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Updated on: 02 Mar 2025, 01:59
Originally posted by GMATinsight on 29 Apr 2021, 00:32.
Last edited by Bunuel on 02 Mar 2025, 01:59, edited 3 times in total.
Last edited by Bunuel on 02 Mar 2025, 01:59, edited 3 times in total.
Corrected the OA
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C
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:
33% (02:07) correct
67%
(01:18)
wrong
based on 52
sessions
History
Date
Time
Result
Not Attempted Yet
Is \(x^2-x\) even?
1) x is not divisible by 2
2) \(x^2\) is divisible by 2
1) x is not divisible by 2
2) \(x^2\) is divisible by 2
Source: https://www.GMATinsight.com
30 Apr 2021, 00:10
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GMATinsight
The OA is of course wrong. It should be C and not E. Please amend it as I cannot think of any case where \(x^2-x\) is even.
Now \(x^2-x=x(x-1)\)
(a) If x is integer, x(x-1) means product of two consecutive integers, one of which will surely be even. Hence the product will always be even.
(b) If x is a fraction, then x(x-1) means product of two fractions having a difference of 1. Hence the product will almost always be a fraction, and, therefore, not even.
1) x is not divisible by 2
If x is odd, then answer is yes. Say \(x=3...3^2-3=6=even\)
If x is a fraction, then answer is no.
Insufficient
2) \(x^2\) is divisible by 2
If x is an integer, then yes.....\(x^2=4\).
If x is a square root, then it can be no....\(x^2=2\)
Insufficient
Combined.
\(x^2\) is even but x is not, so \(x^2-x = even \ - \ (not \ \ even) \ = \ not \ even\)
Sufficient
C
30 Apr 2021, 08:09
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GMATinsight
When asking for the parity of an expression, we are allowed to reduce the powers of any terms as that does not change the parity. Hence instead of asking for if \(x^2 - x\) is even, we could ask if \(x - x = 0\) is even. Yet 0 is always even, hence as long as x is an integer this expression is even. The question did not actually tell us whether x was an integer, so we're looking for evidence that x is an integer.
Statement 1:
x might not be an integer. Insufficient.
Statement 2:
If \(x = \sqrt{2}\) then x doesn't have to be an integer. Insufficient.
Combined:
Again let's use \(x = \sqrt{2}\). It satisfies both statements yet x is still not an integer. On the other hand, we need to check if x is allowed to be an integer. If x was an integer it would have to be odd due to statement 1, if it was odd then there is no possible way if satisfies statement 2. Hence x cannot be an integer. Sufficient.
Ans: C
