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23 Aug 2021, 08:37
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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:
75%
(hard)
Question Stats:
47% (02:00) correct
53%
(02:21)
wrong
based on 113
sessions
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Not Attempted Yet
The variables x, y, and z are all integers. If yz = −15, is xz an odd negative integer?
(1) xy = 12
(2) x + y = 7
(1) xy = 12
(2) x + y = 7
23 Aug 2021, 14:50
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ajaanrkhan
Note 1:
ODD*ODD = ODD
ODD*EVEN = EVEN
EVEN*EVEN = EVEN
Note 2:
ODD + EVEN = ODD
ODD + ODD = EVEN
EVEN + EVEN = EVEN
In accordance with the blue rule above:
xz will be ODD only if x and z are both ODD.
Since yz = ODD, y and z must both be ODD.
Thus:
Since z is ODD, xz will be ODD only if x is also ODD.
Question stem, rephrased:
Is x ODD?
Statement 1, in accordance with the red rule above:
Since xy = EVEN and y is ODD, x must be EVEN.
Thus, the answer to the rephrased question stem is NO.
SUFFICIENT.
Statement 2, in accordance with the green rule above:
Since x+y = ODD and y is ODD, x must be EVEN.
Thus, the answer to the rephrased question stem is NO.
SUFFICIENT.
24 Aug 2021, 12:41
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ajaanrkhan
Since y*z results in an odd integer, we know y and z must be both odd, and one of them is negative. We can either have \(- 3*5 = 15\) or \(-1*15 = 15\).
Statement 1:
Since y has to be odd, we must have \(|y| = 3\) or \(|y| = 1\), as other values don't give us an integer x. Multiply both sides by yz to get \(xy^2z = 12*(-15)\). We know \(y^2\) has to be 9 or 1, so \(xz = 12*(-15)/9 = -20\) which is even, or just \(12*(-15)\) which is also even. Sufficient.
Statement 2:
y is odd, so x must be even. Then xz must be even. Sufficient.
Ans: D
