranging we get
|y|<x-\(x^2\) ?
Statement 1: y>x
we certainly know that if y>x then for any value of x
|y| > x - \(x^2\)
Hence Statement 1 is sufficient
Statement 2: x<0
|y|< x - \(x^2\)
We know that |y| will be a positive number hence
0< x - \(x^2\)
Taking x common we get
x(1-x)>0
So the conditions are x>0 and 1-x>0 which implies x<1
hence the range of values is between 0<x<1. For values of x between 0 and 1
but since it's give x<0
Statement 2 is sufficient to say that |y| > x-\(x^2\)
Hence both the statements alone are sufficient
In both the cases |y| is not less than x - \(x^2\). Hence D
Kudos if it helps
16 Sep 2020, 08:02
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Genoa2000
Rearrange the terms in the question so that y and x are isolated, is \(| y | < x- x^2\) ?
Statement 1:
Our goal is to assemble the inequality from the question. If \(y > x\), then \(y > x - x^2\) must be true as \(x^2\) is nonnegative. Then note if y is nonnegative, \(|y| = y\) and \(|y| > x - x^2\) is true. If y is negative, then we would have \(|y| > y > x- x^2\). Either way, \(|y| > x - x^2\) is true. Therefore the question can be answered with "no" and this is sufficient.
Statement 2:
\(x - x^2 = x * (1 - x)\). If x is negative, then \(x * (1 - x) = Negative * Positive = Negative\). Therefore |y| is always greater than \(x*(1-x) = x - x^2\). Again we answer "no" so this is sufficient.
Ans: D
16 Sep 2020, 10:30
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We want to know if |y| < x - x^2.
If y > x, then certainly |y| > x. But x^2 is a square, so it is positive (or zero), and it is always true that x > x - x^2. So if Statement 1 is true, then |y| > x > x - x^2, and we can be certain the answer to the question is 'no'.
From Statement 2, if x < 0, then x - |y| is certainly negative (we're subtracting a positive thing from a negative thing). Since x^2 is positive, it could never be smaller than the negative number x - |y|, so again we can be certain the answer is 'no' and the answer is D.
If y > x, then certainly |y| > x. But x^2 is a square, so it is positive (or zero), and it is always true that x > x - x^2. So if Statement 1 is true, then |y| > x > x - x^2, and we can be certain the answer to the question is 'no'.
From Statement 2, if x < 0, then x - |y| is certainly negative (we're subtracting a positive thing from a negative thing). Since x^2 is positive, it could never be smaller than the negative number x - |y|, so again we can be certain the answer is 'no' and the answer is D.
