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05 Jan 2020, 08:06
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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x is a positive integer. Is it odd or even?
(1) \(x^2+x\) is even
(2) \((x + 2) (x – 4)\) is even
(1) \(x^2+x\) is even
(2) \((x + 2) (x – 4)\) is even
Updated on: 15 Jul 2020, 10:47
Originally posted by BrushMyQuant on 05 Jan 2020, 09:24.
Last edited by BrushMyQuant on 15 Jul 2020, 10:47, edited 1 time in total.
Last edited by BrushMyQuant on 15 Jul 2020, 10:47, edited 1 time in total.
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STAT 1:
\(x^2+x\) is even
This can be simplified as x*(x+!).. Now product of two integers will always be even as one of them will be even and other one will be odd
(Related Theory: Product of n consecutive integers is always a divisible by n!)
So, STAT 1 is insufficient
STAT 2:
\((x + 2) (x – 4)\) is even
so STAT 2 is product of two numbers (x+2) and (x-4)
Notice that in both of them we are either adding or subtracting an even numbers from x
So, if x+2 or x-4 or both is/are even (As the product is even so at least one has to be even) then
x also has to be even (As Even + Even = Even and Even - Even = Even)
So, STAT 2 is sufficient to prove that x is even.
So, Answer is B
Hope it helps!
\(x^2+x\) is even
This can be simplified as x*(x+!).. Now product of two integers will always be even as one of them will be even and other one will be odd
(Related Theory: Product of n consecutive integers is always a divisible by n!)
So, STAT 1 is insufficient
STAT 2:
\((x + 2) (x – 4)\) is even
so STAT 2 is product of two numbers (x+2) and (x-4)
Notice that in both of them we are either adding or subtracting an even numbers from x
So, if x+2 or x-4 or both is/are even (As the product is even so at least one has to be even) then
x also has to be even (As Even + Even = Even and Even - Even = Even)
So, STAT 2 is sufficient to prove that x is even.
So, Answer is B
Hope it helps!
05 Jan 2020, 09:42
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Asad
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.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Since \(x^2+x = x(x+1)\) is a product of two consecutive integers, it is always an even number whatever the integer \(x\) is.
Thus, condition 1) doesn't tell anything about the parity of \(x\).
Since condition 1) does not yield a unique solution, it is not sufficient.
Condition 2)
\((x+2)(x-4)\) is even
⇔ \(x+2\) is even or \(x-4\) is even.
⇔ \(x\) is even since both "\(x+2\) is even" and "\(x-4\) is even" are equivalent to "\(x\) is even" since they are the expressions of adding or subtracting even numbers \(2\) and \(4\) from \(x\).
Since condition 2) yields a unique solution, it is sufficient.
Therefore, B is the answer.
