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Difficulty:
65%
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59% (02:30) correct
41%
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wrong
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If x is a negative integer, is x < −3?
(1) x^2 + 6x < 7
(2) x^2 + |x| ≤ 2
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(1) x^2 + 6x < 7
(2) x^2 + |x| ≤ 2
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
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Posts: 1,001
1) x^2 + 6x -7 <0 ==> (x+7)(x-1) < 0 ==> -7<x<1. Insufficient!
2) For negative integers, this condition is only true for x = -1. Sufficient!
Answer B.
2) For negative integers, this condition is only true for x = -1. Sufficient!
Answer B.
rahul16singh28
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Bunuel
Statement I: Take X = 0, - 4 ------------ Not Sufficient
Statement II: Here, X can be only 0,-1,1... Sufficient.
DavidTutorexamPAL
Dear David,
Please confirm my understanding. In statement 2 you can say that x=-1 because of the following reasoning.
\(IxI<= 2 - x^2\)
we have 2 cases:
1) \(x>0\) --> \(x<=2-x^2\)
2) \(x<0\) --> \(x >= x^2-2\)
However we know that x is negative hence we pick only case 2.
\(x >= x^2-2\)
\(x^2-x-2<=0\)
\((x-2)(x+1)<=0\)
so \(x>=-1\) & \(x<=2\)
We know that x is a negative integer so within this range only \(x=-1\) is acceptable.
Am I correct?
GMAT 1: 760 Q51 V42
Posts: 10,057
Bunuel
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have 1 variable (\(x\)) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.
Condition 1)
\(x^2 + 6x < 7\)
⇔ \(x^2 + 6x - 7 < 0\)
⇔ \((x+7)(x-1) < 0\)
⇔ \(-7 < x < 1\)
⇔ \(-6 ≤ x ≤ -1\) since \(x\) is a negative integer.
The possible values of \(x\) are \(-6, -5, -4, -3, -2\) and \(-1\).
Since condition 1) does not yield a unique solution, it is not sufficient.
Condition 2)
\(x^2 + |x| ≤ 2\)
⇔ \(x^2 + |x| - 2 ≤ 0\)
⇔ \(|x|^2 + |x| - 2 ≤ 0\) since \(|x|^2 = x^2\)
⇔ \((|x| - 1 )( |x| + 2 ) ≤ 0\)
⇔ \(-2 ≤ |x| ≤ 1\)
⇔ \(0 ≤ |x| ≤ 1\) since \(0 ≤ |x|\)
⇔ \(|x| ≤ 1\)
⇔ \(-1 ≤ x ≤ 1\)
Then \(x = -1\) is the unique solution since \(x\) is a negative integer.
Since condition 2) yields a unique solution, it is sufficient.
Therefore, B is the answer.
Posts: 53
Can someone please explain me how statement 2 is sufficient ?
i did not understand any of the above solutions.
i did not understand any of the above solutions.
Harshjha001
Notice that we are given than x is a negative integer (-1, -2, -3, -4, ...). Next, notice that if x is -2 or less (-3, -4, -5) than x^2 + |x| is more than 2. So, for negative integers, x^2 + |x| ≤ 2 hold true only if x = -1. Thus the answer to the question whether x < -3 is NO.
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B satisfies +1 and -1 since negative hence -1 suff.
A satisfies all negative nos
hwnce IMO B
A satisfies all negative nos
hwnce IMO B
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